*Hi we are here to do some extra examples on measuring angles and converting back and forth between degrees and radians.*0000

*I hope you had a chance to try this out on your own a little bit. *0008

*There are common values that you use a lot in all kinds of trigonometric functions and situations.*0012

*It is worth at least working them out once on your own and memorize them after that.*0020

*The common values are 0 degrees, 30 degrees, 45degrees, 60degrees, and 90 degrees.*0027

*What we are going to do is find the complementary and supplementary angles for each one in both degrees and radians.*0037

*The reason these angles are so important is because 90 degrees is a right angle*0047

*What we are doing is chopping a right angle up into either two equal pieces which gives us 45 degrees or three equal pieces which gives us the 30 degrees and 60 degrees angles.*0054

*Those are very common ones that come up very often.*0065

*It is worth knowing what these are in both degrees and radians.*0068

*Knowing their complements and supplements and knowing what the complements and supplements are in degrees and radians as well.*0073

*Let me make a little chart here, we are starting out in degrees.*0082

*We have the 0 degrees angle, 30 degrees, 45 degrees, 60 degrees, and 90 degrees.*0087

*Let me convert those into radians first.*0098

*The 0 degrees angle is still 0 in radians.*0102

*Well remember that a 90 degrees angle is pi over 2 radians.*0107

*A 30 degrees angle is 1/3 of that, it’s pi over 2 divided by 3, that is pi over 6.*0114

*45 degrees is 90 degrees divided by 2, that is pi over 2 divided by 2 which is pi over4.*0123

*60 degrees is twice 30 degrees, 60 degrees is 2 x pi over 6, 2 pi over 6 is pi over 3.*0131

*Those are pretty convenient fraction if you write them in radians.*0143

*The complementary angles we will do it in terms of degrees first.*0147

*Remember complementary angles add up to 90 degrees. *0154

*If you know what the angle is, you will do 90 degrees minus that number to get the complementary angle.*0158

*If you start with the 0 degrees angle, the complementary angle is 90 degrees because 90 minus 0 is 90.*0165

*30 degrees angle the complementary angle is 60 because those add up to 90.*0173

*45 degrees angle is its own complement because 45 and 45 is 90.*0178

*60 degrees angle its complement is 30.*0185

*A 90 angle its complement is 0.*0187

*Let us do those in terms of radiance*0192

*The 0 radiant angle its complement is going to be pi over 2 because those add up to pi over 2.*0198

*Remember that is the same as a 90 degree angle.*0207

*Pi over 6 + pi over 3 is pi over 2.*0210

*You can work out the fractions there or you can remember that 30 + 60 equals 90.*0216

*45 degree angle or pi over 4 is its own complement. *0222

*Pi over 3 we already figured out that its complement is pi over 6.*0228

*Pi over 2 its complement is 0 because those add up to pi over 2 or 90 degree angle.*0235

*Let us figure out the supplementary angles.*0243

*Remember supplementary angle means that they add up to 180 degrees.*0246

*In each case we are looking for what adds up to pi over 2 or 90. *0251

*We start with 0, that means the supplement is 180 itself.*0256

*If we start with 30 it is 150.*0262

*45 a 180 - 45 is 135.*0265

*180 - 60 is 120.*0271

*180-90 is 90 itself.*0275

*Finally, if we find the supplementary angles in terms of radians. *0280

*Remember we are looking here for, before they add it up to a 180 degrees in terms of radians they should add it up to pi.*0286

*0 + pi adds up to pi.*0295

*Pi over 6 + 5 pi over 6 adds up to pi.*0299

*Pi over 4 + 3 pi over 4 adds up to pi.*0306

*Pi over 3 + 2 pi over 3 adds up to pi.*0313

*Pi over 2 + pi over 2 adds up to pi.*0320

*All of these are common values. *0326

*These are all conversions back and forth between degrees and radians that you should now very well as a trigonometry student just because these angles come up so often.*0328

*It is probably worth understanding the pictures behind each of these numbers that I have written down.*0337

*For example, when we look at complementary angles.*0344

*Here is a 30 degree angle and its complement is a 60 degree angle.*0349

*In terms of radians, that is pi over 6 radians.*0355

*The 60 degree angle is pi over 3 radians.*0362

*Add those together, you will get pi over 3 + pi over 6 adds up to pi over 2.*0367

*Here is another right angle and I’m dividing it in 2 into 45 degrees, which is pi over 4 radians.*0377

*We see that that angle is its own complement 45 degrees is pi over 4 radians.*0385

*I will do the supplements in blue.*0394

*If we start out with a 30 degree angle or pi over 6 radians.*0399

*Then its supplement is a 150 degree angle which is 5 pi over 6.*0411

*If we start out with a 60 degree angle which is pi over 3 radians then its supplement.*0423

*Remember to put them together and they are supposed to make 180 degree or pi radians is 120 degrees which is 2 pi over 3 radians.*0436

*Finally if you start out with a 45 degree angle which is pi over 4 radians then its supplement is 135 degrees which is 3 pi over 4 radians.*0452

*All of those are angles that you should know very well both in terms of degrees and radians because we will be seeing a lot of them in our trigonometry lessons.*0475

*Finally, our example here is for each of the following angles.*0000

*I want to find out what quadrant it is in.*0005

*I want to find the coterminal angle between 0 and 360 degrees over 0 and 2 pi radians.*0009

*We got 4 angles here and I’m going to give you in both degrees and radians.*0018

*We are going to start out with -5 pi over 4 radians and that will do 735 degrees, -7 pi/3 radians and -510 degrees.*0027

*If you want you can try those on your own.*0047

*I will help you out starting with -5 pi/4 radians.*0051

*Ok, that is not between 0 and 2 pi radians.*0057

*What we are going to is add a 2 pi to it, 2 pi is 8 pi/4, (8 pi/4 – 5 pi/4) is 3 pi/4.*0060

*Let us graph that, 0 pi/2 pi, 3 pi/2, and 2 pi, 3 pi/4 is between pi/2 and pi. *0076

*It is right there, that is in the second quadrant.*0094

*Our answer here is 3 pi/4 and it is in the second quadrant.*0107

*The way you can understand how that came from the original -5 pi/4.*0115

*If you went in the other direction 5 pi/4 in the other direction from the traditional direction because it is negative. You will end up at that angle.*0121

*Its coterminal angle is 3 pi over 4 in quadrant 2.*0131

*735 degrees that is way bigger than 360.*0136

*Let us drop down multiples of 360, we can actually take out two multiples of 360 right away.*0141

*I’m going to subtract 720 degrees and we get 15 degrees.*0148

*If you write things in terms of degrees here, 0 is 0 degrees, Pi/2 is 90 degrees, Pi is 180 degrees, 3 pi/2 is 270 degrees, 2pi is 360 degrees.*0154

*15 degrees is between 0 and 90. That is right about there*0174

*What that means is 735 would actually go out in circle twice and end up right at the same place that 15 degrees ended up.*0182

*So, the coterminal angle is 15 degrees and that is in the first quadrant.*0192

*Let us look at -7pi/ 3, that is less than 0 so I’m going to add 2 pi to that, plus 2 pi, well 2 pi is 6 pi/3.*0203

*That would give us – pi/3, that is still less than 0, let me add another 2 pi. *0217

*That is again 6 pi/3 – pi/3, would give us 5 pi/3, that is between 0 and 2 pi.*0227

*Now, we just have to figure what out quadrant it is in.*0238

*5 pi/3 is a little bit bigger than 3 pi/2, 3 pi/2 is 1 ½, 5 pi/3 is 1.67.*0241

*So, 5 pi/3 is a little bit bigger than 3 pi/2.*0252

*Let me erase some of these extra angles, they are getting in our way.*0260

*I will show you where 5 pi over 3 is.*0273

*If you go between pi and 2 pi, there is 4 pi/3, there is 5 pi/3.*0277

*And so, 5 pi/3 is right there, that is clearly in the fourth quadrant.*0283

*Finally, -510 degrees, where does that one end up? That is definitely less than 0.*0295

*So, I will add 360 degrees and we will end up with -150 degrees, still less than 0 there so I will add another 360 degrees, that gives us 210 degrees.*0303

*That is between 0 and 360 degrees, we know we found our coterminal angle, 210 degrees is just past 180 degrees.*0322

*In fact, it is 30 degrees past 180 degrees, it is right there, that is in the third quadrant.*0333

*Just to recap here, finding these coterminal angles and finding out what quadrant they are in.*0346

*To find the coterminal angle, you take the angle that you are given in degrees or radians and you add or subtract multiples of either 2 pi radians or 360 degrees.*0351

*Then you add or subtract these multiples until you get them in to the range that you want, 0 to 2 pi radians, 0 to 360 degrees.*0357

*Once you get it into those ranges then you can break it down finally and ask are you between 0, pi over 2, pi, 3 pi over 2, or 2 pi.*0380

*That tells you which quadrant you are in or in terms of degrees that would be 0, 90 degrees, 180 degrees, 270 degrees, or 360 degrees.*0394

*That will tell you what quadrant you are in if you are in terms of degrees.*0407

*That is the first or our lessons on trigonometry for www.educator.com.*0412

*Here we talked about angles, we have not really got in to the trigonometric functions yet.*0417

*In the next lessons we will start talking about sine and cosine, all the different identities and relationships,how you use them in triangles?*0421

*We will talk about tangents and secants.*0429

*That is all coming in the later lectures on www.educator.com.*0431

*Hi, This is Will Murray and I'm going to be giving the trigonometry lectures for educator.com.*0000

*We're very excited about the trigonometry series.*0006

*In particular, for me, trigonometry is the class that got me excited about math.*0008

*I'm really looking forward to working with you on learning some trigonometry.*0013

*We're going to start right away here, learning about angles.*0018

*The first thing you have to understand is that there's two different ways to measure angles.*0021

*People use degrees which you probably already heard of, and radians which you may not hear about until you start to take your first trigonometry class.*0029

*They're just two different ways in measuring.*0037

*You can use either one but you really need to know how to use both, and convert back and forth.*0041

*That's what I'll be covering in this first lecture.*0044

*We'll start with degrees.*0048

*Degrees are unit of measurement in which a circle gets divided into 360 degrees*0050

*If you have a full circle, the whole thing is 360 degrees.*0059

*That's 360 degrees.*0066

*Then if you have just a piece of a circle, then it gets broken up into smaller chunks.*0067

*For example, an angle that's half of a circle here that's 180 degrees, because that's half of 360, a quarter of a circle which is a right angle, that would be 90 degrees, then so on.*0072

*You can take a 90-degree angle and break it up into two equal pieces.*0093

*Then each one of those pieces would be a 45-degree angle.*0098

*Or you could break up a 90-degree angle into three equal pieces, and each one of those would be a 30-degree angle.*0104

*We'll be studying trigonometric functions of these different angles.*0114

*In the meantime, it's important just to get comfortable with measuring angles in terms of degrees.*0118

*The second unit of measurement we're going to use to measure angles is called radians.*0124

*That's a little bit more complicated.*0129

*You probably haven't learned about this until you start to study trigonometry.*0131

*The idea is that, you take a circle, and remember that the circumference of a circle is equal to 2Π times the radius, that's 2Π r, it's one of those formulas that you learned in geometry.*0135

*What you do with radians is, you break the circle up, and you say the entire circle is 2Π radians.*0151

*2Π radians.*0159

*What that means is that a one-radian angle, well, if the entire circle is 2Π radians, then 1 radian, use a little r to specify the radians, cuts off an arc that is 1 over 2Π of the whole circle.*0161

*One radian, an angle that is 1 radian cuts of a fraction of the circle that is 1 over 2Π.*0184

*Since the radius is 2Π times r, sorry, the circumference is 2Π times r, if you have 1 over 2Π of the whole circumference, what you get is exactly the length of the radius.*0194

*That's why they're called radians is because if you take one-radian angle, it cuts of an arc length that is exactly equal to the radius.*0214

*That's the definition of radians.*0226

*It takes a little bit of getting used to.*0228

*What you have to remember that's important is that the whole circle is 2Π radians.*0230

*That means a half circle, a 180-degree angle, is Π radians.*0240

*A right angle, a 90-degree angle, or a quarter circle is Π over 2 radians, and so on.*0250

*Then you can break that down into the even smaller angles like we talked about before.*0262

*If you take a right angle and you cut it in half, so that was a 45-degree angle before, that's Π over 4 radians because it's half of Π over 2.*0268

*If you take a right angle and you cut it into three equal pieces, so those are 30-degree angles before, in terms of radians, that's Π over 2 divided by 3, so that's Π over 6 radians.*0281

*You want to practice going back and fort between degrees and radians, and kind of getting and into the feel of how big angles are in terms of degrees and radians.*0299

*We'll practice some of that here in this lecture.*0307

*Remember that Π is about 3.14, so 2Π is about 6.28.*0310

*That means we're breaking an entire circle up in the 2Π radians, so the circle gets broken up into about 6.28 radians.*0317

*That means one radian is about one-sixth of a circle.*0326

*What I've shown up here is pretty accurate that one radian is about one-sixth of a circle.*0332

*It's about 60 degrees but it's not exact there because it's not exactly 6 it's 6.28 something.*0339

*That's about roughly a radian is, about 60 degrees, but we really don't usually talked about whole numbers or radians.*0347

*People almost always talk about radians in multiples of Π the same way I was doing here, where I said the circle is 2Π radians, the half circle is Π radians, the right angle is Π over 2.*0355

*People almost always talk about radians in multiples of Π and degrees in terms of whole numbers.*0366

*Sometimes, they don't even bother to write the little r.*0374

*It's just understood that if you're using a multiple of Π , then you're probably talking about radians.*0376

*Let's practice going back and forth between degrees and radians.*0385

*Remember that 360 degrees is a whole circle.*0389

*That's 2Π radians.*0394

*What that means is that Π radians is equal to 360 degrees over 2, which is 180 degrees.*0398

*Pi radians is 180 degrees.*0412

*That gives you the formula to convert back and forth between degree measurement and radian measurement.*0415

*If you know the measurement in degrees, you multiply by Π over 180 and that tells you the measurement in radians.*0421

*If you know the measurement in radians, you just multiply by 180 over Π , and that tells you the measurement in degrees.*0429

*We'll practice that in some of the examples later on.*0439

*We got a few more definitions here.*0442

*Coterminal angles, what that means is that their angles that differ from each other by a multiple of 2Π radians,*0446

*remember that's a whole circle, or if you think about it in degrees, 360 degrees.*0456

*For example, if you take a 45-degree angle, and then you add on 360 degrees, that would count as a 360 plus 45 is 405 degrees.*0462

*Forty-five and 405 degrees are coterminal.*0484

*In the language of radians, 45 degrees is Π over 4.*0491

*If you add on 2Π radians, if you add on a whole circle to that, you would get...*0497

*This should end up here.*0511

*If you add on 2Π plus Π over 4, well 2Π is 8Π over 4, so you get 9Π over 4.*0512

*Then Π over 4 and 9Π over 4 are coterminal angles.*0522

*The reason they're called coterminal angles is because we often draw angles starting with one side on the positive x-axis.*0528

*We start with one side on the positive x-axis.*0539

*I'll draw this in blue.*0544

*Then we draw the other side of the angle just wherever it ends up.*0545

*Coterminal angles are angles that will end up at the same place, that's why they're called coterminal.*0550

*If they differ from each other, if one is 2Π more than the other one, or 360 degrees more than the other one, or maybe 720 degrees more than the other one, then we call them coterminal because they really end up on the same terminal line here.*0561

*Couple other definitions we need to learn.*0580

*Complementary angles are angles that add up to being a right angle, in other words, 90 degrees or Π over 2.*0583

*If you have two angles, like this two angles right here , that add up to being a right angle, 90 degrees or Π over 2, those are complementary.*0591

*Supplementary angles are angles that add up to being a straight line, in other words, Π radians or 180 degrees.*0608

*Those two angles right there are supplementary.*0620

*That's all the vocabulary that you need to learn about angles, but we'll go through and we'll do some examples of each one to give you some practice.*0629

*Here's our first example.*0640

*If a circle is divided into 18 equal angles, how big is each one in degrees and radians?*0642

*Let me try drawing this.*0648

*We've got this circle and it's divided into a whole bunch of little angles but each one is the same.*0653

*We want to figure out how big each one is, in terms of degrees and radians.*0664

*Let's solve this in degrees first.*0668

*Remember that a circle is 360 degrees.*0671

*If it's divided into 18 parts, then each part will be 20 degrees.*0676

*Each one of those angles will be 20 degrees.*0682

*We've done the degree one, how about radians?*0686

*Remember that an entire circle is 2Π radians.*0691

*If we divide that by 18, then we get Π over 9 radians will be the size of each one of those little angles.*0695

*You can measure this angle either way, we say 20 degrees is equal to Π over 9 radians.*0706

*Second example here, we want to convert back and forth between degrees and radians.*0718

*Let's practice that.*0721

*We want to convert 27 degrees into radians.*0724

*Well, let's remember the formula here, the conversion formula, is Π over 180.*0727

*So we do 27 times Π over 180.*0733

*That's our conversion formula from degrees into radians.*0738

*I'm just going to leave the Π because it doesn't really cancelled anything.*0743

*The 27 over 180 does simplify.*0747

*I could take a 9 out of each ones.*0752

*That would be 3.*0753

*If we take a 9 out of 180, then there'd be 20.*0755

*What we end up with is 3Π over 20 radians, as our answer there.*0761

*Converting back and forth between degrees and radians is just a matter of remembering this conversion factor, Π over 180 gets you from degrees into radians.*0777

*For the second part of this example, we're given a radian angle measurement, 5Π over 12. *0789

*We want to convert that into degrees, 5Π over 12 radians.*0794

*We just multiply by the opposite conversion factor, 180 over Π.*0804

*Let's see here.*0813

*The pis cancel.*0814

*One hundred eighty over 12 is 15.*0818

*That's 5 times 15 degrees.*0826

*That gives us 75 degrees.*0833

*The same angle that you would measure, in radians is being 5Π over 12, will come out to be a 75-degree angle.*0838

*Converting back and forth there is just a matter of remembering the Π and the 180, and multiplying by one over the other to convert back and forth.*0848

*Third example is some practice with coterminal angles.*0859

*In each case, what we want to do is, we're given an angle and we want to find out what quadrant it's in.*0863

*That's assuming that all the angles are drawn in the standard position with their starting side on the positive x-axis.*0872

*We want to start on the positive x-axis.*0882

*We want to see which one of the four quadrants the angle ends up in.*0885

*Then we want to try to simplify these angles down by finding a coterminal angle that's between 0 and 360, or between 0 and 2Π radians.*0893

*Let's start out with 1000 degrees.*0904

*A thousand degrees is going to be, that's way bigger than 360. *0908

*Let me just start subtracting multiples of 360 from that.*0912

*If I take off 360 degrees, what I'm left with is 640 degrees.*0916

*That's still way bigger than 360 degrees.*0926

*I'll subtract off another 360 degrees and what I'm left with is 280 degrees.*0928

*That's between 0 and 360.*0940

*I found my coterminal angle there.*0942

*I wanna figure which quadrant it's gonna end up in.*0945

*Now, remember, if we start with 0 degrees being on the x-axis, that would make 90 degrees being on the positive y-axis.*0947

*Then over here on the negative x-axis, we'd have 180 degrees.*0959

*Down here is 270 degrees, because that's 180 plus 90.*0964

*Then 360 degrees would be back here at 0 degrees.*0971

*Two hundred and eighty degrees would be just past 270 degrees.*0976

*That's a little bit bigger than 270 degrees.*0981

*It would be about right there.*0983

*That's 280.*0986

*That puts it in the fourth quadrant.*0990

*Next one's a radian problem.*1001

*We have -19Π over 6 radians.*1005

*That's one, I'll do this one red.*1011

*That's one that goes in the negative direction.*1014

*We start on the positive x-axis but now we go in the negative direction.*1016

*Instead of going up around past the positive y-axis, we go down in the negative direction and we go -19Π over 6.*1021

*If you think about it, 19Π over 6 is bigger than 2Π.*1031

*Let me start with 19Π over 6 and subtract off a 2Π there.*1039

*Well, 2Π is 12Π over 6, so that gives us 7Π over 6.*1044

*If we do, -19Π over 6 plus 2Π, that will give us, -19Π over 6 plus 12Π over 6 is -7Π over 6.*1055

*That's still not in the range that we want, because we want it to be between 0 and 2Π radians.*1070

*Let me add on another 2Π plus 2Π gives us positive 5Π over 6.*1076

*The trick with finding these coterminal angles with degrees, it was just a matter of adding or subtracting 360 degrees at a time.*1084

*With radians, it's a matter of adding or subtracting 2Π radians at a time.*1092

*Remember, 2Π is a whole circle.*1098

*We end up with 5Π over 6.*1101

*That is between 0 and 2Π so we're done with that part but we still have to figure out what quadrant it's in.*1104

*Well now 5Π over 6, where would that be?*1111

*Well if we map out our quadrants here, 0 is right there on the positive x-axis just as we had before, 90 degrees is Π over 2 radians, 180 degrees is Π radians, and 270 degrees is 3Π over 2 radians.*1114

*Then 360 degrees is 2Π radians, a full circle.*1135

*Where does 5Π over 6 land?*1143

*Well that's bigger than Π over 2, it's less than Π, so, 5Π over 6 lands about right there.*1145

*That's in the second quadrant.*1155

*OK, we have another degree one.*1165

*Negative 586 degrees, and what are we going to do with that?*1168

*It's going in the negative direction so it's going down south from the x-axis.*1173

*Negative 586 degrees.*1180

*Well, 586 is way outside our range of 0 and 360.*1183

*Let's try adding 360 degrees to that.*1187

*That gives you -226 degrees, which is still outside of our range.*1193

*Let's add another 360 degrees.*1203

*We're adding and subtracting multiples of a full circle 360 degrees.*1208

*That gives us positive 134 degrees.*1212

*Now, positive 134 degrees, that is in our allowed range between 0 and 360.*1218

*So we finished that part of the problem.*1227

*Where would that land in terms of quadrants?*1229

*Let me redraw my axis because those are getting a little messy.*1230

*That's 0, 90, 180, 270 and 360.*1237

*Where's 134 going to be?*1246

*A hundred and thirty-four is going to be between 90 and 180, almost exactly halfway between.*1250

*It's about right there.*1255

*That's in the second quadrant.*1258

*The answer to that one is that that's in the quadrant number two there.*1262

*Finally, we have 22Π over 7, again given in radians.*1272

*The question is, is that between 0 and 2Π?*1281

*It's not, it's too big.*1284

*It's bigger than 2Π.*1285

*Let me subtract off a multiple of 2Π.*1287

*I'll subtract off just 2Π, which is 14Π over 7.*1291

*That simplifies down to 22Π over 7 minus 14Π over 7, is 8Π over 7.*1296

*Now, 8Π over 7 is between 0 and 2Π.*1303

*We found our coterminal angle.*1309

*Where will that land on the axis?*1311

*Well, remember 0 degrees is 0 radians, 90 degrees is Π over 2 radians, 180 degrees is Π radians, and 270 degrees is 3Π over 2 radians, and finally, 360 degrees is 2Π radians.*1312

*Eight pi over 7 is just a little bit bigger than 1.*1334

*That's a little bit, or 8 over 7 is a little bigger than 1.*1336

*Eight pi over 7 is just a little bit bigger than Π.*1341

*Let's going to put it about right there which will put it in the third quadrant.*1346

*Let's recap how we found this coterminal angles.*1360

*Basically, you're given some angle and you check first whether it's in the correct range, whether it's in between 0 and 2Π radians,*1363

*or if it's given in degrees, whether it's between 0 and 360 degrees.*1372

*If it's not already in the correct range, if it's negative or if it's too big, then what you do is you add and subtract multiples of 360 degrees or 2Π radians until you'll get it into the correct range, *1378

*the range between 0 and 360 degrees or 0 and 2Π radians.*1393

*Once you get it in that range, if you want to figure out what quadrant it's in, well in degrees, it's a matter of checking 0, 90, 180, 270, and 360;*1402

*in radians,it's a matter of checking 0, Π over 2, Π, 3Π over 2, 2Π.*1417

*Which one of those ranges does it fall into?*1426

*That tells you what quadrant it's in.*1428

*Ok we are trying some more examples of computing values of the inverse trigonometric functions.*0000

*Here we have to find the arctangent of the following common values.*0006

*I will start with my unit circle.*0013

*I remember that when arctangent I am always looking for values between –pi/2 and pi/2 because arctangent is always between pi /2 and –pi/ 2.*0029

*I wrote less than or equal to there, but it should really be less than because arctangent never actually hits – pi/2 or pi/2.*0050

*That is because tangent itself cannot be evaluated on pi/2 or –pi/2 because you are basically trying to divide by 0 there.*0059

*I’m going to make a list here of x and arctan *_{z} 1 and root 3.0071

*It is really useful if you can remember some common values of the tangent function.*0098

*For example tan (pi/0) is 0.*0105

*Tan (pi/6) is sine of pi/6 divided by cosine of pi/6, which works out to root 3/3*0113

*Tan (pi/4) is sine of 4 divided by cosine.*0126

*Well in both are 2/2 so it is one.*0132

*Tan (pi/3) is sin/cosin of pi/ 3, that is root 3.*0136

*That tells me what the values of tangent are at these common values, pi/6, pi/4, and pi/3.*0143

*There is another way to remember the values of the tangent functions, which is to draw this line at x=1.*0155

*The tangent function is actually where these lines hit that line. That is 1, root 3/3, and root 3.*0164

*So if you remember your common values of tangent, it is easy to figure out the values of arctangent.*0178

*Let me start from the bottom because that is the easiest.*0186

*Root 3 here, what angle has tangent root 3?*0190

*Well, we just figured that out, that is pi/3.*0194

*Also, we get that from here.*0199

*Pi/3 is the angle that has tangent root 3.*0201

*What angle has tangent 1?*0206

*Well, there it is right there, it is pi/4. *0210

*What angle has root 3/3? It is pi/6.*0214

*From getting these common values over here.*0220

*What angles has tangent 0? Well, that is 0.*0223

*The negative ones are little trickier but if you just remember that the values in the fourth quadrant down here.*0227

*They are just the negatives of the same values we had up in the first quadrant.*0235

*That is –root 3, -1, and –root 3/3.*0242

*For –root 3, what is an angle that has that as its tangent? it is –pi/3.*0252

*For -1, what is an angle that has that as its tangent? It is –pi/4.*0259

*Finally for –root 3/3, what is an angle that has that as its tangent? It is –pi/6.*0264

*There are really two things that are keys to being able to find our tangents.*0274

*The first is knowing your tangents of the common values.*0278

*Again, if you can not remember those, just remember that tangent is sine over cosine.*0282

*If you remember the sine and cosine of the common values, it all comes back to those triangles.*0295

*Then you can figure out the tangents.*0299

*It is good to remember the common values that you do get from tangent namely root 3, 1, and root 3/3.*0302

*It is good to remember those numbers. But if you can not remember them, you can work them out from sine and cosine.*0311

*Once you figure those out, then you can figure out the common values of tangent of the negative angles.*0316

*You have to remember that arctangent always takes values between –pi/2 and pi/2.*0324

*Looking for angles between –pi/2 and pi/2 that have the tangents that we have been given.*0334

*When we are given these tangents, we just remember values between –pi/2 and pi/2 that have those particular tangents.*0341

*On our last example here, we are given various angles.*0000

*We have to find the arcsine of their sine or the same for cosine and tangent.*0005

*It is very important that you would be able to graph this on the unit circle.*0012

*That is really the way you start the problem like this.*0015

*There is my unit circle and that is 0, pi/2, pi, 3 pi/2, and 2 pi.*0021

*Now, -7 pi/6. The negative means you are going the downward direction from 0.*0040

*If you go a little bit past pi and -7 pi/6 puts you right there, -7 pi/6.*0046

*So, arcsine of sine of -7 pi/6 *0056

*The sine of that angle is the y coordinate and it is positive there.*0068

*So, it is arcsine and that is a 30, 60, 90 triangle.*0076

*I know the common values there are ½. It is positive because the y coordinate is positive.*0083

*It is arcsine of 1 ½ but remember that arcsine is always between – pi/2 and pi/2.*0088

*I want an angle between –pi/2 and pi/2.*0105

*Whose sign is ½. Well, sign is the y coordinate.*0108

*There is an angle who is y coordinate is ½ and that angle is pi/6.*0117

*My x are, there is pi/6.*0126

*It is an angle between –pi/2 and pi/2 who is sign is ½.*0132

*Let us try the next one, arcosine of 4 pi/3.*0138

*Well, I find 4 pi/3, that is between pi and 2 pi.*0148

*In fact, that is down here.*0153

*Its cosine is x coordinate, its cosine is x coordinate and there again we have a 30, 60, 90 triangle.*0158

*Its cosine of 4 pi/3 is ½.*0174

*But since we are on the left side of the page, it is negative because the x coordinate is negative there.*0183

*Now I’m trying to find the arcosine of – ½.*0189

*Remember in arcosine, you will always look for an angle between 0 and pi.*0196

*I’m looking for an angle between 0 and pi whose cosine is – ½.*0200

*There it is right there.*0209

*That is 2 pi/3.*0212

*It is between 0 and pi, and its cosine is – ½.*0215

*So, our answer there is 2 pi/3.*0219

*Finally, we have the arctangent of the tangent of -5 pi/4.*0229

*Let us figure out where that angle is on the unit circle.*0239

*It is negative so we are going around in the downward direction from 0.*0242

*5 pi/4 is just past pi.*0246

*We go down on the downward direction and we end up over here at 5 pi/4.*0250

*So, -5 pi/4. I’m sorry that is -5 pi/4. Let me go back carefully.*0260

*Its tangent is sine/cosine.*0270

*That is a 45, 45, 90 triangle.*0276

*Its tangent is sine/cosine.*0285

*Sine is root 2/2 because it is positive.*0288

*Cosine is –root 2/2 that simplifies down to -1.*0293

*We want the arctangent of -1.*0299

*Now, arctangent always takes values between -pi/2 and pi/2 arctangent.*0302

*We are looking for an angle between –pi/2 and pi/2 whose tangent is -1.*0317

*So, I’m looking for an angle whose tangent is -1.*0327

*There it is at that 45, 45, 90 triangle.*0339

*There is my angle whose tangent is -1.*0344

*And that angle is –pi/4.*0348

*To recap, what really made it possible to this problem were finding sine, cosine, and tangent to start with.*0361

*The first thing you need to is your common values of sine, cosine, and tangent.*0369

*You need to be able to graph these things on the unit circle.*0374

*You need to identify the 30, 60, 90 triangles and the 45, 45, 90 triangles and know the common values.*0378

*Remember the common values for the 30, 60, 90 triangles are always ½ and root 3/2.*0388

*The common values for a 45, 45, 90 triangles are root 2/2 and root 2/2.*0398

*You need to know those common values and identify which triangle you are working with.*0414

*And figure out what quadrant you are in to figure out whether the sine and cosine are positive or negative.*0423

*The way you remember that is by all students take calculus.*0430

*In the first quadrant, they are all positive.*0434

*In the second quadrant, sine is positive only.*0437

*In the third quadrant, tangent is positive.*0440

*And in the fourth quadrant, cosine is positive.*0443

*You figure out which quadrant all these things are positive or negative and then you can find your sine and cosine.*0448

*And then what we have to do is find arcsines, arcosines, and arctangents.*0456

*The key to that was remembering the common values but also remembering these ranges.*0462

*Arcsine, it has to be between –pi/2 and pi/2.*0470

*Arcosine, between 0 and pi.*0474

*And arctangent, between –pi/2 and pi/2.*0477

*What we are doing in this step for all three problems was we had the sine or the cosine of the tangent.*0482

*We were trying to find an angle in the appropriate range that had the correct sine, cosine, or tangent.*0490

*That is where our arcosine, arsine, and arctangent are.*0498

*We are looking for an angle whose sign, cosine, and arctangent is the value that you are given.*0501

*We are looking for an angle inside this range.*0510

*In each case, we have a value and we found angles inside that range that have the right sine, cosine, or tangent.*0514

*That is what the answers are.*0521

*That is the end of our lecture on computations of inverse trigonometric functions.*0524

*These are the trigonometry lectures for www.educator.com.*0529

*Hi these are the trigonometry lectures for educator.com and today we're talking about computations of inverse trigonometric functions.*0000

*In the previous lecture, we learned the definitions and we practiced a little bit with arcsin, arccos, and arctan, you might want to review those a little bit before you go through this lecture.*0009

*I'm assuming now that you know a little bit about the definitions of arcsin, arccos, and arctan, and we'll practice using them and working them out for some common values today.*0020

*The key thing to remember here is where these functions are defined and what kinds of values you're going to get from them.*0032

*Arcsin, remember you start out with the number between -1 and 1, and you always get an answer between -π/2 and π/2.*0040

*It's very helpful if you remember the unit circle there.*0052

*Arcsin always gives you an angle in the fourth and the first quadrant between -π/2, 0, and π/2.*0055

*You're looking for angles in that range that have a particular sine.*0068

*Arccos, also between -1 and 1, produces an answer between 0, and π.*0075

*Again, it's helpful to draw the unit circle and keep that in mind.*0081

*There's 0, π/2, and π.*0088

*You're trying to find angles between 0 and π that have a given cosine.*0095

*Arctan, you can find the arctan of any number.*0101

*Again, you're trying to find an angle between -π/2 and 0, and π/2 that has a given tangent.*0110

*I say exclusive here because you would never actually give an answer of -π/2 or π/2 for arctan because arctan never actually hits those values.*0129

*If you think of that coming from the other direction, we can't talk about the tangent of π/2 or -π/2, because those involve divisions by 0.*0143

*When we're talking about arctan, we'll never get -π/2 or π/2 as an answer.*0153

*Let's practice finding some common values.*0161

*Here's some common values that we should be able to figure out arc sines of.*0166

*Let me start by drawing my unit circle.*0172

*There's -π/2 and 0, and π/2, remember, our answers always going to be in that range.*0183

*Let me just graph those common values and see what angles they correspond to.*0191

*I'll make a little chart here.*0197

*x and arcsin(x), so we got -1, negative root 3 over 2, negative root 2 over 2, -1/2, 0, 1/2, root 2 over 2, root 3 over 2, and 1.*0198

*Remember, sin(x), sine is the y-coordinate.*0222

*I'm looking for an angle that has y-coordinate of -1 to start with, so I want to find an angle that has y-coordinate down there at -1, and that's clearly -π/2.*0227

*That's the answer.*0238

*Negative root 3 over 2, that's an angle down there, so the angle that has sine of negative root 3 over 2, must be -π/3.*0240

*Negative root 2 over 2, that's the one right there, so that's a -π/4.*0258

*-1/2, the y-coordinate -1/2, is right there, that's -π/3.*0267

*Arcsin(0), what angle has sin(0)?*0277

*Well, it's 0.*0280

*What angle has sin(1/2)?*0283

*Well, what angle has vertical y-coordinate 1/2?*0285

*That's π/3.*0288

*Root 2 over 2, that's our 45-degree angle, also known as π/4.*0294

*Root 3 over 3, that's our 60-degree angle, also known as π/3.*0303

*Finally, we know that the sin(π/2) is 1, so the arcsin(1)=π/2.*0313

*In each case, it's a matter of looking at the value and thinking, "Okay, that's my y-coordinate, where am I on the unit circle?"*0321

*"What angle between -π/2 and π/2 has sine equal to that value?"*0332

*Of course, if you know your values of sine very well, then it's not too hard to figure out the arcsin function.*0339

*You really don't need to memorize this.*0345

*You just need to know your common values for sin(x) very well, and to know when they're positive or negative.*0348

*Then you can figure out the values for arcsin(x).*0354

*In our second problem here, we're asked to find which of the arcsin, acrccos, and arctan functions are odd, even or neither.*0360

*It's really, the key to thinking about this one is probably to think about the graphs and not so much to think about the original definitions of odd or even.*0369

*Let me write down the important properties to remember here.*0377

*An odd function has rotational symmetry around the origin.*0382

*The way I remember that is that 3 is an odd number and x*^{3}, y=x^{3} has rotational symmetry around the origin.0406

*Even functions have mirror symmetry across the y-axis and the way I remember that is that 2 is an even number, and the graph of y=x*^{2} has mirror symmetry across the y-axis.0423

*That's how I remember the pictures for odd or even functions.*0458

*Now, let me draw the graphs of arcsin and arccos, and arctan, and we'll just test them out.*0464

*Arcsin, remember you take a piece of the sine graph, there's sin(x) or sin(θ).*0473

*Arcsin, I'll draw this one in blue.*0482

*It's the reflection of that graph in the y=x line, that's arcsin(x) in blue.*0485

*Now, if you check that out, that has rotational symmetry around the origin.*0497

*So, arcsin(x) is odd.*0507

*Let's take a look at cosine and arccos.*0517

*Cosine, remember, you got to snip off a piece of cosine graph that will make arccos into a function.*0522

*There's cos(θ), and now in blue, I'll graph arccos.*0530

*There's arccos(x) in blue.*0546

*Now, that graph is neither mirror symmetric across the y-axis nor is it rotationally symmetric around the origin.*0552

*So, it's not odd or even, which is a little bit surprising, because even though if you remember cos(θ) is even.*0565

*It turns out that arccos(x) is not odd or even, even though cos(θ) was an even function.*0590

*Finally, arctan, let me start out by drawing the tangent graph, or at least the piece of it that we're going to snip off.*0597

*It kind of looks like the graph of y=x*^{3}, but it's not the same as the graph of y=x^{3}.0609

*One big difference is that tan(x) has asymptotes at π/2 and -π/2, and of course y=x*^{3} has no asymptotes at all.0617

*What I just graphed here is tan(θ) and then in blue I'll graph arctan(θ).*0627

*I'm flipping it across the line y=x.*0638

*It also has asymptotes neither horizontal asymptotes.*0642

*That blue graph is arctan(x), and if you look at that, that is rotationally symmetric around the origin.*0648

*So that's also an odd function.*0666

*This problem is really kind of testing whether you know the graphs of arcsin, arccos, and arctan look like.*0675

*If you don't remember those, then you go back to sine, cosine and tangent, and you snip off the important pieces of those graphs, and you flip them around y=x to get the graphs of arcsin, arccos, and arctan.*0685

*Those are the graphs that I have in blue here, arcsin, arccos, and arctan.*0700

*The other thing that this problem is really testing is whether you remember the graphical characterizations of odd and even functions.*0705

*If you know that odd functions have rotational symmetry around the origin, even functions have mirror symmetry across the y-axis, it's easy to check these graphs to just look at them and see whether they have the right kind of symmetry.*0717

*Of course, what you find out is that arcsin(x) has rotational symmetry, arccos(x) doesn't have either one, arctan(x) also has rotational symmetry.*0732

*For our third example here, we're trying to find arccos of the following list of common values.*0746

*Again, it's useful to start with a unit circle here.*0753

*Once you start with a unit circle, remember that with arccos, you're looking for values between 0 and π.*0766

*Arccos is always between 0 and π.*0779

*We're looking for angles between 0 and π that have cosines equal to this list of values.*0786

*I'll make a little chart here.*0796

*-1, negative root 3 over 2, negative root 2 over 2, -1/2, 0, 1/2, root 2 over 2, root 3 over 2 and 1.*0805

*Remember, cosine is the x-value.*0823

*I'm going to draw each one of these values as an x-value, as an x-coordinate, and then I'll see what angle has that particular cosine.*0826

*So, -1, the x-coordinate of -1 is over here, clearly that's π, that's the angle π.*0834

*Negative root 3 over 2, I'll draw that as the x-coordinate, and that angle is 5π/6.*0845

*Negative root 2 over 2, I'll draw that as the x-coordinate, I know that's a 45-degree angle, so that's 3π/4, that's the arccos of negative root 2 over 2*0863

*-1/2, if we draw that as the x-coordinate, that's a 30-60-90 triangle, that's 2π/3.*0879

*What angle has cos(0)?*0894

*That means what angle has x-coordinate 0, that's π/2.*0896

*What angle has cos(1/2)? *0903

*Again, a 30-60-90 triangle, that must be π/3.*0909

*Root 2 over 2, that's a 45-degree angle, so that's π/4.*0917

*Root 3 over 2, that's a 30-60-90 angle again, that's π/6.*0927

*Finally, what angle has x-coordinate 1?*0939

*That's just 0.*0943

*The trick here is remembering your common values of cosine on the unit circle.*0947

*I know all the common values of cosine on the unit circle very well because I remember my 30-60-90 triangles, and I remember my 45-45-90 triangles.*0952

*I know which ones are positive and which ones are negative.*0965

*Finally, I remember that arccos is always between 0 and π.*0968

*So, I'm looking for angles between 0 and π that have these cosines.*0973

*These are the angles that have the right cosines and are in the right range.*0981

*We'll try some more examples of these later.*0985

*Ok we are here to try more examples of the addition and subtraction formulas.*0000

*This time we are going to use the formula for cos(A-B) and the co function identities to derive the other three addition and subtraction formulas.*0007

*If you remember back in the previous set of examples, we proved the formula for cos(A-B).*0016

*We did it without using the other addition and subtraction formulas.*0024

*We are not getting trapped in any circular loops of logic.*0028

*We really did prove the cos(A-B) from scratch.*0031

*And now that we got that available to us, we are going to start with that formula and we are going to try to derive all the others.*0035

*Hopefully, it would be easier than the original proof of the cos(A-B) formula.*0044

*Let us remember what that formula is because we are allowed to use it now.*0050

*The cos(A+B) is equal to cos A cos B + sin A sin B, we are allowed use that.*0055

*I want to derive the other three formulas.*0070

*I'm going to start with cos(A+B) and I'm going to write that as cos((A-(-B)).*0073

*I'm gong to write in addition, in terms of a subtraction.*0079

*The point of that is now I can use my subtraction formula.*0091

*So, this is cos A. I'm just going to invoke this formula above except whenever I see a B, I will change it to (-B).*0095

*I have cos A cos(-B) + sin A sin(-B).*0107

*Remember, cosine is not even a function.*0125

*That means cos(-x) is the same as cos(x).*0128

*Sine is an odd function, sin(-x) is equal to -sin(x), I got cos A and cos(-B), but cos(-B) is the same as cos B.*0137

*Now sin A and sin -B, sine is odd so sin(-B) is -sin B.*0156

*But look, now I got cos(A+B) is equal to cos A cos B - sin A sin B, that is the formula for cos(A+B).*0169

*I was able to do that much more quickly than we were able to prove the original formula for cos(A-B).*0180

*Let us see how that works for sin(A+B).*0188

*Now, I'm going to have to bring in the co function identities, let me remind you what those are.*0194

*Those say that cos(pi/2)-x is the same as sin(X), sin(pi/2)-x is equal to cos(x).*0198

*Somehow we are going to use those to derive the sin formulas from the sin formulas.*0215

*The way we do that is I have sin(A+B), I'm going to use the first co function identity and write that as cos((pi/2-(A+B)).*0221

*That is by the first co function identity.*0238

*Now, that is cos(pi/2-A). I am going to group those two terms together and then -B, because it was minus the quantity of A+B.*0241

*I'm going to use my cos subtraction formula, this one that we started with.*0255

*cos, I am going to substitute n instead of A-B, I have (pi-2)-(A-B).*0263

*So, this is cos of the first term, cos(pi/2-A), cos of the second term is B + sin of the first term x sin of the second term.*0273

*But now, cos(pi/2-A) again using the co function identity is just sin A, sin A cos B.*0293

*Now, sin(pi/2-A) using the co function identity at the second co function identity is cos A x B.*0305

*Now we got the addition formula for sin, because we started with sin(A+B) and we reduced it down to sin A cos B + cos A sin B.*0318

*That is where the addition formula for sin comes in.*0330

*Finally, sin(A-B) we are going to do the same trick that we did for cos(A+B).*0334

*We will write this as sin, instead of writing it as a subtraction, we will think of it as adding a negative.*0344

*This is sin A - (-), I'm sorry A + (-B).*0355

*The point of that is that we can then invoke the sin formula that we just proved, we got sin(something +something).*0365

*According to the sin formula that we just proved, it is the sin of the first one x the cos of the second one which is (-B) + cos of the first one x sin of second one which is (-B).*0372

*Now again, we are going to use the odd and even properties.*0390

*This is sin A, cos (-B), cos does not even function so that is cos B + cos A. *0394

*Actually I should have said plus because look we have sin (-B) and sin (-x) is -sin(x). This is -cos A sin B.*0406

*But now, we started with sin(A-B) and we derived sin A cos B - cos A sin B.*0418

*That is exactly the subtraction formula for sin.*0429

*In each one of those identities, we did not use anything external.*0434

*We just started with the identity for cos(A-B) and then we made some clever substitutions to figure out cos(A+B), sin(A+B), sin(A-B).*0438

*Just making little substitutions into the one formula that we started with to get the formulas for the other three expressions.*0454

*Remember, it was a lot of work to prove that original formula for cos(A-B).*0464

*But once we have that one we can sort of milk it over and over again to get the other three formulas.*0469

*Extra example 2, which is to convert 75 degrees and -15 degrees to radians and we will use the addition and subtraction formulas to find the cos and sin.*0000

*So, 75 degrees, we will start out with that one.*0014

*Remember, the conversion formula is pi/180, that simplifies down to 5pi/12, that is not a common value. *0020

*I do not know the cos and sin of 5pi/12.*0034

*I‘m going to write that as a combination of two angles that I do know, that is (pi/4 + pi/6).*0040

*That is because pi/4 is 3pi/12 and pi/6 is 2pi/12 and you put them together and you got 5pi/12.*0044

*The key point of that is the pi/4 and pi/6 are common values.*0060

*I know the sin and cos(pi/4 and pi/6), I have memorized them and hopefully, you have memorized them as well.*0066

*Once I use my addition and subtraction formulas I can figure out the sin and cos of 5pi/12.*0074

*Let me remind you the addition and subtraction formulas we will be using.*0081

*Here, we are going to find cos(A+B) which is(cos A cos B) – (sin A sin B).*0085

*I’m going to go ahead and write the formula for sin(A+B).*0103

*It is equal to (sin A cos B) +(cos A sin B).*0110

*What is invoked those here we are trying to find the cos(5pi/12) which is the same as the cos(pi/4 + pi/6).*0122

*I’m going to use the cos addition formula cos(pi/4) cos(pi/6) – sin(pi/4) sin(pi/6).*0138

*All of those are common values, I have got those all computed to memory.*0159

*This will be very quick to finish from here.*0163

*This is cos(pi/4) I remember is square root of 2/2, cos(pi/6), I remember is square root of 3/2 – sin(pi/4) is root 2/2, sin(pi/6) is just ½.*0166

*If I put those together the common denominator there is 4, (root 2 x root 3 is 6) – (root 2 x 1).*0182

*That is my cos(5pi/12) which is the same as the cos of 75 degrees.*0195

*Let us find the sin now, sin(5pi/12) is equal to sin(pi/4) + pi/6, which by the addition formula for sin is sin of the first one pi/4, (cos of the second one pi/6) + (cos of the first one x sin of the second one).*0202

*And now again those are common values, I remember them all.*0234

*Sin(pi/4) is root 2/2, cos(pi/6) is root 3/2, cos(pi/4) is also root 2/2, sin(pi/6) is just ½.*0237

*I put these together over common denominator 4 and I get (root 6 + root 2/4).*0257

*What to remember those two values because we are actually going to use them in the next part.*0271

*The next part is to figure out -15 degrees, we want to start out by converting that to radians.*0277

*-15 degrees we multiply that by our conversion factor pi/180, that gives us 15/180, simplifies down to 112, so we get –pi/12 radians.*0283

*Now, there are two ways we could proceed from here. We can write –pi/12 as (pi/6 – pi/4) and that is because pi/6 is 2pi/12, pi/4 is 3pi/12. You subtract them and you will get –pi/12.*0307

*We could do at that way or we can write –pi/12 as (5pi/12 – pi/2- 6pi/12).*0334

*I want to do it that way because I want to practice that plus I think the sin and cos of pi/2 are a little bit easier to remember, I want to practice that method.*0341

*Let me write the formulas for sin and cos because we are going to be using those.*0351

*Cos(A-B) is(cos A cos B) + (sin A sin B) and sin(A-B) is equal to (sin A cos B) – (cos A sin B).*0356

*I’m going to be using those subtraction formulas the cos(-pi/12).*0390

*If we use the second version that is cos(5pi/12-pi/2).*0399

*And now by the subtraction formula that is (cos(5pi/12) x cos(pi/2)) +(sin(5pi/12) x sin(pi/2)).*0408

*Now look at this, the cos(pi/2), remember that is cos 90 degrees, the x coordinate of 90 degrees angle that is 0.*0431

*That whole term drops out, sin(pi/2) is 1.*0441

*This whole thing simplifies down to sin(5pi/12) and we worked that out on the previous page.*0447

*The sin(pi/12) we did this work before, that was the (square root of 2) + (square root of 6)/4.*0460

*We are invoking previous work there, this would be something that I would not have remembered but because I just work that out in the previous problem I remember the answer now.*0470

*We are going to try to figure out the sin(-pi/12) the same way. *0485

*So, sin(-pi/12) is the same as sin(5pi/12) – pi/2, because it is (5pi/12 – 6pi/12).*0489

*Using the subtraction formula for sin, that is sin of the first one, which is sin(5pi/12) x cos of the second one (pi/2) – cos of the first one (5pi/12) x sin of the second one (pi/2).*0506

*The point of that is that the pi/2 values are very easy.*0526

*I know that the cos, just like before is 0 and the sin is 1.*0529

*This whole thing simplifies down to –cos(5pi/12).*0539

*Again, I worked out the cos(5pi/12) on the previous page, the cos(5pi/12) in the previous page was (root 6 – root 2)/4.*0550

*Or we want the negative of that this time, I will just switch those around and I will get root 2 – root 6 divided by 4.*0565

*The key to doing that problem, well first of all, converting those angle to radians, that is a simple conversion factor of pi/180, that part was fairly easy.*0583

*Once we figured out how to convert those angle to radians, it was a matter of writing them as either sums or differences using addition or subtraction of common values, pi/6, pi/4, pi/3.*0593

*Things that you already know the sin and cos of by heart.*0608

*75 degrees 5pi/12, the key there was to figure out that was pi/4 + pi/6 and then know that you remember the common values, the sin and cos(pi/4) and pi/6.*0614

*So you can work out the sin and cos of 5pi/12, the -15 degrees converted into –pi/12 and then we can write that as pi/6 – pi/4, that would be one way to do it.*0630

*Or since we already know the sin and cos of 5pi/12, it is a little bit easier to write it as 5pi/12 – pi/2.*0646

*Then we can use the addition and subtraction formulas which because of the pi/2, essentially reduced it down to knowing the sin and cos of 5pi/12, which we figured out on the previous page.*0654

*So, that is how you use the addition and subtraction formulas to find the values of sin and cos of other angles when you already know the sin and cos of the common values.*0668

*That is the end of the lecture on addition and subtraction formulas.*0681

*We will use these formulas later on to find the double and half angle formulas that is coming next in the trigonometry lectures on www.educator.com.*0684

*Hi this is Will Murray for educator.com and we're talking about the addition and subtraction formulas for the sine and cosine functions.*0000

*The basic formulas are all listed here.*0009

*We have a formula for cos(a-b), cos(a+b), sin(a-b), and sin(a+b).*0012

*Unfortunately, you really need to memorize these formulas but it is not quite as bad as it looks.*0021

*In fact, if you can just remember one each for the cosine and the sine, maybe if you can remember cos(a+b) and sin(a+b), we'll learn later on in the lecture that you can work out the other formulas just by making the right substitution into those starter formulas.*0026

*If you remember what cos(a+b) is then you can substitute in -b in the place of b, and you can work out what the cos(a-b) is.*0047

*The same for sin(a+b), if you can remember the formula for sin(a+b), you can substitute in -b for b and find out the formula for sin(a-b).*0058

*You do have to remember a couple of formulas to get started, but after that you can work out the other formulas by some basic substitutions.*0070

*It's not as bad as it might sound in terms of memorization here.*0078

*There's a couple of cofunction identities that we're going to be using as we prove and apply the addition and subtraction formulas.*0082

*It's good to remember that cos(π/2 - x) is the same as sin(x).*0092

*The similar identity sin(π/2 - x) is equal to cos(x).*0098

*Those aren't too hard to remember if you kind of keep a graphical picture in your head.*0104

*Let me show you how those work out.*0111

*Let me draw an angle x here.*0113

*Then the cosine and sine, remember the x and y coordinates of that angle.*0117

*That's the cosine, and that's the sine.*0125

*And π/2 - x, well π/2 - x, remember of course is a 90-degree angle, so π/2 - x, we just go back x from π/2.*0128

*There's x and then that right there is π/2 - x.*0139

*If we write down the cosine and sine of π/2 - x, this is the same angle except we just switch the x and y coordinates.*0147

*When you go from x to π/2 - x, you're just switching the sine and cosine.*0157

*That's kind of how I remember that cos(π/2 - x)=sin(x) and sin(π/2 - x)=cos(x).*0164

*We'll be using those cofunction identities, both to prove the addition and subtraction formulas later on, and also to figure out the sines and cosines of new angles as a quicker in using these addition and subtraction formulas.*0172

*Let's get some examples here.*0189

*The first example is to derive the formula for cos(a-b) without using the other addition and subtraction formulas.*0191

*There's a key phrase here, it says, without using the other formulas.*0199

*The point of that is that once you figure out that one of these formulas, you can figure out a lot of the other formulas from the first one.*0204

*If you can figure out one formula, you need one formula to get started because otherwise you kind of get in a circular logic clue.*0213

*You need one of these formulas to get started and we'll have to go a bit of work to prove that.*0221

*Figuring out the other formulas from the first one turns out not to be so difficult.*0227

*What we'll do is we'll work out the formula for cos(a-b).*0233

*Then in our later example, we'll show how you can work out all the others just from knowing the cos(a-b).*0238

*This is a bit of a trick, it's probably not something that you would easily think about.*0246

*It really takes a little bit of ingenuity to prove this.*0250

*We'll start with a unit circle.*0254

*There's my unit circle.*0269

*I'm going to draw an angle a and an angle b.*0272

*I'm going to draw an angle a over here, so there's a, this big arc here.*0274

*I'll draw a b a little bit smaller, so there's b.*0283

*Then (a-b) is the difference between them.*0293

*This arc between them is going to be (a-b).*0296

*That's (a-b) in there.*0299

*Now, I want to write down the coordinates of each of those points.*0302

*The coordinates there, I'll write them in blue, are cos(a), the x-coordinate, and cos(b), the y-coordinate.*0307

*That's the coordinates of endpoint of angle a.*0318

*In red, I'm going to write down the coordinates of the endpoint of angle b, cos(b), sin(b).*0324

*Now, I'm going to connect those two points up with a straight line.*0338

*I want to figure out what the distance of that line is, and I'm going to use the Pythagorean formula.*0345

*Remember, the distance formula that comes from the Pythagorean formula is you look at the differences in the x-coordinates.*0350

*So, (x*_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}, you add those together and you take the square root of the whole thing.0365

*That's the distance formula.*0379

*Here, the x*_{2} and the x_{1} are the cosines, so my distance is cos(b)-cos(a).0384

*Actually, I think I'm going to write that the other way around, this cos(a)-cos(b).*0402

*It doesn't matter which way I write it because it's going to be squared anyway.*0411

*Plus [sin(a)-sin(b)]*^{2}, then I'll have to take the square root of the whole thing.0416

*To get rid of the square root, I'm going to square both sides.*0429

*I get d*^{2}=(cos(a)-cos(b))^{2}+(sin(a)-sin(b))^{2}.0435

*That's one way of calculating that distance.*0452

*Now, I'm going to do something a little different.*0455

*I'm going to take this line segment d and I'm going to move it over, move it around the circle so that it starts down here at the point (1,0).*0458

*There's that line segment again.*0475

*Remember, the line segment was cutting off an arc of the circle exactly equal to (a-b), exactly equal to an angle of the size (a-b), which means that that point right there has coordinates (cos(a-b),sin(a-b)).*0481

*That point has coordinates (cos(a-b),sin(a-b)).*0513

*Now, I'm going to apply the distance formula, again, to the new line segment in the new place.*0519

*That says, again, the change in the x coordinates plus the change in the y coordinates, square each one of those and add them up and take the square root.*0525

*So, d is equal to change in x coordinates, that's cos(a-b).*0535

*Now, the old x-coordinate is just 1 because I'm looking at the point (1,0).*0543

*That quantity squared plus the change in y coordinates, sin(a-b) minus, the old y-coordinate is 0, squared.*0547

*Then I take the square root of the whole thing.*0562

*I'm going to square both sides, d*^{2}=(cos(a)-1)^{2}+(sin^{2}(a-b)).0566

*What I'm going to do is look at these two different expressions here for d*^{2}.0590

*Well, they're both describing the same d*^{2}, they must be equal to each other.0597

*That was kind of the geometric insight to figure out to get me an algebraic equation setting a bunch of things equal to each other.*0608

*From here on, it's just algebra, so we're going to set these two equations equal to each other.*0616

*The first one is (cos(a)-cos(b))*^{2}+(sin(a)-sin(b))^{2} is equal to the second one, (cos(a-b)-1)^{2}+(sin^{2}(a-b)).0622

*Now, I'm just going to manipulate this expression expanded out, cancel few things and it should give us the identity that we want.*0652

*Remember, the square formula (a-b)*^{2}=a^{2}-2ab+b^{2}.0659

*We're going to be using that a lot because we have a lot of squares of differences.*0671

*On the first term we have cos*^{2}(a)-2cos(a)cos(b)+cos^{2}(b)+sin^{2}(a)-2sin(a)sin(b)+sin^{2}(b)=cos^{2}(a-b)-2×1×cos(a-b)+sin^{2}(a-b).0679

*Now, there's a lot of nice ways to invoke the Pythagorean identity here.*0727

*If you look at this term, and this term, cos*^{2}(a) and sin^{2}(a), that gives me 1-2cos(a)cos(b).0735

*Now I have a cos*^{2}(b) and a sin^{2}(b), so that's another 1-2sin(a)sin(b), is equal to, now look, cos^{2}(a-b) and sin^{2}(a-b), that's another 1.0749

*It looks like I forgot one term on the line above when I was squaring out cos(a-b)-1*^{2}, I got cos^{2}(a-b)-2cos(a-b), then there should be +1^{2}, there's another 1 in there.0772

*There's another 1 in there, -2cos(a-b).*0793

*That's it because we already took care of the sin*^{2}(a-b) that got absorbed with the cos(a-b).0800

*There's a lot of terms that will cancel now.*0807

*The 1s will cancel, 1, 1, 1 and 1, those will cancel.*0809

*We're left with -2, I'll factor that out, cos(a)×cos(b)+sin(a), because I factored out the -2, sin(b), is equal to -2cos(a-b).*0814

*Now, if we cancel the -2s, look what we have.*0838

*We have exactly cos(a)×cos(b)+sin(a)×sin(b)=cos(a-b).*0840

*That's the formula for cos(a-b).*0860

*That was really quite tricky.*0864

*The key element to that is that we did not use the other addition and subtraction formulas.*0867

*We really derived this from scratch, which means that we can use this as our starting point.*0873

*Later on, we'll derive the other addition and subtraction formulas but we'll be able to use this one to get started.*0878

*The others will be a lot easier.*0884

*This one was trickier because we really had to later on geometric ideas from scratch.*0887

*What we did was we graphed this angle a and angle b.*0892

*We connected them up with this line segment d, and we found the length of that line segment using the Pythagorean distance formula.*0897

*Then we did this very clever idea of translating and moving that line segment d over, so that it had a base of one endpoint at (1,0).*0905

*We found another expression for the length of that line segment or that distance, also using the Pythagorean distance formula but starting and ending at different places.*0916

*We get these two expressions for the length of that line segment d, and then we set them equal to each other in this line.*0929

*Then we got this sort of big algebraic and trigonometric mess, but there was no more real geometric insight after that.*0937

*It was just a matter of sort of expanding out algebraically using the Pythagorean identity to cancel some things that kind of collapse together, sin*^{2}+cos^{2}=1.0944

*It all reduced down into the formula for cos(a-b).*0957

*Now, let's try applying the addition and subtraction formulas to actually find the cosines and sines of some values that we wouldn't have been able to do without these formulas.*0964

*In particular, we're going to find the cosine and sine of π/12 radians and 105 degrees.*0975

*Let's start out with cosine of π/12 radians.*0982

*Cos(π/12), that's not one of the common values.*0987

*I don't have that memorized, instead I'm going to write π/12 as a combination of angles that I do have the common values memorized for.*0992

*Here's the trick, remember π/12=π/4 - π/6, that's because π/4 is 3π/12 and π6 is 2π/12.*1003

*You subtract them, and you get π/12.*1020

*The reason I do it like that is that I know the sines and cosines for π/4 and π/6.*1022

*I can use my subtraction formulas to figure out what the cosine and sine of π/12 are in terms of π/4 and π/6.*1030

*I'm going to use my subtraction formula cos(a-b)=cos(a)×cos(b)+sin(a)×sin(b).*1042

*Here, the (a-b) is π/12, so a and b are π/4 and π/6.*1060

*This is cos(π/4-π/6), which is cos(π/4)×cos(π/6)+sin(π/4)×sin(π/6).*1066

*Now, π/4, π/6, those are common angles that I have those sines and cosines absolutely memorized so I can just come up with those very quickly.*1095

*The cos(π/4) is square root of 2 over 2.*1105

*The cos(π/6) is square root of 3 over 2.*1109

*The sin(π/4) is root 2 over 2.*1113

*The sin(π/6) is 1/2.*1115

*Those are values that I have memorized, you should have them memorized too.*1117

*Now, we simply combine these, root 2 times root 3 is root 6.*1123

*I see I'm going to have a common denominator of 4 here, and root 2 times 1 is just root 2.*1130

*That gives me the cos(π/12) as root 2 plus root 6 over 4.*1139

*I'm going to work out sin(π/12) very much the same way, it's the sine of (π/4) - (π/6).*1145

*I remember my formula for the sin(a-b), it's sin(a)×cos(b)-cos(a)×sin(b).*1160

*I'll just plug that in as sin(π/4), cos(b) is π/6, minus cos(π/4)×sin(π/6).*1177

*So, sin(π/4) is root 2 over 2, cos(π/6) is root 3 over 2, minus cos(π/4) is root 2 over 2, and sin(π/6) is 1/2.*1196

*Again, I have a common denominator of 4, and I get root 2 times root 3 is root 6, minus root 2.*1212

*What we did there was we just recognized that (π/12) is (π/4)-(π/6), and those are both common values that I know the sine and cosine of.*1227

*I can invoke my cosine and sine formulas to figure out what the cosine and sine are of (π/12).*1238

*Now, let's do the same thing with a 105 degrees.*1246

*We'll do everything in terms of degrees now.*1250

*I know that 105, well, to break that up to some common values that I recognize, that's 45+60.*1254

*I'm going to be using my addition formulas now.*1264

*I'll write those down to review them.*1267

*cos(a+b)=cos(a)×cos(b)-sin(a)×sin(b), and when I'm at it, I'll remember that the sin(a+b)=sin(a)×cos(b)+cos(a)×sin(b).*1269

*The cos(105), that's the same as cos(45+60).*1300

*Using the formulas with a as 45 and b as 60, I get cos(45)×cos(60)-sin(45)×sin(60).*1310

*Again, 45 and 60 are both common values, I've got the sines and cosines absolutely committed to memory, and hopefully you do too by the time you've gotten this far in trigonometry.*1328

*Cos(45) is square root of 2 over 2, cos(60) is 1/2, sin(45) is square root of 2 over 2, and the sin(60) is root 3 over 2.*1338

*I'll put those together.*1353

*Common denominator is 4, and I get square root of 2 minus the square root of 6, as my cos(105).*1355

*Sin(105) works very much the same way.*1366

*We'll write that as sin(45+60), which is sin(45)×cos(60)+cos(45)×sin(60).*1371

*Now, I'll just plug in the common values that I have committed to memory, root 2 over 2, cos(60) is 1/2, plus cos(45) is root 2 over 2, and sin(60) is root 3 over 2.*1391

*Common denominator there is 4.*1408

*This is root 2 over 2 plus root 6 over 4.*1413

*That was a matter of recognizing that 105 degrees.*1422

*It's not a common value itself but we can get it from the common values as 45+60.*1426

*Those both are common values, so I know the sines and cosines, so I can figure out what the sine and cos of 105 is, by using my addition and subtraction formulas.*1434

*I'll mention one more thing there which is that we could write 105.*1443

*If we convert that into radians, that's 7π/12 radians.*1450

*Remember, the way to convert back and forth is you just multiply by π/180.*1455

*Then, 7π/12, well that's the same as 6π/12, otherwise known as π/2 + 1π/12.*1462

*We figured out what the sine and cosine of π/12 were on the previous page.*1473

*Once you know the sine and cosine of π/12, you could work out the sine and cosine of 7π/12 by doing an addition formula on π/2 + π/12.*1482

*This is really an alternate way we could have solved this problem.*1494

*Given that we had already figured out the sine and cosine of π/12.*1504

*Let's try another example there.*1509

*We're going to use the addition and subtraction formulas to prove a trigonometric identity sin(5x)+sin(x) over cos(5x)+cos(x) is equal to tan(3x).*1512

*It really may not be obvious how to start with something like this.*1525

*The trick here is to write 5x, to realize 5x is 3x+2x, and x itself is 3x-2x.*1530

*If we start with a=3x and b=2x, then 5x=a+b, and x itself is a-b.*1544

*That's what the connection between this identity and the addition and subtraction formulas is.*1561

*We're going to use the addition and subtraction formulas to prove this identity.*1566

*Let me write them down now and show how we can combine them in clever ways.*1571

*I'm going to write down the formula for sin(a-b).*1574

*Remember, that's sin(a)×cos(b)-cos(a)×sin(b).*1580

*Right underneath it, I'll write the formula for sin(a+b) which is the same formula sin(a)×cos(b)+cos(a)×sin(b).*1590

*Now, I'm going to do something clever here.*1608

*I'm going to add these two equations together.*1613

*The point of that is to make the cos(a)×sin(b) cancel.*1617

*If we add these equations together, on the left-hand side we get sin(a-b)+sin(a+b).*1623

*Remember, you're thinking in the back of your head, a is going to be 3x and b is going to be 2x.*1638

*On the left side, we really got now sin(x)+sin(5x), which is looking good because that's what we have in the identity.*1644

*On the right side, we get 2sin(a)×cos(b), and then the cos(a)×sin(b), they cancel.*1650

*That was the cleverness of adding these equations together.*1667

*We get 2sin(a)×cos(b).*1669

*If I plug in a=3x and b=2x, I will get sin(a-b) is just sin(x), plus sin(a+b) which is 5x.*1674

*On the right-hand side, I'll get 2sin(a) is 3x, cos(b) is x.*1692

*That seems kind of hopeful because that's something I can plug in to the left-hand side of my identity and see what happens with it.*1703

*Before we do that though, I'm going to try and work out a similar kind of formula with the addition and subtraction formulas for cosine.*1710

*Let me write those down.*1718

*Cos(a-b) is equal to cos(a)×cos(b) plus, cosine is the one that switches the positive and the negative, plus sin(a)×sin(b).*1720

*I wanted to figure out cos(a+b).*1743

*It's just the same thing changing the positives and negatives, so cos(a)×cos(b)-sin(a)×sin(b).*1749

*I'm going to do the same thing here, I'm going to add them together in order to make them cancel nicely.*1760

*On the left-hand side, I get cos(a-b)+cos(a+b)=2cos(a)×cos(b).*1767

*That's it, because the sin(a) and sin(b) cancel with each other.*1785

*I'm going to plug in a=3x and b=2x, so I get cos(x) plus cos(a+b) is 5x, is equal to 2 cosine, a is 3x, and b is 2x.*1791

*Let's keep this in mind, I've got an expression for sin(x)+sin(5x), and I've got an expression for cos(x)+cos(5x).*1810

*I'm going to combine those and see if I can prove the identity.*1823

*I'll start with the left-hand side of the identity.*1832

*I'll see if I can transform it into the right-hand side.*1836

*The left-hand side is sin(5x)+sin(x) over cos(5x)+cos(x).*1841

*Now, by what we did on the previous page, I have an expression for sin(5x)+sin(x), that's sin(3x)×cos(2x).*1860

*That's by the work we did on the previous page.*1881

*Also on the previous page, cos(5x)+cos(x)=2cos(3x)×cos(2x).*1890

*That was also what we did on the previous page.*1901

*But now look at this, the cos(2x) is cancelled, and what we get is 2sin(3x) over 2cos(3x).*1906

*The 2s cancel as well and we get just tan(3x), which is equal to the right-hand side.*1920

*We finished proving it.*1929

*The trick there and it really was quite a bit of cleverness that might not be obvious the first time you try one of these problems, but you'll practice more and more and you'll get the hang of it, is to look at this 5x and x, and figure out how to use those in the context in the addition and subtraction formulas.*1933

*The trick is to let a=3x and b=2x, and the point of that is that (a-b), will then be x, and a+b will be 5x.*1951

*That gives us the expressions that we had in the identity here.*1967

*Once we see (a-b) and (a+b), it's worthwhile writing down the sine and the cosine each one of (a-b) and (a+b), and kind of looking at those formulas and kind of mixing and matching them, and finding something that gives us something that shows up in the identity.*1974

*Once we get that, we start with the left-hand side of the identity, we work it down until we get to the right-hand side of the identity.*1997

*We'll try some more examples of that later*2004

*Hi we are trying some examples of the double angle formulas for sin and cos.*0000

*We are going to try to find the sin and cos of 4pi/3.*0005

*We are going to use all three cos formulas and check that they agree.*0010

*We are also going to use our common values to find the sin and cos of 4pi/3 to check our answers.*0014

*Let me write down the double angle formulas that we are going to be using.*0021

*We are going to use sin 2X = 2 sin X cos X.*0025

*These are probably worth remembering, but if you do not remember them, you can work them out from the addition formula for sin and cos.*0034

*Cos(2X) is cos x *^{ 2 } - sin x^{ 2 }.0043

*Of course, the X here would have to be 2pi/3 because what we are really trying to find is the sin and cos of 4pi/3.*0051

*So, sin(4pi/3) according to our double angle formula is equal to 2 x sin(2pi/3) x cos(2pi/3).*0061

*Now, 2pi/3 that is a common value, I remember its sin and cos, its sin is root 3/2 and its cos is -1/2.*0078

*It is negative because 2pi/3 is in the second quadrant, so its X coordinate is negative.*0089

*Remember, cos is the X coordinate and this simplifies down to cancel and we will get –root 3/2.*0094

*cos(4pi/3) is cos(2pi/3)*^{2} - sin(2pi/3)^{2}.0106

*So, plug those common values in the cos(2pi/3) is -1/2 and sin(2pi/3) is positive root 3/2.*0122

*I will get ¼ negative goes away because they got squared – root 3 squared is ¾ and so I get -1/2.*0137

*That was the first of the three formulas for cos(2x).*0151

*Let me remind you what the other two formulas are.*0154

*cos(2x) is equal to 2 cos X*^{ 2 } - 1 and the other version we have of that formula is 1 – 2 sin X^{2}.0156

*These are all different formulas for cos(2x) and we will try each one now.*0177

*The first one there is cos(4pi/3) is equal to 2 cos(2pi/3) *^{2} - 1.0183

*The cos(2pi/3) is -1/2 because it is in the second quadrant -1, that is (2 x ¼ - 1), which is ½ - 1, which is – ½.*0201

*If we use the other version of the formula we will get cos(2pi/3) is (1- 2 sin(2pi/3)*^{2}) which is (1-2 x (root 3/2)^{2}).0225

*2pi/3 is our common value, I remember its sin, (1 – 2 x root 3 squared is 3), 4 in the denominator so 1 – 3 (1/2), and again we will get -1/2.*0243

*That is very reassuring because if you look at the three different formulas for cos, we got the same answer for all three of them.*0262

*That was the first point we wanted to check, but now let us check using our common values, there is 0, pi/2, 3pi/2, 2pi.*0272

*Now, 4pi/3 is bigger than pi, it is down here.*0294

*4pi/3 is 2/3 around the unit circle to 2pi.*0300

*That is one of our common triangles, that is the 30, 60, 90 triangles.*0306

*I know that the length of the sides there are root 3/2 and 1/2, I can figure out the sin and cos from that.*0311

*I just have to figure what are the positive or negative.*0320

*Well, the cos(4pi/3) is negative because the X value is negative, so it is – ½.*0323

*The sin(4pi/3) is also negative because the Y value is also negative there, it is – root 3/2.*0334

*Those are the answer we get using the common values on the unit answer.*0346

*But if you look, that is also the answers we got using the double angle formula breaking 4pi/3 up into ((2 x (2pi/3)).*0351

*We got sin, was – root 3/2, and cos was – ½.*0363

*It indeed, in fact, agrees with the values that we got from the unit circle.*0367

*Finally we are going to use the double angle formulas to prove another trigonometric identity.*0000

*We are going to prove that sec(2x) is equal to (sec X*^{ 2 }) / (2 – sec X ^{ 2 }).0005

*We are going to start with the right side because it looks more complicated.*0013

*The right hand side and we will try to manipulate it to the left hand side.*0017

*Let me start with the right hand side, (sec X*^{2}) / (2 – sec X ^{ 2 }).0022

*Again, we do not know what to do with the trigonometric identity, it is often good to start with the more complicated side.*0035

*Secondly, convert everything to sin and cos.*0042

*Here I got a lot of sec, I am going to convert it to the definition of sec is (1/cos), this is (1/cos X *^{ 2 }).0046

*My denominator, I have 2 – 1/ (cos X *^{ 2 }).0055

*I see a lot of cos*^{ 2 } in the denominator.0064

*I think I’m going to try to clear that by multiplying top and bottom by cos X *^{ 2 }.0068

*On the top, I will just get 1, on the bottom I get 2 cos X *^{ 2 } - 1.0077

*But look at that, 2 cos X *^{ 2 } - 1. 0090

*That is one of the formulas that I remember for cos(2x), this is 1/cos(2x).*0094

*Now, let us remember by definition, one of our cos is exactly the same as sec.*0104

*So, this is sec(2x) and that is the left hand side of the identity that we are trying to prove.*0110

*We proved that we started with the right hand side, we derived the left hand side. *0120

*The key things to notice in there, the way it worked was, first of all the right hand side is a little more complicated, so we are going to work on that one.*0124

*When I see a bunch of sec, I try to convert it into sin and cos because I know how to manipulate sin and cos.*0131

*I got more formulas for them than for sec, tan, cosec, and cot.*0138

*I converted into sin and cos.*0144

*I see some cos in the denominator, I decided to multiply by cos X *^{ 2 } to clear away those denominators.0147

*I'm multiplying that thru and there is really some pattern recognition here knowing your identity formulas.*0155

*When I see that 2 cos X *^{ 2 } - 1, a little bell goes of in my head, “wait I have seen that somewhere before, oh yes that is equal to cos(2x)”.0162

*Now I got 1/cos(2x), that is by definition sec(2x) and so I converted into the left hand side.*0174

*That is how you can use the double angle identities to prove more complicated trigonometric identities.*0184

*That is the end of our lecture on double angle identities.*0192

*These are the trigonometry lectures for www.educator.com.*0196

*Hi, welcome back to the trigonometry lectures on educator.com.*0000

*Today, we're going to learn about the double angle formulas, so here they are.*0004

*The first one is sin(2x)=2sin(x)×cos(x).*0008

*You may think there's so many formulas to remember in trigonometry.*0014

*This one, if you have trouble remembering it, you can work it out from the addition formula.*0019

*You do have to remember something, but if you can remember the sin(a+b)=sin(a)×cos(b)+cos(a)×sin(b).*0024

*If you remember that one, then you don't really need to learn anything new here because you can work it out so quickly.*0041

*Just take a and b, both to be x in the addition formula.*0046

*If a is x and b is x, then what you get here is sin(2x)=sin(x)×cos(x)+cos(x)×sin(x).*0052

*What you get is just 2sin(x)×cos(x).*0068

*If you can remember the addition formula, the double angle formulas are really nothing new to remember here, same goes for the cos(2x) formula.*0074

*If you remember the addition formula for cosine, you might want to try just plugging in x for each of the a's and b's, and you'll see that what you get is exactly cos*^{2}(x)-sin^{2}(x).0082

*Now, there's two other ways that you often see this formula written as 2cos*^{2}(x)-1, and 1-2sin^{2}(x).0096

*Those might look different but actually you can figure them out very quickly, or check them very quickly, because 2cos*^{2}(x)-1 is 2cos^{2}(x) minus, now remember 1 is the same as sin^{2}(x)+cos^{2}(x).0106

*If you work with that a little bit, you have 2cos*^{2}-cos^{2}.0126

*That's just a single cos*^{2}(x)-sin^{2}(x), and so all of a sudden this goes back to the original formula for cos(2x).0132

*You could do this, you can check the second formula the exact same way, if you convert the 1 into sin*^{2}+cos^{2}, you'll see that it converts back into this original formula for cos(x).0143

*Even though it looks like there's 4 new formulas to remember here, really the basic sin(2x) and cos(2x), you can work both of those out from the additional formulas.*0158

*The other two formulas for cosine, you can just work them out if you remember the original formula for cosine and then the Pythagorean identity, sin*^{2}+cos^{2}=1, which certainly any trigonometry student is going to remember the Pythagorean identity.0172

*It's really not a lot of new memorization for these formulas.*0185

*The more interesting question here is how are you going to use them.*0190

*Let's try them out on some examples.*0194

*Our first example here is we're just going to get some practice using the sine and cosine of 2x formulas, the double angle formulas.*0197

*To find the sine and cosine of 2π/3.*0206

*Even though 2π/3 is a common value, hopefully you can work out the sine and cosine of 2π/3 without using the double angle formulas.*0211

*We're going to try them out using the double angle formulas, and then we'll just check that the answers we get agree with the values that we know coming from the common values.*0219

*We'll use that as a check, we won't use that at the beginning.*0232

*We're also going to use all three of the formulas for cosine and just check and make sure that they all work out, that they all agree with each other.*0235

*Let's start out by remembering those, actually, four formulas, sin(2x) is 2sin(x)×cos(x), and cos(2x) is cos*^{2}(x)-sin^{2}(x).0243

*Here, we're being asked to find the sine and cosine of 2π/3.*0264

*We're going to use x=π/3, that way 2x is 2π/3.*0267

*So, sin(2π/3), using x=π/3, it's 2sin(π/3)×cos(π/3).*0277

*I remember that the sin(π/3), that's a common value, so the sin(π/3) is root 3 over 2, cos(π/3) is 1/2, the 2 and that in 1/2 cancel, and what we'll get is root 3 over 2.*0294

*Now, let's try the cosine, cos(2π/3), is cos*^{2}(π/3)-sin^{2}(π/3) according to our formula, but we're going to check it out and see if it works.0314

*Now, cos(π/3) is 1/2, so (1/2)*^{2} minus the sin(π/3) is root 3 over 2, we'll square that out.0333

*1/2 squared is 1/4, root 3 over 2 squared is, root 3 squared is 3, 2 squared is 4, we get 1/4-3/4=-1/2.*0344

*Now, there were two other formulas for cos(2x), we want to check out each one of those, cos(2x)=2cos*^{2}(x)-1.0357

*It was also supposed to be equal to 1-2sin*^{2}(x).0370

*We're going to check out each one of those.*0376

*Cos(2π/3), using those other formulas, is equal to 2cos*^{2}(π/3)-1, which is 2.0379

*Now, cos(π/3), that's a common value, that's 1/2, (1/2)*^{2}-1, which is 2×1/4-1, which is 1/2-1, is -1/2.0393

*Let's use the other version, 1-2sin*^{2}(π/3), we'll use the last cosine formula there.0413

*That's 1-2, now, sin(π/3), I remember that's a common value, root 3 over 2.*0425

*We're going to square that out, that's 1-2 times, root 3 squared is 3, and 2*^{2} is a 4.0433

*That's 1-3/2=-1/2.*0443

*The first thing we noticed is that these 3 different formulas for cos(2x) they all gave us the answer -1/2.*0452

*They do check with each other, that's reassuring.*0460

*Now, let's work out the sine and cosine of 2π/3 just using the old-fashioned common values.*0463

*Let me draw my unit circle.*0471

*There's 0, π/2, π, and 3π/2.*0482

*2π/3 is 2/3 the way from 0 to π.*0490

*There it is right there.*0493

*That's my 30-60-90 triangle, so I know the values there are root 3 over 2 and 1/2.*0496

*I just have to figure out the sine and cosine, which ones are positive and which ones are negative.*0505

*I know that the cos(2π/3) because that's the x-value, and the x-value is negative, that's -1/2.*0512

*The sin(2π/3) is the y-value, which is positive, that's root 3 over 2.*0524

*We worked those out just looking at the unit circle and remembering the common values but that checks out with the values we got from the formulas there sin(2π/3) and each one of the formulas for cos(π/3).*0530

*What we're doing there is working out each one of the formulas for sin(2x) and cos(2x) with x=π/3.*0545

*That separates it out into expressions in terms of sines and cosines of π/3, which I remember so I just plug those in and I get the sine and cosine of 2π/3.*0556

*All the cosine formulas agree with each other and they all check with the values that I can find just by looking at the unit circle.*0567

*Our next example is to use the double angle formulas to prove a trigonometric identity.*0577

*It's not so obvious how to start with this one.*0584

*We're actually going to start with the right-hand side because it looks more complicated.*0588

*I'm going to start with the right-hand side and that's 2tan(x)/1+tan*^{2}(x).0592

*I'm evaluating the right-hand side, I'm going to work with it a bit and hopefully I can simplify it down to the left-hand side, but we'll see how it goes.*0605

*First thing, I'm going to do is to change everything into sines and cosines.*0615

*That's a good rule when you're not sure what to do with the trigonometric identity is to change everything into sines and cosines.*0619

*If you got a tangent or a secant, or a cosecant or a cotangent, convert it into sines or cosines.*0626

*It will probably make your life easier.*0631

*I'll write this as 2, tangent, remember is sin/cos, and 1+tan*^{2}, that's 1+sin^{2}(x)/cos^{2}(x).0633

*Now, I see a lot of cosines in denominators here, I think we're going to try to clear those out.*0651

*We multiply top and bottom by cos*^{2}(x) and see what happens with that.0655

*That's multiplying by 1, so that's safe.*0662

*On the top, I have 2sin(x), now I had a cos(x) in the denominator but I multiplied by cos*^{2}, that gives me cos(x) in the numerator.0665

*In the bottom, I have cos*^{2} times 1+sin^{2}(x) over cos^{2}, that gives me 1×cos^{2} is cos^{2}(x), plus the cos^{2}(x) cancels with the denominator sin^{2}(x).0677

*Now, look at this, the top is exactly 2sin(x)×cos(x).*0695

*I remember that, that's my formula for sin(2x).*0701

*Now the bottom, that's the Pythagorean identity, so that's just 1, cos*^{2}+sin^{2}(x) is 1.0709

*This converts into sin(2x), but that's equal to the left-hand side of what we were trying to prove.*0716

*We started with the right-hand side because it looked a little more complicated there.*0724

*I see a bunch of tangents, I am not so sure what to do with those, I convert them into sines and cosines.*0729

*I see a lot of cosines in the denominator, so I multiply top and bottom by cos*^{2}(x).0736

*Then I start noticing some formulas that I recognized, 2sin(x)×cos(x) is a double angle formula, and cos*^{2}(x)+sin^{2}(x), that's the Pythagorean identity.0745

*It reduces down into the right-hand side.*0754

*Let's try another example here, we're going to use the addition and subtraction formulas to derive a formula for tan(2x) in terms of tan(x).*0760

*Remember, we have formulas for sin(2x) and cos(2x), we're going to find a formula for tan(2x) just in terms of tan(x).*0771

*When we get that, we're going to check the formula on a common value π/6, because I know what the tangent of that is, and I know what the tan(2x) is, so we can check whether our formula works.*0780

*Let me start out with, tan(2x), don't know much about that except that the definition of tan(2x) is sin(2x)/cos(2x).*0793

*Now, I'm going to use, well, it's the addition and subtraction formulas but it's really the double angle formulas.*0811

*Of course, those come from the addition and subtraction formulas.*0817

*Now, sin(2x) is 2sin(x)×cos(x), that's the double angle formula for sine.*0820

*Of course, you find that out from the addition formula.*0829

*Cos(2x) is cos*^{2}(x)-sin^{2}(x), that was the first double angle formula for cosine.0832

*Now, it's not totally obvious how to proceed next, but I know that I'm trying to get everything in terms of tan(x).*0844

*Right now, I've got a bunch of cosines lying around, I'd like to move those down into the denominator.*0852

*The reason is because tangent is sin/cos, so I would like to be dividing by cosines.*0858

*What I'm going to do is I'm going to divide the top by cos*^{2}(x), and I'll divide the bottom by cos^{2}(x).0866

*We're dividing top and bottom by cos*^{2}(x), that's dividing by 1, so that's legitimate, we'll see what happens.0876

*Now, in the numerator, we get 2sin(x), we had a cos(x) before, we divided by cos*^{2}, we get 2sin(x)/cos(x).0882

*In the bottom, we're dividing everything by cos*^{2}(x), we get 1-sin^{2}(x)/cos^{2}(x).0896

*That's really nice because now we have sin/cos everywhere and that's tangent.*0908

*We are asked to find everything in terms of tan(x).*0913

*What we get here is 2sin/cos is tan(x) over 1-sin*^{2}(x)/cos^{2}(x) is tan^{2}(x).0918

*Our formula, our double angle formula for tangent is tan(2x)=2tan(x)/(1-tan*^{2}(x)).0931

*Now, I didn't list this at the beginning of the lecture as one of the main formulas that you really need to memorize.*0944

*It kind of depends on your trigonometry class, in some classes they will ask you to memorize this formula, this formula for tan(2x).*0949

*I don't think it's worth memorizing.*0957

*In my trigonometry classes, I don't require my students to memorize these formulas for tan(2x).*0960

*I do require them to memorize sin(2x) and cos(2x) and I figure they can work out the other ones from that.*0965

*You may have a teacher who requires you to memorize the formula for tan(2x).*0973

*If so, here it is, here is the formula that you want to remember.*0979

*Let's check that out on a value that I already know the tangent of, let's try x=π/6.*0984

*The tan(2π/6), according to this formula, would be 2×tan(π/6)/(1-tan*^{2}(π/6)).0994

*Now, π/6 is a common value, tan(π/6), I remember that, I've got that one memorized, it's root 3 over 3.*1011

*If you don't have that one memorized, it probably is a good one to memorize, but if you don't have it memorized, you can work it out as long as you remember sine and cosine of π/6.*1023

*You just divide them together and get the tan(π/6).*1034

*This is 2 times root 3 over 3, over 1 minus root 3 over 3 squared.*1037

*Let's do a little over that, that's 2 times root 3 over 3, over 1 minus root 3 over 3 squared, is 3, over 3 squared is 9.*1048

*That's 3/9 which is 1/3.*1063

*This is 2 root 3 over 3, divided by 2/3.*1067

*Remember how you divide fractions, you flip it and multiply, 3/2, that cancels off the 2 and the 3, this whole thing boils down to just a root 3 as tan(2π/6).*1076

*Of course, 2π/6 is just π/3.*1092

*π/3 is another common value that I know the tangent of.*1101

*tan(π/3), I remember, is root 3, that's a common value.*1105

*Again, if you don't remember that, remember the sine and cosine of π/3, divide them together and you'll get root 3.*1112

*Look at that, our answers agree.*1120

*That confirms our formula for tan(2x).*1122

*To recap the important parts of that problem, we have to figure out tan(2x).*1126

*We wrote it as sin/cos of 2x.*1132

*We expanded each one of those using the double angle formulas that we learned at the beginning of the lesson.*1135

*Then, I was trying to get this in terms of tan(2x).*1140

*I wanted to get some cosines in the denominator, that's why I divided top and bottom by cos*^{2}(x).1144

*That converted the thing into something in terms of tan(x).*1150

*Then we checked that out by plugging in x=π/6, that's something that I know the tangent of, worked through the formula, and we got an answer square root of 3.*1156

*That checks the common value that I also know tan(π/3) is square root of 3.*1170

*We'll try some more examples of that later.*1176

*Ok we are going to try some examples of the half angle formulas.*0000

*Now, we are going to find the sin and cos of pi/8 and we are going to check that our answer satisfies the pythagorean identity sin*^{2} + cos^{2} = 1.0005

*Let me remind you of the half angle formulas, we have sin(1/2x) is equal to + or – the square root of ½, 1 – cos X, we will be using that and cos(1/2x).*0015

*Same formula just for the plus, square root of 1/2 , 1 + cos X.*0033

*In this case, we want to find the sin and cos of pi/8, now that is not one of the common values that we need to memorize.*0045

*I do not remember the sin and cos of pi/8, what I do know is that pi/8 is ½ of pi/4.*0052

*And pi/4 is a common value, I know the sin and cos of pi/4, that is my starting point. *0065

*I’m going to take the X to be pi/4 and I’m going to plug into the half angle formulas.*0073

*Sin(pi/8) is equal to + or – the square root of ½, of 1 – cos(pi/4), because my X is pi/4.*0080

*Now, cos(pi/4) is a common value that I remember, I know that by heart it is square root of 2/2.*0099

*I’m going to put this over common denominator, so ½ is 2/2, (2 – root 2/2), (+ or – the square root of 2 – root 2/4), (+ or – the square root of 2 – root 2), the square root of 4 is just 2 there.*0111

*Now, I need to deal with this + or -, figure out whether it is positive or negative.*0135

*Let me draw a quick unit circle here, I know that pi/4 is over here, pi/8 is half of that so it is down here.*0140

*Definitely in the first quadrant there, both sin and cos will be going to be positive, remember sin and cos, and the x and y coordinates. *0157

*I know that I want to take the positive square root here in quadrant 1, sin(pi/8) must be positive.*0167

*So, sin(pi/8) then must be equal to the + square root of (2 – root 2/2).*0185

*Now let us figure out the cos(pi/8), I will do that on the new page here.*0201

*cos(pi/8) is equal to + or – the square root of 1/2 ((1 + cos(pi/4)), the X is pi/4 and then we are finding the cos(x/2) here.*0207

*This is + or – the square root of ½ (1 +… now cos(pi/4), we remember that very well, that is a common value, root 2/2).*0228

*Put this over the common denominator so we get (2 + root 2)/2, combine those fractions so we get (2 + root 2)/4.*0241

*Finally, we can split up the square root into numerator and denominator, (2 + root 2)/4.*0255

*Again, we already said that pi/8 is in quadrant 1, the first quadrant, so cos(pi/8) must be positive, that is the X value.*0264

*We have cos(pi/8) equal to the + square root of (2 + root 2).*0284

*I simplify slightly wrong here the square root of 4 of course is 2, I accidentally wrote it that down again as 4.*0300

*Cos(pi/8) is the + square root of (2 + root 2)/2.*0311

*Now we figured out the sin and cos of pi/8 using our half angle formulas.*0321

*The last thing the problem asks us to do is to check that those sin and cos verify the pythagorean identity sin*^{2} + cos^{2} = 1.0336

*Let us work those out, sin*^{2}(pi/8) + cos^{2}(pi/8).0337

*The sin was (root 2 – root 2)/2, we need to square that, the cos was the same thing as the facts (2 + root 2)/2.*0351

*If we square that out in the numerator region, we will get square root*^{2}, so they cancelled each other we get ((2 – root 2)/denominator 2^{2} is 4) + ((the numerator is (2 + root 2)/4).0367

*If we combined those the minus root 2 and plus root 2 cancel, so we get 4/4, of course simplifies down to 1.*0384

*That does check then that sin*^{2}(pi/8) + cos^{2}(pi/8) = 1, it checks the pythagorean identity.0395

*They key to that problem was really just remembering the half angle formula for sin and cos, sin(1/2x) and cos(1/2x).*0402

*To get pi/8 we just noticed that it is half of pi/4 and we plugged it into the formulas and we get this answer which has a + or – square root.*0414

*We got to remember that pi/8 is in quadrant 1, its sin and cos, its x and y values are both going to be positive, so we take the positive square root.*0424

*So, let us try one more example here. We have to prove the trigonometric identity involving some half angle formulas, tan (x) × tan (1/2x) = sec (x) – 1.*0000

*This one is a little tricky but it is going to be a lot easier if you remember an example we did in the earlier part of the lecture.*0013

*What we figured out the trigonometric for tan (1/2x), we figured out a half angle formula from the earlier example.*0020

*We proved that tan(1/2x) = 1 – cos(x) / sin(x), that was a little bit of work to prove.*0039

*Actually, that involves removing some + or – and some square roots, that was tricky.*0054

*But having done that work, it would be pretty easy to prove this trigonometric identity.*0059

*Let me start with the left hand side because that one looks a little more complicated and I will try to manipulate it into the right hand side.*0066

*The left hand side is tan(x) × tan (1/2x), now this tan(x), I’m not sure exactly what to with that so for a lot of better option. *0075

*I will convert that in to sin(x) / cos(x), that is the definition of tan, you can always convert tan in to sin / cos.*0088

*Tan(1/2x) that is not so obvious but I remember this previous example where we derived this identity 1 – cos (x) / sin(x).*0097

*I’m going to convert that in to 1- cos(x) / sin(x).*0108

*Now the sin(x) is cancelled and we are left with 1 – cos(x) / cos(x), if I put that up in to 1 / cos(x) – cos(x) / cos(x) *0116

*Then 1 – cos by definition is sec(x) and cos / cos is of course 1.*0135

*And look we transformed it in to the right hand side.*0142

*Again, as with most trigonometric identities it is not always obvious to proceed with them but a general rule of the thumb is try to attack the more complicated looking side first.*0149

*In this case, that was the left hand side, tan(1/2x) I remember my half angle identity for tan(1/2x) that we proved earlier on in the earlier part of the lecture.*0160

*I expand that out, I do not know what to do with the tan so I just convert it in to sin and cos, that is something you do if you can not think of anything else to do, convert everything to sin and cos.*0173

*Then it starts to cancel nicely, separates out, and suddenly converts in to the right hand side once I remember the definition of sec(x).*0185

*The general rule is there, go for the more complicated side, convert into sin and cos, invoke any half angle or double angle or that you already know.*0194

*That is the end of the half angle formulas as part of the trigonometry lectures on www.educator.com, thanks for watching.*0207

*Hi, these are the trigonometry lectures for educator.com and today we're going to talk about the half-angle formulas.*0000

*The main formulas that we're going to be using today, we have a formula for sin(1/2 x) and cos(1/2 x).*0007

*They're a little bit cumbersome, sin(1/2 x) is equal to plus or minus the square root of 1/2 of 1-cos(x).*0017

*Cos(1/2 x), same formula, except there's a plus in it.*0027

*They're a little bit cumbersome but we'll practice using them and you'll see that they're not so bad.*0033

*What makes them difficult is the plus or minus and the square root signs.*0037

*That's probably the confusing part.*0042

*Actually, what's inside the square root sign isn't bad at all, the 1-cos(x), 1+cos(x) aren't too bad.*0044

*Let's try them out right away with some examples.*0050

*Our first example is to find the sine and cosine of 15 degrees and then we'll check that our answers satisfy the Pythagorean identity sin*^{2}+cos^{2}=1.0054

*The first thing to notice here is 15 degrees is not a common value, it's not one that where we've memorized the sine and cosine.*0065

*We'll have to use the half-angle formulas here.*0076

*We'll start with 15 is 1/2 of 30, so we're going to use the sine formula.*0079

*Remember, sin(1/2 x) is equal to plus or minus the square root of 1/2(1-cos(x)).*0089

*The x in question here is 30, we're trying to find the sin(15), sin(15) is equal to plus or minus the square root of 1/2(1-cos(30)).*0102

*That's plus or minus the square root of 1/2.*0120

*Now, 30 degrees is a common value, that's π/6, and I've got all the sines and cosines of the common values memorized.*0125

*Cos(π/6), cos(30) is root 3/2.*0134

*I'm just going to do a little bit of algebra with this expression here, plus or minus the square root of 1/2.*0140

*I'm going to put 1 in root 3 over 2 over a common denominator, that will be 2-3/2.*0148

*I'm just writing 1 there as 2/2.*0154

*If I combine these, I get 2 minus root 3 over 4.*0158

*I still have this plus or minus which is not very good because I want to give a single answer, I don't want to give two different answers.*0168

*Let's remember where 15 degrees is.*0175

*Here's 0 degrees, and here's 90 degrees, 15 degrees is way over here right in the first quadrant.*0180

*Since sine and cosine are the x and y values, actually sine is the y-value, and cosine is the x-value.*0188

*They're both positive in the first quadrant.*0197

*15 degrees is in quadrant 1.*0201

*The sine is its y-value, it's positive, sin(15) is greater than 0, it's positive.*0212

*Our answer then, is sin(15) must be the positive square root there, it's the square root of 2 minus root 3 over 4.*0220

*Now, if you've been paying really close attention to the educator.com trigonometry lectures, you'll know that we've actually solved this problem before.*0236

*15 degrees if you convert it to radians, is actually π/12, and we worked out the sin(π/12) before not using the half-angle identities but using the addition and subtraction formulas.*0246

*We worked out the sine and cosine of π/12 by realizing it as π/4-π/6.*0264

*When we worked it out, sin(π/12) using the subtraction formulas, we got the answer square root of 6 minus the square root of 2 over 4.*0273

*There's a little bit of a worry here, because it's seems like we did the same problem using two different sets of formula and we got two quite different looking answers.*0290

*We got the square root of 6 minus the square root of 2 over 4 last time we did it.*0302

*This time we have the square root of 2 minus the square root of 3 over 4.*0306

*Actually, that can be simplified a little bit into, if we just take the square root of the top part, then the square root of 4 is just 2.*0314

*We have this two different answers here, or at least they seem to be different.*0324

*Let me show you that these two answers can actually be reconciled.*0329

*How do these answers agree?*0336

*Let me start with the old answer, the one we did in a previous lecture on educator.com.*0346

*Our old answer was root 6 minus root 2 over 4.*0353

*What I'm going to do is square the numerator, root 6 minus root 2 squared.*0361

*To pay for that, I have to take a square root later.*0370

*Remembering an algebra formula (a-b)*^{2} is a^{2}-2ab+b^{2}, I remember that algebra formula.0373

*I've got a quantity that I'm squaring here, root 6 squared is 6 minus 2 root 6 root 2 plus root 2 squared is just 2, all over 4.*0385

*This simplifies down to 8 minus 2 root 12 over 4.*0405

*But root 12, I could pull a 4 out of that, and it turns into a 2 on the outside.*0413

*But I already had a 2 on the outside, you combine those and you get 4 root 3 over 4.*0424

*Now, I can factor 4 out of the numerator.*0427

*When it comes outside it becomes a 2, left on the inside will be, the 8 turns into a 2, and the minus 4 root 3 becomes a minus root 3 over 4.*0432

*That simplifies down to 2/4 cancel into 1/2, square root of 2 minus root 3 over 2.*0446

*Look at that, that's our new answer that we've just arrived using the half-angle formula.*0454

*We could do that problem using the addition and subtraction formulas as we did a couple of lectures ago or we could do it using our new half-angle working it out from what we now about 30 degrees.*0465

*Either way, we get down to the same answer.*0481

*We still have to find the cos(15).*0486

*We're going to use the half-angle formula for cosine, cos(1/2 x) is equal to plus or minus the square root of 1/2 times 1 plus cos(x).*0490

*The cos(15), 15 is 1/2 of 30, this is square root of 1/2 times 1 plus cos(30).*0506

*Cos(30) is a common value that I remember very well, 1 plus root 3 over 2.*0520

*If I put those, combine those over a common denominator, I get 1/2 2 plus root 3 over 2.*0529

*That simplifies down to 2 plus root 3 over 4, or if I take the square root of the bottom 2 plus root 3 over 2.*0541

*I still have that plus or minus but remember 15 degrees is safely there in the first quadrant.*0556

*It's sine and cosine, it's x and y values.*0570

*They're both positive, 15 degrees is in quadrant 1, cos(15) is positive.*0571

*The cos(15) must be the positive square root 2 plus root 3 over 2.*0590

*There's my answer.*0600

*Last thing we were supposed to check there was that the answer satisfies the Pythagorean identity sin*^{2}+cos^{2}=1.0602

*Sin*^{2}(15)+cos^{2}(15), sin^{2}(15) was square root of 2 minus root 3/2.0613

*We're going to square that.*0627

*Plus cos*^{2}(15) is 2 plus root 3, square of that, over 2.0630

*We'll square that one out.*0636

*Now, in the top, the square root and the square will cancel each other away.*0642

*We get 2 minus root 3.*0647

*In the bottom, we have 2, squared is 4, plus 2 plus root 3/4.*0650

*When we add those together, the root 3s cancel.*0657

*We just get 4 over 4 which is 1.*0662

*When we worked out sin*^{2}(15)+cos^{2}(15), we did indeed get 1, showing that it does confirm the Pythagorean identity.0668

*The key to that problem was really just recognizing that 15 is 1/2 of 30, and then invoking the sine and cosine half-angle formulas plugging in x=30 working them through doing a little bit of algebra, and getting our answers there.*0677

*The only other step that was a little bit tricky was recognizing whether we wanted to use the positive or the negative square root.*0694

*That's the matter of recognizing that 15 degrees is in the first quadrant, in both cases sine and cosine are both positive.*0702

*For our second example here, we're asked to use, to prove a trigonometric identity, (cos(1/2 x)+sin(1/2 x))/(cos(1/2 x)-sin(1/2 x))=sec(x)+tan(x).*0712

*That's a pretty complicated identity.*0727

*It's not really obvious where to start.*0730

*You might want to jump into the half-angle formulas because you see cos(1/2 x), sin(1/2)x.*0733

*I'm going to say, let's try to avoid the half-angle formulas here if we can.*0741

*Here's why.*0745

*Remember that cos(1/2 x) is equal to plus or minus the square root of something or other, so is sin(1/2 x) is equal to plus or minus the square root of something or other.*0746

*If we start putting those in, we're going to have plus or minuses, or lots of square roots, it's going to get complicated.*0761

*I'm going to try to avoid those.*0766

*Instead, I have another strategy which we've seen before in proving trigonometric identities.*0769

*If you have (a+b)×(a-b), remember from algebra, that's the difference of squares formula.*0775

*That's a*^{2}-b^{2}.0781

*That can be really useful if you have an (a+b) in the denominator or an (a-b) in the denominator.*0783

*You multiply both sides by the conjugate, by the other one, and then you get the difference of squares.*0790

*Let's try that out on this one.*0796

*The left-hand side, I'm going to work with the left-hand side because I see that (a-b) in the denominator.*0799

*That's cos(1/2 x)+sin(1/2 x)/(cos(1/2 x)-sin(1/2 x)).*0811

*Now, I'm going to multiply top and bottom by the conjugate of the denominator.*0823

*That means where I saw a minus before, I'm going to multiply by the same expression with a plus in it, sin(1/2 x).*0826

*Of course, I have to multiply the top by the same thing, (cos(1/2 x)+sin(1/2 x).*0837

*Let's see where we'll go with that.*0845

*In the numerator, we actually have cos((1/2 x)+sin(1/2 x))*^{2}, so that's cos^{2}(1/2 x)+2sin(1/2 x)×cos(1/2 x)+ sin^{2}(1/2 x).0848

*We can invoke this difference of squares formula.*0873

*We get cos*^{2}(1/2 x)-sin^{2}(1/2 x).0877

*There's several good things that are going to happen right now but they will only happen if you remember the double angle identities.*0886

*Let me write those down for you.*0894

*I'm going to write them in θ instead of x.*0896

*Remember that sin(2θ)=2sin(θ)×cos(θ), cos(2θ)=cos*^{2}(θ)-sin^{2}(θ).0901

*Now, look at what we have here.*0922

*There's several good things that are going to happen.*0923

*First of all, cos*^{2} and sin^{2}, those combined, and those give me a 1.0926

*Now, we have 2 sine of something, cosine of something, and the something is 1/2 x.*0933

*If you look back at our sin(2θ), 2 sine of something and cosine of something is equal to sine of 2 times that thing.*0941

*We have sin(2×1/2 x).*0951

*Now, I have cos*^{2} of something minus sin^{2} of something, and I know that cos^{2} of something minus sin^{2} of something is equal to cosine of 2 times that something, so cos(2×1/2 x).0958

*You can simplify this a little bit, this is (1+sin(x))/cos(x).*0979

*I'll split that up into 1/cos(x) + sin(x)/cos(x).*0990

*Those are expressions that I recognize, 1/cos(x) is sec(x), sin(x)/cos(x) is tan(x).*0998

*Look, now we've got the right-hand side of the original trigonometric identity.*1008

*That was the right-hand side right there.*1014

*That was a pretty tricky one.*1018

*There were several key steps involved there.*1020

*The first is looking at the left-hand side and noticing that we have something minus something in the bottom, so we're going to use this difference of squares formula.*1024

*We're going to multiply top and bottom by the conjugate.*1036

*Once we multiply top and bottom by the conjugate, we get something that looks pretty messy, but we start invoking these identities all over the place.*1039

*First of all, sin*^{2}+cos^{2} gives you 1.1046

*Secondly, 2 sine of something cosine of something, that's the double angle formula for sine.*1051

*Then, cos*^{2} of something minus sin^{2} of something, that's the double angle formula for cosine.1058

*That simplifies it down to (1+sin(x))/cos(x).*1065

*Those split apart and convert easily into secant and tangent, and all of a sudden we have the right-hand side.*1070

*You may have to experiment a bit with different techniques when you're proving these trigonometric identities.*1077

*The ones that I'm using for example, these are ones I've worked out ahead of time, so I know right away which technique I'm going to use.*1083

*Even when I'm working on this, I'll try multiplying a few different things together, maybe splitting up things differently and invoking different half-angle formulas, double angle formulas, and finally I find the sequence that works.*1090

*When you're asked to prove this trigonometric identities, go ahead and experiment a little.*1103

*If it seems like it's getting really complicated, maybe go back and try something else.*1108

*Eventually, you'll find something that converts to one side of the equation into the other.*1113

*For our next example, we have to prove a half-angle formula for tangents, and we're told to be careful about removing plus or minus signs.*1119

*The reason for that is we're going to be using the sine and cosine half-angle formulas, and those both have plus or minus in them.*1129

*Let me remind you what those are.*1139

*The cos(1/2 x) is plus or minus the square root of 1/2 times 1+cos(x).*1144

*The sin(1/2 x) is plus or minusthe square root of 1/2 times 1-cos(x).*1157

*Those are the formulas we're going to be using.*1173

*We're given the tan(1/2 x).*1177

*Let me start with that, I'll call that the left-hand side.*1179

*Left-hand side is tan(1/2 x).*1186

*Now, I don't have a formula yet for tan(1/2 x), I'm going to split that up into sin(1/2 x)/cos(1/2 x) because I do have formulas for those.*1192

*Those are my half-angle formulas.*1204

*In the numerator, I get plus or minus the square root of 1/2.*1207

*Sin(1/2 x) is 1-cos(x).*1213

*In the denominator, cos(1/2 x), the same thing except that I get 1/2 (1+cos(x)).*1219

*Here's the thing, you might think that you can cancel plus or minus signs, but you really can't.*1228

*The reason is that plus or minus signs means that both the top and bottom could be positive or could be negative.*1234

*When you divide them together, you don't know if the answer's going to be positive or negative.*1241

*I'm going to put one big plus or minus sign on the outside but I can't just cancel those away.*1246

*I'm also going to combine everything under the square root here.*1252

*I get (1/2 (1-cos(x)))/(1/2 (1+cos(x))).*1257

*The obvious thing to do there is to cancel the (1/2)'s, so you get (1-cos(x))/(1+cos(x)).*1270

*It's not so obvious where to go from here but remember that we've been practicing this rule (a-b)×(a+b)=a*^{2}-b^{2}.1282

*That comes up all over the place for trigonometric identities and with other algebraic formulas as well.*1295

*The trick is when you have either one of those in the denominator, you multiply by the conjugate.*1302

*Here, we have 1-cos(x) in the denominator, I'm going to multiply by 1+cos(x).*1308

*Of course, I have to multiply the numerator by the same thing.*1317

*Sorry, I have 1+cos(x) in the denominator, so the conjugate would be 1-cos(x).*1322

*Multiply top and bottom by 1-cos(x).*1327

*The point of that is to invoke this difference of squares formula in the denominator.*1331

*This is all taking place under a big square root.*1337

*1-cos(x), I'll just write that as (1-cos(x))*^{2}.1342

*I don't need to multiply that out.*1350

*In the bottom, I got 1-cos(x) times 1+cos(x), that's the difference of squares formula, that's 1-cos*^{2}(x).1351

*Now, I'm going to separate out the top and the bottom part here, because in the top, I've got a square root of a perfect square.*1367

*On the top, I'm just going to write it as 1-cos(x), because I had the square root of (1-cos(x))*^{2}.1382

*In the bottom, I still have a square root, 1-cos*^{2}(x), that's something that should set off some warning bells in your brain.1389

*Certainly doesn't mind, because I remember that sin*^{2}+cos^{2}=1, the Pythagorean identity.1398

*If you move that around, if you see 1-cos*^{2}, that's equal to sin^{2}.1407

*So, 1-cos*^{2}(x)=sin^{2}(x), that cancels with the square root.1415

*I've already got a plus or minus outside, I don't need that another one, (1-cos(x))/sin(x).*1423

*I almost got what I want, I've almost got the right-hand side (1-cos(x))/sin(x).*1435

*The problem is this plus or minus.*1444

*The directions of this exercise said we have to be very careful about why we can remove any plus or minus signs.*1446

*Let me write the big question here, "Why can we remove this?"*1455

*That actually takes a bit of explanation.*1464

*I'm going to go on a new slide to explain that.*1467

*From the last slide, we figured out that the left-hand side is equal to plus or minus (1-cos(x))/sin(x), but we aren't sure if we can remove the plus or minus from the right-hand side.*1470

*Let's think about that.*1497

*First of all, I know that cos(x) is always less than 1.*1499

*That's because cos(x), remember, is the x values on the unit circle so it's always between -1 and 1.*1509

*1-cos(x) is always greater than 0, the numerator here, the 1-cos(x), the numerator 1-cos(x) is always positive.*1517

*That part isn't really affected by the plus or minus.*1543

*What about the sin(x)?*1547

*I know that that is not always positive.*1548

*What about sin(x)?*1555

*Let me draw a unit circle, because this really depends on where x lies on the unit circle.*1560

*There are several different cases depending on where x lies on the unit circle.*1575

*Let me write down the four quadrants, 1, 2, 3, and 4.*1580

*Let's try and figure out where x could lie on the unit circle.*1589

*There's sort of 4 cases.*1593

*If x is in quadrant 1, then remember, sin(x) is its y-value, so sin(x) is positive.*1600

*And x/2, if x is in quadrant 1, if that's x right there, then x/2 will also be in quadrant 1.*1620

*Tan(x) will also be positive.*1636

*Remember All Students Take Calculus, they're all positive in the first quadrant, second quadrant, only sine is, third quadrant only tangent is, and fourth quadrant only cosine is.*1640

*If x is in quadrant 1, then they're both positive.*1652

*Both sides here, the tan(1/2 x) and the sin(x), Oops, I said tan(x) and I should have said the tan(1/2 x) or x/2, they're both positive.*1654

*Let's check the second quadrant, if x is in quadrant 2, then sin(x) will still be positive, and x/2 ...*1666

*Well if x is over here in quadrant 2, then x/2 will be in quadrant 1 because it's half of x.*1686

*It's tangent, will still be positive.*1699

*Again, both the sin(x) and the tan(x) will both be positive.*1704

*Third case is if x is in quadrant 3, then sin(x) is less than 0 because x is down here in quadrant 3.*1709

*Where will x/2 be?*1730

*If x is in quadrant 3, that means x is bigger than π, x/2 is bigger than π/2.*1734

*x/2 will be over here in quadrant 2.*1748

*Tan(x/2), in quadrant 2, only the sine is positive, the tangent is negative.*1760

*Sine is negative because x is in quadrant 3, and tan(x/2) is also negative.*1769

*Finally, if x is in quadrant 4, then sin(x) is less than 0 because it's still below the x-axis, its y-coordinate is negative.*1775

*If x is somewhere over here in the fourth quadrant, x is between π and 2π, x/2 will be between π/2 and π.*1797

*So x/2 is still in quadrant 2, tan(x/2) is still negative.*1812

*Now, there's 4 cases there.*1825

*In the first 2 cases, sine was positive and tangent was positive, tan(x/2) is positive.*1826

*In the second 2 cases, in the last 2 cases, sine was negative and tan(x/2) was also negative.*1834

*Sin(x) and tan(x/2), they're either both negative or both positive, that means they have the same plus or minus sign, have the same sign in terms of positive and negative not as any have the same sign.*1844

*We can drop the plus or minus, and finally say tan(x/2) or tan(1/2 x)=(1-cos(x))/sin(x).*1872

*In all 4 cases, the tan(x/2) has the same plus or minus as sin(x).*1897

*Then remember the 1-cos(x) is always positive, the left-hand and the right-hand side will always have the same plus or minus, we don't need to attach another plus or minus.*1905

*That was a pretty tricky one.*1916

*The secret to that was starting with the tan(x/2), expanding it out using the formulas for cos(1/2 x) and sin(1/2 x) or x/2.*1918

*We worked it down, we did some algebra, simplifying a square root.*1934

*Then we still have that plus or minus at the end.*1938

*What we had to do was this sort of case by case study of each of the four quadrants to say when is sine positive or negative, when is tan(x/2) is positive or negative.*1940

*The last term, 1-cos(x) was always positive.*1951

*Finally, we figured out that sin(x) and tan(x/2) could be positive or negative, but they'll always be the same so we don't need to put in a plus or minus there.*1956

*We'll try some more examples later on.*1966

*The examples we're going to be using later, we're going to be using this formula for tan(x/2), so it's worth remembering this formula for tan(x/2).*1969

*Later on, we'll be using that to solve some more trigonometric identities.*1982

*We are trying some examples of right angle in trigonometry, we are finding lengths of sides and angles in right triangles.*0000

*The master formula we are using here is SOHCAHTOA, and remember that SOHCAHTOA only works in triangles where one of the angles is a right angle, so in right triangles.*0012

*SOHCAHTOA does not work in other kind of triangles, we are going to learn some complicated rules later on called the law of cosines and the law of sines, that will work in any kind of triangle.*0026

*Right now we are just looking at right triangles and we are using SOHCAHTOA to figure out the relationships between the length of the sides and the measure of the angles.*0037

*What we are given here is a right triangle with a short side of length = 3 and hypotenuse of length = 7.*0046

*We want to find all the angles in the triangle so let me try to graph this out.*0056

*We know that it is a right triangle, one angle is a right angle, short side has length = 3 and the hypotenuse has length = 7, we want to find all the angles in the triangle.*0067

*I label the angles as theta and phi, I think it will be probably useful to find the third side of the triangle.*0079

*we know that the third side satisfies if I call it x for the time being.*0091

*x*^{2} + 3^{2} = 7^{2}, so x^{2} + 9 = 49.0096

*X*^{2} is 40 and x = square root of 40, we can factor 4 out of that so 2 square root of 10.0108

*That one did not come out to be very neat as some of other triangles have then.*0121

*I want to find this angles now, theta if I use SOHCAHTOA, I know that sin (theta) is equal to the opposite / hypotenuse.*0125

*That one I deliberately picked one where I would not have to use the square root to make my life a little bit simpler.*0137

*So, opposite is 3, the hypotenuse is 7, theta itself is arcsin(3/7).*0142

*Now remember here, if you calculator is set to radian mode then you will get a very strange answer looking here.*0154

*If you are looking for degrees, you have to set your calculator to degree mode.*0161

*On my calculator that is a matter of pushing the mode button and then moving it over and selecting the degree option.*0166

*It has two options, degree and radians, you want to set the degree options if that is the kind of answer you are looking for.*0172

*In my case, I’m going to do the inverse sin of 3 / 7, and it tells me that it is 25.4 degrees.*0180

*That tells me what theta is, I would like to figure out now what phi is, if I use the sin = opposite / hypotenuse.*0202

*If I use that for phi then I’m going to have to look at this ugly number 2 square root of 10.*0211

*Instead, I’m going to use cos(phi) = adjacent/ hypotenuse, phi is right there so the adjacent is 3.*0218

*Cos(pi) is 3/7, phi is arcos(3/7) and if I work that out on my calculator I get 64.6 degrees approximately.*0230

*I’m rounding here to the nearest decimal place, so now I have phi.*0252

*That gives me all the angles of the triangle because the last angle is a right angle, but I want to check these by seeing whether these angles adds up to 180 degrees.*0258

*25.4 + 64.6 those add up to 90 and the last angle is a 90 degree angle, it does in fact give me a 180 degrees.*0270

*That is very satisfying to see that does checked, the key there was to remember the SOHCAHTOA formula which works in all right triangles.*0286

*And then when we figured out that we are given two sides of the triangle, we are able to figure out the angles using sin(theta) =( opposite / hypotenuse), cos(phi) = adjacent / hypotenuse.*0294

*For our last example of right triangle trigonometry, we are given a right triangle that has one angle of 65 degrees and a hypotenuse of length=3.*0000

*I want to find the lengths of all the sides of the triangle, let me draw this out.*0012

*We got one angle on the corner, I will call it (theta) and we are given that 65 degrees, we got a right angle and we know that the hypotenuse is length = 3.*0022

*The question is to find the lengths of the other sides of the triangle and I’m going to use my standard formula SOHCAHTOA, (Some Old Horse Caught Another Horse Taking Oats Away).*0033

*I’m going to use the sin portion of that formula, so sin(theta) = (opposite/hypotenuse).*0048

*I do not know what the opposite is but I know that the hypotenuse is 3 and I know that (theta) is 65 degrees.*0064

*The opposite is equal to 3 x sin(65 degrees), remember to set your calculator to degree mode.*0076

*If you have it set in radian you will get a very confusing looking answer here.*0083

*I typed in 3 × sin(65) and I get that, that is approximately equal to 2.7.*0090

*That tells me that the side opposite theta is approximately 2.7 units long.*0100

*Finally, I am going to use the cos part of SOHCAHTOA, cos(theta) = adjacent/hypotenuse, and the hypotenuse is still 3 units, the cos(theta) is cos(65 degrees).*0107

*If we solve that for the adjacent side, we get ((3 cos(65 degrees)) and the calculator tells me that it is approximately equal 1.3 units.*0129

*That tells me that the adjacent side to (theta) is 1.3.*0137

*Now again, I figured out each one of those sides using SOHCAHTOA, I’m going to check it using the Pythagorean theorem.*0159

*As I check here, I will do 1.3*^{2} + 2.7^{2} and that should be approximately equal to 1.3^{2} + 2.7^{2}.0168

*If you work that out on the calculator that actually gives you 8.98.*0187

*There is a little bit of rounding when I found those values that is very close to 9 which is 3 squared.*0192

*That tells us that the Pythagorean theorem is satisfied by these lengths which means that we almost certainly did those right.*0200

*Again, that came back to writing down all the information we had in the triangle and labeling one of the angles.*0206

*We labeled the angle we were given as (theta=65 degrees) and then using the SOHCAHTOA relationships to set up some equations.*0212

*And then solving for the length of the sides that we did not know, the sin gave us the length of the opposite side.*0221

*The cos SOHCAHTOA relationship gave us the length of the adjacent side.*0230

*And then it was an easy matter to check that those actually satisfy the Pythagorean theorem.*0235

*That is the end of the lecture on trigonometry in right triangles.*0240

*We will come back later and talk about trigonometry in triangles that do not necessarily have a right angle.*0245

*We will be using the law of sines and law of cosines to analyze those.*0251

*Thanks for watching the trigonometry lectures on www.educator.com.*0255

*Hi, these are the trigonometry lectures for educator.com.*0000

*Today, we're going to talk about trigonometry in triangles that have a right angle.*0005

*These are called right triangles.*0009

*The master formula for right triangles, we've seen it before it's the SOH CAH TOA.*0011

*That's the word that I remember to know that the sin(θ) ...*0017

*Let me draw a right triangle here.*0022

*If θ is one of the small angles, not the right angle, then the sin(θ) is equal to the length of the opposite side over the length of the hypotenuse.*0026

*Cos(θ) is equal to the adjacent side over the hypotenuse.*0042

*The tan(θ) is equal to the opposite side over the adjacent side.*0050

*For shorthand, sine is equal to opposite over hypotenuse, cosine is equal to adjacent over hypotenuse, tangent is equal to opposite over adjacent.*0053

*It's probably worth saying SOH CAH TOA often enough until it sticks into your memory, because it really is useful for remembering these things.*0063

*If you have a hard time remembering that, the little mnemonic that is also helpful for some students is Some Old Horse Caught Another Horse Taking Oats Away.*0071

*That spells out SOH CAH TOA for you.*0083

*One key thing to remember here is that SOH CAH TOA only works in right triangles.*0085

*You have to have one angle being a right angle.*0093

*If you have a triangle that is not a right triangle, if you don't have a right angle, don't use SOH CAH TOA because it's not valid in triangles that don't have right angles.*0096

*That's if you have no right angle.*0116

*We're going to learn in the next lectures on educator.com, we'll learn about the law of sines and the law of cosines.*0122

*Those work in any triangle where you don't need a right angle, but when you have a right angle, it's definitely easier and better and quicker to use SOH CAH TOA.*0128

*Let's try that out with some actual triangles.*0138

*On the first example, we have a right triangle with short sides of length 3 and 4, and we want to find all the angles in the triangle.*0140

*Let me draw a triangle here.*0149

*We're told that the short sides have length 3 and 4.*0154

*Of course, one angle is a right angle, so I don't need to worry about that.*0158

*I'll call these angles θ, and I'll call this one φ.*0162

*First thing we're going to need to know is what the hypotenuse of this triangle is.*0166

*h*^{2}=3^{2}+4^{2}, Pythagorean theorem there, which is 9+16, which is 25.0172

*So, h is the square root of 25, h is 5.*0184

*Let me draw that in there.*0187

*Now, I want to figure out what θ and φ are.*0191

*I'm going to use SOH CAH TOA.*0197

*Let me write that down there for reference, SOH CAH TOA.*0200

*I'm going to figure out what θ is by using the SOH part of the SOH CAH TOA.*0205

*Sin(θ) is equal to the opposite over the hypotenuse.*0214

*The opposite angle to θ is 3 and the hypotenuse is 5.*0225

*So, θ=arcsin(3/5).*0231

*I'm going to work that out on the calculator.*0238

*My calculator has an arcsine button, it actually writes it as sin*^{-1}, which I don't like that notation because makes it seem like a power.0241

*In any case, I'm going to use inverse sine of 3/5.*0251

*There's a very important step here that many students get confused about which is that, if you're looking for an answer in terms of degrees, which in real world measurement, it sometimes easier to use degrees than radians.*0257

*You have to set your calculator to degree mode.*0270

*Most calculators have a degree mode and a radian mode.*0274

*In fact, all calculators that do trigonometric functions have a degree mode and a radian mode.*0278

*The default is probably radian mode.*0284

*If your calculator is set in radian mode and you try to do something like arcsin(3/5), you'll get an answer that doesn't look right.*0287

*You may be confused if you're checking your answers somewhere, it may not agree with what the correct answer it.*0297

*What you have to do is set up your calculator in degree mode, if you want an answer in degrees.*0304

*My calculator is a Texas Instruments.*0311

*It's got a mode button, I just scroll down, it say's RAD and DEG to convert it from radians to degrees.*0315

*I'm going to convert it into degree mode and then I'll get an answer in terms of degrees.*0323

*That's a step that many students forget and they get kind of confused when they get an answer which is in terms of radians, but it doesn't agree with the degree answer they were looking for.*0330

*Now, I've got my calculator set in degree mode.*0338

*I'll do the arcsin(3/5) which is 0.6.*0342

*It tells me that that is approximately equal to 36.9 degrees.*0348

*I found one of the angles in the triangle.*0358

*Of course, another one is a right angle, so it's 90 degrees.*0359

*For φ, I think for φ, I'm going to practice the cosine part of the SOH CAH TOA.*0362

*I know that cos(φ) is equal to adjacent over hypotenuse.*0369

*That's the adjacent side to φ, so φ is over here, it's adjacent side is 3, hypotenuse is still 5.*0378

*So, φ=arccos(3/5).*0390

*Again, I'll do that on my calculator.*0396

*The calculator tells me that that's 53.1 degrees, approximately equal to 53.1 degrees.*0404

*Now, there's a little check here you can do to make sure that you work this out correctly.*0412

*We know that the angles of a triangle add up to 180 degrees.*0416

*If we check here, 36.9+53.1 plus the last angle was a 90-degree angle, a right angle, if you add those up, 36.9+53.1 is 90 plus 90, you get 180 degrees.*0421

*That tells me that my answers are right.*0439

*The key formula here to remember is SOH CAH TOA.*0443

*Everything comes down to drawing the angles in the triangle, and just using sine equals opposite over hypotenuse, cosine equals adjacent over hypotenuse, tangent equals opposite over adjacent.*0446

*Now, we're given a right triangle and we're told that one angle measures 40 degrees.*0460

*Let me call that angle right here, that would be 40 degrees.*0472

*The opposite side has length 6.*0480

*I want to find the lengths of all the sides in the triangle.*0486

*Of course, finding the angles is no big deal because one side is a right angle, we're told that it is a right triangle.*0491

*I can find the other angle just by subtracting from 180.*0498

*In fact, 180-40 is 140, minus 90, is 50.*0502

*I know that other angle is 50.*0511

*The challenge here is to find the lengths.*0513

*We're going to use SOH CAH TOA.*0516

*I'm going to apply SOH CAH TOA to 40.*0521

*I know that sin(40) is equal to 6 over the hypotenuse, because that's opposite over hypotenuse.*0526

*The hypotenuse, if I solve this, that's equal to 6/sin(40).*0535

*Remember to convert your calculator to degree mode before you do this kind of calculation.*0545

*6/sin(40) is approximately equal to 9.3.*0551

*That tells us the length of the hypotenuse, 9.3.*0561

*Now I'd like to find the length of the other sides.*0567

*I'm going to use the cosine part of SOH CAH TOA.*0572

*I know that cos(40) is equal to the adjacent over the hypotenuse.*0574

*If I solve that for the adjacent side, that's equal to hypotenuse times the cos(40).*0585

*I know the hypotenuse now.*0592

*If I multiply that by cos(40).*0596

*What I get is approximately 7.2 for my adjacent side.*0601

*The sides of my triangle are 6, 7.2 and 9.3.*0611

*Again, there's an easy way to check that.*0618

*We'll check that using the Pythagorean theorem.*0620

*I want to check that 6*^{2}+7.2^{2} gives me 87.8, which is approximately equal to 9.3^{2}.0623

*I know that I got those side lengths right.*0656

*The third example here, we have the lengths of the two short sides of a right triangle, are in a 5:2 ratio.*0662

*Let me draw that out.*0668

*We're given that it's a right triangle.*0676

*We've got lengths in a 5:2 ratio.*0679

*I don't actually know that these lengths are 5 and 2, but here's the thing, if I expand this triangle proportionately, it won't change the angles.*0683

*If I blow this up to a similar triangle that actually has side lengths of 5 and 2, I'll get a similar triangle with the same angles.*0695

*I can just assume that this triangle has actually side lengths of 5 and 2.*0706

*I want to find all the angles in the triangle.*0711

*I'll label that one as θ, that one is φ.*0715

*I know that sooner or later, I'm going to need the hypotenuse of the triangle, I'll go ahead and find that now.*0719

*h*^{2}=2^{2}+5^{2}, that's equal to 4+25 which is 29.0722

*My hypotenuse is the square root of 29.*0738

*I'd like to practice all parts of SOH CAH TOA, and we've used sine and cosine in the previous problems.*0747

*I'm going to use the tangent part.*0757

*Remember, tangent is opposite over adjacent.*0760

*Tan(θ) there, the opposite side is 5, and the adjacent side has length 2.*0765

*I'm going to find arctan(5/2), θ is arctan(5/2).*0772

*Remember that you want to have your calculator in degree mode here, because if you have your calculator in radian mode, you'll get an answer in radian which would look very different from any answer in degrees that you were expecting.*0783

*I calculate arctan(5/2), and it tells me that that is approximately equal to 68.2 degrees.*0800

*Now, I found θ, I've got to find φ now.*0815

*I could find φ just by subtracting θ from 90 degrees, because I know θ+φ adds to 90 degrees.*0819

*Remember, one angle of the triangle is already 90 degrees, the other two must add up to 90 degrees.*0826

*I could just find the other angle by subtracting but I want to avoid that.*0833

*I want to practice using my SOH CAH TOA rules.*0839

*The other reason is if I find it using some other method, then I can add them together at the end and use that to check my work.*0841

*I'm going to try using a SOH CAH TOA rule.*0848

*I'm going to find it using angle φ.*0853

*Looking at φ, I know that sin(φ) is equal to the opposite over hypotenuse.*0854

*Sin(φ) is equal to the opposite over hypotenuse.*0866

*The opposite side to φ is 2, and the hypotenuse is the square root of 29.*0872

*φ is equal to the arcsine of 2 over the square root of 29.*0881

*That's definitely something I want to put into my calculator.*0886

*I'll figure out inverse sine of 2 divided by square root of 29 ...*0891

*The inverse sine of 2 divided by square root of 29 ...*0911

*It tells me that that's approximately equal to 21.8 degrees.*0917

*That's what I got using SOH CAH TOA.*0925

*Again, I'm going to check it by checking that the three angles of this triangle add up to a 180 degrees.*0928

*I've got 68.2 degrees plus 21.8 degrees, those add up to 90, plus the last 90-degree angle, the right angle.*0937

*Those do add up to 180 degrees.*0951

*That tells me that my work must probably be right.*0954

*That came back to looking at SOH CAH TOA, and figuring out what the angles were based on the SOH CAH TOA.*0958

*We know that tangent is opposite over adjacent, and we also used that sine is opposite over hypotenuse.*0968

*We'll try some more examples later.*0974

*We are learning about the law of science today, we are trying some more examples of solving triangles completely using the law of science.*0000

*Remember, that means you are given some data about the triangles, some information about the lengths and some of the sides of the triangle, and the measure of some of the angles.*0008

*What you have to do, is first of all figure out if there is a triangle that satisfies that data or maybe if there is more than one.*0019

*And then solve for all the other lengths and sides, and all of the other measures of the angles in the triangle.*0027

*In this example, we are given a triangle- it looks like we are given two angles and a side, let me draw the triangle.*0036

*Capital letters for the corners and the sides you will use lower case letters, it has to be opposite of the capital of the same letter.*0052

*That tells you where the orientation of all the information that you are given, given that (A is 40 degree), angle A=40 degrees, angle B=110 degrees.*0063

*Certainly the way I drew it is not to scale because 110 would be end up two’s angle, it is bigger than 90 degrees.*0074

*We are given that (a) has side length=7, side (a) has length=7, I fill in the information that I have.*0081

*I want to find out first of all, are we going to have a solution here? We are given here a side and then two angles of a triangle.*0089

*This is (side, angle, angle) the side is not between the angles that is why (side, angle, angle) not (angle, side, angle).*0098

*The thing you want to check there is whether the angles you have been given are legitimate, did they add up to less than 180 degrees.*0109

*We talked about that in the beginning of the lecture, when you are given certain pieces of information how do you know if there is one solution or no solutions or two solutions?*0121

*In (side, angle, angle), you just check whether the angles add up to 180 degrees.*0130

*In this case, 110 + 40 = 150, that is less than 180 so we have not already exceeded the angle limit for the triangle.*0137

*In that case, it has a unique solution, it has one solution, that would be exactly one triangle satisfying the data we have been given.*0150

*That answers the first question, the trickier part is use the law of science to find all the missing quantities here.*0166

*Let me write down the law of science, sin(A)/a = sin(B)/b = sin(C)/c.*0173

*What do we know there? We know (A), so we can find its sin very easily, we know (B) and we know (a).*0189

*We can figure out (C) very quickly because we know two of the angles and it is very easy to find the third one.*0200

*So, (C) = 110 – 40, sorry it is 180 – the two angles we are given (40+110), so that is 180-150=30 degrees, we can fill that in very quickly, (C)=30 degrees.*0207

*We got sin(C), we can figure that out quickly, we need to find (b) and (c), we need to use a lot of sin to find them.*0230

*Sin(A)/a=sin(B)/b and trying to solve for (b) here, I will cross multiply and get b sin(A)=a sin(B) or (b)=a sin(B)/(A).*0241

*Actually, I’m going to fill in before I cross multiply, I will fill in the quantities that I know, sin(A)=40, (a)=7, (b)=?, (B)=110.*0264

*Now, I’m going to cross multiply, and I get (b) = (7)sin(110)/sin(40).*0281

*Remember to set your calculator to degree mode so you do not get strange answers because your calculator is thinking in terms of radians.*0291

*I will work out (7)sin(110)/sin(40) what I get there is approximately, I’m going to round it to 10.23, so that is the length of side (b).*0304

*We found all three angles, two of the sides, we just need to find side (c).*0327

*I’m going to use the law of science. Sin(A)/a=sin(C)/c, angle A=40, a=7, C=30, c= we do not know, that is what we are solving for, so I will cross multiply.*0333

*I will get c=(7)sin(30)/sin(40), there is actually a common value, I do not what the sin is but it is not really that useful here.*0359

*Certainly I do not know sin(40), so I’m plugging these numbers into my calculator anyway.*0376

*(7)sin(30)/sin(40) is approximately equal to 5.45.*0381

*Let us recap what we needed to do there, we were given two angles and one side of a triangle, I filled in everything I knew, that gave me a (side, angle, angle) configuration.*0408

*It is not (angle, side, angle) because of the orientation of where everything was, the side was not in between the two angles, it was outside of the two angles.*0421

*We have (side, angle, angle) and remember from the list at the beginning of the lecture, if you have (side, angle, angle).*0432

*You just have to check that the angles add up to less than 180 degrees, which they did and then you know you has exactly one solution.*0440

*Once you determine that, you know you are going to find one solution.*0448

*It is quick to find the third angle because you know the angles add up to 180 and then you have to find the two missing sides.*0453

*That takes a little more time but they both come from the law of science.*0460

*You just write down the law of science, you fill in the information that you do know and then you solve for the missing side.*0465

*That worked to find both of those missing sides, then we know all three sides of the triangle and all three angles, we completely solved the triangle.*0472

*On our last example here, we are given a side, and other side, and an angle.*0000

*Let me draw out the information that we have here.*0006

*Capital letters on the angles (A, B, C), lower case letters on the sides, opposite development angles.*0020

*Let us fill in what we have, Side a=10, side b=8, angle B=20 degrees, that is what we are given. *0030

*We are given two sides and an angle, the angle is not in between them so this is (side, side, angle) not (side, angle, side) because the angle is not between them.*0050

*That is a little worrisome because (side, side, angle) is the ambiguous case, it could have no solutions, one solution, or two solutions.*0063

*We do not really know yet until we do some more work, we do not know whether there is going to be one triangle that fits this data, or no triangle fits this data, or maybe two triangles that fits this data.*0089

*We are going to work with the laws of science and try narrowing this down, sin(A)/a=sin(B)/b=sin(C)/c.*0104

*Let me see which of these we know, I’m going to fill in what we know (a), (b), we are told the measure of angle (B) so I can figure out the sin(B) very quickly.*0116

*The reasonable thing is to try and figure out first of all what is the sin(A), let me go ahead and try to calculate that.*0134

*Sin(A)/a=sin(B)/b, I will fill in what I know there, I do not know sin(a) yet, but I do know angle (A), but I do know that (a)=10, (B)=20, (b)=8.*0142

*So sin(A), if I cross multiply there I will get (10) sin(20)/8.*0168

*Let us see what that comes up to be, 10 x sin(20)/8 turns out to be approximately 0.43.*0178

*Ok, I’m looking for an angle whose sin is 0.43, let me draw my unit circle again.*0196

*The top half of the unit circle, 0.43 is the y value, remember sin is y value, there is 0.43.*0207

*There are two angles here, there is one and there is another one, they both have sin of 0.43.*0219

*One is (theta) but then there is angle that is actually 180-(theta), they both have sin(0.43).*0226

*My calculator does not really know that, if I work out from the calculator, if I just type arcsin(0.43), let me type that in.*0237

*After we calculate (10) sin(20)/8 and there is 0.43.*0254

*I if I just type in arcsin in my calculator it tells me that it is 25.3 degrees.*0275

*But, that is angle (theta) there is 25.3 degrees, we already know that there is another angle that also have sin(0.43) and that is this other angle 180 – (theta)=154.7 degrees.*0294

*Angle (A) could be either one of those two possibilities, 25.3 degrees or 154.7 degrees.*0324

*That means we have two different possibilities for angle (A), each one of them gives us a set of data that we can solve out the rest of the triangle with.*0335

*We get two possible triangles as our solutions, we have to solve for each one separately.*0349

*We got two solutions here and we are going to solve each one of them separately, I will do that on the next slide.*0365

*Here we found that we have two solutions depending on what angle (A) is.*0372

*I will solve for each one separately, let me draw my triangle.*0383

*(A, B, C) and on the first one we figured out that angle (A) = 25.3 degrees.*0393

*Then we also have some given data a=10, b=8, we do not know what c is, angle (B)=20 degrees, we have not figure out the rest of that.*0409

*Let us go ahead and start figuring out the rest of it, angle (C)=(180-20-25.3)=134.7 degrees.*0428

*In this triangle, 134.7 and finally we have to figure out what side c is, that is (c), we are going to use the law of science on that.*0448

*sin(C)/c=sin(B)/b since those is the simplest values that I can see.*0464

*Sin(134.7) that is angle (C)/we are solving for (c), is the sin(20)/8, if we cross multiply and solve for (c) there, *0475

*(c)sin(20)=(8)sin(134.7), (c)=(8)sin(134.7)/sin(20) you definitely want to go to the calculator for that.*0497

*So I type in (8)sin(134.7)/sin(20).*0515

*It tells me that it is approximately 60.63, what I am seeing there is approximately 60.63.*0530

*I have solved that for all three angles and lengths of that triangle but remember we have another completely different triangle which is based on finding a different value for angle (A).*0549

*We have to solve for those as well, let me draw that one, (A, B,C) (a, b, c) and I will fill in the value that were given.*0561

*Angle (B)=20, (a)=10, (b)=8, and we figured out the other possible value for angle (A) was 154.7 degrees, that is the other possible value.*0580

*And then we want to solve out the rest of the triangle which really means finding angle (C) and (c).*0613

*Angle (C) is pretty easy to find which is (180-20-154.7), which is (180-174.7) = 5.3 degrees.*0618

*Definitely not drawn in the scale here since this angle the way I drawn it is larger than 5.3 but that is ok.*0638

*Now we want to find (c) and we are going to use the law of cos just like we did on the other triangle it is the same arithmetic but with different numbers.*0656

*Sin(C)/(c)=sin(B)/(b), now this time angle (C)=5.3 degrees, (c)=we still do not know, (B)=20 degrees, (b)=8.*0664

*Cross multiply to solve for (c), (c)=(8)sin(5.3)=(c)sin(20), so (c) is (8)sin(5.3)/sin(20).*0687

*I will go to the calculator to figure that out, that tells me that, that is approximately 2.16.*0706

*Now, we have solved out all three angles and sides of that triangle, we are done with that.*0734

*To recap a little bit, let us see what we are given in this triangle.*0741

*We are given a side, another side, and then an angle, this really is a (side, side, angle) triangle.*0746

*Unfortunately, that is the ambiguous case where it could have no solution, one solution, or two solutions.*0756

*We do not really know until you start going to the law of science to figure out what is happening.*0762

*When we apply the law of science, we got a value for sin(A)=0.43, the problem is that there are two angles that have sin(0.43).*0767

*Your calculator will only give you one of them, if you take in there sin(0.43) it will give you this one 25.3 degrees.*0786

*But we know, if you look at the unit circle that sin(180)-(theta) is the same as sin(theta).*0795

*At the same time as looking at 25.3, we have to look at this other possible value 154.7 degrees for (A).*0807

*So we get two different possible angles for (A), that leads us to two different triangles and we have to solve out each one completely.*0816

*In that point, you really can not overwrap the work anymore, you have to split up this two triangles into two different problems.*0824

*In each one you draw it out and you see what pieces of information you are missing.*0833

*In this case, it was the third angle and the third side.*0838

*The third angle, you find out just by subtracting the two angles you have so that was pretty easy and then you use the law of science to solve for the length of the third side.*0845

*In the second triangle we do the same thing, we find the third angle by subtracting from 180 degrees and then we use the law of science to find the third side.*0855

*We got two different triangles, they both satisfy the initial data but using this rule for science, this co function identity sin(180-theta)=sin(theta).*0865

*We figured out the two possible angles and then we can completely solve the two possible triangles separately and we got a whole set of sides and a whole set of angles on each one.*0877

*That was our practice on using the law of science, in the next lecture we will come back and we will look at the law of cos which is a different way to solve out a triangle completely.*0887

*To find out all the angles and all the sides and you use the law of cos when you have a different set of initial data for the triangle.*0897

*We will learn about that in the next lecture, these are the trigonometry series on www.educator.com.*0906

*Hi, these are the trigonometric lectures for educator.com, and today we're going to learn about one of the really big rules of trigonometry which is the law of sines.*0000

*In a later lecture, we're going to learn another rule called the law of cosines.*0010

*They kind of go hand in hand.*0013

*The idea is that these rules help you figure out what the angles and sides are in any triangle.*0015

*I really mean any triangle here.*0024

*Remember a rule we learned before was SOH CAH TOA, that was kind of the old rule.*0028

*The thing about SOH CAH TOA is, it only works in right triangles, in other words, triangles where one angle is a right angle.*0035

*That was our old rule.*0051

*Our new rule is the law of sines, and we're going to be learning the law of cosines in the next lecture.*0053

*The law of sines works in any triangle.*0059

*You can always use it.*0063

*If you happen to have a right triangle, it's probably easier to use SOH CAH TOA.*0064

*If you don't have a right triangle, you want to use the law of sines or the law of cosines, which we'll learn about next.*0069

*Let me draw a picture to get it started here.*0076

*Here's a triangle.*0085

*It doesn't look like a right triangle.*0086

*It's traditional in trigonometry problems to kind of label the sides and corner, the sides and the angles of a triangle, so that you can keep track of what's opposite what.*0089

*Normally, you would label these things as A, B and C, the corners with capital letters A, B and C.*0102

*Then you label the sides with lowercase letters a, b and c.*0111

*You do it in such a way that lowercase, side labeled lowercase a, is opposite angle labeled uppercase A.*0112

*That puts the a over here, the b over here, and c over here.*0122

*What we're going to be doing is looking at a lot of different triangles where you'll be given some information about some of the sides and some of the angles.*0126

*You won't be given all of the information.*0134

*You'll be told the lengths of a couple of sides or maybe one of the angles, or maybe the measure of two of the angles, but not the third angle and the measure of one of the sides, things like that.*0139

*The trick here is you want to use the law of sines and whatever else you can, to find out all the angles and all the sides of the triangle.*0151

*That's called solving the triangle completely, when you figured out what all the angles and what all the sides are.*0162

*There's several different ways that the information could be given to you.*0168

*You'll hopefully remember some of these from your geometry class.*0176

*The first one is Angle Side Angle.*0179

*What that means is that ...*0183

*Let me draw a picture of this.*0185

*Kind of as you walk around the edge of a triangle, you'll know what one angle is and you'll know what one side is, and you'll know what the next angle is.*0187

*You'll be given two angles and a side.*0198

*It's important that they'd be in that order, the angle comes first and then the following side, then the next angle.*0201

*When you're given an angle side angle set up, you know that it always has a solution.*0208

*There's always a solution.*0217

*There's always a triangle that has those properties.*0218

*It's always unique, meaning that there's only one triangle that has those properties.*0221

*You're always looking for just a single triangle.*0226

*There is one thing that you need to check, which is that the two angles that you're given must add up less than 180 degrees.*0230

*That's not so surprising because, of course, if you're given two angles that are bigger than 180 degrees, collectively bigger than 180 degrees, they can't be the two angles of a triangle.*0239

*We all know that three angles of a triangle sum up to 180 degrees.*0248

*The next situation that you might given is Side Angle Angle.*0254

*Let me draw that one.*0260

*This one was angle side angle.*0261

*You might be given side angle angle.*0266

*That means, again, walking around the edge of the triangle.*0272

*You're told what one side is, and then you're told what one angle is, then you're told what the next angle is.*0275

*Assuming that the angles sum up to less than 180 degrees, because if they sum up to more than 180 degrees, you'll never going to get a triangle.*0283

*Assuming that the angles sum up to less than 180 degrees, this again will have always exactly one solution.*0293

*You know you're going to find a triangle and you're just going to have to worry about one triangle.*0301

*The more complicated case, it's called the ambiguous case.*0306

*You might see something in your trigonometry book called the ambiguous case.*0310

*It's Side Side Angle, where you're told the length of one side, then another side, then one angle.*0314

*This one gets a little tricky.*0331

*When you're told two sides in a row, and then an angle that's not the angle in between them, it's one of the other two angles, this does not always have a solution.*0333

*It might have no solution.*0346

*You might try to solve it out and you'll get into some kind of problem.*0348

*We'll see some examples of that so you'll see how it works, or fails to work.*0350

*It might have exactly one triangle that has those properties, or there might be two triangles.*0356

*That gets a little confusing.*0362

*We'll try some examples of that in the examples.*0364

*Angle Angle Angle, you could potentially be given.*0369

*The thing about angle angle angle, if you're just given the three angles of a triangle, well, you could take that angle and blow it up or take that triangle and blow it up or shrink it as much as you want, you'll get similar triangles that have the same three angles.*0376

*That never has unique solution.*0392

*It always has infinitely many solutions, assuming that those three angles sum up to 180 degrees.*0395

*Those triangles, the triangles that you get, are all going to be similar.*0400

*They're all going to be proportionate to each other.*0407

*You could get larger or smaller versions of the same triangle if you're only given angle angle angle.*0410

*Let me fill this in.*0422

*This was the side angle angle, and this was the side side angle case.*0424

*There's two other cases that you might be given, but we're not really going to talk about them in this lecture.*0431

*I'll mention them now and then we'll start solving those triangles in the next lecture.*0436

*The reason is that they really work better for the law of cosines.*0441

*After we learn the law of cosines in the next lecture, then we'll study these two cases.*0444

*Those two cases are side angle side, where you're given two sides and the angle between them, so side angle side.*0450

*That's different from side side angle because of the position of the angles.*0464

*In side angle side, the angle is between the two sides are given.*0469

*In side side angle, it's one of the other angles.*0473

*The other case we'll study in the next lecture is side side side, where you're given all three sides of the triangle.*0477

*Both of those cases don't really lend themselves very well to solutions by the law of sines.*0493

*We'll use the law of sines for these first three cases, angle side angle, side angle angle, and side side angle.*0500

*We'll use the law of sines for those three cases.*0512

*In the next lecture, we'll learn the law of cosines and we'll study side angle side, and side side side.*0515

*We'll get to some examples now.*0523

*First example, we're given an angle, another angle, and a side.*0524

*First thing we need to do is draw a picture of this triangle and then see where we can go from there.*0530

*I've got A, B, C.*0541

*Remember, we'll use capital letters for the angles, and then lowercase letters for the sides.*0545

*You'll always orient them so that lowercase a is opposite angle A.*0550

*That puts B down here, and C over here.*0556

*We're given that c has length 4.*0560

*We're given that angle A measures 50 degrees and angle B measures 60 degrees.*0562

*What we have here is we're given two angles and a side between them, this is angle side angle.*0572

*With angle side angle, that's one has a unique solution if the angles add up to less than 180.*0581

*In this case, the angles sum up to 110, which is less than 180, there is a unique solution.*0592

*We're going to try to solve the triangle, remember, that means finding all the angles.*0617

*Find all the measures of all the angles and the lengths of all the sides.*0625

*There's one that's very easy to start with, which is angle C, that's just 180-a-b, which is 180-110, which is 70 degrees.*0628

*I know that angle C is 70 degrees.*0648

*Looks like my triangle is not really drawn into scale here, because that angle is a little bit smaller than 70 degrees.*0652

*That's okay, we can still do the trigonometry on this.*0658

*I didn't know ahead of time that it was going to be 70 degrees because I haven't solved that part yet.*0662

*That's okay, we'll still work out the rest of the trigonometry.*0667

*Now, we're going to use the law of sines to solve the rest of this.*0672

*Let me write down the law of sines.*0673

*Remember, it says sin(A)/a=sin(B)/b=sin(C)/c.*0675

*Let's see, we've got to figure out what little a and little b are.*0690

*Those are the only things that are missing in our triangle.*0696

*I'm going to use sin(A)/a is equal to, I think I'm going to use sin(C)/c because I already know what little c is.*0697

*If I use B, then I wouldn't know what little b was, I would have too many unknowns.*0714

*Sin(A) is sin(50)/a = sin(70)/4.*0720

*If I cross multiply there, I get 4sin(50)=asin(70), or a=4sin(50)/sin(70).*0735

*Now, it's a matter of plugging those values into my calculator.*0752

*A very important concept that sometimes trigonometry students forget about, I mentioned this in the previous lecture but I want to emphasize it again, is that your calculator needs to be in degree mode if you're working these problems out using degrees.*0757

*If your calculator is in radian mode, then it will know what to do with sin(50) or sin(70) but it won't be what you want.*0772

*You need to convert your calculator into degree mode before you solve any of these problems if you're using degrees.*0783

*I've converted my calculator, I've used the Mode button to convert it from radians into degrees.*0788

*Now, I'm going to work out 4×sin(50)/sin(70), what I get is approximately, I'm rounding this a little bit, 3.26.*0793

*That tells me that a is approximately 3.26.*0816

*So, I've sold for a, now I'm going to solve for little b the same way.*0825

*Sin(B)/b = sin(C)/c, I'll fill in what I know here.*0831

*I know that capital B is 60 degrees, don't know little b yet.*0841

*Capital C is 70 degrees, and little c is 4.*0849

*Again, I'll cross multiply, bsin(70)=4sin(60), b=4sin(60)/sin(70).*0856

*I'll plug that into my calculator, 4sin(60)/sin(70).*0870

*That gives me approximately 3.69.*0880

*Now, I've solved the triangle.*0895

*I've found the lengths of all three sides, and I've found the measures of all three angles.*0897

*The key to solving this problem is to identify which quantities you're given.*0902

*We're given two angles and a side.*0907

*Once we knew that it was angle side angle, we knew that we had a unique solution because the angles didn't add up too big.*0910

*We could find the third angle and then we kind of worked our way around using the law of sines to solve down and find the missing side lengths.*0918

*We'll try some more examples to that.*0927

*Now, we're given triangle ABC.*0931

*Let's see.*0937

*We're given side a, side b, and angle A.*0938

*Let me draw that out.*0942

*My drawing might not be to scale, that's alright.*0944

*So ABC, I'll use capital letters for the angles, and lowercase letters on the opposite sides, a, b and c.*0952

*We're given that angle A is 40 degrees, we're given that side a has length 3, side b has length 4, and we want to solve for the other quantities here.*0961

*The first thing is to identify what kind of information we've been given.*0975

*Two sides and an angle that is not in the middle, this is a side side angle configuration because the angle is not in the middle, that would be side angle side.*0984

*Side side angle, if you remember back to beginning of the lecture, it might have ...*0993

*Side side angle is the ambiguous case.*1006

*It might have no solutions, one solution, or two solutions.*1008

*That's a little disturbing, that this thing could have more than one solution or it might not have solution.*1030

*We have to solve it out and see what we can find.*1035

*We're going to use the law of sines, that says sin(A)/a=sin(B)/b=sin(C)/c.*1038

*We'll solve that out and we'll see what we can find.*1053

*Let's start with ...*1056

*I see that I know angle A and side a, I know side b but not angle B, and I don't know anything about the C.*1059

*I'm going to start out with the a's and b's.*1067

*I'll start out with sin(A)/a=sin(B)/b.*1068

*Now, sin(40)/3=sin(B)/4.*1080

*I'm solving this thing for sin(B), if I multiply the 4 over, I get sin(B)=4sin(40)/3.*1098

*If I work out what 4sin(40)/3 is, it tells me that it's about 0.86.*1113

*I'm looking for an angle whose sine is around 0.86.*1139

*By the way, I've got my calculator in degree mode since the angles were given in degrees.*1144

*I'm looking for an angle whose sine is about 0.86.*1149

*Here's the thing.*1153

*If I just type arcsin(0.86) on my calculator, it will tell me that B is approximately equal to 59.0 degrees.*1154

*This is where it gets really tricky because, remember, if you're looking for an angle whose sine is 0.86, remember that means its y-value, sine is the y-value, is 0.86 ...*1175

*There are two angles that have that value as its sine, there's a θ and then there's this big angle which is 180-θ.*1204

*So, sin(180-θ) is the same as sin(θ).*1222

*Since we know that sin(B) is 0.86, B could be 59 degrees, angle B could also be 180-59 degrees, in other words, 121 degrees.*1233

*We've got two different possibilities for angle B.*1259

*That means we're going to have two different triangles, we have two solutions here.*1265

*We have two solutions depending on which angle B is.*1276

*For each one, we're going to have to solve around and find the other information in the triangle using that value for angle B. That's a little disconcerting.*1282

*Let me go to a new slide and we will draw out each one of those triangles with each one of those solutions.*1290

*We already figured out with this problem that there are two solutions.*1297

*Let me solve each one of those separately.*1310

*Here's A, B and C.*1317

*We're given that A was 40 degrees, and side a has length 3, side b has length 4.*1320

*In the first solution that we've figured out on the previous page, b had measured 59 degrees.*1329

*Then we want to figure out the values of the other angles and the lengths in the triangle.*1342

*First of all, angle C is 180-40-59, that works out to 81.0 degrees.*1348

*Now, let's figure out what the length of side c is, we'll use the law of sines for that.*1376

*Sin(C)/c=sin(A)/a, sin(81.0)/c=sin(40)/3, if we cross multiply that, we'd get c=3sin(81.0)/sin(40).*1384

*Now, we'll work that part out on my calculator, remember, put your calculator in degree mode for this.*1427

*3sin(81.0)/sin(40), I get an approximate answer of 4.61 for side c there.*1437

*Now, I've solved that triangle completely.*1457

*I've found all three sides and all three angles, but there's another solution, remember, with a different angle for B.*1460

*I have to start all over with that possible triangle.*1468

*Let me redraw the triangle from scratch because I don't want to get confused with any of my earlier work.*1471

*Even though I'm drawing it in a similar fashion, remember this is not drawn to scale, this triangle will actually look quite different if we drew it to scale.*1477

*On the other solution, angle B measured 121 degrees.*1490

*A was still 40 because that was given in the original problem.*1502

*Now we have to find the values for this new triangle.*1505

*It's the same solution techniques, we'll be going through the same kinds of calculations but we're using completely different numbers now.*1509

*C=180-40-121.0, that comes out to be 140-121=19.0 degrees.*1517

*Now, we've found C is 19.0, but we still have to find side little c, we'll use the law of sines for that.*1542

*Sin(C)/c=sin(A)/a, if we fill in what we know, Sin(19.0)/c=sin(40)/3.*1553

*If we cross multiply, we'd get c=3sin(19.0s)/sin(40).*1573

*Got my calculator set on degree mode, so 3sin(19)/sin(40), my calculator tells me that that's approximately 1.52.*1584

*Now, I've solved that triangle completely.*1612

*I've got three angles and three side lengths, totally different from the angles and side lengths that we've solved for in the first triangle, even though they both satisfy the initial data given in the problem.*1615

*That's kind of the curse of the side side angle case.*1627

*It is ambiguous and you can get more than one solution.*1634

*The way we figured out there were two solutions was when we initially tried to solve for angle B.*1638

*We've figured out sin(B)/b=sin(A)/a, then we solved for sin(B) was equal to a particular number there but I think it was around 0.8.*1645

*The problem is there was more than one angle that has that sine, and we have to investigate the possibility of both of them.*1664

*That's what led us to the two solutions.*1670

*We went with each one of those angles, we find the one by the arcsine button on my calculator or the inverse sine, the other one we found subtracting that angle from 180 because those angles have the same sine.*1673

*With each of those angles, those set us on two completely different roads to solve down the triangles using the law of sines and find the other values of the triangles, and get us two completely different answers.*1693

*Both of which are valid, both of which satisfy the initial data given in the problem.*1705

*For our next example, we're given a triangle ABC.*1714

*We have, again, two side lengths and angle not in between them.*1719

*Let me draw this out.*1724

*Angles are always capital letters A, B, C.*1731

*The sides are always lowercase letters opposite the angle with the same letter, there's little a, b, and c.*1735

*This time, we're told that side a has length 7, side b has length 12, and angle A measures 45 degrees.*1742

*It says, determine how many triangles there are with these conditions and solve them completely.*1753

*The first thing is to figure out which of those situations were in, side side angle, side angle side, angle side angle.*1757

*What we want to do is look at this and we've got two sides and an angle, but the angle is not in between them, so this is side side angle.*1770

*Side side angle, unfortunately, that's the ambiguous case.*1779

*It could have, no solutions, no triangles at all, it could have one solution or it could have two solutions.*1791

*We don't really know yet without doing a little extra examination here.*1811

*We're going to start out using our law of sines and see what happens.*1818

*Let me remind you what the law of sines is sin(A)/a=sin(B)/b=sin(C)/c.*1821

*Which of those pieces of information have we been given?*1834

*We know what capital A is, so I can find its sine.*1836

*I know what little a is.*1840

*I know what little b is.*1842

*That's it.*1844

*It makes sense to solve for sin(B).*1846

*Let me rewrite that, sin(A)/a=sin(B)/b, so sin(40)/7=sin(B)/12.*1850

*If you cross multiply that and then solve for sin(B), we get sin(B)=12sin(45)/7.*1872

*I'll plug that into my calculator, 12sin(45)/7 comes out to 1.21 approximately.*1886

*This is very significant because, let's look at this, 1.21=sin(B).*1906

*Remember, the sine of any angle is always between -1 and 1.*1917

*Here we've got the sine of an angle equal to 1.21, that's bigger than 1.*1925

*That's a real problem.*1933

*Sin(B) here is greater than 1, that's a contradiction of what we know about sines and cosines, which means that there can not be any triangle satisfying these conditions.*1937

*As soon as we get to sin(B) being bigger than 1, we know that no such triangle exists.*1956

*If you try to find the arcsine on your calculator of something bigger than 1, my calculator gives me an error, because it knows that sine should always be between -1 and 1.*1968

*Immediately, I know that something's gone wrong, and what's gone wrong is that we must have been given bogus initial data.*1982

*We know that no such triangle exists.*1989

*We know that there are no solutions here, which in a sense is fortunate because it means we don't have to do a lot of work to go ahead and try to find the other sides and angles, because we know there isn't any such triangle in the first place.*1992

*We'll try some more examples like this later.*2007

*Hi! We are doing some more examples on using the law of cos and Heron’s formula.*0000

*We are given one now where the side lengths triangle (a, b and c) are (16, 30, 34).*0005

*We are asked to determine how many triangles satisfy these conditions and to solve the triangles completely.*0013

*What we are given here are three side lengths that is a (side, side, side) situation.*0020

*Because it is (side, side, side), it has a unique solution if it satisfies that check where each side is less than the sum of the other two.*0026

*Let us check that out for these three sides, let us check if (16 < 30+34) that is certainly true.*0059

*Is (30 < 16+34)? Clearly true, Is (34 < 16+30)? That is certainly true.*0068

*There is a unique triangle satisfying these three lengths of sides.*0083

*Let us try to find out what the angles would be in that triangle because we know what the side lengths are, let me draw the triangle.*0099

*(16, 30, 34) I am going to label an angle here, I will label this angle C and then we are going to use the law of cos to find the angles in the triangle.*0112

*That is why I started out by labeling angle C because I know the way I remember cos is with an angle C in it.*0126

*Let me write that down, c*^{2}=(a^{2}) +(( b^{2} - (2ab)cos(C)).0133

*The way I have labeled angle C, that makes the opposite side (c) and then these sides must be (a) and (b) here.*0145

*I can fill everything into the law of cos and then I can solve for angle (C), I will fill that in 16*^{2}= (30^{2}+34^{2}) – 2(30)(34) x cos(C).0154

*(16*^{2})=256, (30^{2}=900)+(34^{2}=1156), ((2 x 30 x 34=2040 cos(C)).0178

*The angles are a little bit messy here, I’m going to move cos(C) over the other and I will get 2040 cos(C)=900+1156-256 that is 1800 because we moved the 256 over on the other side.*0196

*Cos(C) = 1800/2040, (C)=arcos of that horrible fraction (180/204) but I will go ahead and plug that straight into my calculator.*0220

*(180/204) I’m using degree mode for this, you have to be careful because otherwise it would give you an answer in radians, but that gives me answer of about 28.1 degrees.*0253

*That fills in one of my angles for my triangle there 28.1 degrees.*0271

*We got one of the angles, we will find out the other two exactly the same way, let me go ahead and work them out for you, I will redraw my triangle.*0281

*(16, 30, 34) this was 28.1, we already figured that out.*0293

*In order to find the other two angles, what I’m going to do is draw angle (C) in a different place now.*0302

*I am relabeling which angle is which, the point is that is I do not have to rewrite my law of cos switching around the (a, b, c) there.*0306

*This time I’m going to draw my angle (C) there and that makes side (c) the one opposite and (a), (b) other one next to it.*0317

*I will write down my law of cos, (c*^{2})=(a^{2}+b^{2})-((2ab cos (C)).0328

*I will work through and I will solve for (C), that is (30*^{2})=(16^{2}+34^{2})-((2(16)(34)cos(C)).0338

*Now, it is just a little algebra to simplify this, 900=256+1156, now (2×16×34)=1088, and we still have cos(C).*0355

*I want to simplify this, I want to move the cos(C) on the other side, 1088 cos(C)=(256+1156-900), that simplifies down to 512.*0374

*Cos(C) =512/1088, I will take the inverse cos of that in my calculator.*0399

*I get 61.9 degrees approximately, that tells me one more of my angles, I only have one to go.*0418

*I could find the third angle by adding up the two angles and subtracting it 180, but I think I would like to practice the law of cos again.*0440

*At the end we can use that adding up to 180 as to check if we did the law of cos right.*0448

*We are going to find the third angle by using the law of cos again and in order to use my law of cos with the same (a,b, c) formula, I’m going to rotate the (a, b, c) on the triangle.*0453

*I’m going to cross out my old (a, b, c) and I will re-label the angle that I do not know as (C) which means its opposite side is 34 (c).*0468

*(a)(b) are now the two sides next to it, I will plug into the law of cos to solve for (C).*0481

*c*^{2} is (34^{2})=a^{2}is (16^{2})+b^{2} is now (30^{2})- 2(16)(30) cos(C).0491

*34*^{2} is 1156, 16^{2} is 256, 30^{2} is 900, 2×16×30 is 960, cos (C).0510

*Now an interesting thing happens because I am going to subtract 256+900 for both sides, that is equal to exactly 1156.*0527

*What do I get is 0 is -960 cos (C) and if I divide it by -960, I will get cos(C)=0.*0537

*What angle has cos(0)? Well that is exactly a right triangle. (C) is exactly a 90 degree angle, let me fill that one in.*0547

*We could have noticed that this was in fact a right triangle because 16*^{2} + 30^{2} = 34^{2}.0558

*It does not matter because the law of cos works in all triangles.*0568

*It works just as well in right triangles as in other triangles but of course the rules for right triangles SOHCAHTOA do not work in other triangles.*0571

*Now we solved the triangle completely, we got all three sides and we got all three angles but I want to check and see if those three angles actually do add up to 180 degrees.*0581

*To check here, I look at (28.1+61.9+90) if you add those together you will indeed get 180 degrees which means we must have done the problem right.*0594

*Let us just recap what we did there, we are given a (side, side, side) presentation of a triangle.*0612

*The first thing that we did was check that each one of those sides was less than the sum of the other two.*0621

*If that check have not work out, we would have stopped right there and said there is no such triangle, but that check did worked out.*0626

*Then we go on to using the law of cos to find the each of the angles of the triangle.*0632

*We take the law of cos here and we fill in the lengths of the three sides, and then we solve down to find the cos of a missing angle.*0641

*Once we know the cos of the missing angle, we can use our cos to find the angle itself.*0652

*We did that three times just applying it to each angle and term, we got each of the three angles and then we check in the end if they added up to 180.*0656

*A triangle has two sides of length 8 with an included angle of 45 degrees, we want to find the length of the third side and the area of the triangle.*0000

*Let me try drawing that, there is 8, there is 16, that is about a 45 degree angle but it is not really intended to be drawn on the scale.*0010

*Notice here that we are given here a (side, angle, side) presentation of a triangle, (side, angle, side) and the angle is less than 180 degrees.*0026

*We know that there is a unique triangle satisfying this data.*0036

*We want to find the length of the third side, now this is tailor made for the law of cos.*0041

*Let me write down the law of cos to get started, the law of cos says that (c*^{2})=(a^{2}+b^{2}) – ((2ab cos(C)).0046

*Let me call the missing side (c), which means that its opposite angle is (C) and (a) and (b) will be the sides that we know.*0061

*Now, we can put all that information into the law of cos and we can solve for the missing side (c).*0069

*I plug that in, (c*^{2}=(8^{2} is 64), (16^{2} is 256) – ((2(8)(16) cos(C)).0077

*(64+256=320) – (2x8x16=256), (C) is the given angle, that was given as 45 a degree angle and I know what is the cos of 45 degree is.*0094

*That is one of the common values that we learned earlier on the trigonometry lectures, the cos of that is the same as pi/4, the cos of the is square root 2/2.*0114

*That simplifies down to 320-128 square root of 2, (C) is equal to the square root of 320-128 square root of 2.*0127

*That is probably something worth checking out on the calculator, I work out the square root of 320-128 square root of 2.*0144

*It tells me that it is approximately 11.8, we have that third side is approximately 11.8 units long.*0157

*That is the first problem of the example here, now we are asked to find the area of the triangle.*0173

*I want to do a little more trigonometry to find that area, I am going to drop altitude from this top angle here and I want to try to find the length of that altitude.*0182

*The reason I’m trying to find is that is I remember the area formula, area=1/2 base x height.*0192

*I know the base is 16, that is even labeled with side length (b), do I have it labeled on my triangle.*0201

*The height is the length of that altitude, I have got to solve for that length right there.*0211

*I’m going to use SOHCAHTOA here because I have the hypotenuse of the triangle and I have the angle here.*0217

*I know that by SOHCAHTOA, sin(theta)=opposite/hypotenuse, sin(45)=opposite/8 and so the opposite=(8)sin(45) is equal to 8.*0224

*sin(45) is something I know because it is a common value, it is pi/4.*0257

*That is square root 2/2, that is 4 square root 2=length of the opposite there, that is 4 square root 2.*0261

*My area, which is ½ base x height which is ½(16)(4 square root of 2) is what I figured out the height was.*0275

*That is (8 x 4), that is 32 square root of 2, is my area.*0290

*If you want that to be a decimal, we can approximate that in the calculator as about 45.3.*0297

*That gives us the area of the triangle based on the ½ base x height calculation, we really done with this one.*0311

*I like to check it using another formula we learned from trigonometry which is heron’s formula.*0319

*Let me remind you what heron’s formula is.*0326

*Heron’s formula says that the area of the triangle, if you know all three sides, which we did figure out, that is the square root of s(s-a)(s-b)(s-c), that was heron’s formula.*0331

*(a, b, c) are the lengths of the sides of the triangle but this mysterious quantity (s) is the semi perimeter, that means ½ of the perimeter which is the sum of the sides.*0350

*Let us work that out first, we know that two of the sides were 8 and 16, and we worked out on the previous side that the third side is approximately equal to 11.8.*0363

*We put that together, 8 + 16 = 24 +11.8 =35.8 and half of that is 17.9, that is the semi perimeter.*0381

*Let me drop that in to heron’s formula now, the area is equal to (17.9 x 17.9 – a(8)) ((17.9-b(16))((17.9-c(11.8)).*0395

*Now it is just a matter of simplifying that, that is 17.9–8 = 9.9, 17.9-16=1.9, 17.9-11.8=6.1.*0430

*I’m going to multiply those together on my calculator. *0452

*I get 2053.9, take the square root of that and I get approximately 45.3, which is what we figured out on the previous side.*0466

*That is a very useful check that we are doing everything right on the previous side.*0483

*Let us recap what we did there, we are given a triangle with two sides and an included angle (8, 16, and the included angle which is 45).*0488

*We use the law of cos to find the length of the third side of the triangle.*0498

*The law of cos is tailor made if you have (side, angle, side), you use the law of cos to find the length of the third side.*0504

*We then have to find the area of the triangle, I did that the first time by dropping an altitude of the triangle.*0512

*Find the length of the altitude and use the old geometry formula ½ base x height to find the area.*0519

*The other way we could possibly find the area was to use heron’s formula, which is useful when you know all three sides of the triangle.*0526

*You find the semi perimeter which is what we did here and then you take that and you drop it into heron’s formula for the area and you drop all three sides in there.*0534

*It is just a matter of working through some arithmetic to find the area.*0544

*That is the end of our lecture on the law of cosines and solving triangles and using heron’s formula.*0550

*These are the trigonometry lectures on www.educator.com.*0555

*This is Will Murray for educator.com and we're here today to talk about the law of cosines, which is the second of the two big trigonometric rules.*0000

*Remember, last time we talked about the law of sines.*0007

*You kind of put those together, and together those enable you to find the length of any side and the measure of any angle in a triangle, if you're given enough information to start with.*0011

*Let's start with the formula here.*0022

*The law of cosines is c*^{2}=a^{2}+b^{2}-2abcos(C).0024

*Let me draw you a a triangle so we can see how that applies.*0031

*Remember, the convention is that you use lowercase a, b, and c for the sides of the triangle, and uppercase A, B, and C for the angles.*0037

*You use the same letter for the angle and the side opposite it.*0048

*My uppercase A goes here, and my B goes here because it's opposite of side b, here's angle C.*0053

*The point of the law of cosines is it relates the lengths of the three sides a, b, and c, little a, little b and little c, to the measure of one of the angles which is capital C here.*0061

*The point is that, first of all, you can use this in any triangle.*0077

*It's not just valid in right triangles.*0080

*Remember the big rule we had, SOH CAH TOA, is only valid in right triangles.*0084

*The law of cosines is valid in any triangle.*0088

*It's a generalization of the Pythagorean theorem, in a sense that, remember the old Pythagorean theorem was just c*^{2}=a^{2}+b^{2}, that only works in a right triangle.0093

*If you look at the law of cosines, if angle C is a right angle, then the cos(π/2) or the cos(90) is zero.*0105

*If angle C is a right angle, then this term 2ab-cos(c), drops out,the law of cosines just reduces down to the Pythagorean theorem, c*^{2}=a^{2}+b^{2}.0115

*You can kind of think of the Pythagorean theorem as just being a consequence of the law of cosines.*0128

*The law of cosines is the more general one that applies to any triangle.*0134

*The Pythagorean theorem is the more specific one that just applied when angle C happens to be a right angle.*0137

*Let's see how it's used.*0142

*The law of cosines is really used in two situations.*0146

*First of all, it's used in a side angle side situation.*0151

*That means where you know two sides of a triangle and the included angle.*0155

*The reason it's useful ...*0160

*Let me write the law of cosines again c*^{2}=a^{2}+b^{2}-2abcos(C).0162

*The point here is that if you label this sides as little a and little b here, that makes this angle capital C and little c is down there.*0173

*If you know the side angle side, in other words, if you know, little a, little b, and capital C, then you know all of the right-hand side of the law of cosines.*0185

*You can then solve for little c.*0196

*That's why the law of cosines is useful for side angle side situations, it's because you can fill in everything you know on one side of the law of cosines, then you can solve for little c.*0199

*It's also useful for side side side situations.*0211

*Let me draw that.*0218

*Side side side means you know all three sides of a triangle, but you don't necessary know any of the angles yet.*0219

*The point is, if you know little, little b, and little c, then you know all of these parts of the law of cosines, so you can solve for the cosine of capital C, and you can figure out what that angle capital C is.*0227

*Then you can figure out what one of the angles is.*0245

*You can just kind of rotate the triangle, and relabel what a, b, and c are to find the other two angles.*0248

*If you know all three sides of a triangle, the law of cosines is very useful for finding the angles, one at a time.*0254

*Remember, there's a couple other ways that you can be given information for triangles.*0261

*You can be given, angle side angle, or side angle angle, or side side angle.*0266

*Those two don't really lend themselves very well to solution by the law of cosines.*0273

*If you're given one of those situations then you really want to use the law of sines which we learned about in the previous lecture.*0280

*There's one more formula we're going to be using in this lecture which is Heron's formula.*0289

*The point here is that, if you know all three lengths of sides of a triangle, I'll call them a, b, and c, as usual, then you have a nice formula for the area.*0294

*It's got one more variable in it, this s.*0309

*S is 1/2 (a+b+c), that's the semi perimeter.*0312

*Remember, the perimeter is the distance around the edge of a triangle, that's a+b+c.*0317

*The semi perimeter is just 1/2 of a+b+c.*0323

*We worked that out ahead of time and we call it s.*0327

*We plugged that into this formula, that's a fairly simple formula just involving s and then a, b, and c, and it spits out the are of the triangle for us.*0331

*That's very useful if you know the lengths of the sides.*0339

*You never really have to look at any angles, and you don't have to get into any sines or cosines, no messy numbers there, hopefully.*0342

*Let's try out some examples here for law of cosines.*0350

*First example, we're given a triangle ABC.*0353

*Let me go ahead and draw that.*0355

*I'm going to put a here, and b here, and c here, which forces the angles, remember the angles go opposite the sides of the same letter.*0362

*We're given the a=3,b=4, and angle c measures 60 degrees. *0374

*We want to first of all determine how many triangles satisfy these conditions and then we want to solve the triangles completely.*0383

*To answer the first question, we have a side angle side situation.*0392

*What we know is that side angle side always has a unique solution assuming the angle is less than 180.*0400

*In this case, the angle is 60 which is less than 180, so there's a unique solution.*0408

*There's exactly one triangle satisfying these conditions.*0419

*That answers the first question, how many triangles satisfy those conditions, exactly one.*0431

*Now we have to solve the triangle completely, this is where the law of cosines is going to be useful.*0438

*Let me copy down the law of cosines c*^{2}=a^{2}+b^{2}-2abcos(C).0444

*This is very useful because we know a and b, and capital C.*0453

*I can just plug all those in and solve for little c.*0460

*Let me do that, a*^{2} would be 9, b^{2} would be 16 because 4^{2} is 16, minus 2×3×4=24, cos(60) ...0464

*This is 25-24, now cos(60), that's one of my common values, that's π/3, I remember that the cos(60) is 1/2.*0483

*This is 25-24×1/2, 25-12, c*^{2} is 13, c is equal to the square root of 13.0493

*I can get an approximation for that on my calculator, that's about 3.61.*0513

*That's approximately equal to 3.61.*0521

*I'll fill that in on my triangle.*0530

*Now, I've got the third side of the triangle.*0536

*The only thing that's left is it says to solve the triangles completely.*0540

*I need to find the other two angles A and B.*0544

*To do that, I'm going to use the law of sines.*0547

*Let me write down the law of sines to refresh your memory, sin(A)/a=sin(C)/c.*0551

*I know what side a, side c, and capital C are.*0569

*I'm going to cross multiply this and I get csin(A)=asin(C).*0574

*I'll fill in the values that I know.*0583

*I know little a is 3, sin(C)=sin(60), sin(A), I don't know that yet, and little c, I figured out, is 3.61.*0585

*I'm solving for sin(A)=3sin(60)/3.61.*0604

*I'll just work that out on my calculator.*0616

*What I get is approximately 0.72.*0629

*By the way, it's very important that your calculator be in degree mode if you're using degrees here.*0637

*I gave angle C as 60 degrees.*0642

*It's very important that you set your calculator to degree mode.*0646

*If your calculator's in radian mode, then it will interpret that 60 as a radian measure, so your answers will be way off, so set your calculator to be in degrees before you try this calculation.*0650

*A is arcsin(0.72), I'll work that out on my calculator.*0661

*That tells me that a is just about 46.0 degrees.*0673

*Now, I've got a measure for angle A.*0683

*I'm going to use the law of sines to find the measure for angle B, but I need a little more space.*0688

*Let me redraw my triangle.*0694

*We've got a, b, and c.*0704

*I figured out the c was 3.61, a was given as 3, b was given as 4, C was given as 60.*0711

*I figured out that A was 46 degrees.*0721

*I'm just trying to find the measure of angle B now.*0727

*I'm going to use again the law of sines.*0729

*Sin(B)/b=sin(C)/c, I'll fill in what I know here, I know that little b is 4, sin(B), I don't know, sin(C)=sin(60), little c is 3.61.*0734

*I'll cross multiply that, 3.61sin(B)=4sin(60), sin(B), that's what we're solving for is equal to 4sin(60)/3.61.*0760

*Let me work that out on my calculator, 4sin(60)/3.61=0.96.*0784

*B is arcsin(0.96) and I'll work that out, that's 73, just about 74 degrees, rounds to 74 degrees.*0801

*Now I figured out angle B, 74 degrees.*0828

*Now, I've solved for all three sides of the triangle, and all three angles of the triangle.*0837

*It's nice at this point, even though we're done with the problem to get some kind of check because we've done lots of calculations here, we could have made a mistake.*0841

*What I'm going to do is add up all three angles in the triangle, and make sure that they come up to be 180 degrees.*0848

*To check on my work here, I'll add up 60+46+74, that does indeed come out to be 180 degrees.*0856

*That suggests that we probably didn't make a mistake in solving all those angles.*0877

*Just to recap this problem here, we're given a side angle side situation, that's a definite tip-off that you're going to be using the law of cosines.*0882

*I filled in my side, the included angle, and a side.*0891

*The first thing I did was I used the law of cosines to find the missing side.*0895

*To solve the triangle completely, I still had two angles that I didn't know, I used the law of sines after that to find the two angles that I didn't know based on knowing the other sides and the other side, and angle.*0903

*Let's try another one now.*0918

*In this one, we're given the side lengths of the triangle, a, b and c are 5, 7 and 10.*0922

*Let me draw a possible triangle like that, 5, 7 and 10.*0926

*We want to find out how many triangles satisfy this conditions and solve the triangles completely.*0935

*The first thing to do with this problem is to identify what we're given.*0941

*We're given a side side side configuration.*0945

*That usually gives you a unique triangle.*0950

*What you have to do is check that each side is less than the sum of the other two.*0952

*Let's check that out.*0960

*Unique if each side is less than the sum of the other two.*0963

*That'll be a quick real check.*0980

*We're comparing 5 with 7+10, 5 is certainly less than 17, so that works.*0983

*7 should be less than 5+10, that certainly works, 7 is less than 15.*0989

*10 should be less than 5+7, that certainly works.*0995

*If any of those checks had failed, for example 10 if it had been 13, 5, and 7, instead of 10, 5, and 7, then the last check would have failed because 13 is not less than 5+7.*1001

*At that point, we would have stopped and said, "This is invalid. There is no triangle that satisfies those conditions."*1015

*Since all those three conditions checked, it does mean that there is a unique solution.*1022

*There's exactly one triangle with those three lengths of sides.*1036

*We've found all the side lengths.*1043

*We need to find the angles, this is where the law of cosines is really handy.*1046

*Let me write that down, the law of cosines says c*^{2}=a^{2}+b^{2}-2abcos(C).1051

*You can label the sides and angles whichever way you want.*1061

*I'm going to label the top one c.*1067

*Let me write that outside of the triangle.*1075

*That makes the bottom side c and then the two other sides a and b.*1077

*What I can do here is I can plug in little a, little b and little c, and then I can solve for the cosine of capital C, then in turn solve for what angle capital C is.*1085

*Let me do that.*1098

*If c is 10, that's 10*^{2} equals, a and b is 5^{2}+7^{2}-2×5×7×cos(C).1100

*Now it's just a little bit of arithmetic, 100=25+49-70cos(C).*1115

*25+49=74, if we pull that over to the other side, we get 26, -70cos(C).*1128

*Cos(C)=-26/70.*1145

*I'm going to figure out that C is arccosine or inverse cosine of -26/70, I'll do that part on my calculator.*1153

*Remember, you have to be in degree mode for this.*1180

*What I get is that C is approximately equal to 111.8 degrees.*1186

*That's 111.8 in that corner.*1196

*Now I'm going to go over in the next page.*1201

*I'm going to keep going to find out the other two angles.*1204

*We'll find them exactly the same way.*1206

*Let me go ahead and redraw my triangle, 5, 7, 10.*1207

*We've already figured out that that angle is 111.8.*1216

*What I'm going to do is relabel the sides and the angles because I still want to use that law of cosines c*^{2}=a^{2}+b^{2}-2abcos(C).1220

*I'm going to relabel everything here so that I can label C as a new angle, and I can solve for a new angle.*1233

*Relabel that angle as capital C, then my a and b will be 7 and 10.*1242

*Sorry, small C will be 7, and a and b will be 5 and 10.*1256

*We'll go through and we'll work that one out.*1263

*Little c*^{2}, that's 49 is equal to a^{2}, that's 100, plus b^{2} is 25, minus 2ab, that's 2×10×5×cos(C).1267

*Let me work this out, this is 125, 49 equals 125 minus 2×10×5, that's 100, cos(C).*1285

*If I subtract 125 from both sides, I get -76 is -100cos(C), divide both sides by -100, we get 76/100=cos(C).*1300

*C=arccos(76/100), I'll plug that into my calculator.*1324

*It tells me that that's approximately 40.5 degrees for that angle right there.*1342

*Now, there's one angle left to find.*1362

*Again, I'm going to relabel the sides and the angles so that I can continue to use the law of cosines in its standard form.*1364

*I don't have to change around what a, b and c are in the law of cosines.*1373

*I'll do this in red.*1375

*In red, I'm going to call this angle capital C, and that means I have to relabel my sides, that means little c is equal to 5, a and b are 7 and 10.*1380

*Now I'm going to plug those values into the law of cosines.*1399

*c*^{2} is 25, equals 49, 7^{2}, plus b^{2} is 100, minus 2×a×b, 7×10, times cos(C).1403

*Now it's a matter of solving that for capital C again.*1422

*I get 25 equals 149, minus 2×7×10, that's 140, cos(C).*1424

*If I subtract 149 from both sides, I get -124, is equal to -140cos(C).*1440

*Cos(C)=124/140, C=arccos(124/140), I'll go to the calculator on that.*1455

*I get 27.7 degrees.*1481

*Let me fill that in, 27.7 degrees.*1490

*Now we've solved the triangle completely.*1494

*We started out with all three side lengths and we found all three angles in the triangle.*1497

*We're really done but it's always good to find a way to check your work.*1503

*Let me check my work in blue here.*1506

*Again, I found all three angles.*1508

*I'm going to add them together and see if I get 180 to check that out.*1511

*I'm going to add up 111.8+40.5+27.7, what I get is exactly 180 degrees.*1514

*That tells me that I must have been right in getting those three angles for the triangle.*1535

*Just to recap here, what we were given was three sides of a three angle.*1540

*We were given three side lengths, that was a side side side situation.*1546

*We had to check that the three lengths, that we didn't have a situation where two sides added up to be less than the third side.*1554

*We had to check that each side was less than the sum of the other two.*1560

*Once we did that, we knew we have exactly one solution.*1564

*Then we filled in the side lengths of the triangle, and we used the law of cosines.*1568

*The law of cosines lets you fill in three side lengths and then solve for the cosine of one of the angles.*1574

*You work it through, solve for the cosine, then you get each one of the angles by taking the arccosine, then you just go through a separate procedure like that for each one of the angles.*1580

*In our third problem, we're trying to find the area of a triangle whose side lengths are 5, 7 and 10.*1595

*Now, this one is really a set up for Heron's formula, because Heron's formula works perfectly when you know the three sides of a triangle.*1603

*We're going to use Heron here.*1612

*Heron says that the area is equal to the square root of s times s minus a, s minus b, and s minus c.*1618

*You've got to figure out what s is.*1632

*S is the semi-perimeter of a triangle which means 1/2 of the perimeter, 1/2 of the sum of the three side lengths.*1635

*That's 1/2 of 5+7+10, which is 1/2 of 22, which is 11.*1646

*I'm going to plug that in to Heron's formula, wherever I see an s, that's 11×(11-5)×(11-7)×(11-10).*1658

*Then I'll just work on simplifying that, that's 11×6×4×1, that's the square root of 264, 11×24.*1677

*That problem was really pretty quick if you remember Heron's formula.*1697

*Heron's formula is very useful.*1700

*If you know three side lengths of a triangle, then what you do is you work out the semi-perimeter, you just drop the side lengths into this formula for the semi-perimeter, then you drop the semi-perimeter and the three side lengths for the Heron's formula for the area.*1703

*It simplifies down pretty quickly to give you the area of the triangle.*1720

*We'll try some more examples later on.*1722

*Hi we are trying more examples of problems where you have to find the area of a triangle given various data, this one we are given two sides (7, 13, and an included angle of 32 degrees).*0000

*Let me draw out the triangle there, (7, 32, and 13), I want to find the area of this triangle.*0015

*I’m going to drop perpendicular and altitude here and I’m going to use SOHCAHTOA to find the length of that perpendicular.*0034

*Remember, sin(theta) = opposite/hypotenuse and sin(32) = the altitude of the triangle is the opposite side that I am looking for and the hypotenuse is 7.*0042

*Opp=(7)sin(32), I will check that on my calculator and it says that it is approximately equal to 3.7, that is this length right here, 3.7.*0064

*The area, the old fashioned area formula is ½ base x height, which is equal to ½, now the base is 13, we were given that, x 3.7.*0083

*What we get is 24.1.*0111

*Let us recap what we did there, I drew out what I knew on the triangle.*0122

*I knew two sides and an included angle, I wanted to find an area.*0126

*I wanted to find base and height of the triangle so I dropped this altitude down from the top corner and then I found out the length for the altitude using SOHCAHTOA and the altitude and hypotenuse that I already knew.*0130

*Remember SOHCAHTOA does not work in evry triangle, it only works in right triangles.*0144

*What I did here, by dropping the altitude, the perpendicular, was I made a little right triangle inside the triangle that we are given so it is ok to use SOHCAHTOA there.*0149

*Once I found the length of that altitude, I just used area is 1/2 base x height.*0160

*I knew the base, height, and I plugged down it into the formula and simplified it down to 24.1.*0167

*Now I know the area of the triangle.*0173

*Alright our last example here, we are given two sides of a triangle and an included angle.*0000

*We want to find the area of the triangle, let me draw out what we know.*0007

*We got side of length 5, side of length 6, and the included angle we are told is 60 degrees.*0012

*I want to find the area of the triangle, what I would like to do is find the length of the third side of the triangle.*0030

*I can do that since I know (side, angle, side), I can find the length of the third using the law of cos so I will do that first.*0037

*Once I know the length of all three sides, I’m going to use heron’s formula to find the area of the triangle.*0047

*Let me write down the law of cos, c*^{2} = a^{2} + b^{2} - ((2ab) cos (C)).0052

*That is really useful if you know two sides and an included angle because if you call those two sides (a and b), that makes (c) the third side.*0065

*(C) will be the angle in between (a) and (b) because it should be opposite (c).*0076

*We can plug all those into this formula and solve for (c), c*^{2} = 25 that is 5^{2} + 6^{2} is 36 – (2)(5)(6) cos(C) which is 60 degrees.0082

*(25 + 36 = 61) – (2×5×6=60) cos(60) that is a common value, I know that one, that is pi/3.*0101

*I know the cos of pi/3 is ½.*0114

*This is 61-30=31 that was c*^{2} and c=square root of 31 which is approximately equal to 5.6.0118

*My third side here is approximately equal to 5.6.*0141

*I’m going to the next slide to finish this so let me fill in what I just figured out.*0149

*I know that I had (5 and 6) and I just figured out that the third side was approximately equal to 5.6.*0158

*That angle was 60 but we are not going to be using that anymore.*0166

*Now I want to find the area of the triangle since I know the three sides already, this will going to be quick using heron’s formula.*0170

*Heron says that the area = square root(s)(s-a)(s-b)(s-c).*0181

*You have to know what (s) is, (s) is the semi perimeter so ½ x the full perimeter (a+b+c).*0196

*Let us work that out first, then we will drop that into heron’s formula.*0206

*This is ½ of (5+6+5.6), 5+6=11+5.6=16.6, ½ of 16.6 is 8.3.*0211

*Now I’m going to drop that into heron’s formula and also the lengths of the three sides because I took some trouble to work out the length of the third side.*0232

*My (s) is 8.3 then I have (8.3- the first side=5), (8.3 -the second side=6) and finally (8.3- the third side =5.6).*0245

*It is just a matter of simplifying the numbers,(8.3 x 3.3 x 2.3) and (8.3 – 5.6=2.7).*0271

*I will multiply those numbers together, what I get is 170.1.*0287

*To take the square root of that I get just about 13.04 as my area.*0303

*Let us recap what we needed to do for that problem.*0318

*We were given two sides and an included angle so that is (side, angle, side) and we wanted to find the area of the triangle.*0321

*We used some different methods in the previous problems, this time what we used was the law of cos to find the third side.*0329

*We used the law of cos there and that was really useful because the law of cos is perfect when you have (side, angle, side).*0338

*You just drop the (side, angle, side) into your law of cos and it will tell you what the third side is.*0349

*Once you have all three sides, then you use heron’s formula which gives you this nice formula for the area.*0355

*It does not look at the angle at all, it just uses the three sides and this quantity (s).*0362

*But then (s) just comes back to popping in the three sides into the semi perimeter formula.*0366

*We worked out the semi perimeter, it came to 8.3, drop that in for the (s) in heron’s formula, drop the three sides in and then just simplify it down to find the area of the triangle.*0372

*What I hope you taken away from this is that there are several different ways you can find the area of a triangle.*0387

*You can use the law of cos, heron’s formula, we also used SOHCAHTOA, and ½ base x height.*0392

*Different methods work better in different situations but they all help you find the area of a triangle.*0399

*These are the trigonometry lectures for www.educator.com, thanks for watching.*0406

*We're working on the trigonometry lectures, and today we're talking about finding the area of a triangle.*0000

*We've learned several important formulas over the past few lectures.*0007

*Today, we'll be combining them all and learning a few different methods to find the area of a triangle depending on what kind of initial data you're given.*0009

*I want to remind you about some very important formulas, first of all the master formula that works for right triangles is SOH CAH TOA.*0019

*Let me remind you how that works.*0028

*It only works in a right triangle.*0030

*You have to be very careful of that.*0032

*You have to have one right angle.*0034

*If you talk about one of the other angles, θ, then you label all the sides as the hypotenuse, the side opposite θ and the side adjacent θ.*0036

*SOH CAH TOA stands for the sin(θ) is equal to the length of the opposite side over the hypotenuse, the cos(θ) is the length of the adjacent side over the hypotenuse, the tan(θ) is equal to the opposite side over the adjacent side.*0049

*Remember, we have a little mnemonic to remember that, if you can't remember SOH CAH TOA, you remember Some Old Horse Caught Another Horse Taking Oats Away.*0065

*Now, that'll help you spell out SOH CAH TOA.*0075

*Remember, SOH CAH TOA only works in right triangles, you have to have a right angle to make that work.*0078

*The law of cosines works in any triangle, I'll remind you how that goes.*0083

*We're assuming here that your sides are labeled a, b, and c, little a, little b, and little c, then you label the angles with capital letters opposite the sides with the same letter, that would make this capital A, capital B, and this, capital C.*0090

*The law of cosines relates the lengths of the three sides, little a, b and c, to the measure of one of the angles, c*^{2}=a^{2}+b^{2}-2abcos(C).0107

*This works in any triangle, it does not have to be a right triangle.*0121

*In fact, if it happens to be a right triangle then the cos(C), if C is a right angle, the cosine of C is just 0, that whole term drops out and you end up the Pythagorean theorem for right triangle, c*^{2}=a^{2}+b^{2}.0124

*Finally, the important formula that we're going to be using for areas is Heron's formula.*0141

*That's very useful when you know all the three sides, a, b and c, of a triangle ABC.*0148

*First, you work out this quantity s, the semi-perimeter, where you add up a, b and c, that's the perimeter, divide by 2, so you get the semi-perimeter, then you drop that into this area formula, and you drop in the lengths of all three sides.*0158

*Heron's formula gives you a nice expression for the area of a triangle without ever having to look at the angles at all.*0174

*That's very useful as well.*0181

*We'll practice combining all those formulas in different combinations and see how we can calculate the areas of triangles in various ways.*0184

*The first example, we're given a triangle that has two sides of length 8 and 12 with an included angle of 45 degrees.*0193

*Let me set that up.*0200

*That's 8, that's 12, and this is 45 degrees.*0203

*I'll draw in the third side there.*0210

*I'm not going to try to use Heron's formula yet, I'm going to try to use old-fashioned SOH CAH TOA here.*0211

*The thing is, remember, SOH CAH TOA only works in right triangles.*0217

*I don't necessarily have a right triangle here.*0221

*What I'm going to do is draw an altitude, drop a perpendicular from the top corner of that triangle.*0222

*Now I do have a right triangle.*0233

*I'm going to try and find the length of that altitude, I'm going to use SOH CAH TOA.*0234

*Sin(θ) is equal to the opposite over the hypotenuse.*0240

*Sin(45) is equal to the opposite, now the hypotenuse of that little triangle is 8, so root 2 over 2 is equal to the opposite over 8.*0247

*That's because I know the sin(45), that's one of my common values, that's π/4.*0262

*I memorize the sine, cosine and tangent of all the common values way back earlier in the course.*0267

*If you haven't memorized that, you really should commit all those common values 45 and multiples of 30 to memory.*0273

*If I solve for the opposite here, I get that the length of the opposite is equal to 8 root 2 over 2, that's 4 root 2.*0280

*That means that that altitude is 4 root 2.*0291

*Now we can use the old formula from geometry for the area of a triangle, just 1/2 base times height.*0300

*The h stands there for height instead of hypotenuse.*0309

*I know that's a little confusing to be using h for two different things, but we're kind of stuck with that in English that hypotenuse and height both start with the same letter.*0311

*One-half the base here is 12, and the height we figured out was 4 square root of 2.*0320

*We multiply those out, that's 6 times 4 square root of 2, that's 24 square root of 2 for my area there.*0329

*That one came down to drawing an altitude in the triangle and then using SOH CAH TOA.*0341

*We didn't really have to use anything fancy like the law of cosines or Heron's formula, although we could have.*0347

*You'll see some examples later where we use the law of cosines and Heron's formula instead.*0352

*In this one, we just drew this altitude, we used SOH CAH TOA to find the length of the altitude, then we used the old-fashioned geometry formula, 1/2 base times height, to get the formula of the triangle.*0358

*In the next example, we're given a triangle with side lengths 10, 14 and 16.*0372

*Let me draw that.*0377

*10, 14, and 16 ...*0382

*We're asked to find the area.*0384

*If you have all three sides of a triangle and you're asked to find the area, it's kind of a give-away that you're going to use Heron's formula.*0386

*Because Heron's formula works very nicely based on the side lengths of the triangle only.*0394

*You never even have to figure out what the angles are.*0399

*That's what we're going to use.*0402

*Heron, remember, says that you have to start out by finding the semi-perimeter.*0404

*That's (1/2)×(a+b+c), that's (1/2)×(10+14+16).*0410

*(10+14+16) is 40, a half of that is 20.*0422

*Now I know what s is.*0425

*Heron says I plug that into my area formula which is this big square root expression of s times s minus a, s-minus b, and s minus c.*0427

*My s was 20, a is 10, b is 14, and c is 16.*0444

*Now this is just a very easy simplification, this is 20 times 10, times 20-14 is 6, and 20-16 is 4, so 20×10 is 200, times 24 is 4800.*0460

*I can simplify that a bit.*0485

*I can pull off, let's see, I can pull a 10 out of there, and that's 10 square root of 48.*0488

*Now, I can pull a 16 out and make that 40 square root of 3 for my area.*0498

*Let's recap what we're given there.*0506

*We're given a triangle and we were told the three side lengths.*0508

*If you know the three side lengths and you're going for the area, you almost certainly want to use Heron's formula.*0513

*It's a quick matter of finding the semi-perimeter, and then dropping the semi-perimeter and the three side lenghts into the square root formula.*0518

*Then you just simplify down and you get the area.*0526

*In this third example, we're asked to find the area of a triangle whose side lengths are 7, 9 and 14.*0531

*Ordinarily, I'd say that's a dead give-away that you want to use Heron's formula, but unfortunately, this example they specifically say without using Heron's formula.*0539

*I'd really like to use Heron's formula on that but it has asked me to find the area of the triangle without using Heron's formula.*0549

*I'm going to do a little bit of extra work here.*0558

*I'll start out by drawing my triangle.*0561

*There's 7, 9, and 14.*0565

*I think I'm going to try to find that angle right there.*0572

*I'll call that angle C.*0575

*The reason I called it angle C is because I'm planning to use the law of cosines to find that angle.*0577

*Remember the law of cosines is very useful if you know all three sides.*0581

*You can quickly solve for an angle.*0587

*Let me remind you what the law of cosines is.*0589

*It says c*^{2}=a^{2}+b^{2}-2abcos(C).0592

*If you know little a, b and c, you can just drop them into that formula, and you can solve for capital C.*0602

*That's what we're going to do here.*0607

*I wrote 19 for that side, of course, that's supposed to be 9.*0610

*My c is the side opposite angle C, a is 7 and b is 9.*0617

*I'm going to plug these into the law of cosines, and it says, 14*^{2}=7^{2}+9^{2}-2×7×9×cos(C).0625

*I'll do a little algebra on that.*0646

*14*^{2} is 196, 7^{2} is 49, 9^{2} is 81.0648

*I was looking ahead to this next calculations, 7×9 is 63, 2 times that is 126.*0662

*We get 126cos(C).*0671

*49+81 is 50+80, that's 130.*0674

*If I move that to the other side, I get, let's see, 130 from 196 is 66, is equal to -126cos(C), so cos(C), solving for that part, that's what we don't know, is -66/126.*0682

*If I take the arccosine of that on my calculator, of course I've got to use degree mode if I'm planning to use degrees for this problem that I am, I'll take the arccos(-66/126).*0705

*What I get is 121.6 degrees approximately for angle C.*0723

*Normally, I would just fill that into my drawing, but in my drawing, I've shown it as an acute angle and that really isn't appropriate because 121.6 is bigger than 90 degrees.*0731

*I think I have to modify my drawing based on what C came out to be, and draw that part as an obtuse angle.*0743

*I'll redraw there.*0756

*That's angle C which we figured out was 121.6 degrees.*0762

*That's side c which is 14, that's side a which is 7, and this is side b which is 9.*0769

*What I'm going for is I'm trying to find the area of the triangle ultimately, but I can't use Heron's formula because the problem specifically told me I wasn't allowed to use Heron's formula.*0777

*I want to try to use the old-fashioned 1/2 base times height, but I don't know the height of the triangle.*0791

*I'm going to try to find the height of the triangle, that's why I had to find that missing angle.*0799

*To find the height of the triangle, I'm going to drop an altitude here.*0804

*Since it's an obtuse angle in the triangle, this altitude is actually outside the triangle.*0809

*There's that altitude.*0817

*I want to find the height of that altitude.*0820

*In order to find it, I have to find that angle right there, and that is the supplement of 121.6.*0821

*So θ=180-121.6, which is approximately equal to 59.4 degrees.*0832

*That tells me what angle θ is.*0845

*Based on that and this a=7, I can find the missing height of the triangle.*0848

*I'm going to use SOH CAH TOA for that.*0857

*Sin(θ) is equal to the opposite over the hypotenuse.*0860

*Sin(59.4) times the hypotenuse, is 7 ...*0869

*We don't know what the opposite length is.*0880

*I'll leave that there.*0884

*I get that the opposite is equal to 7sin(59.4).*0885

*I'll work that out on my calculator.*0894

*What I get is that that's approximately equal to 6.03.*0910

*That length right there that we found, 6.03.*0919

*I'm going to need a little more space to work this out, I'm going to go to the next slide here and just redraw my triangle, and remind you what we know about that.*0923

*We were given this triangle with side lengths 7, 9, and 14.*0935

*The first thing we did was we used the law of cosines to find this missing angle.*0942

*That was 121.6 degrees.*0947

*Then we dropped a perpendicular, an altitude, from the top angle there.*0953

*To find that, we had to figure out that θ there was 59.4 degrees.*0960

*Then using that value of θ and the hypotenuse of 7, we figured out that this was 6.03 units long.*0968

*Now we're in good shape to find the area of the triangle using old-fashioned geometry.*0978

*It's just 1/2 times the base times the height.*0983

*The base here is 9.*0986

*That's the base.*0992

*You might be worried a little bit that the top of the triangle sticks out a bit over the edge of the base, that actually doesn't matter.*0994

*That doesn't make the formula invalid.*1002

*It is still ...*1004

*You count the base as being 9, even though the top of the triangle sticks out over the edge of the base.*1005

*The height we've just worked out is 6.03.*1011

*Now I'm going to multiply that by 9 and by 1/2, and I get approximately 27.1 for the area of the triangle.*1019

*Let's recap what happened there.*1033

*I was given three sides for a triangle.*1036

*Normally when you're given three sides and you want to find the area, that's a dead give-away that you want Heron's formula.*1038

*Unfortunately, this example asked us to find it without using Heron's formula.*1043

*That's why we used law of cosines to find that angle, and then we used SOH CAH TOA to find the length of an altitude of a triangle, in other words, the height of a triangle.*1049

*SOH CAH TOA right there.*1065

*Then we used the old-fashioned area formula using the base and the height to give us the area of the triangle.*1067

*We'll try some more examples of these later.*1073

*We are trying some more examples of applied trigonometry to solve word problems.*0000

*In this first one, we are given a child flying a kite on 200ft of string and the kite’s string makes an angle of 40 degrees with the ground.*0006

*We want to figure out how high is the kite of the ground.*0014

*Remember, whenever you have these word problems, the important thing is to draw a picture right away.*0019

*That helps you convert from lots of words into some triangles or geometric picture where you can invoke the equation that you know.*0023

*Let me draw this now, the child is flying a kite on 200ft of string and the kite’s string makes an angle of 40 degrees with the ground, that is a 40 degree angle.*0034

*I want to figure out how high the kite is off the ground so let me draw in something to mark the height here.*0050

*There is the kite up there and here is the child, we are not going to worry about how tall the child is because that really will not make a difference.*0066

*We are trying to find this distance right here, that is the height of the kite off the ground.*0077

*This is a right triangle so we can use our formulas for right triangles which is SOHCAHTOA.*0086

*Remember SOHCAHTOA only works for right triangles, if you do not have a right triangle then you will going use something like the law of sin or law of cos, or heron’s formula.*0092

*If you have a right triangle, SOHCAHTOA is usually faster.*0102

*Let us see what I know here, I do not really know anything except the hypotenuse of this triangle and one angle, and then this is the opposite side to the angle that I know.*0108

*I’m going to use this part of SOHCAHTOA, sin=opposite/hypotenuse, sin(theta)=opposite/hypotenuse that is because I know the hypotenuse and I’m looking for the opposite side there.*0122

*Sin(40)=opposite/200 and if I solve that by for the opposite side, I will get the opposite side as equal to (200) sin (40).*0140

*Let me check on my calculator what that is, since that is not a common value, that is not one that I just remember.*0162

*What I get is an approximate value of 128.6 and we are told that the unit of measurement here is feet, so, 128.6ft.*0171

*That one was a pretty quick one, the reason it was quick because we really had a right angle there so we are able to use SOHCAHTOA.*0199

*If we did not have a right angle and we had to use the law of cos or law of sin, it probably would have been longer.*0205

*Let us recap what it took to do that problem.*0212

*First of all, you read the word and try to make a sense of it then you draw a picture.*0215

*It is very important that you draw a picture to illustrate what is going on.*0219

*We drew a picture with a child here, the kite here, and we know that the child is flying a kite on 200ft of string.*0223

*We fill that in, we are told that the kite’s string makes a 40 degree angle with the ground and we draw that in.*0233

*We are trying to find how high is the kite? We draw in the height of the kite.*0240

*We have a right angle, definitely a candidate for SOHCAHTOA and the reason I picked sin is because I want to find the opposite side to the angle and I know the hypotenuse.*0247

*That is why I worked with sin, sin=opposite/hypotenuse.*0258

*I filled in the values that I know 40 and 200, and I just solved it for the opposite.*0263

*I got a number and then I checked back to see that we are using feet.*0268

*That tells me that my unit for measurement for the answer should be feet.*0274

*Our last example here is another pretty wordy one, two straight roads lead from different points along the coast to an in land town.*0000

*Surveyors working on the coast measure that the roads are 12 miles apart and make angles of 40 degrees and 110 degrees with the coast.*0009

*I want to find out how far is it to the town along each one of the roads?*0019

*This is a pretty complicated one, when you get like one like this with lots of words, it is absolutely essential that the first thing you do is draw a picture.*0025

*Let me draw a picture, that is supposed to be the coastline there.*0034

*Those are supposed to be fish swimming in the ocean and we have got two different points along the coast.*0047

*Two roads leads from these points to an in land town and these roads make angles of 40 degrees and 110 degrees with the coast.*0055

*That is about 110 and that is about 40, we know that straight roads go off and they go to some town somewhere in land.*0069

*The last thing we are told is that the roads are 12 miles apart, I’m going to write in my third side as 12 miles there.*0086

*We want to figure out how far it is to the town along in each of the roads.*0099

*First of all, let us look at this triangle we got here.*0104

*What I have been given is two angles of the triangle and the side in between them, so we have been given an (angle, side, angle) situation.*0108

*When you are given an (angle, side, angle) situation, the thing you want to check is whether those angles are really legitimate.*0120

*In other words, whether they add up to less than 180 degrees.*0127

*The angles sum up to, in this case (110+40=150) which is less than 180, which is good, there is a unique solution.*0132

*I’m going to label everything I know here, I’m going to label that unknown angle as (C) and the others are (A and B).*0160

*I will label my sides (a, b – remember you put the sides opposite the angles of the same letter and that side that we know is 12 is (c).*0172

*Now I have labeled everything I know and I want to find the length of the roads, that is (a and b), so how can I find those?*0189

*This is an (angle, side, angle) situation, (angle, side, angle) if you remember that is the one you want to solve using the law of sin.*0202

*That one you want to solve with the law of sin, it does not work very well with the law of cos.*0218

*Certainly, it does not work with SOHCAHTOA because we do not have a right angle here.*0223

*We are going to use the law of sin, let me remind you what that is, that says sin(A)/a=sin(B)/b=sin(C)/c.*0227

*That is the law of sin and we want to use that to solve for (a) and (b).*0245

*Let me work on (a), I will write down sin(A)/(a)=sin(C)/c, the reason I’m going to that is because I know what (c) is and I do not know what any of the others are.*0251

*I can find (C) very easily, (C) is 180-(A)-(B), that is because the three angles of a triangle add up to 180.*0271

*That is 180-110-40, 180-150=30, so (C) is 30 degrees, that is not my full answer to the problem but that is going to be useful to me.*0284

*Sin(A)=sin(110), sin(a)= I do not know yet, sin(C)=sin(30), sin(c)=12.*0307

*I’m going to cross multiply to solve for (a), 12 sin(110)=(a)sin(30), if I solve that for (a), (a)=(12)sin (110)/sin(30).*0325

*I’m going to reduce that using my calculator, remember to set your calculators on degree mode if that is what you are using.*0349

*If you use radian mode, your calculator will interpret this as a 110 radians and 30 radians and will give you answers that do not make sense.*0355

*I’m going to simplify (12)sin(110)/sin(30), it tells me that it is approximately 22.6.*0364

*The unit of measurement here is miles be we were told here that the measurement on the coast was given as 12 miles.*0388

*That tells us how long the A road is, 22.6 miles.*0398

*Now let us figure out how long the B road is, again we are using the law of sin.*0405

*Sin(B)/(b)=sin(C)/sin(c), I fill in what I know there.*0409

*I know that (B) is 40 degrees, I do not know (b), that is what I’m solving for, sin(C) is sin(30), (c) is (12).*0421

*I will cross multiply there, I get (12)sin(40)=(B)sin(30), I want to solve for (b), (b)=(12)sin(40)/sin(30).*0441

*I will plug that into my calculator.*0470

*Make sure that it is in degree mode and I get an approximate answer of 15.4 miles for (b).*0477

*Now I have the lengths of those two roads to the town in land.*0493

*Let us recap what we did there, we were graded with this long problem and lots of words.*0498

*The first thing to do is draw a picture and try to identify everything you are being given in the problem.*0502

*We have different points along the coast, we have an in land town, we measure the roads are 12 miles apart.*0509

*I filled that 12 miles into my distance along the coast, I filled in the two angles, and that was really all the problem gave me.*0517

*But that was enough to set up a triangle and to notice that I have (angle, side, angle).*0529

*Once I know that I have (angle, side, angle) I can the measure of the angles, make sure it is less than 180 and that tells me that it has a unique solution.*0535

*Now with (angle, side, angle) it is really not a good one to work with cos, certainly it does not work with SOHCAHTOA because we do not have a right angle.*0545

*Remember that SOHCAHTOA only works with right angles.*0554

*We are stuck using the law of sin, which actually works very well for (angle, side, angle).*0558

*First thing we had to do is find that third angle, the way we did that was by noticing that the angles add up to 180 degrees.*0564

*We used that to find the value of the third angle, that the value of angle C is 30 degrees.*0572

*We were able to plug that in to these incarnations of the law of sines, sin(A)/(a)=sin(C)/(c).*0578

*We plug in everything we know and the only thing missing is this (a).*0588

*We were able to solve this down and get (a) equal to 22.6 and we figured out that the unit of measurement is miles. *0594

*That is why we said that the answer was in term of miles.*0603

*We used the law of sines again, sin(B)/(b)=sin(C)/(c) and fill in everything we know, reduced it down, solve for (b) and give our answer in terms of miles.*0607

*That tells us the length of those two roads that finishes off this problem.*0620

*That also finishes the lecture on word problems and applications of triangle trigonometry.*0624

*Thanks for watching this trigonometry lectures on www.educator.com.*0631

*Hi, these are the trigonometry lectures for educator.com, and today we're going to look at some word problems and some applications of triangle trigonometry.*0000

*We're going to be using all the major formulas that we've learned in the previous lectures.*0009

*I hope you remember those very well.*0014

*The master formula which works for right triangles is SOH CAH TOA.*0016

*You can remember that as Some Old Horse Caught Another Horse Taking Oats Away.*0021

*Remember, that only works in a right triangle.*0025

*If you have an angle θ, that relates the sine, cosine and tangent of θ to the hypotenuse, the side opposite θ, and the side adjacent to θ.*0030

*You interpret that as the sin(θ) is equal to opposite over hypotenuse, the cos(θ) is equal to adjacent over hypotenuse, and the tan(θ) is equal to opposite over adjacent.*0043

*The law of sines works in any triangle.*0056

*Let me draw a generic triangle here, doesn't have to be a right triangle for the law of sines to work, so a, b, and c ...*0060

*Generally, you'd label the angles with capital letters, and label the sides with lowercase letters opposite the corner with the same letter.*0071

*That makes this a, this is b, and this is c.*0082

*The law of sines says that sin(A)/a=sin(B)/b=sin(C)/c.*0086

*That's the law of sines.*0106

*Law of cosines also works in any triangle.*0108

*Let me remind you what that one is.*0112

*We had a whole lecture on it earlier, but just to remind you quickly, it says that c*^{2}=a^{2}+b^{2}-2abcos(C).0114

*That's useful when you know all three sides, you can figure out an angle very quickly using the law of cosines.*0129

*Or if you know two sides and the angle in between them, you can figure out that third side using the law of cosines.*0134

*Remember, law of cosines works in any triangle, doesn't have to be a right triangle.*0142

*It still works in right triangles.*0147

*Of course, in right triangles, if C is the right angle, then cos(C) is 0, so the law of cosines just boils down to the Pythagorean formula.*0149

*You can think of the law of cosines as kind of a generalization of the Pythagorean theorem to any triangle, doesn't have to be a right triangle anymore.*0158

*Finally, Heron's formula.*0166

*Heron's formula tells you the area of a triangle when you know the lengths of the three sides.*0173

*Heron says that the area is equal to the square root of s×(s-a)×(s-b)×(s-c).*0178

*The a, b, and c are the lengths of the sides, you're supposed to know what those are before you go into Heron's formula.*0193

*This s I need to explain is the semi-perimeter.*0199

*You add a, b, and c, you get the perimeter of the triangle, then you divide by 2 to get the semi-perimeter.*0205

*That tells you what the s is, then you can drop that into Heron's formula and find the area of the triangle.*0212

*We'll be using all of those, and sometimes it's a little tricky to interpret the words of a problem and figure out which formula you use.*0218

*The real crucial step there is as soon as you get the problem, you want to draw a picture of the triangle involved, and then see which formula works.*0225

*Let's try that out on a few examples and you'll get the hang of it.*0237

*The first example here is a telephone pole that casts a shadow 20 feet long.*0241

*We're told the sun's rays make a 60-degree angle with the ground.*0247

*We're asked how tall is the pole.*0250

*Let me draw that.*0252

*Here's the telephone pole, and we know it casts a shadow, and we know that that shadow is 20 feet long.*0254

*That's a right angle.*0262

*The reasons it casts a shadow is because of these rays coming from the sun.*0265

*There's the sun casting the shadow.*0274

*We want to figure out how tall is the pole.*0278

*We're told that the sun's rays make a 60-degree angle with the ground.*0281

*That means that angle right there is 60 degrees.*0284

*We want to solve for the height of the telephone pole, that's the quantity we want to solve for.*0288

*That's the side opposite the angle that we know.*0295

*We also are given the side adjacent to the angle we know.*0303

*I see opposite and adjacent, and I see a right angle.*0305

*I'm going to use SOH CAH TOA here.*0309

*I know the adjacent side, I'm looking for the opposite side.*0315

*It seems like I should use the tangent formula here.*0325

*Tan(60) is equal to opposite over adjacent.*0330

*Tan(60), 60 is one of those common values, that's π/3.*0339

*I know what the tan(60) is, I've got that memorized and hopefully you do too, square root of 3 is the tan(60).*0346

*If you didn't remember that, I at least hoped that you remember the sin(60) and the cos(60), that tan=sin/cos.*0353

*You can always work out the tangent if you don't remember exactly what the tan(60) is.*0364

*The sin(60) is root 3 over 2, the cos(60) is 1/2, that simplifies down to square root of 3.*0368

*That's what the tan(60) is.*0377

*The opposite, I don't know what that is, I'll just leave that as opposite, but the adjacent side I was given is 20.*0380

*I'll solve this opposite is equal to 20 square root of 3.*0387

*Since this is an applied problem, I'll convert that into a decimal, 20 square root of 3 is approximately 34.6.*0394

*The unit of measurement here is feet, I'll give my answers in terms of feet.*0411

*That tells me that the telephone pole is 34.6 feet tall.*0420

*Let's recap there.*0428

*We were given a word problem, I don't know at first exactly what it's talking about.*0430

*First thing I do is I draw a picture, so I drew a picture of my telephone pole, I drew a picture of the shadow then I noticed that's a right triangle, I could use SOH CAH TOA to solve it.*0433

*I tried to figure out which quantities do I know, which quantities do I not know.*0447

*I knew the adjacent side, I knew the angle, but I didn't know the opposite side and that seemed like it was going to work well with the tangent formula.*0451

*I plugged in what I knew, I solved it down using the common value that I knew, and I got the answer.*0461

*In the next one, we're trying to build a bridge across a lake but we can't figure out how wide the lake is, because we can't just walk across the lake to measure it.*0469

*It says that these engineers measure from a point on land that is 280 feet from one end of the bridge, 160 feet from the other.*0481

*The angle between these two lines is 80 degrees.*0490

*From that, we're supposed to figure out what the bridge will be, or how long the bridge will be.*0494

*Lots of words here, it's a little confusing when you first encounter this because there's just so many words here and there's no picture at well.*0498

*Obviously, the first thing we need to do is to draw a picture.*0504

*I had no idea what shape this lake is really but I'm just going to draw a picture like that.*0513

*I know that these engineers are trying to build a bridge across it.*0523

*Let's say that that's one end of the bridge right there and that's the other end.*0529

*They measure from a point on land that is 280 feet from one end of the bridge and 160 feet from the other.*0532

*It says the angle between these two lines measures 80 degrees.*0551

*If you look at this, what I have is a triangle, and more than that I have two sides and the included angle of a triangle.*0557

*I have a side angle side situation, and I know that that gives me a unique solution as long as my angle is less than 180 degrees.*0566

*Of course, 80 is less than 180 degrees.*0576

*I know I have a unique solution.*0580

*I'm trying to find the length of that third side.*0585

*If you have side angle side and you need the third side, that's definitely the law of cosines.*0590

*I'm going to label that third side little c, and call this capital C, label the other sides a and b.*0595

*Now, the law of cosines is my friend here, c*^{2}=a^{2}+b^{2}-2abcos(C).0605

*We know everything there except for little c.*0623

*I'm just going to plug in the quantities that I know and reduce down and solve for little c.*0626

*Let me plug in c*^{2}, I don't know that yet, a^{2} is 160^{2}, plus b^{2} is 280^{2}, minus 2×160×280×cos(C), the angle is 80 degrees.0634

*I don't know exactly what that is but I can find that on my calculator.*0658

*160*^{2}=25600, 280^{2}=78400, 2×160×280, I worked that out as 89600, the cos(80), I'll do that on my calculator ...0664

*Remember to put your calculator in degree mode if that's what you're using here.*0692

*A lot of people have their calculator set in radian mode, and then that gives you strange answers, because your calculator would be interpreting that as 80 radians.*0694

*It's very important to set your calculator to 80 degrees.*0704

*Cos(80)=0.174, let's see, 25600+78400=104000, 89600×0.174=15559, that's approximate of course, if we simplify that, we get 88441.*0710

*That's c*^{2}, I'll take the square root of that, c is approximately equal to 297.4.0761

*Our unit of measurement here is feet, so I'll give my answer in terms of feet.*0778

*Let's recap what made that one work.*0791

*We're given this kind of long paragraph full of words and it's a little hard to discern what we're supposed to be doing.*0794

*First thing we see is, okay, it's a lake, so I drew just a random lake, it's a bridge across a lake, so I drew a picture.*0800

*That's really the key ideas to draw a picture.*0808

*I drew my bridge across the lake here.*0810

*That's the bridge right there.*0813

*It says we measure from a point that is 280 feet from one end of the bridge, and 160 feet from the other.*0817

*I drew that point and I filled in the 160 and the 280.*0825

*Then it gave me the angle between those two lines, so I filled that in.*0831

*All of a sudden, I've got a standard triangle problem, and moreover, I've got a triangle problem where I know two sides and the angle between them, and what I want to find is the third side of the triangle.*0835

*That's definitely a law of cosines problem.*0847

*I write down my law of cosines, I filled in all the quantities that I know then I simplified down and I solved for the answer on that.*0851

*We'll try another example here.*0862

*This time, a farmer measures the fences along the edges of a triangular field as 160, 240 and 380 feet.*0864

*The farmer wants to know what the area of the field is.*0873

*Just like all the others, I'll start out right away by drawing a picture.*0876

*It's a triangular field, my picture's probably not scaled, that doesn't really matter, 160, 240 and 380.*0885

*I have a triangle and I want to figure out what the area is.*0895

*If you have three sides of a triangle and you want to find the area that's pretty much a dead give-away that you want to use Heron's formula.*0900

*Let me remind you what Heron's formula is.*0907

*Heron's formula says the area is equal to the square root of s×(s-a)×(s-b)×(s-c).*0910

*The a, b, and c here just the lengths of the three sides but this s is the semi-perimeter of the triangle.*0929

*That's 1/2 of the perimeter a+b+c.*0939

*The perimeter is the distance around if you kind of walk all the way around this triangle.*0945

*That's (1/2)(160+240+300), 160+240=400, plus 300 is 700, 1/2 of that is 350, so s is 350.*0949

*I just drop the s and the three sides into Heron's formula.*0968

*That's 350×(350-160)×(350-240)×(350-300), (350-160)=190, (350-240)=110, and (350-300)=50.*0976

*I can pull 100 out of that immediately, so this is 100 times the square root of 35×19×11×5.*1014

*I'm going to go to my calculator for that, 35×19×11×5=36575, square root of that is 191, times 100 is 19.1, rounds to 25.*1026

*This is the area of a field, the units are squared here, and we were talking in terms of feet, so this must be square feet.*1060

*Let's recap how you approach that problem.*1081

*First of all, you read the words and right away you go to draw a picture, I see that I have a triangular field, the edges are 160, 240 and 300 feet.*1086

*I'm asked for the area of the field.*1099

*Now, once you have the three sides and you want the area, there's pretty much no question that you want to use Heron's formula on that.*1101

*That's the formula that tells you the area based on the three sides very quickly.*1109

*You write down Heron's formula, that's got a, b and c there.*1115

*It's also got this s, the s is the semi-perimeter.*1118

*You figure that out from the three sides.*1122

*You plug that into Heron's formula and you plug in the three sides.*1126

*The a, b and c are 160, 240 and 300.*1133

*Then, it's just a matter of simplifying down the numbers until you get an answer and figuring out that the units have to be square feet, because the original measurements in the problem were in terms of feet, and we're describing an area now, it must be square feet.*1137

*We'll try some more examples later.*1153

*We are learning about vectors and we are trying some more examples.*0000

*We are trying to find the components of a vector of magnitude 8 that points 30 degrees West of South.*0003

*First thing to do is to draw a picture here, let me put my North, East, South, West (Never Eat Shredded Wheat).*0009

*I want a vector that points 30 degrees West of South and that would be down there.*0023

*That is 30 degrees but I like to think of these things in terms of a reference angle from a positive x axis.*0029

*That would be a 60 degree angle there, the reference from the positive x axis would be a 180 + 60 that would 240 degrees.*0037

*Now I want to find the components of that vector, I know its magnitude is 8 so r=8.*0050

*Remember we got the components, these useful formulas x=arcos(theta), y=arcsin(theta).*0058

*We are going to use those, x=r(8)cos(theta) the reference angle is 240 degrees.*0069

*That is 8(cos) is the x coordinate of that angle and the x coordinate is -1/2.*0080

*Negative because its on the left hand side where the x value is negative so that is -4.*0089

*The y component is (8)sin(240degrees), again common value, I memorized the answers for multiples of 30 degrees so I do not need to check the calculator for this.*0097

*This is (8)(-square root of 3/2) that is -4 square root of 3.*0112

*The components there are xy=-4 square root of 3.*0122

*That one was one of the simpler ones, we are given a vector, magnitude, direction.*0136

*The magnitude of course is the r, the direction essentially is the theta but we want to draw a picture and make sure what theta we are talking about.*0140

*It says that this direction is given in terms of West of South*0150

*That is why I drew my vector down here, 30 degrees to the West of South and then I worked that out to be 240 degrees from the x axis.*0155

*Then I dropped those into arcos(theta) and arcsin(theta).*0165

*Simplify those down using the common values and I get the components of the vector.*0170

*Finally we have another one, it is a ship this time, sailing at a current that flows at 5km/hr NE.*0000

*The ship’s captain steers the course 30 degrees East of South at 40km/hr relative to water.*0009

*I want to find out the true speed and true course of the ship.*0017

*This is again one or more we are going to have two velocities.*0021

*There is the ship’s velocity and the current’s velocity, I want to figure out each one separately.*0025

*Find the components and add them up and then convert that back into a vector.*0030

*We are given the current first, I’m going to draw a graph of that and find its components.*0037

*The current flows 5km NE so this is the current, I will do this in black.*0046

*NE so that makes that this is N, this is E, that is a 45 degree angle.*0057

*The components we can get using arcos(theta) and arcsin(theta).*0064

*The current is arcos(theta), (phi)cos(45), (phi)sin(45) which is a common value that I recognized.*0072

*Remember that 45 is pi/4 and its cos is square root/2, so it its sin.*0092

*Now, I’m going to add up the vectors and the numbers are not going to be very nice.*0103

*I’m going to have to throw that in to my calculator.*0107

*Let me find the decimal approximation for (phi)(square root of 2)/2 it is about 3.54.*0111

*That is the components of the vector for the component.*0133

*Now, I will do the ship in blue, the ship is going 30 degrees East of South.*0137

*There is South down there, the ship is going 30 degrees East of South so that is a 30 degree angle.*0144

*It is steering a course at 40 km/hr relative to the water so the magnitude is 40 km.*0154

*I want to find the components of that vector.*0165

*I want to get a reference angle from the positive x axis, I know that it is 270 degrees down to the -y axis and then 30 more degrees, 270 + 30 would be 300 degrees.*0169

*To find the components of the ship’s vector we are going to use arcos(theta) and arcsin(theta).*0182

*The ship is (40)cos(300 degrees) and (40)sin(300 degrees).*0190

*300 degrees common value, I recognize that one that is 5pi/3 and its cos is +1/2 because its x coordinate is positive, its sin is (–square root of 3)/2 because its y coordinate is negative.*0204

*That is a common value, I do not need to check it on a calculator.*0228

*That simplifies down to 20 and -20 square root of 3.*0232

*I am going to be adding that together with the component for the current.*0237

*I am going to go ahead and throw that in to my calculator.*0243

*Of course 20 is still 20, 20 square root of 3 is 34.64, those are the components of the vectors for the ship.*0247

*Now, what I want to do is add these two up to find the net components of the ship’s travel.*0266

*Let me move the current vector down, I am going to draw a copy of the current vector, same magnitude and same direction.*0276

*I will draw the net in red here, I add these two vectors together and I will get the net.*0285

*Let me remind you that the blue one is the ship and the red one is the net.*0294

*For the net, I will just going to add those two together.*0302

*I take those two and I add them together, I get 23.54 for the (x).*0308

*For the (y) I get -34.64, so I get -31.10, that is the x and y components of the net travel of the ship.*0316

*We had its travel relative to the water but the water is also moving so the net travel, you add up those two velocities and you will get the component of the ship’s motion.*0343

*We are asked to find the true speed, speed is the magnitude of the velocity.*0353

*Remember velocity is a vector and speed is a number, and the true direction.*0361

*The speed is the (r), the magnitude of the velocity so that is the square root of (x*^{2} + y^{2}).0369

*Where this is my (x), the (x) I am talking about is 23.54 and the y is -31.10.*0378

*I am going to work those out.*0386

*It works out to be just about 39.0 and it looks like our units here are km/hr.*0400

*That is the magnitude of that vector which is the true speed of the ship, that is half of our answer there. *0413

*Remember, we find the direction by finding (theta) is arctan(y/x) and then there is this question of whether you have to add 180.*0425

*I said you have to add 180 depending on which quadrant you are in.*0439

*You have to add 180 if you are in the second or third quadrant.*0444

*But we are in the fourth quadrant here, our (x) coordinate here so we are in the fourth quadrant, we do not have to add 180 here.*0452

*That is a small mercy there this is because we are in quadrant for or alternatively you can think of that as because the (x) is greater than 0, the (x) is positive there so you do not have to add 180.*0464

*We can just figure out the arctan(y/x), tan inverse, make sure you are in degree mode here.*0484

*The arctan(-31.1/23.54) I will work that out.*0492

*It works out to -52.9 degrees, of course I’m rounding there.*0514

*That really tells you the angle from the positive (x) axis, I do not really like giving a direction in terms of a negative angle.*0525

*If this is a real ship I would not want to say it is sailing in the negative angle.*0532

*Let me draw that out, that tells us that, that angle right there is 52.9 degrees.*0537

*A useful way to describe that would be as some angle East of South and so I can find the other complimentary angle by doing 90-52.9.*0553

*I worked that out and what it tells me is that its 37.1 degrees as reference in terms of East of South.*0581

*The direction that this ship is actually moving is 37.1 degrees East of South.*0607

*Just to remind you there, we calculated this angle right here that is where the 37.1 came from.*0621

*That was a pretty tricky problem, we had two different things going on here.*0630

*We had a ship sailing and a current, so we had a velocity vector for the current and then we had another velocity vector for the way the ship is steering relative to the water.*0634

*What we have to do there was to take the current’s vector 5km/hr NE and use arcos (theta) and arcsin(theta) to find the components of the current.*0646

*We took the ship’s velocity vector relative to the water and we use arcos(theta) arcsin(theta) here, the (r) is 40 because that is the magnitude, the speed of the ship relative to the water.*0663

*The 300 we got from the fact that it was 30 degrees East of South.*0677

*We graph that out and we found that it was 300 degrees from the positive (x) axis, that is where we get 300.*0682

*Plot that through arcos(theta) and arscin(theta) and we get a vector for the ship.*0690

*We add those two set of components together and we get the net components of the ship’s motion in terms of (x) and (y).*0697

*But we want to find the speed and direction, so we know the (x) and (y) components to find the speed that is the magnitude of the velocity.*0708

*We use square root (x*^{2} + y^{2}), that came out to be 39.10, and it is km/hr, because we solved it all our distances were in terms of km.0716

*To find the direction, we used arctan(y/x) and there is always this issue of do you have to add 180? We are in quadrant 4 so we do not have to add 180.*0730

*We add 180 when if we were in quadrant 2 or 3, we do not have to add 180.*0740

*Another way to check that is to check whether the (x) is positive or negative, but our (x) is positive so ok no adding of 180.*0747

*We find the arctan (y/x) and it comes out to be -52.9 degrees.*0754

*That means that your 52.9 degrees below the positive (x) axis.*0761

*But if you want to give that in terms of the major compass directions North, South, East, West, we looked down there and discovered that is 37.1 degrees East of the South direction.*0766

*That is how we finally gave our answer as 37.1 degrees East of South.*0780

*That is the end of the vector lecture on trigonometry as part of the trigonometry series on www.educator.com.*0786

*Thanks for watching.*0792

*Hi, these are the trigonometry lectures on educator.com, and today we're going to talk about vectors which is a big topic in most trigonometry classes, and they also come up in a lot of physics applications as well.*0000

*You really want to think of vectors as being a way of measuring physical concepts that have both magnitude and direction, and kind of the graphical way you think about vectors is as arrows.*0012

*Think of vectors as being arrows.*0029

*Every vector has both magnitude and direction.*0033

*Those are the two things that are important about a vector.*0035

*The magnitude is the length of the vector.*0039

*Magnitude, that's the length of the arrow.*0044

*The direction is if you put it in the coordinate system, there's an angle θ that determines the direction of the vector.*0051

*Every vector has both magnitude and direction.*0061

*Of course, if the magnitude is 0, then you have an arrow that just reduces down to a dot, that doesn't really have a direction.*0063

*Most vectors, except for the zero vector, have magnitude and direction.*0071

*You can draw these vectors anywhere you want, you can move them around as long as you don't change the length or the magnitude, and as long as you don't change the direction.*0075

*You can move them around but you aren't allowed to move their direction, and you aren't allowed to stretch them or shrink them.*0087

*That's the idea of vectors.*0095

*They're used primarily to represent lots of physical concepts about things like velocity.*0097

*Why is velocity a vector?*0106

*We think of velocity as being synonymous with speed but that's really not quite the idea.*0108

*Velocity tells you which direction you're going and how fast you're going.*0115

*There's two ideas imbedded in velocity, one is how fast you're going, and one is the direction you're going.*0122

*That's something that's very natural to measure with a vector.*0130

*Another typical example of a physical concept that is measured with a vector is force.*0135

*If you push on something, you're pushing it in a particular direction and with a particular amount of force.*0141

*There's a magnitude that tells how hard you're pushing it, and there's a direction which tells you in which direction you're pushing it.*0148

*These are two examples of physical concepts.*0157

*There are lots and lots that when you try to describe them, you really need to talk about both magnitude and direction, so we keep track of them using vectors.*0161

*There's some equations to be associated with vectors.*0171

*If you think of vectors having components x and y, think of it as the terminal point of a vector where the arrow is going to as x and y in the initial point as the origin, then the magnitude ...*0178

*We often use r for the magnitude, you can figure that out using the Pythagorean theorem, it's just the square root of x*^{2}+y^{2}.0197

*The direction is slightly more tricky.*0206

*To find the direction of a vector, you have to remember that tanθ=y/x, so it's tempting to say that θ=arctan(y/x), that's sometimes true but more subtle than that.*0209

*Let me show you why it's more subtle than that.*0227

*Remember that arctangent always gives you an angle between -π/2 and π/2.*0232

*Arctangent always gives you an angle in the fourth quadrant or the first quadrant, -π/2 to π/2, which means that if you take a vector in the second quadrant or the third quadrant ...*0240

*Let me label my quadrants here.*0262

*If you take a vector in the second quadrant or the third quadrant, and you take y/x, then you take arctan of that, it will give you an answer, an angle in the fourth or first quadrant and that will clearly be wrong.*0265

*The way you fix that is you add 180 to the arctan(y/x).*0282

*You want to do that whenever the vector is in the second or third quadrant, that in turn turns out to be whenever the x-coordinate is negative.*0288

*You use this formula when in the quadrants 2 or 3, which is the same as saying, when x is less than 0.*0306

*This standard formula, θ=arctan(y/x).*0331

*I'm running out of space here.*0338

*Let me make a little space over here.*0342

*Use when in quadrants 1 or 4, when x is bigger than 0.*0349

*That's a little tricky.*0367

*The formula for θ, it's usually arctan(y/x), but if you're in the second or third quadrant, or in other words, if your x-coordinate is less than 0, then it's arctan(y/x)+180.*0369

*That's probably the trickiest formula.*0381

*If you know r and θ, then it's easy to find x and y, this just comes from SOH CAH TOA.*0385

*There's the x-coordinate and the y-coordinate.*0402

*Remember sin(θ) is equal to the opposite, this is from SOH CAH TOA, over the hypotenuse.*0404

*Here, the opposite is y, the side opposite to y, the hypotenuse is r, you can solve that out to y=rsin(θ).*0413

*Similarly, cos(θ) is equal adjacent over hypotenuse, which is equal to x/r, you can solve that out to x=rcos(θ).*0431

*Those just come from old-fashioned right triangle trigonometry and SOH CAH TOA, x=rcos(θ), y=rsin(θ).*0448

*Those are always true no matter what is positive or negative.*0457

*You can be very safe with the x and y formulas.*0462

*The only tricky one is when you know x and y and you're solving for θ, it's a little tricky.*0465

*You have to check the sin(x), and you either have to use the arctan(y/x) or arctan(y/x)+180.*0470

*Let's practice that with some actual vectors and some examples.*0478

*The first problem is find the magnitude and direction of a vector whose horizontal and vertical components are -3 and 4.*0483

*Let me draw that vector.*0490

*X component is -3, y component is 4.*0495

*There's that vector (-3,4).*0505

*We want to find the magnitude and direction, that's r and θ, for that vector.*0508

*I'll use the formulas that we learned on the previous slide, r is equal to the square root of x*^{2}+y^{2}, which is the square root of -3^{2}+4^{2}.0518

*That's 9+16, 25.*0533

*The magnitude of that vector is 5 units.*0540

*The direction of that vector, θ=arctan(y/x), 4/-3.*0547

*Remember that the formula for θ is tricky because it can be arctan(y/x) or, depending on where the vector is, it can be that plus 180.*0563

*This vector is in the second quadrant, we have to do plus 180 here.*0574

*That's because the x-coordinate is negative.*0579

*That's not a common value, let me work that out on my calculator.*0583

*I'm going to set degree mode here because I'm talking in terms of degrees.*0591

*When I said the 180, it's kind of determined that I have to be using degrees here.*0597

*Arctan(4/-3)=-53.1+180.*0603

*Of course, -53.1 degrees would be down there, that's -53.1 degrees.*0613

*That's certainly wrong because that's 180 degrees away from the vector that we're looking for, that's why we have to add the 180.*0623

*When we add the 180 to that, we get 126.9 degrees.*0629

*If you want to give that as an angle from the x-axis, there's your answer, 126.9 degrees.*0644

*In a lot of these applications, we're going to give North, South, East, West compass directions for our directions.*0650

*If you think about that, that's 90 degrees plus 36.9 degrees, that means that angle right there is 36.9 degrees.*0655

*If you want to give your answers in terms of a compass direction, this will be 36.9 degrees west of north.*0673

*It kind of depends on the kind of answer you're looking for.*0686

*If you're looking for an answer as an angle around from the x-axis, there's your answer right there, 126.9.*0691

*If you want a compass direction oriented from the North Pole, then I would call this 36.9 degrees west of north.*0698

*Let's recap there.*0706

*We were given the two components of a vector, the x and y, we find the r just by this Pythagorean formula, we throw those in and we get the magnitude of the vector.*0708

*We find the direction of the vector by using arctan(y/x), but because the vector is in the second quadrant, because its x is negative, we have to add 180.*0718

*We worked that through, we get an angle, and then if we want to describe it in terms of compass directions, we write it as 36.9 degrees west of north.*0732

*Let's look at another example here.*0746

*We're told that a horizontal force of 30 Newtons is applied to a box on a 30-degree ramp, and we want to find the components of the force parallel to the ramp and perpendicular to the ramp.*0749

*Certainly, a problem like this, you want to draw a picture.*0760

*You definitely want to draw a picture.*0763

*Let me draw a 30-degree ramp, there's the ground, there's my 30-degree ramp.*0768

*I've got a box on this ramp.*0779

*Apparently, I'm pulling on this box horizontally with a force of 30 Newtons.*0785

*There's my horizontal force and I know that that's 30 Newtons.*0795

*What I want to do is find the components of the force parallel to the ramp and perpendicular to the ramp.*0802

*I want to break this vector up into two pieces, one of which is parallel to the ramp, one of which is going straight up the ramp, and one of which is perpendicular to the ramp.*0808

*One of which goes straight in to the ramp.*0822

*If you see, I'm finding two vectors that are parallel and perpendicular to the ramp that add up to the force there.*0824

*Then I want to figure out what those vectors are.*0834

*I'm going to translate this angle 30 degrees.*0840

*This is probably a little easier if I get rid of some of the elements of the picture.*0844

*Let me just write this as a right triangle like that, with a 30-degree angle there.*0849

*That's the hypotenuse of the triangle, it's 30 Newtons there.*0862

*And I'm trying to find the lengths of the other sides.*0866

*This is a right triangle, it's okay to use SOH CAH TOA, I don't have to use law of sines or law of cosines because I have a right triangle.*0868

*SOH CAH TOA, that says that sin(θ) is equal to the opposite over the hypotenuse.*0881

*The opposite angle to the angle θ here or the opposite side is that.*0889

*That's the adjacent angle to angle, if we call that angle θ.*0897

*Of course, this is the hypotenuse.*0905

*Sin(30) is equal to the opposite over the hypotenuse, is 30, the opposite is equal to 30sin(30).*0908

*That's one of my common values, I remember the sin(30).*0924

*I don't have to go to the calculator for this one.*0927

*The sin(30), that's π/6, that's equal to the sin(π/6)=1/2, 30×1/2 is 15.*0929

*To find the adjacent side, I'm going to use the cosine part of SOH CAH TOA.*0951

*Cosine is equal to adjacent over hypotenuse.*0955

*Cos(30) is equal to ...*0964

*I don't know the adjacent yet but I know that the hypotenuse is 30.*0969

*If I solve that for the adjacent side, I get 30cos(30).*0972

*Again, 30 is one of my common values, it's π/6.*0980

*I know that the cos(30) is root 3/2, so I get 15 root 3.*0982

*If I fill those in, this is 15 root 3, and this is 15, the answer to the original question is that the parallel force, force parallel to the ramp is 15 root 3 Newtons, and the force perpendicular to the ramp is exactly 15 Newtons.*0994

*Let's recap what made that problem work.*1047

*Again, we started with the word problem, first thing you want to do is draw a picture.*1050

*I drew my picture of my box on a 30-degree ramp, I drew in my horizontal force as a vector, vector 30 units long, because it's a force of 30 Newtons.*1055

*I was asked to find the components both parallel and perpendicular to the ramp.*1065

*What I really had to do is break this arrow up into two arrows, one of which was parallel to the ramp, one of which is perpendicular to the ramp.*1071

*I broke it up into those two arrows, those two vectors then I abstracted into another triangle, then I could use SOH CAH TOA to find the lengths of those two sides of the triangles.*1081

*That's what I did here, I found the lengths of those two sides of the triangles, I plugged them back in to my original picture, then I interpreted them in the context of the original problem.*1096

*We have another word problem here.*1112

*It says a small plane leaves Great Falls heading north, 60 degrees west, that means 60 degrees west of north, an air speed of 160 kilometers per hour.*1116

*Meanwhile, the wind is blowing from north 60 degrees east at 40 kilometers per hour.*1127

*We're told the Great Falls is 180 kilometers south of the Canadian Border.*1134

*We want to know how long until the plane enters Canadian air space.*1140

*This is quite a complicated vector problem.*1142

*As usual with a problem, when you're overwhelmed by lots of words, you want to draw a picture, so let me try to draw a picture here.*1146

*We're told that the plane is heading 60 degrees west of north.*1158

*Let me just write in my compass directions here, North, West, South, East.*1164

*We're told that the plane is heading 60 degrees west of north.*1174

*There it is, that's a 60-degree angle.*1182

*It's heading at an air speed of 160 kilometers per hour, that means that the magnitude of that vector is 160.*1186

*Meanwhile, the wind is blowing from the direction north 60 degrees east at 40 kilometers per hour.*1196

*I think what we're going to be doing is we're going to be adding the contribution of the plane's engines and the contribution of the wind.*1203

*We want to figure those out separately.*1213

*I want to find the components of each one of those vectors.*1216

*Let's find the components of the plane.*1221

*If that's 60 degrees west of north, then that's really 60+90, that's 150 degrees.*1225

*The vector of the plane's motion is, remember x=arccos(θ) and y=arcsin(θ), we know our θ is 150 degrees, so x=160cos(150), and the y component is 160sin(150).*1239

*Let's figure out those first.*1280

*Cos(150), that's one of the common values, that's 5π/6.*1283

*I remember that it's cosine is negative square root of 3 over 2.*1289

*It's negative because the x-coordinate is negative, over there in the second quadrant.*1297

*Sin(150), that's sine of 5π/6, that's 1/2, that's a common value that I remember.*1300

*This is -80 square root of 3, and 80 for the plane's motion.*1312

*Those are the x and y components of the plane's motion.*1320

*Meantime, the wind, let me fork out the wind in blue.*1325

*The wind is blowing from north 60 degrees east.*1329

*The wind is coming from 60 degrees east of nort, it's coming from that direction.*1338

*What I'm going to do is move that vector down so that it starts at the origin.*1350

*There's the wind in blue.*1358

*We know that it's blowing at 40 kilometers per hour.*1361

*We want to figure out the components of the wind.*1365

*Let me figure out the wind.*1369

*Now, that is a 30-degree angle there, because that's 60.*1372

*The whole angle is 210, because it's 30 degrees past 180.*1381

*The angle is 210.*1388

*Using rcos(θ), rsin(θ), we have 40cos(210), and 40sin(210).*1390

*We got to work out what those are.*1406

*The cos(210), again that's a common value, that's 7π/6.*1408

*I can work that out, that's negative root 3/2, negative because we're over there in the third quadrant, the x-coordinate is negative.*1414

*Sin(210)=-1/2, it's negative because again third quadrant and the y-coordinate is negative there too.*1428

*This is -20 root 3, and -20.*1437

*That's the components of the wind's vector.*1447

*We found the plane's vector and the wind's vector, the net travel of the plane will be the sum of those two vectors.*1452

*We have the plane in black.*1468

*We have the wind blowing it in blue.*1474

*In red, I'll show the net travel is the sum of the two vectors.*1480

*I want to add up the x and y components of the plane and the wind there.*1486

*I'm going to do all these in red.*1492

*The net is equal to the plane plus the wind.*1495

*The net velocity of the plane, that's equal to ...*1501

*If I add up -80 square root of 3 and -20 square root of 3, I get -100 square root of 3, then 80-20=60.*1515

*That came from adding up those two vectors.*1527

*The net travel is -100 square root of 3 in the x direction, and 60 in the y direction.*1533

*Let's go back and read some more of the problem.*1547

*It says the Great Falls is 180 kilometers south of the Canadian border, and the border runs due east-west.*1550

*Let me draw a very rough map of the United States and Canada here.*1558

*Here's the US, and here's Canada.*1564

*We're told that this plane, now we know it's travelling according to that vector, we're really only interested when it crosses the Canadian border.*1568

*That is a matter of how long it takes, its y-component, to get us across the Canadian border, and we're told that the vertical distance is 180 kilometers.*1581

*Let me draw my net vector a little bit shorter here to make a little more to scale.*1604

*The y-component there is 60 kilometers per hour.*1611

*The question is how long will it take to go 180 kilometers north if its y-component is increasing at 60 kilometers per hour.*1616

*At 60 kilometers per hour, that's an easy division problem, it will take 3 hours to go 180 kilometers north.*1628

*Let's go back and look at that problem again and recap it.*1657

*Lots of words to start with, we're greeted by this problem that's very long and very wordy.*1660

*First thing to do is draw a picture and try to isolate the different quantities involved.*1667

*First we drew a picture of the plane's motion.*1671

*The plane's motion is this one in black here, this black vector we know it's 160, it has a 160 magnitude.*1674

*We're told that the angle is 60 degrees east of north, but in terms of a reference angle from the x-axis, that's really 150 degrees because it's 90+60.*1683

*That's where the 150 comes in.*1701

*To find its components we used x=rcos(θ), y=rsin(θ).*1702

*We plug those in and we get the components of the plane's motion.*1708

*Then we want to try to find the components of the wind's vector.*1712

*The wind is blowing from north 60 degrees east at 40 kilometers per hour.*1719

*That's where we get this vector is blowing from the northeast, I showed it blowing to the southwest.*1726

*From the northeast is the same as to the southwest.*1730

*It has magnitude 40, I put that in to rcos(θ) and rsin(θ), and I work out the components of the wind's motion.*1733

*The net motion of the plane is the way its flying, it's air speed, or it's velocity relative to the wind, plus the wind's velocity.*1744

*I add those two together, and in red, I get a vector representing the net velocity of the plane.*1757

*That's my vector representing the net velocity of the plane.*1765

*Finally, the question when you blow all the words out of it is asking how long does it take the plane to go 180 miles north.*1769

*Since the border runs east-west, it really doesn't matter how far the plane moves east or west during that time, we only care how long it takes to go 180 kilometers north.*1779

*I think I said miles before but our unit is kilometers.*1789

*We want to go 180 kilometers north, we don't care about the east-west motion which is why I never really did anything with the -100 square root of 3, instead we looked at their vertical motion, the 60 ...*1791

*Ok we are going to work on some extra examples with you, I hope you have time to practice a little bit on your own.*0000

*Here we are given a right triangle that has 1 leg of length 5, and a hypotenuse of length 13.*0006

*What we want to do is find the sin, cos, and tan of all the angles in the triangle.*0024

*I hope you know that the master formula here is going to be SOHCAHTOA.*0029

*That is the one we are going to use but in order to use SOHCAHTOA you have to know the lengths of all the sides of the triangle.*0035

*Let me set a variable for this length, we need to solve x*^{2} + 5^{2} = 13^{2}.0044

*That is x*^{2} + 25 =169, x^{2} = 169-25, that is 144, so x=12.0057

*That tells us our third side length, it looks like my triangle is not really drawn on the scale because 12 is a lot bigger than 5.*0073

*But we can still do the calculations here even if we did not draw the triangle perfectly to scale, you can still figure out the calculations.*0079

*Let me label the angles (theta) and (phi) and we will look at (theta) first.*0088

*Sin(theta) using SOHCAHTO is opposite/hypotenuse, so that is 12 is the opposite angle of (theta) and the hypotenuse is 13.*0094

*Cos(theta)=adjacent of the side is 5/13, the tan(theta) =opposite of adjacent that is 12/5.*0112

*Sin(phi) let us do that one first, is equal to the opposite/hypotenuse.*0134

*The opposite side to (phi) is 5 and the hypotenuse is still 13.*0146

*The cos(phi) is equal to the adjacent side, the side adjacent to (phi) is 12/13.*0155

*The tan(phi) is equal to the opposite of adjacent, the opposite side of (phi) is 5, and the adjacent side is 12.*0165

*Finally, let me label the right angle (alpha), the sin(alpha), I’m not going to use SOHCAHTOA on this but the sin (alpha) is just the sin (pi/2) is 1.*0180

*The cos(alpha) is the cos (pi/2) which is 0 and then the tan(alpha).*0194

*Again we have not really learned about tan yet, but it turns out that the tan of a 90 degree angle or the tan of pi/2 is undefined.*0202

*I learned why that is true when we get to studying the tan function later on.*0211

*To recap here, everything came out of the master formula SOHCAHTOA, memorize that word SOHCAHTOA or memorize the phrase Some Old Horse Caught Another Horse Taking Oats Away.*0225

*That helps you to remember how to figure out all these sin, cos, and tan, once you know the lengths of the sides of the right angle.*0237

*Our last example here is two functions that we need to graph, two modified sin waves and remember that the way to do that is to start out with sin and cos graphs that you know.*0000

*Let me start graphing sin(x), there is pi, 2pi, 1, -1, pi/2, 3pi/2.*0012

*Remember that sin(x) starts at 0, goes up to 1, comes back to 0 down to -1.*0035

*What I graphed there in black is sin(x), now sin(x)-pi/2 I will do that in blue.*0052

*Sin(x)-pi/2 that takes the graph and it shifts it over to the right by pi/2 units.*0060

*I’m going to take this graph in black and shift it over to the right by pi/2 units.*0067

*Let me extend that a little bit so I will know how to shift it, now I’m going to shift it over to the right by pi/2 units.*0077

*Now it starts at 0 to pi/2, comes back to 0 to 3pi/2 down to -1, back to 0.*0082

*What I have drawn there in blue is sin(x) – pi/2, it is just a basic sin graph but moved over by pi/2 units.*0104

*We have to figure out if its odd, even, or neither*0113

*Remember that odd has rotational symmetry, even has mirror symmetry across the y axis.*0116

*Clearly this blue graph that I have drawn here has mirror symmetry across the y axis but not rotational symmetry.*0124

*This is an even function because it has mirror symmetry across the y axis.*0135

*Let us move on to g(x), that is cos, again I will start with the cos graph that I have memorized and hopefully you have two.*0151

*There is pi, 2pi, 3pi/2, pi/2, 0, -pi, -pi/2, 1 and -1, that is a little low, let me draw that a little higher, -1.*0163

*Remember the cos graph starts at 1 goes down to 0, down to -1 at pi and comes back to 0 and back 1 at 2pi.*0189

*What I graphed there in black was cos(x), the basic cos curve, cos(x + pi/2), I will do this in red.*0210

*You want to think about that as cos(x - -pi/2) and I do that to create that negative sign because when you have a shift you always want to have a negative sign there.*0222

*It helps you figure out which way it is shifting, that means shifting –pi/2 units to the right, which means pi/2 units to the left.*0235

*I take this graph and I shift it p/2i units to the left, so that means it will start here go down to 0, down to -1, back to 0 up to 1 and back to 0.*0244

*That was a little bit of too high there, that curve in red is cos(x + pi/2) and if we look at that curve it has rotational symmetry around the origin.*0264

*If you rotate it back graph, 180 degrees around the origin it would look the same.*0286

*It definitely does not have a mirror symmetry around the y axis, so this is an odd function.*0292

*Odd functions have rotational symmetry around the origin, that is an odd function.*0303

*Remember the way to check whether they are odd or even is to check which kind of symmetry they have or maybe they do not have either kind of symmetry.*0310

*That is the end of our lessons on sin and cos, this is www.educator.com.*0320

*Hi, this is educator.com and we are here to talk about sine and cosine functions.*0000

*I'll start out by giving you the definitions and kind of the master formulas.*0007

*Then we'll go through and work on a bunch of examples.*0010

*The definition of sine and cosine of an angle is, you start out with the axis and the unit circle it's important to know that.*0015

*This is a unit circle meaning the radius is 1.*0029

*What you do is you draw that angle in standard position, meaning it has one side on the x-axis.*0033

*There is data right there.*0042

*Then you look at the coordinates of the point, the x and y coordinates on the unit circle.*0044

*The x-coordinate is, I'll go over that in red, that's the x coordinate right there.*0051

*The y-coordinate, that's the one in blue.*0060

*Those coordinates are, by definition, the...*0063

*I want to do the cosine, in red, cos(θ) and then, in blue, sin(θ).*0071

*That's the definition of what an angle is in terms of coordinates on the unit circle.*0083

*If you know what the angle is, you try to figure out what its x-coordinate and its y-coordinate are.*0090

*Let's call these cos(θ) and the sin(θ).*0097

*Now, we'll see some more examples of that later so that you'll know how to actually compute the cosine and sine, but the definition just refers to those coordinates.*0101

*The most common use of sine and cosine probably is in terms of right triangles.*0113

*Let me draw a right triangle.*0120

*Right triangle just means a triangle where one of the angles is a right angle, a 90-degree angle, or in terms of radians, π over 2.*0123

*What you do is, you let θ be one of the angles that is not the 90-degree angle, so one of the other angles.*0130

*Then you label each one of the sides in terms of its relationship to θ.*0139

*The one next to θ is called the adjacent side.*0145

*The one opposite θ is called the opposite side.*0151

*The long side is called the hypotenuse.*0161

*Then we have the master formula for right triangles, which is, the sine of θ is equal to the length of the opposite side divided by the hypotenuse.*0168

*The cosine of θ is equal to the adjacent side divided by the hypotenuse.*0181

*The tangent of θ, which is something we haven't officially defined yet, so we'll learn about tangent in a later lesson.*0185

*I just want to give you the right triangle formula now, because we're going to try to remember them all together.*0193

*The tangent of θ when we get to it will be the opposite side divided by the adjacent side.*0200

*We don't want to worry too much about tangent now because we haven't learned about it in detail yet.*0205

*I'll get to those later.*0209

*If you put all these formulas together, it's kind of hard to remember their relationships.*0211

*So people have come up with this acronym.*0217

*Sine is equal to opposite over hypotenuse.*0221

*Cosine is equal to adjacent over hypotenuse.*0225

*Tangent is equal to opposite over adjacent.*0227

*If you kind of read that quickly, people call it SOH CAH TOA.*0233

*If you talk to any trigonometry teacher in the world, or any trigonometry student in the world, they should have heard the word SOH CAH TOA,*0235

*because that's kind of the master formula that helps you remember all these relationships.*0242

*They're kind of hard to remember on their own, but if you remember SOH CAH TOA, you won't go wrong.*0247

*If you have trouble remembering that, there is a little mnemonic device that people also use.*0253

*Some old horse caught another horse taking oats away.*0262

*If you remember that sentence, if that's easier for you to remember than SOH CAH TOA, then you can remember all these formulas.*0268

*There is another definition that we need to learn which is, that a function is odd if f(-x) is equal to -f(x).*0276

*Let's figure out what that means.*0289

*We're going to talk about odd and even function.*0291

*Let me give you an example here.*0294

*F(x) is equal to x*^{3}.0299

*Well, let's try f(-x) here.*0305

*I'm going to check this definition of odd function here.*0307

*So, f(-x), that means you put -x into the function, so that's (-x)*^{3} which is (1)3 times x^{3},0310

*(-1)*^{3} is -1, so that's just -x^{3}, and that's negative of the original f(x).0326

*For x*^{3}, f(-x) is equal to -f(x), which means it's an odd function.0334

*There is a way to check this graphically.*0342

*If you graph f(x)=x*^{3}, it looks something like this.0348

*That means that if you look at a particular value of x, and you look at f(x) there, and then you look at -x, f(-x)=-f(x).*0355

*That's what it looks like graphically.*0379

*What it means is that the graph has what I call rotational symmetry around the origin.*0383

*If you put a big dot on the origin and if you spun this graph around 180 degrees, it would look the same.*0393

*That's because f(x) and f(-x) being opposites of each other.*0400

*If you spin the graph around 180 degrees and it looks the same, if it looks symmetric with itself, that's called an odd function.*0408

*The way you remember that odd functions have that property is, just remember x*^{3}, x to the third, because 3 is an odd number and x^{3} is an odd function.0417

*Something, kind of, has that property that x*^{3} has, then it's an odd function.0428

*They're companion definition to that is that f is even if f(-x) equals f(x).*0435

*The difference there was that negative sign on the odd definition.*0443

*No negative sign here on the even definition.*0447

*Let me give you an example of an even one.*0450

*Let's define f(x) to be x*^{2}.0456

*Well, f(-x) is equal to, you plugin -x into the function, so (-x)*^{2} well that's just the same as x^{2}, which is the same as the original f(x).0462

*So, f(-x) is equal to f(x), that checks the definition, so it's even.*0478

*Of course you'll notice that x*^{2}, the 2 there is an even number, that's no coincidence.0486

*That's why we call even functions even is because they sort of behave like x*^{2}.0494

*If you graph those, let me graph x*^{2} for you.0500

*That's a familiar parabola that you learned how to graph in the algebra section.*0505

*If you take a value of x, and look at f(x),*0511

*then you take f(-x), f(-x) is not -f(x), it's f(x) itself.*0517

*It's the same value as f(x) itself.*0532

*You get this f(-x) is equal to f(x).*0537

*What that means is that you have a, kind of, symmetry across the y-axis with even functions.*0542

*If a function is symmetric across the y-axis, if it looks like a mirror image of itself across the y-axis, then that's an even function.*0555

*That's why I say it has mirror symmetry across the y-axis.*0567

*That's what an even function looks like.*0572

*There is a common misconception among students.*0575

*People think, well with numbers, every number's either odd or even.*0578

*People think, well, every function is odd or even and that's not necessarily true.*0583

*Just for example, here's a line but that is not either*0590

*that does not have a rotational symmetry around the origin nor does it have mirror symmetry around the y-axis.*0600

*That function, this line, is not an add or an even function.*0607

*It's a little misleading people think every function has to be an odd or an even function.*0612

*That's not true.*0616

*It's just true that some functions are odd, some functions are even, some functions are neither one.*0618

*We'll practice some examples of that.*0624

*First, we're going to look at some common angles and we're going to figure out what the cosines and sines are.*0626

*Let me draw a big unit circle here.*0635

*That's a circle of radius 1.*0646

*Let's remember where these angles are, 0, of course is on the positive x-axis.*0648

*Here's the x-axis, here's the y-axis, there is zero, *0655

*π over 2, remember that's the same as a 90-degree angle, that's a right angle, so that's up there π over 2;*0660

*π is over here, that's a 180 degrees;*0666

*3π over 2, is down here, and 2π is right here.*0671

*We want to find the cosine and sine of each one of those angles.*0676

*Now remember, cosine and sine, by definition, are the x and y coordinates of those angles.*0680

*What are these x and y coordinates?*0689

*The 0 angle, it's x-coordinate is 1, and it's y-coordinate is 0.*0691

*That tells us that cos(0) is 1, and that sin(0) is 0.*0698

*Pi over 2 is up here, and so it's cosine is the x-coordinate,*0712

*well, the x-coordinate of that point is 0.*0726

*The y-coordinate is 1, and so that's the sin of π over 2.*0730

*Pi is over here at (-1,0), so that's the cosine and sine of π.*0739

*Cos(π) is -1, sin(π) is 0.*0748

*Finally, 3π over 2 is down here at (0,-1), so that's the cosine and sine of 3π over 2.*0760

*Cos(0), sin(π/2), is -1.*0772

*And one more, 2π is back in the same place as 0, so it has the same cosine and sine.*0783

*Cos(2π) is 1, sin(2π) is 0.*0790

*That's how you figure out the cosines and sines of angles.*0805

*As you graph them on the unit circle, and then you look at the x and y coordinates.*0808

*The x-coordinate is always the cosine, and the y-coordinate is always the sine.*0813

*By the way, these are very common values, 0, π/2, 3π/2, and 2π.*0819

*You should really know the sines and cosines of these angles by heart.*0826

*They come up so often in trigonometry context that it's worth memorizing these things, and being able to sort of regurgitate them very very quickly.*0830

*If you ever forget them though, if you ever can't quite remember what the cosine of π/2, or the sine of 3π/2 is,*0841

*Then, what you do is draw yourself a little unit circle, and you figure what the x and y coordinates are and you can always work them out.*0849

*It's worth memorizing them to know them quickly, but if you ever get confused, you are not quite sure, just draw yourself a unit circle and you'll figure them out quickly.*0857

*We're going to use these values, so I hope you will remember these values for the next example.*0869

*In the next example, we're being asked to draw the graphs of the cosine and the sine functions, so let's remember what those values are.*0879

*We're to label all the zeros, and the maxima and the minima of these functions.*0888

*Let me set up some axis here.*0892

*I'm going to label my x-axis in terms of multiples of π.*0898

*The reason I'm going to do that is because we're talking about cosines and sines of multiples of π.*0905

*We're really talking about radians here.*0910

*That's π, that's 2π, that's π/2, and that's 3π/2.*0915

*That's 0, of course.*0929

*The y-axis, I'm going to label as 1 and -1.*0931

*I've set up my scale here, remember that π is about 3.14 so it's a little bit bigger than 3.*0940

*I've set up my scale here so that the π is about a little bit beyond 3 units on the graph.*0945

*I'll extend it a little bit on the left here as well.*0952

*We've got -π, I'll draw that around -3 and -π/2.*0955

*I want to graph the sine and cosine function according to those values that we figured out.*0963

*Remember that the sine and cosine function are correspond to the coordinates of angles on the unit circle.*0968

*So sine and cosine,remember, are the x and y coordinates of angles on the unit circle.*0978

*Now, those coordinates will never get bigger than 1 or smaller than -1.*0986

*That's why on my y-axis, I only went up to -1 and 1, because the coordinates will never get bigger than -1 and 1.*0994

*Let me start out with the cosine function.*1004

*I'll do that one in blue, y=cos(x).*1007

*We'll remember the values that we learned in the previous question.*1014

*Cos(x), cos(0) is 1, cos(π/2) was 0, cos(π) is -1, cos(3π/2) is 0 and cos(2π) is 1.*1018

*What you get is this smooth curve.*1044

*After 2π, remember, after you circle 2π radians, then everything starts repeating.*1057

*What happens after 2π is that it repeats itself.*1065

*It repeats itself in the negative direction as well.*1071

*Now we know what the graph of cos(x) looks like.*1080

*I'll do the sine graph in red.*1085

*Remember that sin(0) is 0, sin(π/2) is 1, sin(π) was 0 again, sin(3π/2) is -1, sin(2π) is 0.*1092

*I'm doing this from memory and hopefully you remember these values as well.*1110

*But if you can't remember these values, you know you can always look back at the unit circle and figure them out again just from their coordinates.*1114

*The sine graph, I'm going to connect this up into a smooth curve.*1122

*It repeats itself after this.*1139

*It repeats in the negative direction as well.*1146

*That's what y=sin(x) looks like.*1151

*It actually has the same shape as cos(x) but it's shifted over on the graph.*1153

*Now, we're asked to label all zeros, maxima and minima.*1160

*Let me go through and label the zeros first.*1163

*This is on the cos(x), this is (π/2,0), that's the 0 right there.*1167

*There is one right there, (3π/2,0).*1177

*This one, even though I haven't labeled it on the x-axis, is actually (5π/2,0), because it's π/2 beyond 2π,(-π/2,0).*1182

*Those are the zeros of the cosine graph.*1197

*The maxima, the high points, remember, cosine and sine never get bigger` than 1 or less than -1.*1202

*Any time it actually hits 1, it's a maximum.*1207

*They're the two maxima, at 0 and 2π.*1213

*The minimum value is -1, so there is (π,-1) and the next one would be at (3π,-1), there is one at (-π,-1).*1218

*Now, let me do the zeros of this sine graph.*1240

*There is one (0,0), (-π,0), (π,0), and (2π,0).*1245

*The maximum value would be 1 and that occurs at π/2, and again at 5π/2.*1258

*The minimum value would be at 3π/2, where the sine is -1,*1273

*Remember sine and cosine never go outside that range, -1 to 1,*1278

*and at -π/2.*1282

*All these values you should pretty much have memorized there the sort of simplest values, the easiest ones to figure out of sine and cosine.*1289

*Let's try an example where we're using this trigonometric functions in a triangle.*1301

*What we're told is that a right triangle has short sides of length 3 and 4.*1307

*We're asked to find the sine, cosine, and tangent of all angles in the triangle.*1317

*Remember, I haven't really explained what tangent is yet, but we did learn that formula SOH CAH TOA.*1323

*That's what we're going to be using here.*1327

*The first thing we need to figure out here is what the hypotenuse of this triangle is.*1330

*We have the Pythagorean theorem that says, h*^{2} = 3^{2} + 4^{2}, which is 9 plus 16, which is 25.1334

*That tells us that the hypotenuse must be 5.*1349

*Now we're going to find the sine, cosine and tangent of each one of these angles.*1354

*Let's figure out this angle first, so I'll call it θ, sin(θ), remember SOH CAH TOA,*1364

*let me write that down for reference here, SOH CAH TOA,*1372

*sin(θ) is equal to opposite over adjacent*1381

*That's 4, that's the opposite side from θ over...*1386

*Sorry, I said sin(θ) is opposite over adjacent, of course, sin(θ) is opposite over hypotenuse, and the hypotenuse is 5.*1394

*So, sin(θ) there is 4/5.*1401

*Cos(θ) is equal to adjacent over hypotenuse.*1404

*Well, the side adjacent to θ is 3, hypotenuse is still 5.*1409

*Tan(θ) is equal to opposite over adjacent.*1416

*Again, we haven't really learned what a tangent means yet, but we can still use SOH CAH TOA.*1423

*The opposite over adjacent is 4/3.*1427

*Let me call the other angle here φ.*1435

*Sin(φ) is equal to the opposite over the hypotenuse, so 3/5.*1442

*Cos(φ) is equal to adjacent over hypotenuse, the adjacent angle beside φ is 4.*1451

*Tan(θ) is equal to the opposite over adjacent, so that's 3/4.*1463

*Finally, we have the right angle here, I'll call that α.*1472

*We can't really use SOH CAH TOA on that, but I know that sin(α),*1476

*α is a 90-degree angle, or in terms of radians, it's π/2.*1480

*The sin(π/2), we learned before, is 1.*1489

*Cos(α) is cos(π/2), and we learned that the cos(π/2) before was 0.*1494

*Finally, tan(α) is tan(π/2), and we haven't really learned about tangent yet.*1505

*In particular, we haven't learned what to do with the tan(π/2).*1514

*But we'll get to that in a later lecture, and we'll learn that that's actually not defined.*1518

*So we can't give a value to the tangent of π/2.*1523

*All of these angles were things we worked out just using this one master formula, SOH CAH TOA.*1532

*That tells you the sine, cosine and tangent of the small angles in the triangle.*1541

*The SOH CAH TOA does not really apply to the right angle.*1548

*But we already know the sine and cosine of a right angle, of a 90-degree angle, because we figured them out before.*1550

*We'll try some more examples here.*1558

*I want to try graphing the function sin(x + π/2) and cos(x - π/2).*1562

*Then, I want to determine whether these functions are odd or even, or neither one.*1570

*Well, something that's really good to remember here from your algebra class, or from the algebra lectures here on educator.com,*1576

*is that you have a function, and you try to graph f(x) minus a constant, *1584

*what that does is it moves the graph of the function over by the amount of the constant.*1593

*That's very useful in the trigonometric setting.*1599

*Let me start out by graphing f(x)=sin(x).*1602

*And we did that in the previous example, and I remember what the sine graph looks like.*1608

*It starts at 0, it peaks at π/2, it goes back to 0 at π, it bottoms out at 3π/2 at -1, and then it goes back to 0 at 2π.*1615

*What I graphed there was just sin(x), I have not introduced this change yet.*1635

*What I'm going to do, in blue now, is the sin(x + π/2).*1643

*What that's going to do is going to move the graph over π/2 units.*1651

*But remember there was a negative sign in there that I don't have here.*1656

*This is really like, sin[x - (-π/2)].*1661

*It moves the graph over -π/2 units, which means it moves it to the left π/2 units.*1672

*I'm going to take this graph and I'm going to move it over to the left π/2 units.*1680

*Now it's going to start at -π/2, come back down at π/2, bottom out at π, and come back to 0 at 3π/2.*1691

*So there is -π/2, 0, π/2, and it comes back at 3π/2.*1709

*That's what our graph of sin(x + π/2) looks like.*1719

*Then the question is, is that odd or even, or neither?*1725

*Well, remember there is a graphic way to look at the graph of a function in determining whether it's odd or even, or neither.*1730

*An odd function, remember, has rotational symmetry, and even function has mirror symmetry across the y-axis.*1738

*Well, if you look at this at this graph, it certainly does not have rotational symmetry.*1750

*If you tried to rotate it around the origin, it would end up down here, and that would be a different graph.*1757

*However, it does have mirror symmetry around the y-axis.*1762

*So because it's mirror symmetric around the y-axis, sin(x + π/2).*1772

*F(x) is an even function because it has mirror symmetry, mirror symmetry across the y-axis.*1781

*OK, let us move on to the next one, cos(x - π/2).*1812

*Again, I'm going to start with the basic cosine function that we learned how to graph in a previous example.*1816

*So that my mark's here, there is π, there is 2π, π/2, 3π/2, 0, -π/2.*1829

*Now, cosine had zero at 1, then it comes down to 0 at π/2, bottoms out at -1 at π, comes back to 0 at 3π/2, and by 2π, it's back up at 1.*1841

*What I have just graphed there in black is cos(x).*1862

*I have not tried to introduce the shift yet.*1865

*But what we want to do is to graph cos(x - π/2).*1871

*That's like saying, you see up here is π/2, that's going to shift the graph π/2 units to the right,*1878

*because it's -π/2, it shifts it to the right.*1888

*We take this graph and we move it over to the right π/2 units.*1890

*I drew that a little too high, let me flatten that out a little bit.*1904

*What we have there is the graph of cos(x - π/2), and of course that keeps going in the other direction there.*1910

*We see, actually if you look carefully, cos(x - π/2) actually turns out be the same graph as sin(x).*1921

*That's a familiar function if you remember those graphs.*1930

*Again, we're being asked whether the function is odd or even, or neither.*1935

*For odd, we're checking rotational symmetry around the origin.*1939

*Look at that.*1944

*If you rotate the graph 180 degrees around the origin, what you'll end up with is exactly the same picture.*1946

*g(x) is an odd function because it has rotational symmetry around the origin.*1953

*Is it an even function?*1982

*Does it have mirror symmetry around the y-axis?*1984

*No, it does not because it has this kind of bump on the right hand side, and it does not have the same bump on the left hand side, so it's not an odd function.*1987

*Sorry, it's not an even function.*1996

*It's just an odd function.*1998

*What we did there was we started with the sine and cosine graphs that we remembered.*2000

*it's worth memorizing the basic sine and cosine graphs.*2004

*Then we examined the shift that each one introduced.*2008

*Each one got shifted π/2 units to the right or left.*2013

*Then we drew the new graphs.*2016

*Then we looked back at them and we checked what kind of symmetry do they have.*2019

*Do they have rotational symmetry or mirror symmetry?*2022

*And that tells us whether they are odd or even.*2027

*We are learning about polar coordinates today and we got some points here in polar coordinates which we have to convert to standard from.*0000

*The (r) should be positive and the (theta) should be between 0 and 2pi and then we have to convert them to rectangular coordinates.*0008

*Let me get started graphing this first point, we are given -13pi/6, I do not like that already.*0018

*I’m going to add 2pi to that and that gives me -13pi/6 + 12pi/6 = -pi/6.*0026

*I still do not like that, that is still not in standard range so I’m going to add another 2pi to that and that gives me 11pi/6, that is good.*0039

*I’m going to graph that now, 11pi/6 is just short of 2pi, over there 11pi/6.*0051

*But we are supposed to go negative two units in this direction, that means I want to end up in the opposite direction from 11pi/6.*0063

*Maybe you can tell from the graph that is 5pi/6.*0074

*If you can not tell that immediately from the graph, just subtract pi from 11pi/6 and you will get 5pi/6.*0078

*The answer there is that we want to go positive 2 units in the direction of 5pi/6.*0088

*Now we got to find my rectangular coordinates, remember the master formula we are using here x=arcos(theta), (y)=arcsin(theta), (x)=(2)cos(2pi/6) which is (2 x –root3)/2, that one I got memorized.*0101

*It is negative because the (x) value is negative there, (y) =(2)sin(5pi/6) on both of those which is (2)x(1/2) because the (y) is positive.*0126

*My (xy) collectively, the rectangular coordinates are –root 3 and 1.*0149

*That is the first point, the second one here 6 (–pi/3), let me graph that one.*0166

*I think I’m going to add 2pi to that right away to get it in the proper range, add 2 pi to that, 2pi – pi/3= 5pi/3.*0172

*If I graph that, 5pi/3 is down here in the fourth quadrant and our radius is 6.*0188

*Our standard form there is 6 (5pi/3).*0205

*I will find (x) and (y) using arcos(theta) and arcsin(theta), so 6 x ((cos(5pi/3)) which is 6, now the cos of 5pi/3 is positive because the (x) value is positive ½.*0215

*(y)=6sin(5pi/3) negative because the (y) coordinate is negative so 6 x(- root 3)/2, these are common values that I have memorized.*0234

*The (xy) collectively can be written and simplified down to 3 and –square root of 3.*0247

*The third one here is 7pi/4, let me find that, this is number three now, that was number two.*0267

*Number three, 7pi/4, that is all the way around in the fourth quadrant, just short of 2pi.*0275

*7pi/4 but I want to go -5 units in that direction, that takes me -5 units in the opposite direction.*0283

*I can tell from the graph that is 3pi/4 but if you are concerned about working that out graphically, just subtract pi from 7pi/4 and you will get 3pi/4.*0292

*Our standard form of polar coordinates would be (phi)(3pi/4).*0305

*Let me use that to find x and y, x=arcos(theta) that is (phi)cos(3pi/4) which is (5) x (-square root of 2)/2 cos is negative.*0317

*Y=(5) sin(3/4) which is (5) x (square root of 2)/2.*0337

*The x and y collectively give you the coordinates -5(root 2)/2 and 5(root 2)/2.*0353

*Finally, the fourth point here, we go -5pi/4, I do not like that already and I’m going to add 2pi to that right away and get 3pi/4.*0371

*I’m going to go in the 3pi/4 direction, there is 3pi/4 but I have to go -4 units in that direction.*0389

*That means I actually go in the opposite direction 4 units and to find that angle I add pi and that puts me back down to 7pi/4.*0401

*7pi/4 is my reference angle and I want to go 4 units in that direction so my standard polar coordinates there are 4 and 7pi/4.*0415

*My x and y, I find using arcos(theta) and arcsin(theta), x=4cos(7pi/4) now the cos is positive because we are in the fourth quadrant so that is (root 2)/2.*0429

*(y)=(4)sin(7pi/4) which is 4 x (- root 2)/2, negative because we are in the fourth quadrant, the( y) coordinate is negative.*0451

*The x and y collectively are 2(root 2) and -2(root 2).*0465

*There are several steps involved in that problem, we are given these polar coordinates but they are not necessarily in standard form.*0483

*They all might not be positive and the (theta) might not be between 0 and 2pi.*0491

*The first thing you do in all of these is to draw a graph, figure out where your angle is and if you do not like the angle, if it is not between 0 and 2pi you can add or subtract multiples of 2 pi to get it in between 0 and 2pi.*0497

*What we did we a lot of this is we added and subtracted multiples of 2 pi to get the angle in between 0 and 2 pi. *0513

*Then we plot the radius and for some of these, the radius turned out to be negative which means we are actually walking in the opposite direction from the angle that we expected.*0521

*Example on this first one, we thought we are going down to 11 pi/6, in fact we are going in the opposite direction.*0531

*To figure out what the opposite direction was, I subtracted pi and that is how I got 5 pi/6 as my answer.*0540

*That happened on several of those, we ended up walking in the negative direction and so we have to subtract pi or add pi to get our final answers for the reference angle.*0548

*Once we add or subtract pi, we can make the radius positive and that is how we got the radius and reference angle for each of these.*0559

*Once we find the radius and the reference angle, it was a matter of using the standard component formulas arcos(theta) and arcsin(theta).*0568

*In each case we used x=arcos(theta) and y=arcsin(theta), plug those in, these were all common values.*0580

*I knew the cos and sin for all of them without looking them up on the calculator because they were all multiples of pi/3 and pi/4.*0588

*I was able to just write down the coordinates for x and y on all of those.*0596

*For our last example here of polar coordinates, we have to graph the polar equation r=sin(2 theta)*0000

*I’m going to try and graph that, I’m going to start by graphing this using x and y.*0007

*I’m going to start by graphing y=sin(x) that is a very familiar one, I know that has period of 2pi.*0014

*I graph that, now we practice modifying these equations from the basic sin way to more complicated ones in the earlier lecture.*0031

*You might want to review that if this is not making sense to you.*0045

*I’m going to try and graph y=sin(2x), what that does the way that two changes it, if you remember back from the earlier lecture on www.educator.com that changes the period. *0049

*The period is now 2pi/2, so it is pi, which means this thing has a whole period just between 0 and pi.*0066

*I will do a second period between pi and 2pi.*0078

*Let me draw that a little lighter, it is going to go up and down, come back down on pi/2, go down to -1, come back and finish a whole period at pi.*0086

*Then from pi to 3pi/2, it goes up to 1 again, goes down to -1, and comes back at 2 pi.*0109

*There is 1, -1, and this is pi/2, this is 3pi/2.*0117

*Now I’m going to use this to make a chart of values for (theta) and then finally I’m going to make a polar graph.*0129

*Let me make a chart of values for (theta), I’m not going to fill in every (theta) I know.*0134

*I’m just going to fill in probably the multiples of pi/4.*0141

*Each one I’m going to write down what sin(2) (theta) is, I stated sin(theta)=0, sin(2x0)=sin0 so that is 0.*0151

*Now when (theta) is pi/4, I can remember that sin(2)( theta) is sin(pi/2) which is 1.*0163

*Or I can just look at my graph here, that was the point of making this graph.*0170

*I see that sin goes up to 1 at pi/4, when (theta)=pi/2, I see that it will back at 0, sin(2 theta)=sin(pi), so it is 0.*0176

*(theta)=3pi/4, I see them down at -1, (theta)=pi we are back to 0, (theta)=5pi/4 that is right here so we are back up at 1.*0189

*3pi/2 we are back down to 0, 5pi/4 we are back down to -1, and 2pi we are back up to 0.*0216

*You can work all those out, you can plug each one into sin(2)( theta) and do the calculations or you can just check this graph, it is really helpful if you just check the graph.*0236

*I’m going to graph that as a polar equation and let me fill in my key angles here, so there is 0, pi/2, pi, and 3pi/2.*0248

*We are back to 0 at 2pi, now I made some values for all the multiples of pi/4, I’m going to fill those in.*0268

*There is pi/4, 3pi/4, 5pi/4, and 7pi/4, I just labeled my 7pi/4 in the chart, I labeled it as 5 pi/4 but that is 7pi/4.*0287

*Just right here 7/pi and that is where 5pi/4 is.*0310

*Now I’m going to fill in my dots and I’m going to do this in blue.*0319

*At 0 it is just 0, at pi/4 I have grown out to 1, pi/2 I’m back at 0, 3pi/4 I met -1, but -1 in the 3pi/4 direction means you are actually walking in the opposite direction down to -1.*0325

*That is really 3pi/4, -1 in the 3pi/4 direction, let me write that down.*0352

*-1 (3pi/4) that is why I ended up there, at pi I ended up at 0.*0362

*Let me graph what I have so far, I started at 0, I grew out to 1 at pi/4, by the time I got to pi/2 I was back to 0, and then I went to -1 in the 3pi/4 direction.*0371

*And then by pi I was back to 0, at 5pi/4 I’m back at +1, I go +1 in the 5pi/4 direction, let me write that down 1 (5pi/4) so I come down here.*0395

*3pi/2 I’m back to 0, 7pi/4 I met -1 again, now 7pi/4 is down here but I want to walk -1 unit in that direction.*0425

*I end up walking out to the 1 unit in the 3pi/4 direction.*0447

*By 2pi I have come back to 0. *0457

*That last point, even though it looks like it is on the 3pi/4 axis, it is actually you should think of it as -1 on the 7pi/4 direction.*0466

*We have this interesting four leaf clover comes out as our graph as r=(2)sin(theta).*0478

*Let me recap the steps that I followed there.*0486

*The first thing I wanted to do was really graph r=(2)sin(theta), think about it as a rectangular equation and graph it as a rectangular equation.*0490

*To warm up , I graphed y=sin(x) and then I graphed y=(2)sin(x).*0499

*I remember back from the previous lecture on www.educator.com, that, that two changes the period.*0507

*That takes y=sin(x) and it shrinks the period down to be pi, the thing oscillates up and down twice as fast.*0514

*I get this faster oscillating graph, I’m going to use that as a graph to fill in all the (theta) I’m interested in.*0526

*I just took multiples of pi/4, and I looked at my sin(2)(theta) and I got these 0, 1, -1.*0536

*I plotted each one of those on the axis over here, but the important thing to remember is when you have a negative number you are walking in the opposite direction from what you expected.*0544

*Let me put a little timeline here if you are trying to follow us later on.*0556

*The first thing we did was we graphed 1 in the pi/4 direction that really represented going here, that was the first loop we drew.*0561

*The second thing was going down to -1 in the 3pi/4 direction, -1 in the 3pi/4 direction brought us out in to the 7pi/4 direction that was down there.*0573

*The third thing we did is we went back to 1 in the 5pi/4 direction that brought us down to quadrant 3.*0590

*We got to -1 in the 7pi/4 direction but 7pi/4 is in the fourth quadrant walking -1 in that direction actually puts you up on the second quadrant.*0599

*That is why the fourth loop that we graphed was up over here in the second quadrant even though it came from looking at the values in the fourth quadrant.*0619

*It was a pretty complicated example there but it is a matter of plotting your values and then plotting them on each of this axis and keeping track of one thing is positive and one is negative.*0628

*If they are negative remember to walk in the opposite direction going to the opposite quadrant from what you expected.*0641

*That is the end of these lectures on polar coordinates.*0648

*Thanks for watching, these are the trigonometry lectures on www.educator.com.*0651

*Hi these are the trigonometry lectures for educator.com, and today we're going to learn about polar coordinates which is really a kind of a new way of looking at the coordinate plane.*0000

*It's going to be a lot of fun to check this out.*0010

*I want to remind you how you have been keeping track of points in the plane.*0012

*You've been using ...*0019

*That's a really ugly access.*0021

*You've been using x and y coordinates which are known as rectangular coordinates or Cartesian coordinates.*0024

*The way that works is you graph a point based on its distance from the x-axis and its rectangular distance from the y-axis.*0033

*You look at that distance, that's the y-coordinate, and that distance is the x-coordinate.*0044

*That's all based on rectangles, it's also called Cartesian coordinates because it comes from Descartes, and you've done that before.*0051

*Polar coordinates are the new idea here, and it's really quite different.*0059

*If you have a point somewhere in the plane, instead of trying to orient that point based on rectangular distances from the x and y axis, what you do is you draw a line straight out from the origin to that point.*0066

*You describe that point in terms of how long that line is, r, it's the distance from the origin, in terms of the angle that line makes with the positive x-axis.*0089

*There's the positive x-axis.*0098

*You measure what that angle is from the positive x-axis, and you measure the distance along that line, then give the coordinates of that point in terms of r and θ.*0100

*That's really a new idea.*0113

*We're going to figure out how.*0115

*If you know what the point is in terms of x and y, how do you figure out what the r and θ are and vice versa.*0117

*Let's check that out now.*0124

*If you know the x and y coordinates of a point, the rectangular coordinates of a point, then you can figure out r and θ based on the Pythagorean theorem.*0126

*There's x and there's y, there's θ, and there's r.*0141

*r, just based on the Pythagorean theorem, is the square root of x*^{2}+y^{2}.0149

*That's the same as the magnitude of a vector that we learned in the lecture last time about vectors.*0153

*You don't really have to remember any new formulas here, it's the same formula r=x*^{2}+y^{2}.0160

*Same with θ, we know that the tan(θ)=y/x, that's based on SOH CAH TOA, which means that θ is arctan(y/x).*0167

*Here is where it gets tricky.*0178

*Remember that arctan(y/x) will always give you an angle in the fourth quadrant or in the first quadrant.*0179

*What you do if you have a point that's in the second or third quadrant, or if you have a point in the second or third quadrant, we said before that you add 180 degrees to θ.*0189

*In polar coordinates, you often use radians instead of degrees.*0202

*I've changed that formula slightly.*0207

*It's the same formula except in terms of radians, it's π+arctan(y/x), instead of 180+arctan(y/x).*0209

*You use this formula when your point is in quadrant 2 or 3, is when you have to add on a π to the arctan(y/x).*0217

*In other words, that occurs when the x-coordinate is negative, when x is less than 0.*0239

*This formula, the basic formula arctan(y/x), you use that for quadrants 1 and 4, or when your x-coordinate is positive.*0249

*That's how if you know the x and the y, you can find the r and the θ.*0264

*If you know the r and the θ, you can find the x and the y, by taking rcos(θ) and rsin(θ).*0271

*Those are the same formulas we had when we started with the magnitude and angle of a vector, and we convert it back to find the components of the vector.*0279

*You don't really have to remember any new formulas here.*0289

*It's the same formulas as before, x=rcos(θ), y=rsin(θ).*0292

*Those come from SOH CAH TOA, so it's not very hard to derive those formulas based on the SOH CAH TOA relationships in a right angle.*0299

*That's θ, that's r, that's x and that's y.*0306

*You can use SOH CAH TOA to find x and y equals rcos(θ) and rsin(θ).*0310

*There's some conventions that we tried to follow but it's not absolutely essential.*0316

*The r is usually assumed to be positive.*0322

*Certainly, if you used this formula to find r, you get a square root, so it's always positive.*0325

*But you can also talk about -r's.*0331

*You think about a -r, if you have an angle θ, what would a -r represent?*0336

*Well that just represents going r units in the other direction.*0343

*If r is less than 0, then that just represents going r units on the other side of the origin.*0347

*That's what a -r would represent, but we try to use positive r's if we can.*0355

*Similarly, θ, we try to keep in between the range 0 to 2π.*0360

*Here's 0, π/2, π, 3π/2, and 2π.*0367

*We try to keep our θ in between 0 and 2π.*0373

*If it's outside of that range, then what you can do is you could add or subtract multiples of 2π to try to get it back into that range.*0377

*That's how we try to restrict the values of r and θ.*0385

*You'll see some examples of this as we practice converting points.*0390

*It's probably best just to move on to some examples and you'll see how we generally find r's that are positive and θ's between 0 and 2π.*0394

*But if someone gives us values that are not in those ranges, we can modify them to find different sets of coordinates that do have r's and θ's in those ranges.*0401

*Let's move on to some examples.*0411

*First example is to convert the following points from rectangular coordinates to polar coordinates.*0414

*Remember, we're given here the x and the y, and we want to find the r's and the θ's, because we're given rectangular coordinates that's x and y.*0422

*We want to find polar coordinates, that's r and θ.*0432

*I'm going to draw some little graphs of these points to help me keep track of where they are.*0435

*You can also figure them out just using the formulas that we learned on the previous slide.*0440

*For the first point here, we've got 3 square root of 2 and -3 square root of 2.*0445

*That's positive in the x direction, and negative in the y direction.*0455

*That's a point down there.*0460

*I want to figure out where that is.*0464

*I'm going to use the formulas to figure that out, r is equal to the square root of x*^{2}+y^{2}.0465

*Let me remind you of those formulas up here because we're going to use them quite a bit, x*^{2}+y^{2}.0472

*θ=arctan(y/x), that's the case if x is greater than 0, or π+arctan(y/x) if the case x is less than 0.*0479

*In this case, our r, our magnitude is the square root of 3 root 2 squared, plus negative 3 root 2 squared.*0504

*3 root 2 squared is 9 times 2, that's the square root of 18, plus another 18, which is the square root of 36, which is 6.*0521

*θ is arctan of negative 3 root 2 over 3 root 2, because it's y/x, which is arctan(-1).*0534

*Arctan(-1), that's a common value that I remember.*0556

*That's -π/4, but I would like to get an answer in between 0 and 2π, -π/4 doesn't qualify.*0559

*I'm going to add +2π, and that will give me, 2π is 8π/4, 7π/4.*0570

*My polar coordinates for that point (r,θ) are (6,7π/4).*0583

*That really corresponds with what I can check visually.*0597

*That angle is -π/4 if you go down south from the x-axis, but if you go all the way around the long way, it's 7π/4.*0602

*That's our first set of polar coordinates.*0613

*Moving on to the next one, (-4,-3), that's -4 in the x direction, -3 in the y direction.*0616

*That's somewhere down there.*0625

*So, r is the square root of 4*^{2}+3^{2}.0628

*Since we're squaring them, I'm not going to worry about the negative signs.*0634

*16+9=25, square root of that is 5.*0637

*θ=arctan(-4/-3).*0644

*Now, we know that this angle is in the third quadrant, its x-coordinate is negative, that means we have to add a π to the θ.*0655

*This is arctan(4/3)+π.*0668

*If you plug in arctan(4/3), your calculator would give you an angle up here.*0671

*We have to add π radians to whatever the calculator's answer is.*0676

*Now, arctan(4/3), one thing that's very important here is that I'm going to switch my calculator over to radian mode, that's because I'm looking for answers in terms of radians now.*0681

*I don't want to mix up my radians and my degrees.*0693

*The π is certainly in radians, I didn't say 180 degrees, I said π, I need to make sure that my calculator's in radian mode when I do arctan(4/3).*0697

*I do arctan(4/3), negative's cancelled, that gives me 0.93+π.*0708

*That simplifies down to 4.07.*0726

*My (r,θ) for that second one is (5,4.07).*0734

*Just a reminder that that 4.07 is a radian measure.*0745

*It's not so obvious when you don't have the π in there, when you don't have a multiple of π, but it is a radian measure there.*0750

*That's the radius and the reference angle for that point.*0757

*The third one here, negative square root of 3 and 1.*0762

*Let me graph that one.*0768

*Negative square root of 3 and positive 1, that's the point over there.*0773

*My r is, root 3 squared is 3, plus 1, so that's 2, θ is arctan(y/x), so 1 over negative root 3.*0780

*The x is negative, I have to add a π there.*0798

*That's arctan, if we rationalize that, that's negative root 3/3, plus π.*0807

*Now, negative root 3 over 3, that's a common value.*0818

*I recognize that as something that I know the arctangent of, that's -π/6+π, that's 5π/6.*0821

*My polar coordinates for that point, our (r,θ), is (2,5π/6).*0837

*Finally, we have -2π.*0860

*That's -2 in the x direction, π in the y direction.*0869

*That's a point somewhere up there in the second quadrant, so r is the square root of 2*^{2}+5^{2}, which is the square root of 29, nothing very much I can do with that.0871

*θ, arctan(5/-2), my x is negative and I'm in the second quadrant so I'm using the other formula for the θ, I have to add on a π there.*0888

*My calculator's set to radian mode, I'm going to do arctan(-5/2), and I get -1.19+π, which is 1.95.*0904

*As long as I'm giving decimal approximations, I may as well give a decimal approximation for the r.*0939

*The square root of 29 is approximately equal to 5.39.*0943

*My (r,θ) is (5.39,1.95) radians.*0955

*That's the answer for that one.*0978

*All of these, it really helped me to draw a picture even though I didn't really need that for the calculations, but it was really useful to kind of check that I was in the right place.*0979

*In all four, I drew a picture of where the point was, I figured out which quadrant it was in.*0989

*I worked out the r using the square root of x*^{2}+y^{2}.0995

*I worked out θ using arctan(y/x), but then I had to check which quadrant the point was in.*1001

*If it was in the second or third quadrant, then I had to add π to the answer that the calculator gave me for arctan(y/x).*1008

*In this fourth one, for example, the calculator gave me an answer of -1.19.*1020

*The calculator's answer was down there, that's why I had to add π to it inorder to get the answer.*1027

*I came back to these formulas for r and θ every time, but it's still very helpful to draw your coordinate axis and to show where the point is on those axis.*1034

*For our second example, we have to convert the following points in polar coordinates to standard form.*1048

*That means that the r should be positive and the θ should be between 0 and 2π, then we'll convert these points to rectangular coordinates.*1056

*We're going to start by graphing each one of these points.*1064

*First one, we have -3π/4, that means we're going down 3π/4 in the negative direction.*1071

*That's the same as going around 5π/4 in the positive direction.*1080

*We can write that as (8,5π/4).*1084

*That is legit because it's between 0 and 2π.*1091

*If you didn't like doing that graphically, you could do -3π/4+2π, and that gives you immediately the 5π/4.*1094

*That's how you could figure it out using equations instead of doing it graphically.*1105

*That's the standard form of that point.*1111

*Now, I have to find the rectangular coordinates, and I'm going to use rcos(θ) for x, and rsin(θ) for y.*1113

*My x is 8cos(5π/4).*1127

*The cos(5π/4), that's a common value that I've got it memorized, that's 8 times root 2 over 2, but it's negative because the x is negative there, negative root 2 over 2.*1135

*y is sin(5π/4), the y is also negative there down on the third quadrant, negative root 2 over 2.*1151

*This is why it's so useful to draw a graph where the point is.*1163

*It helps you find the sine and cosine and remember which one is positive or negative.*1165

*(x,y) together simplify down to (-4 root 2,-4 root 2).*1169

*That was my first point.*1190

*The second one, I'm going around 11π/6, that's just short of 2π, but it's in the negative direction, so I go around just short of 2π in the negative direction.*1194

*That's actually the same as going π/6 in the positive direction.*1205

*If you don't like doing that graphically, you can add 2π to 11π/6, and you'll get π/6.*1211

*But then I'm going -6 units in that direction.*1218

*That really takes me 6 units in the opposite direction.*1223

*Really, my reference angle is down here in the third quadrant.*1229

*I can write that as 6, that's π/6, that's an angle of π/6.*1234

*This is π/6 beyond π.*1246

*I have here (6,7π/6) would be the standard version of that point in polar coordinates.*1251

*Now I've got a positive radius and a positive angle between 0 and 2π.*1262

*Now I've got to find the rectangular coordinates for that point.*1269

*I'm going to use rcos(θ) and rsin(θ).*1272

*x=6cos(7π/6), which is 6×cos(7π/6), is negative root 3 over 2.*1274

*y=6sin(7π/6), these are common values that I've gotten memorized which is 6×(-1/2) down in the third quadrant.*1292

*The (x,y) collectively, the coordinates, simplify down to -3 root 3, and -3.*1302

*Okay, next one, -2 and 11π/3.*1316

*Let's figure out where that one is.*1323

*11π/3 is pretty big, I'm going to subtract 2π from that right away.*1325

*11π/3-2π, 2π=6π/3, that's 5π/3.*1331

*That's up there in the second quadrant, 5π/3.*1342

*We'll have to go -2 units in that direction, which takes me down in the negative direction.*1346

*Sorry, I graphed that in the wrong place.*1360

*I was graphing 5π/6 instead of 5π/3.*1363

*Let me modify that slightly.*1366

*5π/3, that's in the fourth quadrant, that's down there.*1370

*5π/3 is down there, not 5π/6.*1376

*I want to go -2 units in that direction.*1381

*That's up there, back in the second quadrant.*1384

*That's really our reference angle.*1389

*We want to go 2 units in that direction.*1391

*Where is that direction?*1394

*That is 2π/3.*1395

*Our final answer for the standard polar coordinates there is (2,2π/3).*1401

*The way we got that just to remind you is we took 11π/3.*1413

*That was really big so I subtracted 2π right away, I got 5π/3.*1415

*Then I graphed 5π/3, but because I had to go -2 units in that direction, I subtracted π off from 5π/3, that's how I got the 2π/3.*1421

*Now, my x and y, rcos(θ) and rsin(θ).*1436

*2cos(2π/3), cos(2π/3)=-1/2, because the x-coordinate is negative, 2×-1/2.*1445

*y is 2sin(2π/3) which is 2 times positive root 3/2, because the y-coordinate's positive in the third quadrant.*1458

*x and y together are, 2×-1/2=-1, 2×root 3/2=root 3.*1470

*I've got x and y for that third point.*1487

*Finally, the fourth point here is (3,-2π/3).*1489

*Let me graph that one.*1495

*-2π/3, we're going down in the opposite direction from the x-axis, that's 2π/3.*1497

*If we make that something in the positive direction, that's 4π/3 in the positive direction.*1507

*If you don't like doing that graphically, just add 2π to -2π/3, and you'll get 4π/3.*1516

*This is the same as (3,4π/3).*1526

*3 is already positive, so we don't have to do anything else clever there.*1529

*That's our standard polar coordinate form.*1533

*The x and y, x=3cos(4π/3).*1537

*Cosine is negative there, -1/2, so 3×-1/2.*1548

*The sine, the y-coordinate, is also negative there.*1557

*3×sin(4π/3), common value, I remember that one, is 3 times negative root 3 over 2.*1558

*The x and y collectively are -3/2 and negative 3 root 3/2.*1568

*This one was a little bit tricky because we've been given r's and θ's that are not in the standard ranges.*1587

*Some of the r's were less than 0, and some of the θ's were either less than 0, or bigger than 2π.*1594

*The trick on all four of these points, is first of all, graph the point.*1602

*Once you graph the point, if it's not in the standard range of 0 to 2π, add or subtract multiples of 2π until you get it into that standard range.*1607

*That's what we did at first, we added or subtracted multiples of 2π to each of those to get it into the standard range.*1615

*After we did that, we looked at the r's.*1625

*If the r's were negative, then we had to go in the opposite direction from the direction we expected.*1628

*That's what happened in this second and third problem with -r's.*1635

*We had to go in the opposite direction from what we expected.*1643

*What we did was we added or subtracted π to get the correct reference angle, and to get a positive value of r.*1648

*That's how we got the r's and θ's into the standard range then.*1658

*After that, we used x=rcos(θ), y=rsin(θ), in each case to give the x and y the rectangular coordinates of the point.*1662

*That's how we did those conversions.*1677

*We'll try some more examples.*1679

*This one we have to graph the polar equation r=2sin(θ), then we're going to check our answers by converting the equation to rectangular coordinates and solving it algebraically.*1681

*What I'm going to do here is make a little chart of all my reference angles.*1693

*I'm going to make a list of all the possible values of θ that I can easily figure out 2sin(θ).*1703

*Then I'll fill in all my 2sin(θ)'s.*1708

*I'll try to graph them and see what happens.*1713

*First of all, let me remind you which angles we know the common values for.*1717

*I'm going to draw a unit circle here.*1720

*The angles that I know, first, the easy ones, 0, π/2, π, 3π/2, and then 2π.*1727

*I also know all the multiples of π/4 and I also know all the multiples of π/3.*1740

*There's π/6, π/3, 2π/3, 5π/6, 7π/6, 4π/3, 5π/3 and 11π/6.*1747

*I know every multiple of π/6 and π/4 between 0 and 2π.*1764

*I'm going to make a chart showing all of these and then we'll try to graph those and see what it turns out to be.*1767

*2sin(θ), first one is 0, sin(0)=0, that's just 0.*1777

*π/6, 30-degree angle, the sin is 1/2, and 2 times that is 1.*1787

*Next one is π/4, 45-degree angle.*1799

*The sine of square root of 2 over 2, this stuff I have memorized, that's square root of 2, which for future reference is about 1.4.*1802

*π/3 is the next one.*1814

*The sine of that is square root of 3 over 2, 2sin(θ) square root of 3, which is about 1.7.*1817

*Next one is π/2, sine of that is 1, 2 sine of that is 2.*1827

*Next one is 2π/3, sine of the square root of 3 again, that's 1.7.*1835

*3π/4, sine is root 2 over 2, twice that is root 2, 1.4.*1845

*5π/6, sine of that is 1/2, 2 sine of that is 1.*1857

*Finally, π, sine of that is just 0.*1866

*Let me go through and figure out the values on the southern half of the unit circle.*1871

*Then we'll get to put all these together and see what kind of graph we get.*1880

*Below the unit circle, the first value after π is 7π/6.*1887

*The sine of that is negative now, it's -1/2, 2 sine of that is -1.*1894

*Then we get to 5π/4, negative root 2 over 2.*1898

*We just multiply by 2 and get negative root 2, which is -1.4.*1906

*Then we get 4π/3, sine of that is negative root 3 over 2, we get -1.7.*1913

*3π/2, sine of that is -1, because we're down here at the bottom of the unit circle.*1927

*This is -2.*1936

*5π/3, the sine of that is negative root 3 over 2, so it's negative root 3 because we're multiplying by 2, about 1.7.*1939

*7π/4, sine of that is negative root 2 over 2, so I multiply it by 2, you get 1.4.*1954

*I forgot my negative sign up above.*1965

*Finally, 11π/6, sine of that is -1/2.*1968

*All of these are values that you should have memorized to be able to figure out very quickly.*1975

*2 times the sine of that is -1.*1980

*Finally, 2π, we're back to 0 again, sin(0).*1981

*We've made this big chart, now I want to try and graph this thing.*1986

*Those are my values of multiples of 0, π/2, π, 3π/2, and 2π.*2006

*Now I'm filling in π/4, 3π/4, 5π/4, and 7π/4.*2019

*This one is π/6 and 7π/6.*2034

*That is π/3 and 4π/3.*2050

*That's 2π/3 and 5π/3.*2067

*Finally, 5π/6 and 11π/6.*2081

*I want to graph each one of this radii on the spokes of this wheel.*2086

*Starting at 0 ...*2094

*I'm at 0, I'll just put a big ...*2096

*I think I better do this in another color or it's going to get obscured.*2101

*I'll do my graph in red.*2103

*I'll put a big 0 on the 0 axis there.*2110

*At π/6, I'm 1 unit out.*2114

*Let's say, that's about 1 unit at π/6.*2116

*At π/4, I'm 1.4 units out.*2122

*It's a little bit farther out, about there.*2126

*At π/3, I'm 1.7 units out, a little farther out.*2130

*At π/2, I'm 2 units out in the π/2 direction.*2138

*2π/3 is 1.7.*2148

*1.7 in the 2π/3 direction.*2155

*1.4 in the 3π/4 direction.*2157

*1 in the 5π/6 direction, it's shrinking back down.*2163

*When we get to π, it's 0.*2169

*If I just connect up what I've got so far, I've got something that looks fairly circular.*2173

*It's a little hard to tell, my graph isn't perfect.*2190

*It's also kind of an optical illusion here because of the spokes on the wheel here, but it looks kind of circular.*2197

*Remember that that was two units out.*2201

*This is one unit out.*2205

*I want to see what happens when I fill in the values from π to 2π.*2207

*I'm going to fill those in in blue.*2211

*It's 7π/6, I'm in the -1 direction.*2215

*I look at the 7π/6 spoke, but then I go -1 in that direction.*2220

*That means I go in the opposite direction for 1 unit.*2225

*I actually end up back in the opposite direction, I end up back here.*2228

*At 5π/4, that's down here, but I go -1.4 in that direction which puts me back there.*2237

*At 4π/3, I go -1.7 which ends me back there.*2247

*3π/2, -2 puts me up there.*2254

*I start out in the 3π/2 direction but I go -2 units, that takes me back up here.*2259

*5π/3 puts me at -1.7.*2267

*7π/4 puts me at -1.4.*2272

*11π/6 puts me at -1.*2277

*At 2π, I'm back at 0 again, and it starts to repeat.*2280

*If you connect up those dots, what you see is another copy of the same graph.*2286

*I'm drawing them slightly separated so that you can see both of them, but it really is a graph just retracing itself again.*2295

*You might have thought that you might get some action down here in the south side of the coordinate plane, but you don't.*2304

*You just get a graph that traces itself over the top side of the coordinate plane, and then retraces itself even when the angles are down on the south side of the coordinate plane.*2315

*Let me recap what happened there.*2327

*We started out with an equation r=2sin(θ).*2331

*First thing I did was I made a big chart of all the θ's that I know the common values of.*2334

*I made a big chart of all my possible θ's.*2340

*Then I filled in what 2sin(θ) is for each one.*2343

*I drew my axis and I drew my spokes with all the angles listed here.*2347

*I plotted the points for each one of those spokes.*2352

*I went out the correct direction for each one of those spokes.*2358

*Except that when I got down to these angles below the horizontal axis, all my signs were negative, which means I'm going in the opposite direction, which actually landed me up on the top side of the axis.*2363

*That's why you have this blue graph that retraces the original red graph.*2378

*That's how we did that one.*2386

*That's really all we need to do to graph that equation, but the problem also asked us to check our answers by converting the equation to rectangular coordinates and solving it algebraically.*2387

*Let's just remember here what this graph look like because I'm going to have to go to a new slide to check it.*2400

*We have what looks like a circle sitting on top of the x-axis.*2407

*It peaks at y=2, and it looks like its center is y=1.*2412

*In terms of x and y coordinates, it looks like we have that circle peaking at y=2 and centered at y=1.*2419

*We're going to check that by working with the equation algebraically.*2426

*Now we're going to work with this equation algebraically, r=2sin(θ).*2434

*What I'm going to do there is I'm going to multiply both sides by r.*2439

*The reason I'm going to do that is because that will give me r*^{2}=2rsin(θ).2444

*I recognized because I remember my transformations of coordinates from rectangular to polar.*2452

*I remember that x=rsin(θ), and y=rcos(θ).*2460

*I also remember that r is equal to the square root of x*^{2}+y^{2}, r^{2}=x^{2}+y^{2}.2468

*With this nice new version of the equation r*^{2}=2rsin(θ), I can write that as x^{2}+y^{2}.2480

*rsin(θ), that's exactly my y.*2490

*I've converted that equation into rectangular coordinates.*2494

*I'm going to solve that algebraically and see if the graph checks out to what I found on the previous slide.*2496

*What I'm going to do is I'm going to write this as x*^{2}+y^{2}-2y=0.2503

*I like to simplify that y*^{2}-2y.2511

*I'm going to complete the square there.*2515

*We're going to complete the square.*2521

*That's an old algebraic technique, hopefully, you'll learn about that in the algebra lectures on educator.com.*2529

*This is y*^{2}-2^{2}.2535

*You take the middle number and you divide it in half and square it.*2538

*Minus 2 divide that by 2, and you square it.*2545

*That's -1*^{2}=1.2550

*Add that to both sides.*2555

*The point of that is that now we have a perfect square.*2558

*We have y*^{2}-2y+1, that's (y-1)^{2}=1.2562

*This is actually an equation I recognize.*2571

*Remember that if you have x*^{2}+y^{2} equals a number, let's say 1, that's a circle of radius 1 centered at the origin.2575

*That's the unit circle.*2601

*This y-1, what that does is, it just takes the original graph of the unit circle and raises it up in the y direction by 1 unit.*2604

*This is a circle of radius 1, that's because of that 1 right there, centered at x=0, that's because we think of that as (x-0)*^{2}, and y=1, that's because of the y-1.2614

*If we graph that, there's 1, there's 2, I'm going to graph a circle of radius 1 centered at the point (1,0).*2644

*There's (1,0).*2661

*There's my circle of radius 1 centered at the point (0,1).*2670

*If you flip back to the previous slide, you'll see that that is exactly the same graph that we got on the previous slide.*2680

*We got it by using polar coordinates and a lot of trigonometry, but the graph that we drew actually really resembled a circle of radius 1 centered at (0,1).*2687

*Let's recap what we did there.*2698

*We started by graphing this thing in polar coordinates.*2700

*I made a big chart of θ's and 2sin(θ)'s, then I graph the r's, their radius, at each one of those values including when the radius turn out to be negative.*2704

*I ended up graphing this.*2715

*I connected them up into a circle.*2718

*That's one way to solve that problem.*2720

*The second way to check our answer was to start with the equation, I see r=2sin(θ).*2722

*Remembering my conversion formulas x=rcos(θ) and y=rsin(θ).*2730

*That's very important not to mix up.*2745

*I multiplied both sides of the equation by r.*2748

*On the left, we get r*^{2}, on the right, we get 2rsin(θ).2755

*The point of that is that the rsin(θ) converts into y, and the x*^{2}+y^{2} is what you get from r^{2}.2760

*That of course comes from this, r*^{2}=x^{2}+y^{2}.2770

*Then it was a matter of doing some algebra to realize that that's the equation of a circle that involve completing a square.*2775

*We figured out that it was equation of a circle, that we can rid of the radius of the circle is 1, the coordinates of the center of the circle are (0,1), so we can graph that equation algebraically.*2783

*It checks out with what we got from the polar coordinates.*2795

*We'll try some more examples later.*2798

*Try them out yourself and then we'll work them out together.*2800

*We are working on some examples involving arithmetic with complex numbers and we have a complicated expression to simplify here 4+2/3-I all quantity squared.*0000

*The first thing we need to do here is perform a division and remember that the way you divide complex numbers is you multiply by the conjugate.*0014

*The point of that is if you have x +y(i) and you multiply it by its conjugate x-y(i), then those multiplied together using the different of squares formula.*0024

*That is x*^{2} - y(i)^{2} but i^{2} is -1 so this is really x^{2} + y^{2}.0036

*That is the way of turning a complex number into a plain old real number.*0046

*Here we are going to write 4 + 2, now I’m going to multiply top and bottom by the conjugate of 3-i.*0053

*At the end we have to square the whole thing, so this is (4 + 2/3-i)(3+i/3+i) and this is 4 + 2.*0070

*The point of multiplying by the conjugate is that (3-i)(3+i) using this difference of squares formula is 3*^{2} + 1 ^{2} .0089

*And then in the numerator 2 x (3+ i) is 6 + (2i), we still have to square it so this is 4 + (6+2i), 3*^{2} + 1 ^{2} that is 9 + 1=10 squared.0104

*I can simplify the fraction a little bit, we can write that as 4 + 3 + (i)/5.*0124

*I’m just taking out 2 out of everything there and now I’m going to put it over a common denominator.*0132

*That is 20 + 3 + i/(5*^{2}) that is 23 + i/(5^{2}).0144

*Now remember that (a + b)*^{2})is a^{2} + 2ab + b^{2}.0166

*23*^{2} is 529 + 2ab 2(23)(i)that is 46i + i^{2} that is -1, then 5^{2} is 25 .0176

*So this whole thing simplifies down to 528 + 46i/25.*0196

*There are several things that made this problem work, the first thing to remember is when you can basically add, subtract, multiply, and divide complex using the same rules as real numbers.*0212

*But when you want to divide a complex numbers, the way you to do it is to multiply the top and bottom by the conjugate of the complex number.*0222

*That is why when have (3-i) in the denominator, we multiplied it by the conjugate, we multiply it by 3 + i.*0232

*That lets you exploit this formula, this x*^{2} + y^{2} formula, that is why I get 3^{2} + 1 ^{2} in the denominator.0243

*Then I got 10 in the denominator, I did a little bit of fraction simplification to get down to this stage 23 + i/5.*0252

*I expanded out the square using (a+b)*^{2} = a^{2} + 2ab + b^{2}.0262

*That is how I get to this line remembering that i*^{2} is -1 and then I simplified it down to get the answer.0270

*Finally we have to find all complex numbers satisfying z*^{2}= 3 + 4(i), this is going to be a little bit complicated.0000

*We are going to try to imagine that we have a complex number (x+y)*^{2}=3+4(i) with (x) and (y) being real.0009

*We will expand that out and we will see what happens.*0037

*(x+y)*^{2} is x^{2} + 2xy(i) + yi^{2} but that is the same as -y^{2} since (i)^{2} is -1.this is equal to 3 + 4i.0039

*If I equate the real and the imaginary parts here, I will get the real part on each side is x*^{2} -y^{2}=3.0062

*The imaginary part I will write this in red is 2(xy)(i)=4(i), that tells me that 2(xy) is equal to 4, we can simplify down a little bit I get (xy)=2.*0076

*This is really two equations and two unknowns, I have x*^{2} - y^{2}=3 and xy=2.0098

*A little bit harder than the two equations and two unknowns that you usually study in algebra class though because this is not linear.*0105

*To be a little more work to solve these, they are not linear equations, they are not equations to form ax + by=c, but we can still solve them. *0112

*I think I’m going to solve the bottom equation for y, so y=2/x and we are going to take that and plug it into the top equation and so I get x*^{2} - (2/x)^{2} = 3.0121

*Now I narrowed it down to one equation and I will try to solve for that on variable x*^{2} - 4/x^{2} = 3.0148

*We got to multiply both sides here by x*^{2} to clear my denominator.0160

*I get( x*^{4} - 4) =3(x) ^{2}, I will move the 3(x) ^{2} over so I get x^{4} - 3(x) ^{2}- 4=0.0170

*That looks a lot like a quadratic equation except that it goes up to the 4th degree.*0185

*But there are no odd powers of (x) in there, I just have x*^{2} and x^{4}.0191

*I’m going to make a little substitution, I’m going to let w=x*^{2} and the point is that is we have not done yet x^{4}=w^{2} - 3w - 4=00197

*That is really nice because that is now a quadratic equation in (w), I know how to solve that.*0217

*I can use a quadratic formula, I can complete the square, but I’m going to factor because I think that is the easiest.*0221

*(w-4) x (w+1) = 0 and (w)=4 or (w)=-1.*0229

*Now I will substitute back in now to get things back in terms of x, I get x*^{2}=4 or x^{2}=-1.0247

*You might think “I know something that has a perfect square equal to -1 that is (i)”.*0260

*But remember we said x and y are real numbers so we can not have x*^{2}= -1 since (x) is a real number.0265

*That x*^{2}=-1 really does not make sense and it must be x^{2}=4.0295

*I’m going to the next slide, we had our solution so far was (z)=x+y(i) with (x) and (y) being real numbers.*0300

*The way what we solved so far was we got down to x*^{2}=4, so x=2 or x=-2.0331

*If you remember back on the previous side, we had solved that y=2/x, for each one of these (x) values we get a corresponding (y) value.*0343

*If x=2 then y=2/2 that would be 1, if x=-2 y would be 2/-2 so we get y would be -1.*0354

*This would give me (z)=2+(i) because (z) was x + y(i), this solution would give me z = -2 – (i), those are the two solutions.*0368

*I like to check that quickly just to make sure it works because there was a lot of work to get through there, I want to make sure I did not do any mistakes.*0393

*We are going to plug that in, z*^{2} would be 2 + (i) ^{2} and actually notice that the second solution is just the negative of the first solution.0403

*If you square a negative number, the negative just goes away.*0421

*We really only have to check one because if the first one is right, then the second one will also be right because it is the negative of the first one.*0427

*This is 2*^{2} + ((2 x 2 x (i)) that is 4(i) + (i) ^{2}.0434

*I’m filling it out or I’m using the old algebra formula (a+b) *^{2} is a^{2} + 2ab + b^{2}.0443

*This is 4 + 4(i), (i) *^{2} is -1, this is 3 + 4(i), that does indeed check there.0453

*Let me recap the strategy that we used there, we guessed z=((x+y(i)).*0466

*We plugged it in to (x+y)(i) *^{2} = the original equation 3 + 4(i).0472

*We expanded out (x+y)(i) *^{2} using foil into something, a real number + a real number x (i).0483

*That is equal to 3 + 4(i), we can equate the coefficients on each sides.*0496

*We said this one must be equal to 3, and this one must be equal to 4.*0501

*That gave us two equations in (x) and (y) and although they are not linear equations it was a little extra work to solve them.*0507

*We could solve them down, I believe we set it to one another and we got a 4th degree equation in (x).*0514

*To solve that we let w=x*^{2}, and we got a quadratic equation in (w).0524

*We are able to factor that into two solutions in terms of (w).*0535

*Each one of those converted into a solution for (x) but when we got x*^{2}= something or x^{2}= something else one of those was -1.0543

*That is not legitimate for a real numbers, that is only legitimate for complex number, (x) was supposed to be real.*0558

*We threw out that one, the other one was x*^{2}=4, that gave possibilities of 2 or -2 for (x).0567

*Each one of those gave me a corresponding lie, remembering our original substitution and so we got our two solutions for (z).*0575

*We could check each one by squaring them out and see if we really get 3 + 4(i), I just squared up the positive one because I know that if you square the negative it will give you the same answer there.*0585

*That is the end of our lecture on complex numbers, these are the trigonometry lectures on www.educator.com.*0595

*Hi this is Will Murray for educator.com and we're here today to learn about complex numbers.*0000

*Up until now, when you have been studying math, you've always learned that you can only take the square root of a positive number.*0008

*Complex numbers, the idea is, we're going to allow ourselves to take square roots of negative numbers.*0015

*We'll start by creating this number that we call i.*0023

*The rule for i is that i*^{2}=-1.0029

*Now, we'll talk about complex numbers which are numbers of the form (x+y)i.*0033

*We're going to talk about adding and subtracting those, and multiplying.*0042

*We'll learn how to multiply and divide those.*0045

*Just remember that x and y are real numbers, and i always satisfies this rule, that i*^{2} is -1.0048

*Now we can talk about the square root of a negative number.*0055

*We'll start off by looking at powers of i because they do follow a pattern i*^{0}, just like anything else to the 0, we define to be 1, we say 1^{0} is 1.0060

*The first power is i, i*^{2}=-1, i^{3}=i^{2}×i, remember i^{2}=-1, so i^{3}=-i, i^{4} remember is i^{2}×i^{2}=-1×-1, i^{4}=1.0072

*Since we've come back to 1, if we multiply on one more power of i, i*^{5} is just i again.0102

*You could see that we get this repeating pattern, 1, i, -1, -i.*0110

*This pattern keeps on repeating, you can figure out any power of i just by remembering this pattern.*0119

*In fact, it keeps going in the other direction too.*0126

*Let me write down some of the negative powers of i, we'll be using them later.*0128

*If you go back in the other direction, i*^{-1}, well it's 1 power before you get to 1, and if you read back from the bottom, i^{-1}, 1 power before you get to 1 would be -i.0131

*i*^{-2} would be -1.0153

*i*^{-3} would be i.0156

*i*^{-4} would be 1.0158

*i*^{-5} would be -i.0163

*i*^{-6} would be -1.0170

*i*^{-7} would be i.0178

*i*^{-8} would be 1 again.0181

*If you just follow the powers then they go in this pattern of fours, 1, i, -1, -i, 1, i, -1, -i.*0186

*Just remember they go by fours and they always follow that pattern 1, i, -1, -i.*0202

*Let's learn how to do operations on complex numbers, addition and subtraction and then the other ones are more complicated.*0210

*The addition one's pretty easy.*0217

*If you want to add x+yi to another complex number a+bi, you just add the real parts together and then the imaginary parts.*0220

*The real parts are the x and the a, you just add those together to get x+a, then the imaginary parts, I'll do those in red, yi+bi, you add those together to get (y+b)i.*0229

*Subtraction is very much similar.*0249

*You just subtract the real parts from each other, then subtract the imaginary parts.*0251

*x+yi-(a+bi), first you subtract the real parts, you get x-a, then you subtract the imaginary parts yi-bi is (y-b)i.*0255

*We'll practice this with actual complex numbers when we get to the examples.*0271

*You have plenty of time to practice these with actual numbers.*0274

*Multiplication is more complicated and it's not what you might think.*0279

*You might think that you just multiply the real parts, and you multiply the imaginary part.*0284

*It's not like that.*0289

*What you have to do here is FOIL it out.*0291

*Let me show how that works a little more slowly.*0293

*If you want to do (x+yi)×(a+bi), remember in the algebra lectures that you have to FOIL it out, first inner, outer last.*0296

*Let me write that out, first inner, outer last.*0309

*The first two terms are x×a, the inner terms are yai, the outer terms are xbi, then the last terms are yb, then i*^{2}.0311

*If you simplify that, you get x×a, the ybi*^{2}, remember i^{2} is -1, that's xa-yb.0333

*I'm going to factor an i out of the other two terms because those are my imaginary terms, (xb+ya)i.*0346

*For the answer, the real part is this xa-yb, then the imaginary part is (xb+ya)i.*0357

*Now you can see where this complicated formula comes from.*0366

*It just comes from Foiling out the four terms there.*0371

*There's another common operation that you need to learn about what complex numbers called conjugation.*0375

*You write the conjugate of a complex number with a bar over it, like this.*0382

*Conjugation means you just take x+yi, and you change it into x-yi, or if it's x-yi, you change it into x+yi.*0389

*Conjugation is very useful because if you multiply a complex number in its conjugate, x+yi times it's conjugate x-yi, you get this difference of squares formula x*^{2}-(yi)^{2}.0401

*But x*^{2}, remember i^{2} is -1, this turns into x^{2}+y^{2}.0421

*That's very useful when you're trying to produce a real number one would know imaginary part, no i part.*0429

*You multiply a complex number by its conjugate, and the answer would always be this real number, in fact, it'll always be a positive real number because x*^{2}+y^{2} will always be positive.0436

*Now we're ready to learn the most complicated operation on complex numbers which is division.*0449

*How do you divide x+yi by a+bi?*0454

*We want to give our answer in a real number plus a real number times i.*0457

*This is quite complicated.*0464

*The trick is to look at the conjugate of the denominator, a+bi conjugate, and multiply top and bottom by the conjugate.*0465

*That's what we'll do here, we multiply top and bottom by a+bi conjugate.*0474

*The conjugate of a+bi is a-bi.*0481

*Remember that the point of multiplying a complex number by its conjugate is that, when you multiply (a+bi)×(a-bi), the answer is just a real number a*^{2}+b^{2}.0485

*That's where the denominator comes from, it comes from that difference of squares pattern we learned from the previous slide, (a+bi)×(a-bi)=a*^{2}+b^{2}.0500

*In the numerator, it doesn't work so nicely,we have to multiply out (x+yi)×(a-bi).*0510

*If we FOIL it out again, we have x×a first terms, outer terms are -xbi, inner terms are +yai, then the last terms are -ybi*^{2}.0517

*If we simplify that and collect the real terms, remember i*^{2} is -1, this is +yb, if we combine that with the xa, that's where this real terms comes from xa+yb.0536

*If we combine this 2 imaginary terms and factor out the i, we get the (ya-xb)i.*0554

*That's where that line comes from.*0560

*You can separate this out to get it into the form we're looking for.*0561

*Remember we wanted to form a real number plus a real number times i.*0566

*If you separate this out, we get (xa+yb)/(a*^{2}+b^{2}), that's a real number, then ((ya-xb)/(a^{2}+b^{2}))i.0572

*We got it into that form we liked.*0583

*That seems extremely complicated.*0587

*The only thing you need to remember here is to divide complex numbers ...*0590

*I'll write this down, to divide complex numbers, the only thing you need to remember is multiply top and bottom by the conjugate of the denominator.*0603

*That's what we did here.*0632

*We multiplied top and bottom by a-bi which came from the denominator being a+bi.*0633

*You just need to remember to multiply the top and bottom by the conjugate of the denominator.*0643

*The point of that is that makes the denominator a*^{2}+b^{2} and the numerator you kind of stuck with whatever mess you end up getting into.0648

*Let's practice now with some actual complex numbers and see how that works.*0658

*We want to solve the following equation for the complex number z.*0663

*Our unknown is z.*0668

*Think of this as being a complex number times z, plus a complex number is equal to another complex number.*0670

*This is sort of like solving with these real numbers az+b=c, then we would solve for z by subtracting b from both sides az=c-b, divide both sides by a, and we get z=(c-b)/a.*0676

*We're going to do the analog of that process but with complex numbers.*0699

*First, we'll subtract from both sides 6+2i.*0702

*We'll subtract that from both sides.*0713

*That will give us on the left (1+2i)×z=(2+9i)-(6+2i).*0714

*That simplifies down to -4+7i.*0728

*Now we'll divide both sides by 1+2i, I'm trying to solve for z.*0736

*I get z=(-4+7i)/(1+2i).*0747

*Remember the way you do division for complex numbers, you multiply top and bottom by the conjugate of the denominator.*0758

*I'll multiply this by 1+2i conjugate.*0766

*This is (-4+7i)/(1+2i).*0776

*The conjugate of 1+2i is 1-2i.*0782

*The point of that multiplication is it makes the denominator very simple.*0791

*It's 1*^{2}-(2i)^{2} but since i^{2}=-1, this is just +2^{2}.0798

*Remember that's because i*^{2}=-1.0810

*The numerator is going to get more messy and there's no way to get around expanding this using FOIL.*0813

*I've got -4, first terms, the outer terms are +8i, the inner terms are 7i×1, the last terms are -14i*^{2}, but i^{2} is -1, that's really +14.0819

*If I simplify this down, my denominator is 5, -4+14, is 10, 8i+7i=15i, this simplifies down to 2+3i.*0843

*My z value there is 2+3i.*0864

*Let's recap what we had to do to make this problem work.*0868

*Essentially, think of each of these complex numbers as real numbers and treat them just as real numbers, and do whatever algebraic operations you need to do to solve the equation.*0872

*In this case, to solve the equation we had to subtract a complex number from both sides.*0883

*I know how to subtract complex numbers now.*0890

*Then we had to divide both sides by the complex number 1+2i.*0892

*That's trickier.*0898

*To divide complex numbers, you multiply by the conjugate.*0899

*We multiplied by the conjugate here, that's 1-2i.*0902

*The point of multiplying by the conjugate is that 1+2i and 1-2i multiply using the difference of squares pattern into 1+2*^{2}.0905

*The reason it's plus is because we have an i*^{2} which makes a minus into a plus.0917

*We get a nice denominator there, the numerator we just had to expand it out using FOIL, and it simplifies down to the complex number that we found.*0923

*For our next example, we have to expand and simplify 1+2i*^{3}.0935

*I'll write that as (1+2i)×(1+2i)×(1+2i).*0941

*I'll just multiply these first two terms together.*0952

*That's (1+2i)*^{2}, I'm using now (a+b)^{2}=a^{2}+b^{2}+2ab.0955

*The 2ab is 2×2i, that's 4i.*0975

*All this times 1+2i.*0981

*Now, (2i)*^{2}, that's 4×-1, that's -4.0984

*This whole thing is (-3+4i)×(1+2i).*0991

*I'll expand that out using FOIL, first terms -3, outer terms is -3×2i, that's -6i, inner term's +4i, the last term's +8i*^{2}.1003

*The 8i*^{2} converts into -8.1024

*So, -3-8=-11, -6i+4i=-2i.*1029

*What we had to do for that problem, we had to cube 1+2i, that's just a matter of multiplying out 1+2i times itself 3 times.*1043

*We just do that in stages, we multiply two of them together using FOIL, simplify it down to a simpler complex number -3+4i, then multiply on the third one using FOIL, then simplify it down to a simpler complex number.*1053

*The third example here, we have to simplify the following powers of i, i to the -6, 19, 33, -13.*1071

*The key thing here is to remember that the powers of i go by 4.*1080

*If we start at i *^{0}, that's 1, i^{1}=i, i^{2}=-1, i^{3}=1, after that they start repeating.1084

*i*^{3}=-i, i^{4}=1.1099

*After that they start repeating in cycles of 4, and it goes back in the other direction too.*1107

*i*^{-1}=-i, i^{-2}=-1, i^{-3}=i, i^{-4}=1.1115

*It goes in cycles of 4 in both directions.*1131

*What we have to do is look at this exponent and try to reduce them down by 4's.*1135

*i*^{-6}, I'll reduce that down by 4, that's the same as i^{-2}, and I check over here, i^{-2}=-1.1141

*i*^{19}, if you're counting by 4's, you got 4, 8, 12, 16 and then 3 more to get to 19, the remainder's 3.1154

*That's the same as i*^{3} because we went down.1168

*Essentially what we did was we went down by 4 at a time, and i*^{3} I remember from my pattern over here is -i.1171

*i*^{33}, if you're counting by multiples of 4, you have 4, 8, 12, 16, 20, 24, 28, 32, 33 is one more, that's i^{1}.1183

*It cycles every 4 powers of i.*1208

*This i*^{1} is just i.1212

*i*^{-13}, counting by powers of 4 you got 4, 8, 12, then 13 is one more than 12, this is the same as i^{-1}.1216

*We know that that's -i.*1229

*The way you simplify powers of i is you just remember that they go in multiples of 4, they go in cycles of 4.*1240

*If you have a really big power of i, you just need to figure out what power is, what its remainder is when you divide by 4.*1250

*For example in the case of 19, 19=4×4+3, you throw out the 4's and just look at that remainder of 3, you get i*^{3}, then you remember i^{3}=-i.1257

*It works for all of them the same way.*1278

*You just have to keep track of what the cycle is for the first four powers of i.*1280

*We are working on some more examples of polar form of complex numbers.*0000

*Remember the equations for polar form of complex numbers are exactly the same as the equations for polar coordinates that we have learned before in the previous lecture.*0004

*If you are having any trouble with these examples, you might want to review the previous lecture on polar forms of polar coordinates.*0015

*Once you understand polar coordinates really well, then the conversions for complex numbers into polar form uses exactly the same equations, even the same special cases.*0025

*They should make more sense for you.*0035

*In this example, we are converting one complex number from rectangular to polar form and another one from polar to rectangular form. *0038

*Let me remind you what the conversion formulas are, r=(square root) of x*^{2} of x^{2} + y^{2}.0048

*(theta)=arctan(y/x) and then just like with polar coordinates, sometimes you have to add on an extra pi.*0057

*The time when you have to do that is if x is less that 0, if x is bigger than 0 then you just stick with the actan(y/x).*0069

*I will go ahead and show you the conversions on the other direction.*0078

*For me the easiest one to remember is e*^{(i)(theta)} = cos (theta) + (i) sin (theta).0082

*If you do not want like that one you can also work everything out from x=arcos(theta) and y=arcsin(theta), either one of those will work for you.*0092

*Let us look at (z) now, (z) is (–root 2) – ((root 2(i)), my(r) is the square root of (root 2)*^{2} is just 2 + (root 2)^{2} is 2 again.0104

*This just simplifies down to the square root of 4 which is 2, (theta) =arctan(y/x), arctan(root 2)/(root 2) because the negative is cancelled.*0122

*But we still have (x) less than 0 so I have to add on a pi, arctan(1)+pi.*0139

*arctan(1) that is a common value, I know that is pi/4 + pi which is 5pi/4.*0145

*My complex number is (re)*^{(i)(theta)}, 2e^{5pi/4(i)}.0166

*It helps to check that graphically, if you graph ( –root 2) – ((root 2 (i)), the circle is a little bit loft sided, but that is not important.*0179

*(-root 2)-((root 2 (i)) that is down on the third quadrant, that is somewhere down here and if you look that really is at an angle of 5pi/4.*0198

*That checks that we probably got the right angle there.*0212

*On the other one, 6e*^{(5pi/6(i)} that is the polar form, I’m supposed to convert it to rectangular form.0218

*W=6, I’m going to use this form e*^{sin(theta)}=cos(theta) + (i) sin (theta).0225

*So 6 x cos(5pi/6) + (i) sin (5pi/6), if you do not like that you can also use x=arcos(theta), y=arcsin(theta), you will get to exactly same formula.*0233

*In fact there would not even me more steps, they would be just about the same number steps.*0251

*Which ever one you is more comfortable to you, feel free to use that one.*0254

*I’m drawing my unit circle and find 5pi/6 on it, 5pi/6 down there, it is just a little bit short of pi and that is a common value.*0261

*I know that is the one with the square root of 3/2 and 1/2 , we just got to figure out which one is positive and which one is negative.*0273

*The (x) one is negative, so this is 6 x (-root3/2) + (i) sin (5pi/6) is +1/2 because the y is positive. *0281

*This simplifies down to 6/2 is 3, -3(root 3) + 3 (i).*0296

*Each one of those was a pretty straight forward application of the formulas, one we had r=square root of (x*^{2} + y^{2}).0312

*And then the arctan formula for (theta), remembering that you put a correction if the (x) is less than 0, and we found our (r), we found our (theta), we did put on the correction.*0321

*By the way,I have been doing all these in terms of radians, if you found arctan(1) in terms of degrees, if your calculator was in degree form it would have given you 45 and you would have to correct that in radians.*0334

*In this case, I did not even use a calculator because arctan(1) is a common value, I remember that was pi/4.*0349

*Add on my correction term of pi and I get 5pi/4 and so (z)=re*^{(i)(theta)} so 2 x 5pi/4(i).0358

*On the other one we have to convert from polar to rectangular form, the polar form was 6e*^{5pi/6(i)}.0370

*You could use the conversion formula x=arcos(theta), y=arcsin(theta) or you can use the formula e*^{(i)(theta)} = cos(theta) + (i)sin(theta). 0378

*I really like that one so I plugged that one in, drew a unit circle to remind me where 5pi/6 is and what is sin and cos are, fill those in and I got the rectangular coordinates for that complex number.*0389

*Ok now we are asked to simplify the expression 1 + (i)*^{7}.0000

*Now that would be a really nasty one if we had to multiply all that out to the 7th power.*0005

*Instead, what we are going to do is convert to polar form and hopefully the exponentiation will be easier in polar form and after we expand it out in polar form we will convert it back to rectangular form.*0009

*Let us see how it goes, remember r=square root(x*^{2} + y^{2}) and (theta) = arctan(y/x).0022

*Sometimes you have to add on an extra pi there, you do that when (x) is less than 0.*0036

*1 + (i) that means that x is 1 and my y is one, r = square root (1+1) which is square root(2).*0043

*(theta) is arctan(1), that is a common value pi/4 and I do not have to introduce the fudge factor this time because the x is positive.*0059

*You can check that on the unit circle 1 + (i) is right there and that does check that the radius is the square root of 2 and the angle is pi/4, that checks my work here.*0074

*What we have here is square root of 2 x e*^{pi/4(i)} that is the polar form of the complex number 1 + (i).0096

*We want to raise that to the 7th power, so we raise both sides to the 7th power.*0117

*Now that looks pretty horrible but this just turns into square root of 2*^{7}, now pi ^{4(i)}.0123

*e to 1 power raised to another power, you just multiply the exponents, that just turns into e*^{7th(i)/4(i)} .0134

*That is the beauty of the polar form is the exponents just multiply or add instead of making it really difficult in multiplying lots of things together.*0148

*The (e) part is already raised to the 7th power, (root 2)*^{7} might take a little bit of work.0158

*Let me look at this, I will write that as 2*^{1/2} to the 7th power and then you could write that as 2^{7 ½}.0165

*2*^{7 ½} is the same as 2^{3½}, 7 ½ is the same as 3½.0178

*That is the same as 2 cubed x 2*^{1½}, law of exponents there and 2 cubed is 8, 2^{1 ½} is root 2 there.0186

*Root (2*^{7 }) is 8 (root 2), so this whole thing turns into 8 (root 2) x e2^{7 pi/4(i)}.0199

*I like to expand out e*^{7 pi/4(i)}, convert that back into rectangular form.0214

*For me, the easiest way to do that is with the formula e*^{(i)(theta)} = cos(theta) + (i)sin(theta).0223

*That one works really well for me, but you can also use x=arcos(theta) and y=arcsin(theta) if you like. *0234

*But I’m going to try the cos(theta) + (i)sin(theta), let me find where 7pi/4 is.*0243

*That is just short of 2pi, it is down there 7pi/4, pi/4 short of 2 pi.*0251

*That is a common value, I know the sin and cos of that.*0261

*There is no (i) there, plus (i) sin (7pi/4).*0272

*That is 8 (root 2) I know the sin and cos of 7pi/4 they are both (square root of 2)/2.*0281

*I just have to figure out which one is positive and which one is negative and since the y coordinate is negative there, we are below the axis.*0288

*The sin(1) is negative, the cos(7pi/4) is +root2/2.*0298

*The sin is –root 2/2, and now I just have to simplify this, 8 x (root 2) x (root 2)/2 that is 4 x (root 2) x root, that is 4 x 2.*0306

*4 x 2 – (i) x 4 x 2, where I am getting 4 x 2 that is 8/2 and then (root 2) x (root 2) so this is 8 – 8(i).*0325

*That was a little bit long but if you think about it, figuring out 1 + (i) *^{7 } directly would also be long because we have to multiply complex numbers together and they get bigger and bigger.0344

*It would have been pretty complicated if we did it in rectangular form.*0355

*Let me recap the steps that we did to solve this problem, we converted it into polar form, first of all.*0359

*We have to find (r) and (theta), we found (r) using square root of (x *^{2} + y ^{2}).0366

*(theta) was just arctan(y/x) and we did not have to introduce the fudge factor because the x was positive.*0373

*Arctan(1) that was a common value, I know that it is pi/4.*0380

*This number converted in to the polar form (root 2) x e*^{pi/4(i)} and we have to raise that up to the 7th power.0385

*We got (root 2) *^{7 } and e^{pi/4(i)} raise to the 7th power, you just multiply the exponents.0395

*That was the real time saving step there was just multiplying the exponents and there was a little work of figuring out what (root 2) *^{7} was.0404

*Right that is 2*^{1/2 to the 7th}, 2 ^{7/2}, 2 ^{3 ½ } and then separate that into 2 cubed is 8 and 2 ½ is root 2.0414

*To figure out e*^{7pi/4(i)}, you could use arcos(theta) and arcsin(theta), but I like to use e^{(i)(theta)} is cos(theta) + (i)sin(theta).0425

*That is what I’m doing here, I found the sin and cos of 7pi/4, that is a common value.*0437

*I figured out which was positive and which was negative and then I just multiplied it through to get my answer.*0444

*Simplifying, the (root 2) is cancelled and I got 8 – 8(i) as my answer there.*0450

*So we really did get some mileage out of converting into polar form.*0455

*That is the end of our lecture on complex numbers in polar form, these are the trigonometry lectures on www.educator.com.*0461

*Hi, these are the trigonometry lectures on educator.com and today we're going to talk about polar form of complex numbers.*0000

*A lot of what we're learning in this lecture is very directly related to polar coordinates.*0007

*If you're a little rusty on polar coordinates, what you might want to do is go back and review what you learned about polar coordinates before we learn about polar forms of complex numbers.*0015

*In particular, the main formulas for converting a complex number into polar form, they're exactly the same formulas that you learned for polar coordinates.*0023

*They should be familiar to you when we go through them now.*0034

*If they're very rusty, you might want to go back and practice those formulas for converting a point into polar coordinates and back, because they'll be really helpful in this section of polar forms of complex numbers.*0037

*Let's start out there.*0050

*Complex numbers can be written in rectangular form, z=x+yi.*0052

*That represents, if you graph it, then you have an x-coordinate and a y-coordinate.*0058

*We write the rectangular formula complex number as x+yi.*0070

*Just like with points, you would give the coordinates as (x,y), with complex numbers, we give the form as (x+yi).*0075

*They can also be written in polar form, z=re*^{iθ}.0085

*That represents the polar coordinates of the same point.*0093

*re*^{i}θ, sometimes people write it as re^{θi}.0098

*That represents the polar coordinates of the point, r is the radius from the origin going diagonally instead of going in a rectangular fashion.*0109

*θ represents the angle that makes with the positive x-axis.*0121

*Just like we've had polar coordinates rθ who have the polar form of a complex number re*^{iθ}.0127

*The conversion's back and forth between those two forms are exactly the same as what we've had for polar coordinates.*0137

*Let's check those out.*0145

*The conversion for r is square root of x*^{2}+y^{2}.0147

*That comes straight from the Pythagorean theorem.*0152

*The conversion for θ is a little more complicated and it's got the same kind of subtleties and nuances that it had with polar coordinates.*0155

*θ is either arctan(y/x) or π+arctan(y/x).*0164

*The way you know which one of these formulas to use is you check the sine of x.*0172

*This is when x is greater than 0.*0178

*This is when x is less than 0.*0183

*Another way to remember that is to ask whether the point is in quadrant 1, 2, 3, or 4.*0187

*Remember arctangent will always give you a value in quadrants 1 or 4.*0200

*If you start out in quadrant 1 or 4, then you just want to use the arctangent function directly.*0206

*If you're looking for point in quadrants 2 or 3, then the arctangent will not give you the right value, that's why you add π to it.*0213

*That's the tricky one.*0235

*x and y, same formulas as we had for polar coordinates before, rcos(θ) and rsin(θ).*0237

*We'll try to use values of r that are positive, but that's not absolutely essential.*0245

*We'll try to use values of θ that are between 0 and 2π, but that's not absolutely essential.*0249

*Let me give you one more formula that's very very useful in working out conversions between rectangular and polar coordinates.*0255

*We write re*^{iθ} as x+yi.0265

*I'll write that as iy.*0277

*Polar form is re*^{iθ}, rectangular form is x+iy.0280

*If you convert that, the x is rcosθ, iy is irsin(θ).*0291

*If you factor out an r there, we get r×cos(θ)+isin(θ).*0304

*If you just take r=1, if you factor out the r from both sides, what you get here is the e*^{iθ}=cos(θ)+isin(θ).0315

*That is an extremely useful formula in converting complex numbers to polar coordinates.*0328

*That one is probably worth memorizing as well.*0337

*c=cos(θ)+isin(θ).*0338

*Let me decorate that a little bit, illustrate how important it is.*0344

*e*^{iθ}=cos(θ)+isin(θ), that's definitely worth remembering.0351

*We'll be using it on some of the examples.*0359

*Let's go ahead and practice doing some conversions here.*0362

*One more thing that I need to show you before we practice that. *0367

*Multiplying two complex numbers in polar form.*0371

*If we have two complex numbers in polar form, r*_{1}×e_{1}^{iθ}, it's got an r and a θ, and r_{2}×e_{2}^{iθ}.0375

*There's a very easy way to multiply them.*0386

*If we multiply these together, what we do is we just multiply the r's together r*_{1}×r_{2}.0390

*Remember the laws of exponents x*^{a}×x^{b}=x^{a+b}.0399

*Here, we have x or e*^{iθ1}×e^{iθ2}, you add the exponents, iθ_{1}+iθ_{1}, just gives you i(θ_{1})+θ_{2}).0409

*You add the exponents there.*0426

*You end up just multiplying the r's and adding the angles θ*_{1}+θ_{2} because they're in the exponents.0429

*Now let's try some examples.*0436

*We're going to convert the following complex numbers from rectangular form to polar form.*0438

*Let's start out with -3+i.*0445

*The -3 is x and y is 1 there.*0449

*We want to find the r and θ, r is the square root of x*^{2}+y^{2}.0454

*Let me write this at the top page so I don't have to keep rewriting it.*0460

*θ=arctan(y/x), that's if x > 0, or we might have to add π to that if x < 0.*0464

*In this case, our r is the square root of x*^{2}, negative root 3 squared is just 3, +y^{2} is 1, that simplifies down to 2.0484

*θ=arctan(y/x), y=1, 1 over negative root 3, which is arctan of negative root 3 over 3.*0498

*That's one of my common values.*0523

*I know what the arctan negative root 3 over 3 is, it's -π/6.*0525

*My x-coordinate was negative there so I haven't actually been using the right formula, I have to add π to each of these, +π.*0536

*You almost always use radians and not degrees here.*0546

*If you do happen to plug this into your calculator, make sure your calculator is in radian mode.*0549

*I didn't have to use my calculator on this one because negative root 3 over 3 is a common value.*0556

*(-π/6)+π=5π/6.*0562

*My polar form for that complex number is r*_{2}e^{iθ}, so e^{(5π/6)i}.0572

*Let's keep going with the next one.*0591

*6 root 2, that's my x, -6 root 2, that's my y, r is the square root of x*^{2}, 6 root 2 squared is 36, times 2 is 72, +y^{2} is 6 root 2 again, 72, square root 144 is 12.0595

*θ=arctan(y/x), that's negative 6 root 2 over 6 root 2, which is arctan(-1) which is -π/4.*0626

*My x in this case was positive so I don't have to introduce that correction term.*0642

*I get w=re*^{iθ}=12e^{(-π/4)}.0650

*I don't really like that negative value of π/4, so what I'm going to do is to make it positive, to get it into the range, 0 to 2π, I'll add 2π to it.*0661

*I'll write that as 12e, I need an i there, ex*^{(7π/4)i}.0672

*You can also understand these things graphically.*0685

*Let me draw a unit circle here.*0693

*Negative root 3 plus i, that means my x is negative root 3 and my y is 1.*0707

*I recognize that as a multiple of root 3 over 2 and -1.*0717

*I recognize that as being over here.*0724

*That's z with radius of 2, because it's 2 times root 3 over 2 and 1/2, I know that that's 5π/6.*0726

*That's the way to kind of check graphically that my z is 2×e*^{(5π/6)i}.0742

*For w, 6 root 2 minus 6 root 2, I know that's 12 times root 2 over 2 root 2 over 2, except the y is negative.*0750

*That value is 7π/4, that's kind of a little graphical check that we have the right polar form for the complex numbers.*0774

*Let's go back and recap what we did for that problem.*0788

*We're converting complex numbers from rectangular form to polar form, really just boils down to these two conversion formulas for r and θ.*0791

*r gives you the magnitude, θ gives you the angle.*0801

*The problem though is that this θ formula is little bit tricky.*0802

*It has this two cases depending on whether x is positive or negative.*0808

*If x is negative then you have to add an extra π to it, that's what we did here, we were adding an extra π to the value of θ.*0811

*Once you find r and θ, you just plug them into this form re*^{iθ}.0820

*That's how we got the answers for each of those.*0825

*For the next one, we're converting from polar form to rectangular form.*0830

*We're given z=4e*^{(-2π/3)i}, and w=2e^{(3π/4)i}.0834

*Let me write down the conversion formulas.*0845

*x=rcos(θ), y=rsin(θ).*0847

*For the first one, x=4cos(-2π/3).*0859

*Let me graph that quickly on the unit circle.*0871

*-2π/3 is down here, it's the same as 4π/3.*0878

*The cosine is -1/2, that's a common value.*0883

*This is 4×-1/2, which is -2.*0890

*The y there is 4sin(-2π/3), the sine of that is negative root 3 over 2.*0894

*This is 4 times negative root 3 over 2, which is -2 root 3.*0913

*We're going for the form x+yi, our z is equal to x=-2, +yi, -2 root 3, i.*0925

*For the second one, we have 2e*^{(3π/4)i}.0943

*I'll graph that on the unit circle to help me find the sine and cosine.*0951

*3π/4 is over there, it's 45-degree angle on the left-hand side.*0954

*I know the sine and cosine very quickly.*0958

*x is equal to r which is 2, cosine of 3π/4, which is 2, cosine of that is negative because it's on the left-hand side.*0962

*2 times negative root 2 over 2, which is just negative root 2.*0977

*y=2sin(3π/4), which is 2 times positive root 2 over 2 because we're in that second quadrant, y-coordinate is positive.*0987

*x+yi is negative root 2 plus root 2i.*1000

*That one wasn't too bad, it was simply a matter of remembering x=rcos(θ), y=rsin(θ), then putting those into x+yi.*1020

*For finding the sines and cosines, it helps if you graph the angle in each case.*1034

*Once you remember those formulas, you just work in through arccos(θ) and arcsin(θ) in each case.*1041

*For the third example, we're going to use polar form in an application.*1050

*We're going to perform a multiplication by converting each one of the complex numbers to polar form, then we're going to check the answer by multiplying them directly in rectangular form.*1054

*-1 plus root 3i, I'm going to figure out my r there.*1065

*My r is equal to square root of x*^{2}+y^{2}.1071

*Let me write these formulas generically, x*^{2}+y^{2}, θ=arctan(y/x), that's if x is bigger than 0, we'll have to add on a π, the fudge factor π if x < 0.1078

*In the first one, r is equal to 1*^{2} plus root 3 squared, that's 3, which is 2 square root of 1 plus 3.1104

*θ is equal to arctan negative root 3 over 1.*1113

*Let me write that as root 3 over -1.*1123

*I have to add on a π because the x is negative.*1129

*Arctan of negative root 3 is negative π/3+π, that was a common value that I remembered there.*1134

*Plus π gives me 2π/3.*1144

*That tells me my r and my θ for the first one.*1151

*Let me go ahead and figure them out for the second one before I plug them in.*1152

*For the second one, we have r is equal to the square root of 2 root 3 squared.*1155

*2 root 3 squared is, 4 times 3, is 12, plus 2 squared is 4, 12+4=16, that gives me root 16 is 4.*1163

*θ is arctan 2 over 2 root 3, but the x-coordinate was negative, I have to add a π, so this is arctan 1 over root 3, is root 3 over 3 plus π.*1177

*Again, that's a common value, so arctan of root 3 over 3, I remember that's a common value, that's (π/6)+π=7π/6.*1205

*If I convert one to each one of these numbers into polar form, this one is 2e*^{(2π/3)i}.1220

*This one is 4e*^{(7π/6)i}.1235

*I want to multiply those, but multiplying numbers in polar form is very easy.*1247

*First, you multiply one radius by the other one, that's 2×4=8, then you add the angles e*^{((2π/3)+(7π/6))i}.1255

*You just add the angles, you multiply their radius by the other one and then you add the angles.*1272

*That's 8e to the, let's see (2π/3)=4π/6, you get (11π/6)i.*1277

*I want to convert that back into rectangular form.*1287

*I forgot to put my e in there.*1293

*I'm going to use this formula e*^{iθ}=cos(θ)+isin(θ).1296

*That one's really useful, definitely worth remembering.*1302

*This is 8cos(11π/6)+isin(11π/6).*1306

*You could also use x=rcos(θ), y=rsin(θ), you'll end up with the same formula at the end, either way works.*1322

*Let me draw on the unit circle to remind where 11π/6 is.*1329

*11π/6 is just short of 2π, it's right there.*1336

*It's a 30-degree angle south of the x-axis.*1341

*The cosine there is root 3 over 2, it's positive because we're on the right hand side.*1345

*The sine is -1/2.*1358

*What we get there that simplifies down to 4 root 3 minus 4i.*1366

*Now we've done it.*1383

*We've converted each number into polar form.*1385

*We multiplied them in polar form which is very easy, then we converted the polar form back into rectangular form to give us our answer.*1388

*It says we have to check our answer by multiplying them directly in rectangular form.*1398

*Let's do the check here, we'll FOIL the multiplication out.*1404

*I'll do the check over here.*1410

*I'll do the check in blue.*1411

*Foiling it out, my first terms give me -1 times 2 root 3, that's positive root 3.*1414

*My outer terms give me -1-2i, so +2i.*1421

*My inner terms give me -2 times root 3 times root 3, that's 6i.*1427

*Those are my inner terms, I'm doing FOIL here.*1437

*First outer, inner, and my last terms are root 3i minus 2i, that's -2 root 3, i*^{2}, but i^{2}=-1, this counts as +2 root 3.1440

*If we simplify that down, we get 2 root 3 plus 2 root 3, 4 root 3, +2i-6i, is -4i.*1459

*That does indeed check with the answer we got by converting into polar form.*1471

*That was kind of a long one.*1477

*Let's recap what we did there.*1478

*We had these two complex numbers.*1480

*We wanted to convert each one into polar form.*1482

*For each one, I found my r, and I used square root of x*^{2}+y^{2}.1484

*I found my θ by using arctan(y/x), [intelligible 00:24:55] each one the x's were less than 0, so I had to add on this fudge factor plus π to get me into the right quadrant.*1491

*I found my r, my θ, another r, my other θ.*1504

*I converted there each one into re*^{iθ} form.1509

*To multiply them together, you multiply the r's but then you add the θ's because they're up in the exponents.*1514

*That's the law of exponents there, so we added the θ's.*1516

*We got a simplified polar form and then we converted back into rectangular form using either the iθ=cos(θ)+isin(θ).*1524

*You could also use x=rcos(θ), y=rsin(θ), you'll get to exactly the same place.*1535

*I know my cosine and sine of 11π/6, that's a common value.*1542

*I get the answer there.*1547

*To check it, I skipped all the polar forms.*1548

*I just multiplied everything out using FOIL, simplified it down and it did indeed check with the answer that I've got using the polar form.*1550

*We'll try some more examples later.*1563

*You should try them on your own first and then we'll work on them together.*1565

*Hi we are working on some more examples of DeMoivre’s theorem.*0000

*Our first extra example here is to convert the complex number 2 (root 2) – 2 (root 2i) in polar form and then use DeMoivre’s theorem to calculate z*^{5}.0004

*Let us first convert that into polar form, remember my equations x*^{2} + y^{2}, square of that used to be (r) and (theta)=arctan( y/x).0016

*Sometimes you have to introduce the extra term plus pi, you have to do that when (x) is less than 0.*0035

*Here, my x=2(root 2), y=-2(root 2), r= x*^{2} + y^{2}.0045

*2 (root 2) *^{2} is 4 x(root 2)^{2} is 2, 4 x 2=8, y=-2(root 2), r= square root of 8 + 8, square root of 16 is 4.0058

*My (theta) is arctan(y/x) is -2(root)2/2(root)2 that is -1. *0077

*There is no fudge factor on this one because the (x) is positive, there is no plus pi.*0089

*Arctan(-1) that is a common value that I remember is –pi/4.*0095

*If you work that out on your calculator in degrees it will say -45, if you have your calculator in radian mode it will say –pi/4.*0102

*But you should not really need a calculator for that because that is a common value.*0109

*That is –pi/4, that is still not in the range that I like which is 0 to 2pi, I think what I’m going to do is I’m going to add 2pi to that and get 7pi/4. *0115

*That is now in the range between 0 and 2pi.*0133

*My z in polar form is 4e*^{7pi/4}, that is the answer to the first part of the problem.0138

*We have converted the complex number into polar form, DeMoivre’s theorem says we want to calculate z*^{5} .0151

*We can do that by z *^{n}=r^{n} x cos(n)( theta) + (i)sin(n)(theta).0159

*That is how DeMoivre’s theorem applies to this one z*^{5} = r^{2}, that is 4^{5} x cos(n)(theta), n=5, theta=7pi/4.0173

*35 pi/4 + (i)sin(35)(pi/4).*0194

*Remember that 35/4 came from n(theta) which was 5 x 7 pi/4.*0205

*I forget to include there the (i) in the polar form, let me put that in, that is 7pi/4i.*0213

*There is supposed to be an (i) in there.*0222

*I do not like the fact that 35 pi/4 is not between 0 and 2 pi.*0226

*Let me see if I can simplify that down, 35 pi/4 if I subtract off 2 pi.*0230

*2pi is 8 pi/4, that would give me 27 pi/4, subtract off another 2 pi that would give me 19 pi/4.*0242

*Subtract off another 2 pi=27pi/4, subtract off another 2pi would give me 11pi/4.*0259

*It is still in the range but if I subtract off 2 pi again, that gives me 3pi/4.*0265

*This whole thing simplifies down a little bit 4*^{5} is (4 x 4 x 4 x 4 x 4)= 1024.0274

*Now the formula of 35pi/4 between 0 and 2pi is 3pi/4, so cos(3pi/4) + (i)sin (3pi/4).*0287

*To simplify that, it is helpful to draw a little unit circle and remember where 3pi/4 is.*0307

*That is right there on the unit circle, 3pi/4 between pi/2 and pi.*0318

*The cos and sin there of both (root 2)/2, you just got to figure out which one is positive and which one is negative.*0326

*But because it is in the second quadrant, the (x) is negative so that is (-root 2)/2, and (y) is positive so (root 2)/2.*0337

*We can simplify this down a little bit, this is -512(root 2) because we divided by 2 + 512 (root 2) and that is my answer.*0352

*Let us recap what we did for that one, we are given a complex number in rectangular form.*0383

*I took those at my (x) and (y) and I plugged them into my formulas for (r) and (theta) to find the polar form.*0388

*For the formula for (theta) the (x) was not zero so I had to include the- sorry, the (x) was positive, I did not have to include the fudge factor.*0396

*When I took the arctan I got –pi/4 so I added on 2pi to get it into the range of 0 to 2pi.*0407

*The polar form for that complex number is 4e*^{7pi/4i}.0415

*To raise it up to the nth power, I’m going to use DeMoivre’s theorem that says z*^{n} is r^{n}, cos(n)(theta) +(i)(sin)(n)(theta).0423

*(theta) was 7pi/4, the n=5, so I get 35pi/4, pretty nasty.*0434

*So I subtract off a bunch off 2 pi and get it down to 3pi/4, I fill in the cos and sin of 3pi/4, which I remember those are common values on the unit circle there.*0441

*Finally I simplify it down and I get my answer there.*0458

*For our last example, we are trying to find all complex 4th roots of the complex number -2-(2 root 3(i)).*0000

*From the beginning I should say we expect 4 answers because we are looking for 4th roots.*0009

*At the end of this, we should have 4 answers and I’m going to convert the number in polar form.*0023

*Let me remind you of the formulas for polar form, x*^{2} + y^{2} and (theta)=arctan(y/x).0031

*If (x) is less than 0 you have to introduce the fudge factor plus pi there and I will go ahead and find that right away.*0042

*R=square root of -2*^{2} is 4, 2 (root 3) ^{2} is 2^{2} x 3, that is 4 x 3 x 12.0051

*That is the square root of 16 is 4, (theta)=arctan(y/x) so that is 2 (root 3)/2 (root 3).*0069

*Positive because we have two negatives, but (x) is negative so I do have to introduce the fudge factor of pi.*0084

*Arctan(root 3) that is a common value for arctan that is pi/3 + pi = 4pi/3.*0092

*I have got my (theta), (z)=re*^{i(theta)}, 4e^{ 4pi/3 x (i)}.0108

*That is the polar form for that complex number, we got the (r) and (theta), I have got polar form for the complex number.*0125

*Now I want to find all complex 4th roots of that, let me remind you what DeMoivre’s theorem says about this.*0135

*It says that z*^{1/n} is equal to r^{1/n} x cos((theta + 2 k(pi/n)) + (i)sin((theta + 2 k(pi/n)).0144

*You figure that out for each value of (k) from 0 to n -1, in this case (n)=4 so (k) goes from 0 to 3.*0177

*I want to make a chart of all the possibilities here, let us work out what (k) could be.*0194

*We already said that could be 0, 1, 2, and 3, now I want to figure out what this angle ((theta + 2k (pi/n)) is.*0211

*((theta + 2k (pi/n)) now (theta) =4pi/3 and n= 4, this is (4pi/3 + 2k pi/4) which simplifies down to 4pi/3/4 is just pi/3 + 2k pi/4 is just k pi/2.*0227

*Let us figure out what that value is for each value of (k), when (k)=0 that is just pi/3, when (k)=1 that is pi/3 + pi/2, the common denominator is 6, that is 2pi/6 + 3pi/6 = 5pi/6.*0270

*When (k)=2 this is pi/3 + 2pi/2 is pi which is 4pi/3, when (k)=3, we have pi/3 + 3pi/2, again common denominator is 6, we have 2pi/6 + 9pi/6= 11pi/6.*0303

*Those are the four angles we will be plugging into the sin and cos formula, if you think about that being alpha.*0338

*Next step is to figure out the sin and cos of alpha is, cos(alpha) + (i) sin (alpha).*0346

*When alpha is pi/3, let me draw a little unit circle to help me figure these values out.*0364

*When (alpha) = pi/3 that is up there, the cos and sin are ½ and root 3/2, they are both positive because we are in the first quadrant.*0379

*For 5pi/6, that is in the second quadrant over there, the sin and cos are root 3/2 and ½ and now the cos is negative, sin is still positive because the (x) is negative and the (y) is positive there.*0406

*4pi/3 is down here on the third quadrant, sin and cos are both negative now and its -1/2 – root 3/2.*0429

*And 11pi/6 is way over there, the sin and cos are root 3/2 and the sin is negative so –(i) x ½ there.*0448

*Those are the sin and cos of those four angles, by the way if you look at that unit circle that I drew.*0477

*Let me highlight where those points are, you will notice that they are exactly evenly spaced around the unit circle.*0483

*It is because we are adding on k pi/2 each time, we are adding on pi/2 each time.*0492

*We get these 4 angles that are exactly spaced out around the unit circle by enables of pi/2 and that is not an accident.*0499

*It is because we started out looking for 4th roots, we divide the unit circle into 4 parts and that is why you go around pi/2 each time.*0509

*That is no accident that those are evenly spaced out by the multiples of pi/2 and have we gone one more step we would have ended up at pi/3, where we started again.*0514

*Let me go back to finding these 4th roots, we find cos(alpha) + (i) sin(alpha).*0533

*The only thing we have to do left is to multiply on the r*^{1/n}, let me figure that out.0540

*R*^{1/n} =4, so ¼, there is a clever thing we can do to the exponents here I know that 4 is 2^{2} to the ¼ so that is 2^{2/4}, that is 2^{1/2} which is root 2.0547

*What we are doing here is we are finding r*^{1/n} x cos(alpha) + (i) sin (alpha).0566

*But the r*^{1/n} is root 2, I’m just going to multiply root 2 by each of the complex numbers in the proceeding column of the chart.0578

*Root 2 x the first complex number is root 2/2 + (i) x root 6/2, that is because root 2 x root 3 is root 6.*0589

*Next one is –root 6/2 + root 2/2 (i) – root 2/2 – root 6/2(i) and finally root 6/2 that is root 2 x root 3 – root 2/2(i).*0604

*I’m just going to box off this last column of the chart and call that my answers.*0638

*What that means is that these are four complex numbers if you take any one of these four complex numbers, those are my four answers.*0649

*If you raise any one to the 4th power, what you should get is the original complex number that we started with -2-(2 root 3(i)).*0659

*That is the end of our last problem, let us go back and recap and see what strategies we used to solve it.*0672

*We started with this number -2-(2 root 3(i)), I want to get that in polar form so I can use it to DeMoivre’s theorem.*0677

*DeMoivre’s theorem only works on complex numbers in polar form so I used my equation for polar form r=square root (x)*^{2} + (y)^{2} and (theta)=arctan(y/x) plus the fudge factor of x is less than 0.0686

*My (x) and (y) I got those from the original complex number and the (theta) I did have to use the fudge factor plus pi.*0702

*Arctan(root 3) is an angle that I know pi/3 and I get (theta) = 4 pi/3. *0712

*I got the polar form of the complex number, (r)(e) *^{(i)(theta)}, r =4, e^{(i)(theta)}=e^{4pi/3} x (i).0719

*Now I used DeMoivre’s theorem, DeMoivre’s theorem says you look at z*^{1/n} is r^{1/n} x cos(alpha) + (i) sin(alpha).0731

*Where the (alpha) is this (theta) + 2k pi/n, and then you plug in different values of (k) going from 0 to n-1.*0744

*Where the n here is 4, so I run (k) from 0 to 3, that is n-1, and here I made a list of my different (k).*0754

*For each one of those I figured out that (theta) + 2kpi/n, that was the second column.*0765

*(theta) + 2k pi/n and that is what I was calling (alpha) so I plugged in (theta) = 4pi/3, n=4 and I plugged different values of (k) each time to get these 4 answers for (alpha).*0772

*To get there 4 answers for (alpha) DeMoivre’s theorem says you look at cos(alpha) + (i) sin(alpha).*0788

*I looked at cos (alpha) + (i) sin (alpha) in this next column of the chart.*0795

*To get those cos and sin I drew a little unit circle here and I read off the cos and sin of each one, those are common values so I do remember those.*0801

*Finally I have to multiply this by r*^{1/n}, what I worked out over here is r ^{1/n}.0811

*(r) was 4 and (n) was also 4, it is 4*^{1/4}, y^{4} is 2 squared.0818

*That gives you a quick way to figure out to simplify down 4*^{1/4}.0825

*Simplify that into square root of 2, so I multiply r*^{1/n} square root of 2, that is here.0831

*Multiply that by each of the values that we got in the proceeding column.*0838

*We multiply this by the square root of 2 and that finally gave me my 4 answers for the 4th roots of -2-(2 root 3(i)).*0844

*These are my 4 answers right here as my four 4th roots.*0856

*That is the end of our lecture on DeMoivre’s theorem to find nth powers and nth roots of complex numbers, in fact that is the end of all our lectures on trigonometry.*0862

*Thank you very much for watching, this is Will Murray for www.educator.com.*0873

*Hi, these are the trigonometry lectures on educator.com.*0000

*We're here today to talk about De Moivre's theorem.*0004

*The De Moivre's theorem is a little bit tricky.*0010

*The idea is that we're going to use the polar form of complex numbers to find nth powers and nth roots of complex numbers.*0013

*We start with a complex number z.*0026

*We write it in polar form re*^{iθ}.0029

*We learned in the previous lecture how to convert a complex number into polar form.*0032

*If that's a little bit unfamiliar to you, what you should really do is go back and review the previous lecture on how to convert a complex number into polar form, and how to convert it back into rectangular form.*0038

*There's several formulas that we're going to be using very heavily here.*0052

*One that we learned in the previous lecture is e*^{iθ}=cos(θ)+isin(θ).0056

*We're going to be using that really heavily.*0064

*Let's see how we can use that to find nth powers of complex numbers.*0067

*For trying to find z*^{n}, we write that as re^{iθ}, to the nth power.0073

*If you think about that, we can distribute this nth power onto the r and onto the e*^{iθn}.0080

*r*^{n}, that just gives you r^{n}.0091

*e*^{iθn}, that's an exponent raised with an exponent.0099

*You multiply the exponents.*0106

*That's where I get iθ×n here, so e*^{i×n×θ}.0107

*If you expand that, e*^{iθ}=cos(θ)+isin(θ), e^{inθ} gives you cos(nθ)+isin(nθ).0116

*That's where De Moivre's theorem comes in handy, and that's where it comes from.*0133

*You can expand this into r*^{n}×cos(nθ)+isin(nθ).0137

*Another way to start out with that is to expand e*^{iθ} into cos(θ)+isin(θ).0145

*The form we're going to be using most often is this form right here z*^{n}=r^{n}×cos(nθ)+isin(nθ).0154

*I know this looks like lots of stuff to remember here.*0166

*The key one that you want to memorize is this one right here, z*^{n}=r^{n}×cos(nθ)+isin(nθ).0168

*Memorize that one and we'll practice it during the examples.*0180

*The next step of using De Moivre's theorem is to, instead of finding nth powers, we're going to find nth roots.*0185

*For example, we'll find square roots and cube roots, and fourth roots of complex numbers.*0194

*This is quite a bit more tricky than it is with real numbers.*0199

*In fact, if you're looking for nth roots of a complex number, you always expect to find exactly n answers.*0204

*If a problem says find all the 8*^{th} roots of a complex number, you better find 8 answers.0211

*Unless of course the complex number happens to be 0, in which case the only roots are 0.*0219

*That's why I say every non-zero complex numbers has exactly n nth roots here.*0224

*Let me show you how to find them.*0231

*It's a little bit complicated.*0232

*First of all, we think about the nth root of z.*0235

*We write that as e*^{1/n}.0236

*That's re*^{iθn}.0240

*Remember, e*^{iθ} is cos(θ)+isin(θ), to the 1/n.0248

*Remember how when we were finding nth powers, we distributed the n into the r, we had r to the n.*0254

*Let me write this again.*0265

*z*^{n}=r^{n}×e^{niθ}, which was r^{n}×cos(nθ), you multiply the angle by n,plus isin(θ).0266

*With 1/n, replacing the n by 1/n, we get r*^{1/n}cos(θ/n)+isin(θ/n).0296

*We start out with just cos(θ/n)+isin(θ/n).*0315

*We look at (θ/n).*0322

*We have to find other nth roots as well.*0324

*Our first nth root is just (θ/n).*0328

*To find the other ones, what we add on is multiples of 2π/n.*0329

*That's why I say 2kπ/n.*0340

*We keep doing that for different values of k.*0343

*The reason we do that is we run all the values of k from 0 to n-1.*0347

*If we plugged in k=n into this formula, we get (θ/n)+(2nπ/n), which would be (θ/n)+(2π).*0354

*In terms of angles, that's the same as (θ/n) again.*0375

*That's why we stop at n-1.*0379

*We don't go to k=n, because when we get to k=n, we're repeating ourselves again.*0383

*Essentially, what we're doing here is we're breaking up the unit circle into multiples of (θ/n).*0388

*θ/n, (θ+2π)/n, (θ+4π)/n, we're just taking all these angles around the unit circle until we get back to θ/n.*0401

*This is a little bit tricky.*0423

*What we do is we find one answer for each value of k.*0426

*One nth root for each value of k.*0432

*Since we run k from 0 to n-1, that's a total of n nth roots.*0445

*That's worth remembering.*0470

*We'll practice these with the examples.*0471

*Anytime you have to find nth roots, the r part is easy, you do r*^{1/n}, but then you have to find this cosine and i-sine formula for each angle, for each value of k from 0 to n-1.0476

*You run this formula separately, n times over, and at the end, you have n complex numbers as your answers.*0494

*We'll check that out with some examples and you'll get the hang of it.*0501

*The first example here, we have to convert the complex number z equals negative root 3 plus i into polar form, then use De Moivre's theorem to calculate z*^{7}.0505

*Remember to convert it into polar form.*0518

*You do r equals the square root of x*^{2}+y^{2}, θ=arctan(y/x).0524

*Sometimes you have to modify that θ formula, sometimes you have to add on a π.*0532

*You know you have to do that when x is negative.*0538

*You do that if x is negative.*0540

*Let's find our r in our θ here.*0542

*Let me graph that thing.*0544

*Graph it just so that we'll be able to check whether our answer's possible.*0557

*Negative square root of 3 on the x-axis, i on the y-axis, that's about right there.*0560

*Let me calculate r and θ to see if it's possible.*0568

*r is the square root of x*^{2}+y^{2}, x^{2}=3, y^{2}=1, square root of 4 is 2.0574

*θ is arctangent of 1 over negative square root of 3 which is negative root 3 of 3, that's a common value.*0585

*The arctangent of that is -π/6.*0598

*There's this fudge factor that I have to include here.*0606

*The x < 0 here, I have to add on a π.*0610

*I add on a π and I get 5π/6.*0615

*That does check with my little graph here because that really is 5π/6, the angle over there, and the radius does indeed look like about 2.*0620

*That's reassuring.*0630

*z=re*^{iθ}, that's 2e^{(5π/6)i}.0632

*We have converted a complex number into a polar form, that was the first part of the exercise.*0642

*The main part here is to use De Moivre's theorem to calculate z to the seventh.*0650

*Let's work that out.*0654

*z*^{7}, the whole point is we're going to use the polar form to find z^{7}, so this is 2e^{(5π/6)i7}.0655

*That's 2*^{7}.0671

*I have each one exponent raised to an exponent.*0675

*I just want to multiply those two exponents e*^{7×5}, is (35π/6)i.0678

*35π/6 is a little cumbersome, that's not in between 0 and 2π.*0696

*I'll work on that a little bit.*0703

*In the meantime, 2*^{7} is 128.0704

*35π/6, how can I simplify that?*0711

*35π/6, let me subtract 2π, 12π/6, that's 23π/6, that's still not in my range between 0 and 2π.*0715

*Let me subtract another 2π, that gives me another 12π/6 off, is 11π/6.*0732

*That is in the range between 0 and 2π.*0743

*This is the same as e*^{(11π/6)i}.0745

*I want to convert that into rectangular form.*0751

*It's very good to remember this formula e*^{iθ} is cos(θ)+isin(θ).0755

*You can also use x=rcos(θ), y=rsin(θ).*0766

*I prefer the e*^{iθ} form.0771

*This is equal to 128cos(11π/6)+isin(11π/6).*0774

*11π/6, where is that on the unit circle?*0797

*That's just π/6 short of 2π.*0801

*That's down there.*0808

*That's a common value, I know what the sine and cosine are.*0811

*It's root 3 over 2 and 1/2.*0814

*Root 3 over 2 is positive, the sine is negative because it's below the x-axis so it's -1/2.*0820

*I have 128/2, that's 64 root 3 minus 64i.*0831

*That's what that simplifies down to.*0842

*Let's review how we did that one.*0848

*We start out with a complex number, and we have to convert into polar form.*0850

*I look at my formulas for r and θ including the fudge factor for θ if x < 0.*0857

*Run that through, my x and y are negative root 3 and 1.*0861

*I get an r.*0866

*I get a θ including the fudge factor.*0867

*That gives me re*^{iθ}.0870

*I've got my polar form.*0872

*To raise it up to the 7*^{th} power, De Moivre's theorem says if you use polar form, then you just put 2^{7}, then nθ.0874

*This is the nθ.*0889

*That reduces down by subtracting 2π at a time, e*^{(11π/6)i}.0890

*This is really nθ here, cos(nθ), sin(nθ), although we reduced down by subtracting off 2π at a time.*0899

*We get 128 times the cosine and sine of 11π/6.*0909

*That's a common value.*0914

*I look at my unit circle to remember my sine and cosine of 11π/6.*0915

*I plug them in and I get my answer.*0923

*Second example here, we find to find all complex 8*^{th} roots of the number 16.0928

*In order to find all complex 8*^{th} root, we have to think about 16 being a complex number.0934

*Of course 16 is just the same as 16+0i, that is a complex number.*0944

*We're asked for 8*^{th} roots.0952

*Let me remind you that because we're asked for 8*^{th} roots, we expect 8 different answers.0956

*We have to find 8 answers here.*0968

*I'm going to try and write 16 in polar form.*0976

*r is equal to the square root of x*^{2}+y^{2}.0981

*θ=arctan(y/x), plus π if the x happens to be negative.*0987

*My r is the square root of 16*^{2}+0^{2}, that's just 16.1001

*θ, my y=0, arctan(0)=0.*1007

*My z=16e*^{0i}.1016

*I could have worked that out, certainly 16e*^{0}=16 just by itself because e^{0}=1.1027

*That's nice to check our work.*1034

*z*^{1/8}.1036

*According to De Moivre's theorem, let me remind you what De Moivre's theorem said about complex nth roots.*1044

*It said that, you do r*^{1/n}×cos((θ+2kπ)/n)+isin((θ+2kπ)/n).1050

*You run this for different values of k.*1086

*k is equal to 0, 1, 2, up to n-1.*1089

*Here, n=8, so z*^{1/8}, r=16^{1/8}.1096

*Cosine is, θ=0, (0+2kπ)/8, plus isin(0+2kπ)/8.*1113

*There's going to be lots of values for k here.*1144

*Maybe I should make a little chart for what k is, and then the different angles that we have for each value of k, the different angles that we're going to be plugging into De Moivre's formula there.*1147

*(0+2kπ)/8 which actually simplifies down to just kπ/4.*1165

*For k goes from 0, 1, 2, 3, 4, 5, 6, you run it to n-1, and n=8, so 0 through 7 there.*1182

*We'll have 0, π/4, 2π/4 is π/2, 3π/4, 4π/4 is π, 5π/4, 6π/4 is 3π/2, and 7π/4.*1194

*For each one of these angles we're going to plug it in and we're going to get an answer.*1224

*I also have to simplify 16*^{1/8}.1229

*Let me see if I can do something with that.*1233

*16*^{1/8}, I know that 16=2^{4}, that's 2^{4/8}, 2^{1/2} is square root of 2.1235

*I'm going to multiply the square root of 2, that's 16*^{1/8} times cos(θ)+isin(θ) for each one of these values of θ.1250

*Let me not reuse the same Greek letter θ, I'll use α.*1264

*For each one of these values of α here ...*1280

*Let me write down what cos(α)+isin(α) is for each one of these values of α.*1288

*Cos(0)+isin(0), cos(0)=1, plus isin(0)=0.*1291

*Cos(π/4)+isin(π/4), cosine and sine of (π/4) are both square root of 2 over 2.*1311

*That's k=1.*1324

*For π/2, the cosine is 0, the sine is 1.*1329

*For 3π/4, we have the cosine is negative root 2 over 2, the sine is positive root 2 over 2.*1337

*I think this will be easy to work out if I draw a unit circle, so that I can easily and quickly find the sines and cosines.*1353

*They're all common values but it helps to draw a unit circle to remember where things are positive and negative.*1362

*What I started with is π/4, there's π/2, there's 3π/4, π, we're going to move on to 5π/4, 3π/2, move on to 7π/4, and we started out as 0.*1369

*Let's see.*1398

*We've already hit 3π/4 moving on to π now.*1400

*Cosine and sine is -1+0i.*1404

*5π/4, cosine and sine are both negative root 2 over 2.*1410

*They're both negative.*1418

*3π/2, the cosine is 0 again, the sine is -1.*1422

*Finally, 7π/4, cosine and sine are root 2 over 2, but the cosine is positive and the sine is negative.*1430

*For our answers, what we have to do is multiply root 2 by each one of these.*1441

*Let me multiply the root 2 by each one of these.*1447

*The first one you just get square root of 2 times 1+0.*1450

*Multiplying root 2 times root 2 over 2, that gives me 2/2, which gives me 1.*1462

*Plus i to the same thing, so just i.*1470

*Multiply root 2 times 0+i, gives me root 2i.*1471

*Multiply root 2 by this, we get -1+i.*1478

*Multiply root 2 here, we get negative root 2.*1485

*Multiply root 2 here, -1-i, remember root 2 times root 2 over 2, is 2/2, which simplifies to 1.*1489

*Multiply root 2 here, we get negative root 2i.*1502

*Multiply root 2 here, we get 1-i.*1507

*We are finally done here.*1514

*We get 8 different answers, 8 different complex numbers here.*1518

*Each one of these complex numbers has the property that if you raised it up to the eighth power, it will come out to be exactly 16.*1529

*Let me write down what we found here.*1541

*Each one satisfies w*^{8}=16.1545

*If you multiply them up by themselves eight times, you'll get back to 16.*1555

*That will be pretty messy I'm not going to check that here, but you can check it on your own if you like.*1561

*Let me recap how we found that.*1569

*We started out with the complex number 16, think about it as 16+0i.*1570

*We wanted to write that in polar form, so I founded r and a θ.*1580

*r=16, and θ=0, z=16e*^{0i}.1585

*Then I use De Moivre's theorem which says that you get complex nth roots by doing r*^{1/n}, that's where the 16^{1/8} came from.1590

*Then cosine plus isine of these angles (θ+2kπ)/n.*1600

*That's why I started to make this chart (θ+2kπ)/8.*1609

*You run the k from 0 to n-1, that's why I ran the k from 0 to 7.*1613

*For each one of those I got an angle that I called α, then I worked out cos(α)+isin(α) for each one of those.*1621

*That's where I got these sectional values here, and for that it was really helpful to plot my α's on the unit circle here.*1630

*Remember what the sines and cosines of each one of them was.*1641

*Those are common values so I didn't need to calculate to look those up.*1645

*Multiply each one by 16*^{1/8}.1648

*A little cleverness with the laws of exponents tells me that that's root 2.*1654

*Finally, I multiply each one of those by root 2, and I get 8 different answers, each one of them is an eighth root of 16 and complex numbers.*1659

*We're now going to find all complex cube roots of -1.*1670

*Cube roots, that's a third root, we expect 3 answers here because we're looking for cube roots.*1675

*I want to put that complex number into polar form first.*1690

*I'm going to use my r equals square root of x*^{2}+y^{2}, and θ=arctan(y/x), plus π if x < 0.1695

*Negative 1, think of that as -1+0i, the x=-1, y=0.*1713

*My r is square root of x*^{2}+y^{2}, that is square root of 1, r=1.1720

*θ=arctan(y/x), that's arctan(0), x < 0, so I have to add π.*1729

*Arctan(0)=0, θ=π.*1741

*z=re*^{iθ}, 1e^{iπ}.1746

*I've got my complex number into polar form, now I'm going to use De Moivre's theorem.*1757

*Let me remind you how that goes, it says r*^{1/n}×cos(θ+2kπ)/n+isin(θ+2kπ)/n.1763

*The key thing here is k=0, 1, 2, up to n-1.*1790

*Here we're finding cube roots, our n=3.*1797

*Let me make a little chart again of the angles.*1803

*n=3, θ=π, I'll make a chart of k and (θ+2kπ)/3 for each value of k here.*1809

*k goes from 0 to n-1, that's 0, 1, and 2.*1837

*(θ+2kπ)/3, when k=0, that's just, θ=π, π/3.*1846

*When k=1, that's (π+2π)/3, which is 3π/3, which is π.*1856

*When k=2, this is (π+4π)/3, which is 5π/3.*1869

*The three angles we're going to be looking at are π/3, π, and 5π/3.*1879

*Let me also work out r*^{1/n}.1887

*r*^{1/n}, r=1, raised to 1/3 power is just 1.1890

*That part is very easy.*1897

*We have cos(α)+isin(α) for each one of these α's.*1899

*Cos(π/3)+isin(π/3) ...*1913

*Let me draw out where that would be. π/3 is about right there.*1922

*π is right there.*1927

*5π/3 is down there.*1931

*Those are the three angles I'm going to be looking at.*1932

*Let me go ahead and include these on my chart.*1936

*Cos(α)+isin(α), cos(π/3)=1/2, isin(π/3) is root 3 over 2, that's because (π/3) is right there.*1940

*π, the cosine is -1, the x-coordinate, isine is zero.*1967

*5π/3,that's down here, cosine is 1/2, the sine is negative root 3 over 2.*1980

*That was cos(α)+isin(α).*1992

*r*^{1/n}×cos(α)+isin(α) is kind of anticlimatic because we already figured out that r^{1/n}=1.1997

*We're just multiplying each of these by 1.*2012

*We get 1/2 plus i root 3 over 2, -1 and 1/2 minus i root 3 over 2 as our three answers.*2015

*Remember we're looking for cube roots, so we did expect to find 3 answers, it's reassuring here that we found our 3 answers.*2029

*Let me remind you how we did that.*2038

*First of all, we were given a complex number.*2040

*We had to convert it inot plar form so I found my r and my θ using the standard formulas.*2042

*I did have to include the fudge factor plus π here because the x < 0.*2049

*Arctan(0) gave me 0, gave me θ=π.*2050

*My θ was π, my r was 1, so I get 1e*^{iπ}.2057

*I go to De Moivre's theroem which says, r*^{1/n}×cos(θ+2kπ)/n, isin(θ+2kπ)/n.2063

*I made a little chart of the different values of k.*2075

*You go from 0 to n-1.*2079

*For each one, I figured out (θ+2kπ)/n.*2081

*That gave me the π/3, π, and 5π/3.*2087

*I found the cosine plus isine of each one.*2090

*I multiplied those by r*^{1/n}.2095

*That gave me my 3 answers.*2098

*Those are the three complex numbers that are cube roots of 1.*2100

*Each one satisfies wr*^{3}=-1.2106

*If you worked out wr*^{3} for any of these complex numbers, you'd get -1.2111

*We've got some more examples for you later.*2116

*Try them out on your own and then we'll work through them together.*2118

*Hi we are working out some examples of common values of sin and cos of common angles.*0000

*Remember what we learned, everything comes back to knowing those two key triangles.*0007

*There is the 45, 45, 90 triangle whose values are (root2)/2, (root 2)/2 and 1.*0014

*And then there is the 30, 60, 90 triangle whose values are 1/2, (root 3)/2 and 1.*0025

*If you remember those set of numbers, you can work out sin and cos of any common value anywhere on the unit circle.*0038

*That is what we are doing here, we have given the values 225 degrees, convert it to radians, let us start with that.*0046

*225 x pi/180 is equal to, well 225/180 simplifies down to 5/4, so that is 5pi/4.*0053

*Let us draw that on the unit circle and see where it lands.*0069

*My unit circle is 0, pi/2, which is the same as 90 degrees, pi radians is equal to 180 degrees, and 3pi/2 radians is 270 degrees, 2pi radians is equal to 360 degrees.*0085

*We have got 225 degrees or 5pi/4, 225 is between 180 and 270, in fact it is exactly half way between there because it is 45 degrees from either side.*0105

*It is right there, if you like that in terms of radians, 5pi/4 is just pi + pi/4, that is the angle we are looking at.*0118

*That is in the third quadrant, we found its quadrant, we converted it to radians, we want to find its cos and sin, that is the x and y coordinates.*0133

*Let me draw those in there, we want to figure out what those x and y coordinates are.*0146

*Look that is a common triangle and I remember what the values of those common triangles are.*0150

*That distance is (root 2 )/2, that distance is (root 2)/2.*0157

*I’m getting that because I remember this common 45, 45, 90 right triangle.*0162

*That means the sin and cos of both (root 2)/2 and we just need to figure out whether they are positive or negative.*0171

*All students take calculus, down there on the third quadrant only the tan is positive, both the sin and cos are negative.*0188

*Another way to remember that is just to remember that in the third quadrant both x and y values are negative.*0201

*Sin and cos of this angle are both negative (root 2)/2.*0207

*Again, what we are using over and over in these examples is these two key triangles.*0220

*The 45, 45, 90 triangle and the 30, 60, 90 triangle, you want to remember the values for those two triangles, (root 2)/2, (root 3)/2, and ½.*0228

*Remember those values and then when you have an angle that is in the other quadrants, it is just a matter of translating one of those triangles over there.*0244

*And then figuring out whether the sin and cos are positive and negative.*0252

*Let us try one more example here, we want to identify all the angles between 0 and 2 pi whose cos is –root 3/2.*0000

*I start by drawing my unit circle that is not quite straight, let me straighten that up a little bit.*0010

*Cos is – root 3/2, now the cos remember is the (x) value so I’m going to go on the (x) axis and I’m going to go to –root 3/2.*0037

*There it is.*0053

*That is –root 3/2 on the (x) axis and then I’m going to draw and see what angles I will get from that.*0058

*It looks like, remember that root 3/2 is one of my common values, that means that the y values are going to be ½.*0073

*I need to figure out which angles those are but that is one of my common values ½ root 3/2 that means that is a 30 degree angle, that is 60 and that is 30.*0083

*I just need to figure out what those angles are, if you remember we started 0, 90, 180, 270, and 360.*0097

*That first angle there is 30 degrees short of 180, the first angle is 150 degrees.*0111

*The second angle is 30 degrees past 180, so that is 210 degrees.*0120

*I have got my angles in degrees I will convert them into radians x pi/180 is equal to 5pi/6 to 10 x pi/180 is 7pi/6 radians.*0127

*I got those two angles in radians now, that is the first one 5pi/6, that is the second one 7pi/6.*0151

*And identify which quadrant each one is in, one of them is in the second quadrant, one of them is in the third quadrant, quadrant 2 and quadrant 3.*0159

*It all comes back to recognizing those common values, ½, square root of 3/2, square root of 2/2.*0184

*Once you recognize those common values, you can put these triangles in any position anywhere on the unit circle.*0191

*You just figure out where is your root 3/2, where is your ½, where is your root 2/2 and then you figure out which one is positive and which one is negative.*0198

*The whole point of this is you can figure out the sin and cos of any angle anywhere on the unit circle as long as it is a multiple of 30 or 45, or in terms of radians if it is a multiple of pi/6, pi/6, pi/4, pi/3.*0209

*You can figure out sin and cos of all these angles just by going back to those 3 common values and by figuring out whether their sin and cos are positive or negative.*0226

*Now you know how to find sin and cos of special angles, this is www.educator.com, thanks for watching.*0238

*Hi, these are the trigonometry lectures on educator.com.*0000

*Today we're going to learn about sine and cosine values of special angles.*0003

*When I say these special angles, there are certain angles that you really want to know by heart.*0009

*Those are the 45-45-90 triangle, and the 30-60-90 triangle.*0014

*Let me talk about the 45-45-90 triangle first.*0022

*I'll draw this in blue.*0028

*Here's a 45-45-90 triangle and I'm going to say that each side has length 1.*0041

*If each of the short sides has length 1, by the Pythagorean theorem, we can figure out that the long side, the hypotenuse, would have length square root of 2.*0052

*I'm going to scale this triangle down a little bit now.*0066

*I wanted to scale it down so the hypotenuse has length 1.*0071

*That means I have to divide all three sides by square root of 2.*0075

*If I scale this down, so the hypotenuse has length 1 that means the shorter sides has length 1 over the square root of 2, because I divided each side by square root of 2.*0079

*Then if you rationalize that, the way you learned in your algebra class, multiply top and bottom by square root of 2. *0095

*You get square root of 2 over 2, and square root of 2 over 2.*0102

*Those are very important values to remember because those are going to come up as sines and cosines of our 45-degree angles on the next slide.*0113

*First, I'd like to look also at the 30-60-90 triangle.*0122

*I have to do a little geometry to work this out for you.*0129

*I'm going to start with an equilateral triangle, a triangle where all three sides have 60 degrees.*0133

*I'm going to have assume that each side has length 2.*0142

*The reason I'm going to do that is because I'm going to divide that triangle in half.*0145

*If we divide that triangle in half, then we get a right angle here and each one of these pieces will have length 1.*0150

*Now, if I just look at the right hand triangle.*0160

*Remember that each one of the corners of the original triangle was 60 degrees.*0169

*That means that the small corner is 30 degrees, and I have a right angle here.*0175

*Now, the short side has length 1, the long side has length 2.*0182

*I'm going to figure out what the other side is using the Pythagorean theorem.*0186

*Let me call that x for now.*0191

*I know that x*^{2} + 1^{2} = 2^{2}, which is 4.0194

*So, x*^{2} = 4 - 1, which is 3.0202

*So, x is the square root of 3.*0209

*That's where I got this relationship, 1, square root of 3, 2.*0213

*There's the 1, there's the square root of 3, and there's the 2.*0219

*Now, I'm going to turn this triangle on its side, and I want to scale it down.*0223

*Originally, it was 1, square root of 3, 2.*0232

*But again, I want to scale the triangle down so that the hypotenuse has length 1.*0238

*To do that, I have to divide everything by 2.*0244

*So the short side now has length 1/2.*0247

*The longer of the two short sides has length square root of 3 over 2.*0251

*Remember that's the side adjacent to the 30-degree angle. *0258

*That's the side adjacent to the 60-degree angle, *0262

*That's the right hand side.*0264

*These two triangles are very key to remember, in remembering all the sines and cosines.*0267

*In fact, if you can remember these lengths of these two triangles, you can work out everything else just from these two triangles.*0273

*Let me emphasize again, the 45-45-90 triangle, its sides have length 1, square root of 2 over 2, and square root of 2 over 2.*0282

*The 30-60-90 triangle, has sides of length 1, 1/2, and root 3 over 2.*0297

*Those are the values that you need to remember.*0310

*If you can remember those, you can work out all the sines and cosines you need to know for every trigonometry class ever.*0312

*Let's explore those a little bit.*0321

*We already figured out, let me draw a unit circle.*0324

*We know that sines and cosines occur as the x and y coordinates of different angles. *0328

*Well, if you're at 0... *0338

*Let me just draw in some key angles here, 0, here's 90, here's 45, here's 30, and here's 60.*0341

*If you're at 0 degrees, which is the same as 0 radians, then the cosine and sine, the x and y coordinates are just 1 and 0.*0359

*We already figured those out before.*0368

*The other easy one is the 90-degree angle up here.*0370

*We figured out that that's π/2 radians, and the cosine and sine are 0, and 1 there.*0374

*Now, the new ones, let me start with 45, because I think that one's a little bit easier.*0380

*The 45-degree angle, there it is right there.*0386

*We want to figure out what the x and y coordinates are because those give us the sine and cosine.*0391

*Well, we just figured out that a triangle that has 1 as its hypotenuse has square root of 2 over 2, as both its x and y sides.*0397

*That's where we get the square root of 2 over 2 as the cosine and sine of the 45-degree angle, also known as π/4 radians.*0407

*For the 30-degree angle, I'll do this one in blue.*0418

*The 30-degree angle, we have again, hypotenuse has length 1.*0422

*Remember, the length of the long side is root 3 over 2.*0431

*And the length of the short side is 1/2.*0436

*That's how you know the sine and cosine of the 30-degree angle or π/6.*0439

*The cosine, the x-coordinate, root 3 over 2, sine is 1/2.*0447

*The 60-degree angle, that's just the same triangle but it's flipped the other way so that the long side is on the vertical part and the short side is on the horizontal axis.*0452

*The short side is 1/2.*0468

*The long side is now the y-axis, that's root 3 over 2.*0473

*That's how we get 1/2 being the cosine of 60 degrees, root 3 over 2 being the sine of 60 degrees.*0478

*These values are really worth memorizing but you remember that you figure all out from those two triangles.*0488

*All you need to know is that one triangle has length 1, has hypotenuse 1, and sides root 2 over 2, that's the 45-45-90 triangle.*0494

*The other triangle has hypotenuse 1, and then the long side is root 3 over 2, short side is 1/2, that's the 30-60-90 triangle.*0509

*Just take that triangle and you flip it whichever way you need to, to get the angle that you're looking for.*0526

*These angles, these sines and cosines are the key ones to remember, root 2 over 2, root 3 over 2, and 1/2.*0534

*From that, what you're going to do is figure out the sines and cosines of all the other angles all over the unit circle.*0546

*Here's my unit circle.*0573

*We figured out all the sines and cosines of all the angles in the first quadrant.*0575

*All we have to do now is figure out all the sines and cosines of the angles in the other quadrants.*0583

*Let me draw them out.*0589

*But it's the same numbers every time.*0590

*All you have to do is figure out whether those numbers are positive or negative, and that just depends on which quadrant you're in.*0593

*All you have to do is remember those key numbers, root 2 over 2, root 3 over 2 and 1/2.*0601

*Then you're going to figure out which ones are positive in which quadrants.*0607

*Let me show you how you'll remember that, 1, 2, 3, 4.*0613

*Remember that the sine is the y value, and the cosine is the x value.*0622

*In the first quadrant, both the x's and y's are positive, and so sines and cosines are going to positive.*0630

*I'm going to write that as an a, which stands for all the values of everything is positive.*0636

*In the second quadrant, over here, the x values are negative, the y values are positive.*0643

*Now, the x values correspond to the cosines, the cosines are negative and the sines are positive.*0653

*I'm going to write as s here, for the sines being positive.*0658

*In the third quadrant, both x's and y's are negative.*0667

*X and y are both negative, that means cosines and sines are both negative.*0669

*Tangent, we haven't learned the details of tangent yet but we're going to learn later that tangent is sine over cosine.*0675

*Both sine and cosine are negative, that means sine over cosine is positive.*0683

*It turns out that tangent is positive down that third quadrant.*0687

*We'll learn the details of tangent later.*0691

*In the meantime, we'll just remember that tangent is positive in the third quadrant.*0693

*In the fourth quadrant here, that was the third quadrant that we just talked about, now we're moving on to the fourth quadrant.*0698

*The x's are positive now, the y's are negative.*0705

*That means the cosine is positive, but the sine is negative.*0708

*I'll list the cosine here because I'm listing the positive ones.*0713

*First quadrant, they're all positive.*0718

*Second quadrant, sines are positive.*0720

*Third quadrant, tangents are positive.*0722

*Fourth quadrant, cosines are positive.*0724

*The way you remember that is with this little acronym, All Students Take Calculus.*0727

*That shows you as you go around the four quadrants, All Students Take Calculus.*0733

*It shows you which ones are positive in each quadrant.*0741

*First quadrant are all positive.*0744

*Second quadrant sines are positive.*0746

*Third quadrant, tangents are positive.*0748

*Fourth quadrant, cosines are positive.*0750

*That's how you'll remember what the signs are in each quadrant.*0753

*That means the positive signs and the negative signs in each quadrant.*0756

*The numbers are just all these values that you've just memorized, root 3 over 2, root 2 over 2, and 1/2.*0758

*We'll do some practice finding sines and cosines of values in other quadrants, based on this table that we remember.*0768

*And these common values in common triangles that we remember.*0775

*Then we'll take those values, introduce some positive and negative signs, and we'll come up with the sines and cosines of angles in other quadrants.*0780

*Let's try some examples.*0787

*First example here is a 120 degrees, you want to convert that to radians, identify it's quadrant, and find it's cosine and sine.*0791

*First things first, let's convert it to radians, 120 times π/180 is 2π/3, because 120/180 is 2/3 so that's 2π/3 radians.*0800

*Identify it's quadrant.*0806

*Well, let me graph out my unit circle here.*0814

*There's 0, there's π/2, π, 3π/2, and then 2π.*0820

*Now, 2π/3 is between π/2 and π.*0840

*In fact, it's closer to π/2.*0845

*It's 2/3 of the way around from 0 to π.*0848

*If you like that in terms of degrees, π is 180 degrees, and so 2π/3 is 2/3 of the way over to 180.*0852

*Now, we're going to find cosine and sine.*0862

*Let me show you how to do this.*0867

*You draw a triangle here. *0868

*Remember, we're looking for the x and y coordinates.*0870

*Draw a triangle here.*0873

*That's a 30-60 triangle.*0875

*This is 60, that's a 60-degree angle.*0877

*That's a 30-degree angle.*0887

*We know what the lengths of these different sides are.*0891

*We know that the long side there is root 3 over 2 and the short side there is 1/2.*0897

*We know, remember, the cosines and sine are the x and y coordinates.*0907

*The cosine of 120 or 2π/3 is 1/2, except that we're going to have to check whether that's positive or negative.*0910

*Remember All Students Take Calculus.*0930

*In the second quadrant, only the sine is positive, so the cosine must be negative.*0935

*The sine of 2π/3, the y value is root 3 over 2.*0942

*In the second quadrant, sines are positive, so that's positive.*0950

*Our cosine and sine are -1/2 and root 3 over 2.*0954

*If you didn't remember the All Students Take Calculus thing, you can also just work it out once you know what quadrant it's in.*0958

*It's in quadrant 2 and we know there that the x coordinates are negative, and the y coordinates are positive.*0964

*The cosine must be negative and the sine must be positive.*0974

*The whole point of this is that you only really need to memorize the values of the triangles, root 2 over 2, root 3 over 2 and 1/2.*0981

*Once you know those basic triangles, you can work out what the sines and cosines are in any different quadrant just by drawing in those triangles and then figuring out which ones have to be positive, and which ones are to be negative.*0991

*Let's try another one.*1005

*This one is converting 5π/3 radians to degrees, identifying it's quadrant, and finding its cosine and sine.*1009

*5π/3 times 180/π, the π's cancel, and the we have 5/3 of 180, 180 over 3 is 80, so this is 5 times 60 is 300 degrees.*1018

*Let's try and find that in the unit circle.*1040

*We have 0, π/2 which is 90, π which is 180, 3π/2 which is 270, and 2π which is the same as 360 degrees.*1053

*Now, 5π/3, that's bigger than π and that's smaller than 2π.*1074

*In fact, that's π + 2π/3.*1080

*That's π, which is right here, plus 2π/3.*1090

*So, 2/3 the way around from π to 2π.*1099

*There it is right there.*1103

*That's in the fourth quadrant.*1104

*So, we figured out what quadrant it's in.*1108

*If you like degrees better, 300 degrees is a little bigger than 270, in fact, it's 30 degrees past 270, and 60 degrees short of 360.*1110

*That's how you know that that angle is in that quadrant.*1121

*Now we have to find its cosine and sine.*1126

*That's the x and y coordinates.*1128

*We set up our triangle there.*1134

*We already know that that's a 60-degree angle, because it's 60 degrees short of 360.*1137

*That's a 30-degree angle and we just remember our common values.*1142

*The horizontal value, that's the short one, that's 1/2, that's the long one, root 3 over 2.*1147

*We know our values are going to be 1/2 and root 3 over 2, we'll just have to figure out which one's positive and which one's negative.*1154

*I know my cosine of 5π/3 is going to be either positive or negative 1/2.*1162

*The sine of 5π/3 is positive or negative root 3/2.*1171

*Remember All Students Take Calculus.*1179

*Down there in the fourth quadrant, the cosine is positive and the sine is negative.*1184

*If you don't remember All Students Take Calculus, you just look that you're in the fourth quadrant, x coordinates are positive, y coordinates are negative.*1189

*You know which one's positive or negative.*1199

*All you have to remember are those key values, 1/2, root 3 over 2, root 2 over 2.*1202

*Remember those key values for the key triangles.*1209

*Then it's just a matter of drawing the right triangle in the right place and figuring out which one is positive, and which one is negative.*1212

*Let's try another one.*1219

*This one is kind of tricky.*1223

*This one's going to be challenging us to go backwards from the sine.*1224

*We have to find all angles between 0 and 2π whose sine is -1/2.*1229

*This is kind of foreshadowing the arc sine function that we'll be studying later on and one of the later lectures.*1242

*In the meantime, the sine is -1/2.*1249

*Remember now, the sine is the y-coordinate.*1252

*We want things whose y coordinates are -1/2.*1256

*I'm going to draw -1/2 on the y-axis, -1/2.*1259

*I'm going to look for all angles whose y-coordinate is -1/2.*1266

*Look, there's one right there.*1272

*And there's one right there.*1276

*I'm going to draw those in.*1277

*I'm going to draw those triangles in.*1280

*I know now that if we have a vertical component of 1/2, the horizontal component has to be root 3 over 2.*1289

*That's because we remember those common triangles, 1/2, root 3 over 2, root 2 over 2.*1298

*We're going to figure out what those angles are.*1307

*I know that's a 30-degree angle.*1312

*I know that that is 180.*1318

*The whole thing is 210 degrees.*1322

*I know that that is 30.*1330

*I know that that must be 60.*1334

*Remember, this is 270 degrees down here.*1337

*We have 270 degrees plus 60 degrees.*1342

*This is getting a little messy, so I'm going to redraw it over here.*1345

*That's the angle we're trying to chase down here.*1357

*We know that's 60.*1360

*That much is 270.*1362

*So, 270 plus 60 is 330 degrees.*1366

*Those are the two angles that we're after, 210 degrees, 330 degrees.*1372

*Let me convert those into radians.*1378

*If you multiply that by π/180 then that's equal to...*1382

*Let's see, that's 7π/6 radians.*1390

*This one times π/180 is equal to 11π/6 radians.*1397

*We've got our two angles in degrees and radians.*1409

*The quadrants, the first one was in the third quadrant, quadrant 3.*1413

*The second one was in the fourth quadrant, quadrant 4.*1420

*Those are the two angles in both degrees and radians that had a sign of -1/2.*1426

*Their y value was -1/2.*1436

*What this comes down to is knowing those common values, 1/2 root 3 over 2, root 2 over 2.*1442

*Once you know those common values, it's a matter of looking at the different quadrants and figuring out whether the x and y values are positive or negative.*1454

*In this case, we had the sine, sine remember is the y-coordinate.*1461

*Since it was negative, we knew that we had to be at -1/2 on the y-axis.*1468

*We found -1/2 on the y-axis, drew in the triangles, recognized the 30-60 triangles that we've been practicing and then we were able to work out the angles.*1473

*We'll try some more examples of that later.*1484

*We are learning about the Pythagorean identity, what we are going to try now is to start with the Pythagorean identity and prove the Pythagorean theorem.*0000

*Remember what we did in the earlier example, was we started with the Pythagorean theorem and we proved the Pythagorean identity.*0009

*The point of this is to show that you can get from one to the other or from the other back to the first one.*0015

*And so that the two factor are equivalent even though one seems like geometric fact and one seems like a trigonometric fact.*0022

*Let us do that, remember that the Pythagorean identity says that sin*^{2}(x) + cos^{2}(x) = 1.0033

*The Pythagorean theorem, that is the theorem about right triangles, so let me set up a right triangle here.*0045

*I will label the sides a, b, and c, that is Pythagorean identity, now I’m going to use my SOHCAHTOA.*0055

*SOHCAHTOA tells us that if we have an angle here, I will this angle (x) in the corner here, the sin(x) is equal to the opposite/hypotenuse.*0065

*The opposite side here is b/c, the cos(x) is equal to a/c, the adjacent/hypotenuse.*0090

*I’m going to plug those in the Pythagorean identity because remember we are allowed to use the Pythagorean identity here.*0105

*That says the sin*^{2}(x) + cos^{2}(x) = 1, if I plug those in I will get b/c^{2} + a/c^{2}= 1. 0112

*Now I’m just going to do a little bit of algebraic manipulation b*^{2}/c^{2} + a^{2}/c^{2} is equal to 1.0127

*I’m going to multiply both sides there by c*^{2} and that will clear my denominators on the left I got b^{2} + a ^{2} and on the right I have c^{2}.0137

*If I switch around the two terms here a*^{2} + b^{2} is equal to c^{2}.0151

*That is the very familiar equation we approved the Pythagorean theorem.*0160

*We started out with the Pythagorean identity from trigonometry and we ended up proving the Pythagorean theorem from geometry a*^{2} + b^{2}=c^{2}.0177

*It is just a matter of writing down the right angle in one corner of the triangle and then working through the algebra, you end up with the Pythagorean theorem from geometry.*0190

*That shows that the Pythagorean identity and the Pythagorean theorem really are equivalent to each other.*0200

*You ought to be able to start with either one and prove the other one.*0206

*Let us try one more example, we are given that sin(theta) is -5/13, and (theta) is in the third quadrant and we want to find cos(theta).*0000

*Let me try graphing out what that might be.*0010

*Ok (theta) is in the third quadrant, that is down here and so its sum angle down there, I do not know exactly where it is but I will draw it down there.*0022

*What I’m given is that sin(theta) is -5/13 and I want to find cos(theta).*0035

*Well I have this Pythagorean identity that says sin*^{2}(theta) + cos^{2}(theta) = 1.0040

*I will plug in sin(theta ) that is -5/13*^{2} + cos^{2}(theta) =10051

*-5/13 when you square, the negative goes away so we get 25/169 + cos*^{2}(theta) is equal to0062

*Well I’m going to have to subtract the 25/169 so I will write 1 as 169/169 then I will subtract 25/169 from both sides.*0074

*I got cos*^{2}(theta) is 144/169, then if I take the square root of both sides to solve for cos(theta), I get cos(theta) is equal to + or – square root of 144 is 12, square root of 169 is 13.0087

*I know the my cos(theta) is equal to either positive or negative 12/13.*0110

*That is all I can get from the Pythagorean identity because it only told me what cos*^{2}(theta) is, I can not figure out from that whether cos(theta) is positive or negative.0116

*But the problem gave us a little extra information, it says that (theta) is in the quadrant.*0127

*Knowing that (theta) is in the third quadrant, I looked down there and I remember that cos is equal to the x coordinate of my angle.*0133

*Cos is the x coordinate, remember all students take calculus, down there in the third quadrant tan are positive but nothing else is positive.*0145

*That means that cos is not positive, it is negative.*0158

*The cos(theta) is equal to -12/13 and must be the negative value because it is down in the third quadrant, that is where the x coordinate is negative.*0162

*The key to this problem is remembering the Pythagorean identity, sin*^{2}(theta) + cos^{2}(theta)=1.0181

*Then you plug the value you are given into the Pythagorean identity and you try to solve for cos(theta).*0189

*Once you work through the arithmetic you get the value for cos(theta) but you do not know if it is positive or negative.*0197

*Then you go over and look whether what quadrant the angle is in, it is in the quadrant and then you either remember the all students take calculus.*0203

*That tells you the plus or minus on the different functions or you just remember that in the third quadrant the x values are negative so the cos value has to be negative.*0212

*Either way you end up with cos(theta) is equal to -12/13.*0222

*That is the end of our set on the Pythagorean identity, this is www.educator.com.*0229

*Hello, this is the trigonometry lectures for educator.com and today we're going to learn about probably the single most important identity in all trigonometry which is the Pythagorean identity.*0000

*It says that sin*^{2}x + cos^{2}x = 1.0011

*This is known as the Pythagorean identity.*0017

*It takes its name from the Pythagorean theorem which you probably already heard of.*0019

*The Pythagorean theorem says that if you have a right triangle, very important that one of the angles be a right angle, then the side lengths satisfy a*^{2} + b^{2} = c^{2}.0024

*You probably heard that already.*0040

*The new fact for trigonometry class is that sin*^{2}x + cos^{2}x = 1.0042

*What we're going to learn is we work through the exercises for these lectures.*0050

*Is if these are really two different sides of the same coin, you should think of this as being sort of facts that come out of each other.*0055

*In fact, we're going to use each one of these facts to prove the other one.*0064

*These are really equivalent to each other.*0069

*Let's go ahead and start doing that.*0071

*In our first example, we are going to start with the Pythagorean theorem, remember that's a*^{2} + b^{2} = c^{2}.0074

*We're going to try to prove the Pythagorean identity sin*^{2}x + cos^{2}x = 1.0084

*The way we'll do that is let x be an angle.*0095

*Let's draw x on the unit circle.*0107

*The reason I'm drawing it on the unit circle is because remember the definition of sine and cosine is the x and y coordinates of that angle.*0112

*If we draw x on the unit circle, the hypotenuse has length 1 and the x-coordinate of that point, remember, is the cos(x), and the y-coordinate is the sin(x).*0124

*Now, what we have here is a right triangle and we're allowed to use the Pythagorean theorem, we're given that and we're going to use that and try to prove the Pythagorean identity.*0148

*The Pythagorean theorem says that in a right triangle, by the Pythagorean theorem...*0160

*Let me draw my right triangle a little bigger, there's x, there's 1, this is cosx, this is sinx.*0174

*By the Pythagorean theorem, one side squared, let me write that first of all as cosine x squared plus the other side squared is equal to 1*^{2}.0186

*That's the length of the hypotenuse.*0205

*If we just do a little semantic cleaning up here, 1*^{2}, of course, is just 1, cosine x squared, the common notation for that is cos^{2}x + sin^{2}x = 1.0207

*We just derived an equation, and look this is the Pythagorean identity.*0228

*What we've done is we started by assuming the Pythagorean theorem and then we used the Pythagorean theorem to derive the Pythagorean identity.*0246

*Let's see an application of that in the next example.*0256

*We're given that θ is an angle whose cosine is 0.47, and θ is in the fourth quadrant.*0260

*We have to find sinθ.*0266

*Let me draw θ, θ is somewhere down there in the fourth quadrant.*0272

*I don't know exactly where it is but θ looks like that.*0279

*Here is what I know, by the Pythagorean identity, sin*^{2}θ + cos^{2}θ = 1.0284

*I'm going to fill in the one that I know, cosθ, cosθ is 0.47.*0295

*This is 0.47*^{2} = 1 + sin^{2}θ.0302

*Now, 0.47, that's not something I can easily find the square of, so I'll do that on my calculator.*0310

*0.47*^{2} = 0.2209, so that's +0.2209, sin^{2}θ +0.2209 = 1, sin^{2}θ = 1 - 0.2209, which is 0.7791.0317

*Sinθ, if we take the square root of both sides, sinθ is equal to plus or minus the square root of 0.7791, which is approximately equal to 0.8827.*0360

*Now, it's plus or minus because I know that sine squared is this positive number, but I don't know whether this sine is a positive or negative.*0382

*We're given more information in the problem, θ is in the fourth quadrant.*0390

*Remember, sine is the y-coordinate, so the sine in the fourth quadrant is going to be negative because the y-coordinate is negative.*0395

*Because θ is in quadrant 4, sinθ is going to be negative, so we take the negative value, sinθ is approximately equal to -0.8827.*0414

*The whole key to doing this problem was to start with the Pythagorean identity sin*^{2}θ + cos^{2}θ = 1.0446

*Once you're given sine or cosine, you could plug those in and figure out the other one except that you can't figure out whether they're positive or negative.*0457

*Their identity doesn't tell you that so we had to get this little extra information about θ being in the fourth quadrant, that totals that the sinθ is negative and we were able to figure out that it was -0.8827.*0463

*Let's try another example of that.*0480

*We're going to verify a trigonometric identity.*0483

*This is a very common problem in trigonometry classes as you'll be given some kind of identity involving the trigonometric functions and you have to verify it.*0486

*For this one, what I want to do is start with the right hand side.*0496

*I'm going to label this RHS.*0503

*RHS stands for right-hand side.*0504

*The right-hand side here is equal to sinθ/(1 - cosθ).*0510

*Now, I'm going to do a little trick here which is very common when you have something plus something in the denominator, or something minus something in the denominator.*0519

*The trick is to multiply the conjugate of that thing.*0529

*Here I have 1 - cosθ in the denominator, I'm going to multiply by 1 + cosθ, and then, of course, I have to multiply the numerator by the same thing, 1 + cosθ).*0533

*The reason you do that, this is really an algebraic trick so you probably have learned about this in the algebra lectures.*0547

*The reason you do that is you want to take advantage of this formula, (a + b) × (a - b*^{2}.0554

*That's often the way of simplifying things using that algebraic formula.*0566

*What we get here in the numerator is (sinθ) × (1 + cosθ), in the denominator, using this (a*^{2} + b^{2}) formula, we get (1 - cos^{2}θ).0570

*Now, let's remember the Pythagorean identity.*0586

*Pythagorean identity says sin*^{2}θ + cos^{2}θ = 1.0590

*That means 1 - cos*^{2}θ = sin^{2}θ.0596

*We can substitute that in into our work here, sinθ×(1 + cosθ).*0603

*The denominator, by the Pythagorean identity, turns into sin*^{2}θ.0613

*We get some cancellation going on, the sine in the numerator cancels with one of the sines in the denominator leaving us just with (1 + cosθ)/sinθ in the denominator.*0623

*That's the same as the left-hand side that we started with.*0639

*We started with the right-hand side and we're able to work it all the way down and end up with the left-hand side verifying the trigonometric identity.*0644

*There were sort of two key steps there.*0654

*One was in looking at the denominator and recognizing that it was a good candidate to invoke this algebraic trick where you multiply by the conjugate.*0657

*If you have (a + b), you multiply by (a - b).*0669

*If you have (a - b), you multiply by (a + b).*0672

*Either way, you get to invoke this identity.*0674

*Here, we had (a - b), we multiplied by (a + b) and then we got to invoke the identity and get something nice on the bottom.*0679

*The second trick there was to remember the Pythagorean identity and notice that (1 - cos*^{2}θ) converts into sin^{2}θ.0686

*Once we did that, it was pretty to simplify it down to the left-hand side of the original identity.*0697

*We'll try some more examples later.*0703

*Hi this is www.educator.com and we are going to try more examples of modified sin waves where we start with the basic equation of sin(x) or cos(x) and the graph of sin(x) or cos(x).*0000

*We introduce these constant which are going to change some of its attributes and we see what that does to the graph.*0013

*Remember the equation we are working with in general is (a)sin arcos(bx + c) + d and then from each of those values we figured out these various attributes amplitude, period, phase shift and vertical shift.*0021

*In this particular equation, the amplitude, remember that is just (a) so you read that as 4, the period is 2pi/b , the b is 2 here, so that is 2pi/2 which is pi.*0041

*The phase shift, that is the strangest one –c/b, that is (–pi/2)/2 or –pi/4.*0066

*Finally the vertical shift, that is the easier one is -1 here.*0085

*Again we will start out with the basic sin way, we will work through introducing these attributes one at a time and see how that moves around and create a new function for us.*0097

*Let us start out with the basic sin way.*0108

*There is pi, 2pi, so I’m just going to graph sin(x) to start with.*0119

*I’m going to graph it flat because I’m looking ahead and noticing that the next step is to increase the amplitude to 4.*0130

*It is really going to be stretched up, I’m going to keep my scale narrow here.*0139

*The next step is to introduce the amplitude, we are going to graph 4sin(x).*0148

*That stretches it up and down by a factor of 4, -4 goes down to -4 goes up to +4, that makes it a lot steeper.*0156

*That is 4sin(x), we got the amplitude incorporated there, remember it is very important to do these in order, amplitude, period, phase shift, vertical shift.*0175

*Next step is to introduce the period, right now the period is 2pi and we want the period to be pi.*0185

*Instead of doing a full cycle in 2 pi units, that is going to do cycle in pi units, that means it shuffles twice as fast.*0197

*Let me try and draw that.*0205

*That is pi/2 there, 3pi/2, ok we got the period incorporated.*0220

*What I really graphed there was 4sin(2x), next we want to incorporate the phase shift.*0230

*I will be graphing 4sin(2x) + pi/2 and this is starting to get a little crowded here, so I’m going to do this one in red.*0241

*That means that the phase shift is –pi/4, that means we move –pi/4 units to the left.*0257

*Instead of starting at 0, I’m going to start at –pi/4 and it came back down before pi/2 so now it is going to come back at pi/4.*0267

*It used to come back up at pi and now it comes back up at 3pi/4.*0287

*That red graph is what we get by incorporating the phase shift and finally we will incorporate the vertical shift and I will do this one on blue.*0301

*We are going to be doing 4sin(2x + pi/2) -1, that means we take the graph and we move it down by one unit.*0316

*It is not the same as changing the amplitude where things stretch up and down.*0328

*Here we are not stretching anything, we are just moving everything directly down by one unit.*0331

*Instead of starting at 0 going up to 4 down to -4, everything is moved down by 1 unit.*0336

*It will peak at three and instead of going down to -4, it will go down to -5 and the middle part will be at -1.*0346

*Let me go and try to graph that in blue.*0355

*It comes back to -1 instead of 0, bottoms out at -5 instead of -4, goes back up peaks at three, back down to -5 and back up to -1 again.*0362

*This blue curve is the final graph there, it is very complicated by the time it is all over but it is a bunch of simple steps.*0382

*The first step is to look at the equation and identify these quantities, amplitude, period, phase shift, and vertical shift.*0395

*We got pretty simple equations for each one of those, the tricky part of the graphing is where you start with sin(x).*0403

*You remember how to graph sin(x), that is very easy and then start incorporating these attributes one by one.*0414

*Amplitude stretches it vertically up and down, period stretches or compresses it horizontally, in this case the period was pi and since it was 2pi/4 we have to compress it by a factor of 2.*0420

*Phase shift moves it over to left and right, vertical shift moves it up or down.*0434

*We finally end up with this blue curve that has been modified according to all those attributes.*0440

*Now we are being asked to find a cos wave and we are given the attributes but we are not given the equations.*0000

*We have to reverse engineer the equation from this attributes.*0007

*Remember the equation we are going to go for is (a)cos(bx+c)+d, but we do not know what a, b, c, and d are, we got to figure them out from these attributes.*0012

*Amplitude is the opposite value of a, we will take a=2 to give ourselves amplitude =2.*0028

*The period is 2pi/b and that is supposed to be equal to 3pi.*0048

*From that we can figure out if we solve that for b, we get (b) is equal to 2/3, that is how we can figure out what (b) should be.*0063

*The phase shift is –c/b which is supposed to be equal to pi/2, now that requires a little bit of solving, so we get –c/b we already figure out that was 2/3 is equal to pi/2.*0073

*That is -3(c)/2 is equal to pi/2, the 2 is cancel and so we get (c) is equal to –pi/3.*0106

*That is how we figured out what (c) is, we reversed engineer this equation –c/b.*0122

*Finally the vertical shift, is (d) which is -2, we figured out what a, b, c, and d are, that means that our equation is a=2cos(2/3x)-(pi/3-2).*0128

*We figured out the first part of the problem which is to find the equation, now we are going to do the tricky part which is graphing it.*0162

*It is not so tricky if you start with the basic cos wave if you remember how to graph that and then you introduce this attributes in the right order.*0168

*I will start with the basic cos wave, cos(x), there is pi, 2pi, 3pi. Cos(x) I know it starts at 1, there is 1, -1, it starts at 1, it goes down to 0.*0179

*Bottoms out at -1 goes back up to 0, goes back up to 1, now I got cos(x) that I pretty much do from memory.*0217

*Next I’m going to do is introduce the amplitude, I’m going to talk about 2cos(x) that just stretches it vertically up and down, it goes up to 2 and down to -2.*0228

*Next I’m going to adjust the period, what I’m actually going to be graphing is 2cos(2/3x).*0257

*I know that makes the period 3pi, our original period is 2pi, this is stretching it out by a factor of 50% instead of doing a full cycle in 2pi, it is going to do a full cycle in 3pi.*0269

*Let me give my self some coordinates here, it still going to start at 0 but it is going to bottom out now.*0286

*It is going to finish up at 3pi instead at 2pi, it is going to bottom out halfway between them that is 3pi/2.*0295

*I took the graph and I stretched it out so that it has a period of 3pi now, phase shift –pi/3.*0324

*What I’m going to be graphing is 2cos(2/3x)-pi/3, sorry the phase shift is pi/2, the c value is –pi/3 but the phase shift is pi/2.*0338

*I’m going to take this graph and move it over pi/2 units to the right.*0358

*It looks like it is getting a little complicated now, I’m going to switch into red here.*0364

*We are going to graph the phase shift in red so that means instead of starting at 0, I’m going to start at pi/2.*0371

*It is going to bottom out at 2pi and it is going to come back up at 3pi + pi/2.*0379

*My orientation points were pi/2, 2pi, 3pi + pi/2, because I just took my orientation points before and move them all over by pi/2.*0413

*That incorporated the phase shift of pi/2, we are almost done here.*0425

*We got on more step to incorporate which is this vertical shift of -2.*0430

*I’m going to do this one in blue so we are going to be graphing 2cos(2/3x)-(pi/3-2) that means we are going to take the whole graph and just move it down 2 units.*0435

*Now remember, before our peaks and values were 2 and -2, we moved that down by 2 units and the peaks and values are now going to be at 0 and -4.*0455

*I’m going to extend my axis down to -4, -3, -4.*0467

*I’m going to take my reference points from before and move them down by 2 units.*0476

*This graph peaks at 0 and it never actually goes north of the x axis because it is going up to 0 and down to -4.*0505

*That is the final graph that we have been asked for, let us recap there.*0520

*We are given all these attributes amplitude, period, phase shift, and vertical shift.*0528

*We know the equation for those 4 things, absolute value of a, 2pi/b, -c/b and d.*0533

*What we do is we plug in those attributes and reverse engineer to figure out what a, b, c, and d are.*0541

*Then we can put those together to get the equation of the function that we are trying to graph.*0550

*To actually graph it, we start with basic cos curve, we introduced the amplitude.*0557

*It is very important to do this in order, introduce the amplitude where you expanded it up, up and down.*0563

*The thing actually stretches out, we introduced the period which collapses it horizontally or stretches it out horizontally.*0570

*In this case, we are changing the period from 2pi to 3pi, so that stretches it horizontally.*0581

*We introduced the phase shift which moves it horizontally to the right or left without stretching it.*0587

*The period was the part where you stretch it, the phase shift just moves it without stretching it.*0593

*Finally, we introduced the vertical shift which moves it up and down without stretching it, in this case it moved down.*0599

*At the end of it, it is quite a complicated process but individually each one of these steps if you keep them in order is not to hard.*0608

*You just start with you original cos graph and then move them around according to these steps until you get the one you want.*0615

*That is the end of our lecture on modified sin waves, this is www.educator.com.*0622

*Hi this is educator.com and today we're going to learn about modified sine waves.*0000

*We're going to learn how to analyze and graph these functions given by, instead of just talking about sin(x) and cos(x), now we're going throw a whole bunch of constants in there.*0007

*We'll talk about asin(bx+c) and then throw a constant outside +d and the same kind of thing with cosine.*0020

*That takes a basic sine or cosine graph and it moves it all around.*0027

*We're going to learn some vocabulary to describe those movements and we're going to learn how to graph those.*0030

*First of all, some vocabulary, remember we're talking about the graph asin(bx+c)+d.*0038

*We're talking about sine waves, these are functions that basically have this shape like sin(x), but they maybe moved in different waves.*0057

*We need some vocabulary to describe the different ways that could be moved.*0073

*The first one that we're going to learn is amplitude.*0079

*The amplitude of the sine waves is the vertical distance from the middle of the waves to the peaks.*0082

*What that represents graphically is this distance right here, that's the amplitude.*0089

*Of course, that's the same as this distance, that's also the amplitude but you can measure it either way.*0102

*If the wave is moved up, if it's floating up above, the x axis, somewhere like that, then the amplitude is still the distance from the middle of the wave to the peaks.*0114

*In terms of equations, it's very easy to spot the amplitude.*0129

*When you're given asin(bx+c)+d, the amplitude is just that number a, or if the a is negative, you just take the absolute value.*0133

*It's just the positive version of that number a that tells you the amplitude right there.*0148

*It tells you how far up away it's going to the peaks, how far down it's going to the valleys.*0158

*The period of a sine waves, I'll show this in red, is the horizontal distance for the wave to do one complete cycle from one peak to the next peak.*0160

*That is the period right there, of that wave, that's the period.*0179

*On that one, that's the period right there.*0189

*Remember, when you're working out the equations, remember that if you have sin(x) that has period 2π, it takes 2π to repeat itself.*0198

*We learned that when we looked into the original sine graphs.*0212

*If you have sin(2x), that makes it wobble up and down twice as fast, the period would be πsin(4x), the period would be π/2.*0215

*The pattern that you noticed here is it that the period is given by 2π over the coefficient of x, 2π/b.*0230

*The b there tells you where the period is, not the b itself but you plug b into that equation and that tells what the period is.*0242

*Two more vocabulary words we need to learn, the phase-shift of a sine wave is the horizontal distance that the wave is shifted from the traditional starting position.*0254

*Let me rewrite the equation here, asin(bx+c)+d.*0269

*The traditional starting position for sine would be at (0,0), and the traditional starting position for cosine would be at (0,1).*0280

*Those are the traditional starting positions but the phase-shift will move the graph to the right or the left.*0302

*In these equations, it's given by -c/b, and that seems a little mysterious and let me explain that a little bit.*0311

*We can write this bx+c, first of all, we can factor b out, we can write that b[x + (c/b)].*0320

*Then, we can write that as b[x - (-c/b)].*0332

*That's where that -c/b comes from.*0340

*If you have trouble remembering that formula -c/b, you can go through this little process to remember that, x-(c/b), that shows you that it moves it c/b units to the right.*0342

*Let me draw that, I'll draw that in blue.*0367

*If you're starting with a sine curve, the phase-shift is the amount that it moves over, that's the phase-shift right there, it moves it over -c/b units.*0370

*If you're starting with a cosine curve, that's that phase-shift right there.*0393

*Finally, the vertical shift is what happens when you take the graph and you just move it up or down vertically without changing anything else.*0407

*Let me start with a sine curve.*0426

*If we apply a vertical shift to that, that amount right there, I'll draw this in red, that's the vertical shift.*0434

*That's a little bit easier to pick out than some of the others, because that's just the d in the original equation.*0455

*If you're moving the graph up or down by an amount of d.*0464

*This can get pretty tricky we're starting with the basic sine and cosine curves, but then we're moving around and stretching them out, and moving them up and down in all different ways.*0469

*It's a little bit tricky but we'll go through some of the examples and you'll get the hang of it.*0479

*In our first example here, we're given an equation 3cos(4x+π)+2.*0485

*We have to identify all these various things, the amplitude, the period, the phase-shift and the vertical shift and then we're going to draw a graph of the function.*0495

*The key thing here is it if you identify these things in order, then it becomes very easy to pick them out using the equations.*0502

*The graph isn't too hard as long as you these things in order.*0510

*The amplitude, remember, that's just the number on the outside, the a in the original equation.*0517

*Let me write down the original equation, acos(bx+c)+d.*0528

*The amplitude is just the a right there, that's the 3.*0539

*The period is 2π/b, the b there is 4, that's 2π/4 which is π/2.*0543

*The phase-shift is -c/b, our c here is π, -π, b is 4, that's -π/4.*0563

*The vertical shift is just that last term d, which is 2.*0580

*Those are the answers to the first part of the question.*0596

*Trickier part is doing the graph and there's a general strategy for doing this graphs that sort of always works, but you really have to follow it closely.*0602

*The strategy is to start with the basic cosine graph, which hopefully you remember how to do it, you start with the basic cosine graph.*0611

*Then move it around according to each one of these parameters.*0620

*The key thing here is you have to do it in order, you have to do amplitude period phase-shift then vertical shift.*0624

*Let's see how that works out.*0630

*Let me draw a basic cosine graph and then we'll try moving it around according to these different parameters.*0634

*Remember, basic cosine graph, there's π, there's 2π, π/2, 3π/2.*0645

*Basic cosine graph starts at 1, goes down to 0 at π/2, -1 at π, up to 0 at 3π/2, and it's 1 again at 2π.*0661

*That's the basic cosine graph, you pretty much have to remember that to get started here.*0676

*First thing we're going to do, is we're going to change the amplitude.*0682

*Let me keep track of this as I go along.*0688

*First one we graphed was cos(x), just y=cos(x).*0692

*Next we're going to graph is 3cos(x), we're going to bring in the amplitude.*0697

*What that does is that it stretches up the peak, then it stretches down the valleys by a factor of 3.*0700

*I'm going to draw the same shift graph but three times as tall, and three times as deep.*0709

*Instead of going from 1 to -1, goes down to -3 and up to 3.*0720

*That second graph I drew there was 3cos(x).*0725

*The next one is to introduce is the period.*0734

*The period is supposed to be π/2, remember, the period is the amount of horizontal distance between one peak and the next peak.*0738

*My current peak is...*0747

*I'm going to change the period to π/2.*0754

*What I'm really graphing here is 3cos4x.*0760

*What that's going to do is, instead of having a period 2π, it shrinks it horizontally, or it compresses it horizontally so that it does a complete period in the space of π/2.*0769

*There's π/2 right there.*0784

*I need to do a whole period between there and there.*0787

*Every π/2, it does a complete cycle.*0800

*What I just drew was 3cos4x.*0811

*The phase-shift I'm going to introduce is -π/4, that means it moves it π/4 to the left and it's getting a little bit messy.*0817

*I'm going to see if I can draw this in red or will see if it's still visible.*0827

*Instead of going from 0 to π/2, I'm going to draw my graph from -π/4 to π/4, because we're moving it to the left by π/4.*0832

*That's the graph in red there, 3cos4x+π because I've introduced the phase-shift in there.*0852

*Now, it's really going to get messy if I try to draw any more on the same axis, so I'm going to set up a new set of axis.*0860

*We had π/2, π, -π/2.*0876

*I'll redraw the red one on this set of axis, -π/2, and that one is going from 3 to -3.*0883

*Remember the red one is the previous graph shifted over by -π/4.*0911

*Finally, we need to introduce the vertical shift of 2, and I'll do this last graph in blue.*0916

*That takes the entire graph and raises it up by 2 units.*0922

*That means instead of going to 3 from -3, it's going to go up in π, and instead of going down to -3, it only goes down to -1.*0927

*Let me label that more clearly, -1.*0942

*I'll draw this one in blue, this is now 3cos4x+π, and I'm introducing the vertical shift of +2.*0947

*I'm taking the graph and I'm moving it up 2 units there.*0965

*That blue graph at the end is our final function.*0972

*This is really a pretty complicated process.*0980

*There's a lot of steps involved but each individual step is not that hard, and if you do them in order and you're careful about each one, it's not too bad.*0983

*Let me just recap there, we started with the original graph of cos(x), that's the starting point.*0993

*Then we introduced the amplitude, and that stretches it vertically by a factor of 3, stretches it up and down.*1000

*We introduced the period which compresses it horizontally.*1013

*We introduced the phase-shift, which takes the whole thing and it moves it to the right or the left.*1020

*Finally, We introduced the vertical shift which takes the whole thing and moves it up or down.*1030

*Remember, it's important to do these in order, amplitude, period, phase-shift, vertical shift.*1041

*If you do those out of order, then they'll mess each other up as you go along.*1050

*We really want to do those in order.*1055

*We want to practice several of these, so let's get moving in other example.*1059

*Here's another example, same questions here, amplitude, period, phase-shift, and vertical shift of the following function.*1061

*Remember we can read this off quickly, just remembering the formula asin(bx+c)+d.*1070

*If we can figure out what a, b, c, and d are, we have formulas for all of these properties.*1080

*Amplitude, that's the a, or if the a is negative, make it positive, that's the absolute value of a, which is just 2 here.*1089

*The period is 2π/b, the b is 2 here, so that's π.*1098

*The phase-shift is -c/b, which is, okay, c is -π/3, so negative of that is (π/3)/2 will give us π/6.*1109

*Finally, the vertical shift is d, which ,in this case, is just 0.*1135

*Finally, the fun part, we get to graph the function*1154

*Remember, you always start with your basic sine or cosine graph and then you start moving it around according to these different parameters but you got to keep these parameters in order.*1159

*Let me start with this one's a sine graph.*1169

*I know the basic shape of the sine graph, I've got that memorized, π and 2π.*1174

*I know sine always starts at 0, goes up to 1, comes back down to 0, to -1, and then back to 0.*1187

*There's 1, -1, there's π, π/2, 3π/2, so that's my basic sine graph.*1199

*That's the first thing I graphed there, sin(x), remember this is the graph of sine and cosine x.*1209

*Now, we start introducing these other attributes and it's important to go in order.*1217

*First of all, we're going to introduce the amplitude.*1222

*The amplitude is 2, but we're really multiplying the graph by -2, -2sin(x) is what I'm going to graph next.*1225

*-2sin(x) that stretches it vertically because of the 2, but it also flips it vertically.*1235

*Instead of starting by going up, it's going to go down, goes down to -2, up to 0, up to 2, and back down to 0.*1246

*That's first one was a little lop-sided, let me just see if I can make that a little more, a little smoother.*1270

*Okay, we've got -2sin(x), look it's got a bigger amplitude than the original graph and it's flipped over because of the negative sign.*1280

*Next thing I'll introduce is the period.*1288

*The period is supposed to be π, so I'm graphing -2sin(2x).*1292

*That speeds the whole thing up, it shortens the period because the period is now π instead of 2π.*1298

*I need to do that entire graph in the space of π instead of 2π.*1306

*There, that one that I just graphed, I shortened the period to be π instead of 2π.*1330

*The period is now π on that new graph.*1337

*Next, we're going to do the phase-shift, that's π/6 units to the right.*1340

*The phase-shift, what I'm about to graph is -2sin(2x)-π/3, so that takes the whole graph and it shifted over π/3 units to the right.*1349

*Let me do this one in red.*1364

*I'm going to take that last graph and shift it over π/3 units to the right.*1370

*Instead of starting at (0,0), it starts at π/3, it's going to come back down to 0, at 5π/6.*1374

*The phase-shift is supposed to be π/6, so I'm going to move everything over by π/6.*1389

*Instead of starting at 0, I'm going to start at π/6 and come back at (π/2)+(π/6) which is actually 2π/3.*1408

*That red curve that I graphed there is -2sin(2x)-π/3.*1439

*The last step is to do the vertical shift which is 0, so we don't have to move the graph at all which means we're done.*1453

*This last graph is the one we want.*1464

*Again, it's a matter of breaking these equations down into their parts.*1470

*It's very complicated if you kind of look at the whole thing but if you look at each steps and you keep each steps in order then it's not to hard.*1475

*Remember the equations for amplitude, period, phase-shift, and vertical shift.*1484

*Once you've got those, you start with your basic sin(x) graph then you change the amplitude which stretches it out up and down, or might flip it.*1489

*You change the period which compreseses it horizontally. *1500

*You do the phase-shift which takes the whole thing and without compressing it, it moves it to the right or the left.*1507

*Finally, the vertical shift moves it up or down.*1511

*You could just keep moving these graphs around until you build up the equation you're looking for.*1515

*This one's a little bit different, we're asked to find the sine wave.*1526

*This time we're told what all the properties are.*1530

*We're given the amplitude, the period, the phase-shift, and the vertical shift.*1535

*We want to find an equation and we want to graph the function.*1540

*I'm going to kind of build this up, the same way we're building the earlier graphs.*1543

*I'm going to start with the basic sin(x), so that's my basic sine wave.*1548

*Now, I'm going to give an amplitude 2, and remember 2 is just a number on the outside, the a.*1554

*Let me rewrite that equation, asin(bx+c)+d.*1559

*Amplitude 2 means a is 2, 2sin(x).*1566

*Now, period 4π, remember our equation for period was 2π/b, that is supposed to be equal to 4π.*1575

*When you solve that out, that tells us that b=1/2, so the b has to be equal to 1/2.*1597

*That means our equation is now 2sin(1/2)x, so we've incorporated the period.*1610

*Phase-shift is supposed to be π/2, but remember the phase-shift, our formula for that is -c/b.*1622

*b is already, we figured out as 1/2, so that's -c/(1/2).*1643

*If we do a little bit of algebra here, we get (1/2)π=2c, so c=-π/4.*1650

*I got that from the equation -c/b=π/2.*1667

*I already figured out my b, now I can figure out my c.*1672

*The next part of that equation is 2sin[(-1/2)x-(π/4)].*1677

*Finally, I want to talk about the vertical shift, which is supposed to be 1 and that's the d.*1692

*Finally, my equation is 2sin[(1/2)x-(π/4)]+1.*1704

*That is the sine wave that I'm looking for.*1715

*Now, I want to graph that thing.*1722

*I start out with the basic sine wave, π, 2π, π/2, 3π/2.*1730

*Basic sine wave starts at 0, goes up to 1, back down to 0 at π, down to -1, back up to 0 at 2π, basic sine wave.*1742

*Now, I'll introduce these properties in order.*1753

*I'll start out with the amplitude.*1755

*Amplitude's supposed to be 2, so I'm going to stretch this thing up instead of going from 1 to -1, it's going to go up from 2 up to 2, and down to -2.*1760

*It'll stretch the thing up, and down.*1774

*Period is 4π, that means the thing stretches out, so that it'll only does one cycle every 4π.*1783

*That means I have to extend my graph quite a bit here, 3π, 4π, so I'll stretch the thing out, so it'll only does the cycle every 4π.*1794

*Now, phase-shift π/2, that means the thing is going to shift π/2 units to the right.*1817

*I better draw a new set of axis here.*1824

*I've got π, 2π, 3π and 4π, 5π.*1835

*What I want to do is take the graph I have above and move it π/2 units to the right because we have phase-shift π/2.*1850

*Instead of starting at 0, I'm going to start at π/2, I go up to there, back down to 0 at 2π plus π/2, back down to -2 between 3 and 4π and back up to 0 there.*1866

*So, connect these up.*1887

*Finally, I have to do one with vertical shift 1, that means I'll take the whole graph and I'll move it up by one unit.*1913

*That means instead of going or peak at 2, it's going to peak at 3 now.*1920

*Instead of going down to -2, it gets moved up by 1 unit so it's going to go down to -1.*1932

*Let me draw this last final curve in red.*1940

*Everything gets moved up by 1 unit.*1948

*I'm going to plot some points here, moving everything up by 1 unit.*1953

*That final curve is the one we want.*1976

*That's 2sin[(1/2)x-(π/4)]+1.*1980

*Again, it's a complicated procedure but if you take it step by step, each one of the steps is not too hard.*1991

*First, we kind of reconstructed the equation from these parameters that we were given.*1997

*Basically we figured out a, b, c and d from these parameters that we were given by sort of reverse engineering the formulas 2π/b, -c/b, and the d.*2004

*Then, we went through step by step.*2020

*We started with the basic sine curve.*2024

*We changed its amplitude, stretched it vertically.*2030

*We changed its period which stretches out horizontally.*2034

*We changed its phase shift which moved it over horizontally not stretching but just moving it without stretching.*2036

*Then, we did the vertical shift, moving it up or down.*2041

*You should practice a few of these curves on your own.*2045

*We'll come back and try some more examples together later.*2047

*We are back with some extra examples of tan and cot functions, we are given here a right triangle with short sides of length 5 and 12.*0000

*Let me try to draw that and what we have to do is find the tan and cot of all the angles in the triangle.*0009

*Our first step there is to figure out what the hypotenuse is, hypotenuse*^{2} is 5^{2} + 12^{2}, which is 25 + 144 which is 169.0022

*That is exactly 13*^{2}, this triangle is rigged up to have a nice hypotenuse of 13 and so now let me label these angles (theta) and (phi), and we will figure out what the tan and cot of each one is.0037

*The key point here, it all comes back to SOHCAHTOA.*0052

*We will be using SOHCAHTOA over and over again in your trigonometry so it is really worth memorizing that one.*0060

*If you have a hard time remembering the word, then remember the acronym Some Old Horse Caught Another Horse Taking Oats Away.*0067

*In particular, the tan is opposite of adjacent, tan(theta), the opposite side is 5, the adjacent side has length 12.*0073

*Cot(theta) is just the opposite of that, so it is 12/5.*0086

*Tan(phi), the opposite side for angle(phi) is 12 and the adjacent side is 5.*0097

*Cot(phi) is just the other way around it is 5/12.*0106

*Finally, the 90 degree angle a bit of special case there, let me give it in radians as pi/2.*0112

*Remember, tan(pi/2) is undefined and cot(pi/2) cos/sin which is 0/1 is just 0.*0119

*Finding cot and tan in right triangles is just a matter of remembering that mnemonics SOHCAHTOA, tan=opposite/adjacent.*0136

*Ok we are given here a bunch of angles and we want to find the tan and cot of each one.*0000

*A good first step is to graph this angles on a unit circle.*0006

*I got a unit circle here, I will label the common values pi/2, pi, 3pi/2, and 2pi and I want to figure out where these angles are.*0019

*2pi/3 is a little bit past pi/2, so we are going to find the tan and cot there.*0031

*7pi/6 is a little bit past pi, and 7pi/4 is in between 3pi/2 and 2pi.*0038

*Those are the three angles and to find their tan and cot, I will draw my triangles in and write down the sin and cos of each one.*0048

*For each angle I’m going to find the sin, cos, and I will use those to find the tan and cot.*0061

*Let us start with 2pi/3, that is over here, I will draw a triangle there, that is a 30, 60, 90 triangle.*0072

*I know that the sin is long one there so that must be (root 3)/2, cos of the short one is ½ except that is negative because the x coordinate is negative.*0082

*For tan, it is (sin/cos) that is just –root 3, cot=(cos/sin) that is -1/root 3, but that rationalizes to -root 3/3.*0094

*Next one is 7pi/6 so that is this angle right there, and that is a 30, 60, 90 triangle.*0110

*7pi/6 the (sin) is ½ but that is negative because the y coordinate is negative.*0121

*(Cos) the big one is root 3/2 but again that is negative because the x coordinate is negative.*0128

*(tan) if you divide those together, sin/cos=1/(root 3)/3 and it is positive because both of those are negative, so that 2 negative cancel.*0134

*(cot) is( cos/sin) that is positive root 3 because they are both negative and so the two negative is cancelled.*0146

*7pi/4 that is this angle down here and that is a 45, 45, 90 triangle, (sin) and (cos) are both root 2/2 for 45, 45, 90 triangle.*0157

*We got to figure out which one is negative and its (sin) because the (y) coordinate is negative here, we are down below the (x) axis.*0177

*(cos) is positive because the (x) coordinate is positive.*0185

*(tan) you divide those together and you get -1, and (cot) you divide (cos/sin) but you still just get -1.*0188

*It is probably not worth memorizing the (tan) and (cot) of those angles but it is worth knowing the common values for the 30,60,90 triangles. *0202

*For 30, 60, 90 it is ½, root 3/2, for the 45, 45, 90 triangles the (sin) and (cos) would be root 2/2 and root 2/2.*0213

*If you remember those common values then you can draw a triangle anywhere else in the unit circle and just figure out which of these values is the (sin) and which one is the (cos).*0226

*You can figure out which one is positive and which one is negative, to get the (tan) you just divide the (sin) and (cos) together, same for the (cot).*0238

*The only tricky part there is figuring the (tan) and (cot) are positive or negative, but you just look back at whether the sin and cos are positive or negative.*0249

*That tells you right away whether the (tan) and (cot) are positive and negative.*0259

*You can also remember the mnemonic all students take calculus which stands for all the values positive in the second quadrant, only the sin is positive, that is why the 2pi/3 (cot) where negative.*0264

*In the third quadrant, (tan) is positive which means the (cot) will be as well.*0283

*In the fourth quadrant, only (cos) is positive which means the (tan) will be negative.*0288

*A lot of different ways to remember that, choose which one works for you.*0294

*We will come back later with more lectures on trigonometry at www.educator.com.*0298

*Hi this is Will Murray for educator.com and we're talking about trigonometry.*0000

*We're finally ready to learn about the tangent and cotangent functions.*0004

*We've been eluding to those in some of the earlier lectures but this is the lecture where we're going to formally define what the tangent and cotangent means and really get some practice with them.*0010

*The definition of tangent is very simple, it's just tangent, by definition, the tangent of an angle is sin/cos.*0020

*You can only talk about that when the cosine of the angle is not 0.*0032

*If cosine of a particular angle is 0, we just say the tangent is undefined.*0039

*The cotangent is just the same thing except you flip them the other way up.*0043

*The cotangent of an angle is cosθ/sinθ.*0047

*If the sinθ happens to be 0, then we say the cotangent is undefined.*0053

*Now, the master formula that I've already mentioned before in a previous lecture for right triangles is SOH CAH TOA.*0061

*You can just remember the words SOH CAH TOA or there's a little mnemonic to help you remember that if you want.*0071

*It's Some Old Horse Caught Another Horse Taking Oats Away.*0076

*If that's easier for you to remember than SOH CAH TOA, then by all means remember that.*0082

*The key point is that this tells you, the SOH CAH TOA formula, tells you how to interpret the sine and cosine and tangent of an angle in terms of the lengths of sides of a right triangle.*0086

*I've drawn a θ down here in one of the angles of a right triangle.*0105

*Then we talked about the side adjacent to θ and the side opposite to θ, then the hypotenuse of the right triangle.*0109

*The master formula that you want to remember is that the sinθ is given by the opposite side over the hypotenuse, the cosθ by the adjacent side over the hypotenuse.*0128

*The tanθ, which we're learning about today, is the opposite side over the adjacent side.*0140

*Let me run over the common values of the tangent and the cotangent function.*0149

*The way to remember these is to remember the common values of the sine and cosine because remember tangent is just sin/cos and cotangent is just cos/sin.*0156

*If you can remember the common values of sine and cosine, you can always work out the common values of tangent and cotangent.*0165

*I've listed the particular, the common values in the first quadrant here, in degrees, we have 0, 30, 45, 60, 90 and those correspond to radians of 0, π/6, π/4, π/3 and π/2.*0173

*You should really have the sines and cosines of these angles memorized to be able to reproduce them very quickly.*0190

*If you can, then you can immediately figure out the tangent and cotangent without really having to memorize anything extra.*0199

*For 0 degrees, the cosine is 1 and the sine is 0.*0205

*Tangent is sin/cos, that's 0/1, that gives you 0.*0212

*Cotangent is cos/sin, which would give you a division by 0, that's why we say it's undefined.*0216

*For 30 degrees, the same as π/6 radians, the cosine and sine are root 3 over 2, and 1/2.*0225

*Tangent, if you divide those together, you get 1 over 3.*0234

*If you rationalize that, that rationalizes into root 3 over 3, so that's the tangent of π/6.*0244

*Cotangent, if you divide them the other way, cos/sin, you just get root 3.*0248

*For 45, the cosine and the sine are both root 2 over 2, so the tangent and cotangent are both 1.*0254

*For 60 degrees or π/3, you just get the opposite you had for 30 degrees or π/6, the tangent turns out to be root 3 and the cotangent turns out to be root 3 over 3.*0263

*For 90 degrees or π/2, the cosine and sine are the opposite of what they were for 0 degrees, and so the tangent is undefined and the cotangent is 0.*0276

*The values that you get for tangent and cotangent, there's 0, 1, root 3 over 3, and root 3, and undefined.*0288

*Those are the common values you get for tangent and cotangent.*0305

*When you have one of these common angles, one of these multiples of π/6 or π/4, in degrees, that's 30 degrees or 45 degrees, it's really.*0311

*When you want to know the tangent or cotangent, you know it's going to be one of these common values and it's just a matter of figuring out which one.*0323

*It's good to remember that these common values come up over and over again because when you're working out values for any particular angle you expect it to be one of these common values.*0333

*There's a couple more facts that I want to talk about, uses for the tangent function.*0347

*One is that the slope of the line is the tangent of the angle that the line makes with the x-axis.*0354

*Let me try to draw that.*0360

*Suppose you have just a random line in a plane like this, what I'm going to do is look at the angle θ that the line makes with the x-axis.*0366

*Now, I'm going to move this line over.*0379

*I'll draw it in blue, the translated version over to the origin.*0382

*That's meant to be a line with the same slope that just moved over to the origin.*0392

*If I make a triangle like this, we still have θ in one corner of the triangle.*0400

*The adjacent side is equal to the amount that the line is running over, so that's the run.*0411

*The opposite side is equal to the rise that the line makes in that triangle.*0421

*If we remember SOH CAH TOA, tanθ is equal to the opposite side over the adjacent side.*0432

*That's the TOA part of SOH CAH TOA, which is equal to the rise over the run, which is equal to the slope of a line.*0443

*That tells you that for any line, the tangent of this angle θ that it makes with the x-axis is equal to the slope of that line.*0455

*Another thing that you want to remember about sines and cosines is that once you know the common values of tangent and cotangent.*0470

*If you remember which quadrants the sine and cosine are positive and negative, you can figure out which quadrants the tangent and cotangent are positive and negative.*0480

*Let me label the quadrants here, 1, 2, 3, and 4.*0491

*Let me label the positive ones.*0500

*Remember sine corresponds to the y value, so sine is positive in the top two quadrants, negative in the bottom two.*0502

*Cosine corresponds to x values, so cosine is positive in the first quadrant and in the fourth quadrant, and negative in the other two quadrants.*0510

*If you want to figure out when if tangent is positive, it's positive either when sine or cosine are both positive or both negative.*0520

*Remember, the tangent is equal to the sin/cos.*0529

*Tangent is going to be positive in the first quadrant, negative in the second quadrant because sine is positive, cosine is negative.*0535

*Tangent is positive in the third quadrant, because both sine and cosine are negative, and tangent is negative in the fourth quadrant because cosine is positive but sine is negative.*0544

*If you put this together, tangent is positive in the first and third quadrants, negative in the second and fourth, which is kind of the origin of this mnemonics that you can use to remember, ASTC.*0557

*The way you remember that is All Students Take Calculus, that stands for all the sine, cosine, tangent functions are positive in the first quadrant.*0573

*In the second quadrant, only sine is, in the third quadrant, only tangent is, and in the fourth quadrant, only cosine is.*0585

*Now, you might wonder how does cotangent fit in to all of this.*0594

*Cotangent, remember, is just the flip, the reciprocal of tangent, tangent was sin/cos, cotangent is cos/sin.*0597

*Cotangent is going to be positive whenever tangent is positive.*0611

*It'll be positive in the first quadrant and in the third quadrant.*0614

*Now we know our common values of tangent and cotangent.*0622

*Remember that 0, 1, square root of 3, and root 3 over 3, and we know which quadrants is going to be positive or negative, we can work out the tangent and cotangent of any common angle around the unit circle.*0625

*Let's try some problems now.*0642

*Let's start out by drawing graphs of the tangent and cotangent functions, and we're going to label the zeros and asymptotes of each and try to figure out what the periods are.*0645

*I'm going to start with the tangent function, and I'll do that one in blue.*0662

*For the tangent function, I know that the tangent of 0 is 0.*0682

*I'm going to put a dot at 0.*0687

*Let me mark my axis a bit, π/2, π, 3π/2, 2π.*0690

*Tangent starts out at 0.*0703

*The tanπ/4 we said is 1, so let me put a dot there.*0710

*When tangent gets close to π/2, remember that's really getting close to dividing by zero so it goes up to positive infinity there.*0717

*Tangent has an asymptote at π/2 and it goes up to positive infinity there, so it looks like that.*0730

*In the second quadrant, tangent is negative, at near π/2 it has an asymptote going down to negative infinity.*0745

*The tangent of π is 0 because it's sin/cos.*0757

*As tangent approaches 3π/2, it has another asymptote.*0764

*That's what the graph of tan(x) looks like.*0790

*That should be in blue because I'm drawing my tangent graph in blue.*0799

*tan(x) has zeros at 0, π, 2π, and so on.*0805

*It has asymptotes, basically, at the places where you're trying to divide by zero, and those are all the places where the cosine is zero.*0823

*Remember tangent is sin/cos.*0835

*Those are at π/2, 3π/2 and so on.*0839

*That's what the tangent function looks like.*0849

*Let me try to graph the cotangent function.*0853

*Now, the cotangent function, remember that's cos/sin, that has asymptotes and zeros exactly the opposite of where tangent did.*0859

*It has zeros wherever cosine is 0, which is π/2, 3π/2 and so on, because cosine is 0 at those places, 3π/2, 5π/2.*0885

*It has asymptotes wherever sine has 0, because we're trying to divide by 0, and that's 0, π, 2π, and so on.*0907

*Cotangent looks kind of just the opposite of tangent, like that.*0922

*That's the cotangent function in red.*0965

*For the next example, we're going to look at right angle triangles, and try to work out some trigonometry there.*0972

*We're given that a right a triangle has short sides of length 3 and 4.*0977

*We're trying to find the tangents and cotangents of all the angles in the triangle.*0989

*The first thing we need to do is figure out what the hypotenuse of this triangle is.*0995

*We know that the hypotenuse squared is 3*^{2}+4^{2}, which is 9+16, which is 25.1000

*The hypotenuse must be 5 units long.*1010

*Let me figure out first of all the tangents of all the angles in the triangle and we're going to use the SOH CAH TOA.*1019

*Remember, tangent equals opposite over adjacent.*1023

*The tanθ is the opposite over the adjacent which is 3/4.*1030

*The tanφ is the opposite side of φ which is 4/3.*1040

*The last angle in the triangle is 90 degrees, that's the tangent of π/2 radians, remember the tangent is undefined for π/2, so I'm just going to leave that as undefined.*1050

*That's because tangent is sin/cos, and cos of π/2 is 0.*1069

*Now, cotangent of each of these angles, cotangent is just the opposite of tangent in the sense that when tan is sin/cos, cot is cos/sin.*1076

*The cotθ, instead of being 3/4, will be 4/3.*1089

*The cotφ, instead of being 4/3, will be 3/4.*1097

*The cot(π/2) which is the cos/sin, that's 0/1, is just 0.*1106

*The secret to working out tangents and cotangents of angles in right triangles, is just to remember the SOH CAH TOA formula.*1117

*In particular, the TOA part, says tangent equals opposite over adjacent.*1125

*That quickly helps you figure out the tangents of angles and right triangles.*1134

*For our next example, we are asked to determine if the tangent and cotangent functions are odd, even, or neither.*1138

*Let's recall the definitions of odd and even functions.*1144

*Odd function is where f(-x)=-f(x).*1149

*Let me draw that a little more clearly, f(-x)=-f(x).*1160

*It's the kind of function that has rotational symmetry around the origin when you look at the graph.*1169

*An even function satisfies f(-x)=f(x) with no negative sign and that has mirror symmetry across the y-axis.*1185

*Those are the two definitions we want to check here.*1209

*Let's try that out for tangent and cotangent.*1213

*Tan(-x), let's see what happens.*1215

*Tangent by definition is sin/cos, so that's sin(-x)/cos(-x).*1222

*Sine remember is an odd function, sin(-x) is -sin(x), cosine is an even function so cos(-x)=cos(x).*1226

*You put this together, you get -sin/cos, you get -tan(x).*1241

*In particular, that tells you that tangent is odd.*1248

*We can check with the graph of tangent, it's probably worth memorizing what the graph of tangent looks like.*1261

*The graph of tangent look like this, it crosses the x-axis right at the origin, and then it has asymptotes at -π/2 and π/2.*1271

*In particular, the graph of tangent has rotational symmetry around the origin.*1283

*If you spun it around 180 degrees, it would look exactly the same.*1290

*It has that rotational symmetry around the origin, the graph of tangent definitely confirms that tangent is an odd function.*1295

*Let's check that out for cotangent.*1304

*Cot(-x), cotangent is just cos/sin, that's cos(-x)/sin(-x).*1305

*Just like before cos(-x)=cos(x), because cosine is an even function, sin(-x)=-sin(x), so what we get is -cos/sin, -cot(x).*1318

*Cotangent is odd also, it's also an odd function.*1336

*Again, we can check that looking at the graph.*1346

*Let me draw a quick graph of cot(x), probably worth remembering but not quite as essential as the tangent function.*1352

*If you're going to remember one of them, remember the graph of the tangent function.*1360

*The cotangent function looks like this, it crosses 0 at π/2, and it has asymptotes at 0 and π.*1366

*The cotangent function, if you draw a dot on the origin, and then rotate the thing, 180 degrees.*1384

*If you rotate the thing 180 degrees around the origin, the graph would look the same.*1396

*It has this mirror symmetry, sorry, not mirror symmetry but rotational symmetry around the origin, so the cotangent function is an odd function.*1402

*It does not have mirror symmetry because if you flipped it across a mirror on the y-axis, it would not look the same so it's not an even function.*1415

*But it does have a rotational symmetry around the origin.*1425

*We got another couple of examples coming up later.*1429

*Why don't you try them out and then we'll work them out together.*1432

*Okay, another example of cotangents and tangents, we're given some common values here 5π/6, 5π/4 and 5π/3.*1437

*Those are all angles in the unit circle.*1447

*We want to figure out what the tangents and the cotangents are.*1451

*A good way to start this is to draw the unit circle and to figure out where those angles are.*1458

*There's my unit circle, and let me label some easy common values, there's 0, π/2, π, 3π/2 and 2π.*1478

*The angles we're being asked to find, 5π/6 is just a little bit short of π, 5π/4 is in between π and 3π/2, and 5π/3 is 5/6 away around the unit circle, it's down there.*1490

*Those are the three angles we're being asked to find.*1514

*It's probably easiest if we just write down the sines and cosines of those angles, because presumably we know those pretty well.*1518

*Then we can figure out the tangent and the cotangent.*1522

*Let's make a little chart.*1527

*We'll find the angle, I'll find the sine and the cosine, then we'll find the tangent just by dividing those together, and the cotangent.*1530

*If we start with 5π/6, that's this angle right here.*1547

*Remember the common values here.*1554

*The sine is the small one, that's 1/2, it's positive.*1558

*Cosine is the bigger one, square root of 3 over 2, and that's negative because the x value is negative there.*1560

*The tangent is sin/cos, so that's 1 over the square root of 3, which rationalizes to square of 3 over 3, and it's negative because the cosine was negative.*1568

*The cotangent is just cos/sin, so that's root 3, that's also negative.*1580

*Now, 5π/4, that's a 45 degree angle, South of the x-axis*1588

*The sine and the cosine, because it's 45 degrees, they're both root 2 over 2, but we've got to figure out whether they're positive or negative.*1599

*Since it's in the third quadrant there, they're both negative.*1608

*That means when you divide them together to get the tangent, you get a positive one.*1613

*When you check the cotangent, you also get a positive one.*1616

*Now 5π/3, that's another 30-60-90 degree angle down here.*1620

*The y value is the big one this time, so it's root 3 over 2, cosine is 1/2.*1629

*The sine is negative because we're still south of the x-axis, but the cosine is positive because we're to the right of the y-axis, meaning the x value is positive.*1634

*Tangent is sin/cos, that's root 3 and that's negative.*1648

*The cotangent is cos/sin, that's 1 over root 3, that rationalizes to root 3 over 3, that's also negative.*1655

*Those are our answers right there.*1671

*Let me just emphasize that the key to figuring this problem out is to remember the sines and cosines using the common values and those common 30-60-90 triangles and the 45-45-90 triangle.*1675

*As long as you remember those common values, then you can work out the sines and cosines, figure which ones are positive and which ones are negative, then divide them together to get the tangents and cotangents.*1687

*Of course it helps that you remember that the common values for tangent and cotangent, their always going to be 1, square root of 3, and square root of 3 over 3.*1699

*Then, it's just a matter of figuring out whether they're positive or whether they're negative, and which is which, for any given angle.*1710

*We are trying some more examples for the sec and cosec function.*0000

*Here we are being asked to draw a graph of the cosec function and we have to label all 0, max, min, and asymptotes, and identify the period.*0005

*The key thing to remember here is the cosec(theta) is 1/sin(theta), really this comes down to understanding the graph sin(theta) very well.*0015

*I’m going to start with a graph of sin(theta), that is probably a graph that you should have already memorized.*0029

*My graph of sin(theta) goes up to 1 and down to -1, starts at 0 goes up to 1 and pi/2, down to 0 at pi, down to -1 that is 3pi/2 and back at 0 at 2pi.*0048

*What I have drawn there is not cosec yet, it is sin(theta), in red I will draw cosec(theta).*0073

*Cosec(theta) is just 1/sin(theta), whenever sin is 1 it is 1, whenever sin is 1- it is -1, whenever sin is 0 where you can not divide by 0, that is where cosec has an asymptote.*0083

*It goes up to infinity whenever sin goes down to 0.*0110

*And down to negative infinity whenever sin goes down to 0.*0119

*It looks a lot like the graph of sec did, it got these U’s going up to infinity and upside down U’s going down to negative infinity depending on where the graph of sin hits 0.*0127

*Let us label everything we have been asked to label here, cosec(theta) never crosses the (x) axis, so it has no 0.*0143

*There are no 0 to label there, max and min it has local min at the bottom of each of these U’s.*0158

*This is at pi/2 and 1, that is a local min.*0165

*We have local max at –pi/2 and -1, that is a local max, and another one at 3pi/2, -1, that is our local max.*0174

*We got the max and min, the asymptotes are places where it goes up to infinity and down to negative infinity.*0197

*Let me label those, there is one right there at –pi.*0205

*Here is another one at 0, another one at pi, and finally here is one at 2pi.*0215

*Basically those are the places where sin(theta) is 0, remember cosec is 1/sin(theta) so whenever you are trying to divide by 0, that is where (cosec) blows up to infinity or drops down to negative infinity.*0228

*We got the asymptotes, the period of the (cosec) function is how long it takes to repeat itself.*0242

*That is one period before it starts repeating itself and that is 2pi and that really comes back to the fact that sin(theta) has a period of 2 pi.*0253

*The period of (cosec) just like (sec) is 2 pi.*0262

*It looks like we answered everything there, the key part there is remember that (cosec) is 1/sin, you probably do not need to memorize the graph of (cosec) itself.*0273

*It is good if you are familiar with this general shape of the U’s going up and the U’s going down but you do not need to memorize the details.*0285

*As long as you remember that it is 1/sin, you can work it out from there.*0295

*Ok for our next example here, we are given a bunch of angles and we have to find the (sec) and (cosec) of these common values.*0000

*Let me start by drawing these angles on a unit circle, that is a squish unit circle, let me see if I can do a little better there.*0009

*This is 0, pi/2, pi/, 3pi/2, and 2pi, it looks like at least one of these angles is given in degrees as well.*0036

*I will label the values in degrees, there is 90, 180, 270, and 360, those are the values in degrees.*0049

*Let us figure out where these angles are in the unit circle.*0060

*5pi/6 is over here, 240 degrees is between 180 and 270, that is down here.*0063

*7pi/4 is between 3pi/2 and 2pi, that is down there.*0075

*Remember (sec) and (cosec), those are that reciprocals of (cos) and (sin).*0082

*The smart thing to do here is to figure out the (cos) and (sin) of those angles and then it would be an easy matter to find the (sec) and (cosec).*0087

*I’m going to make a little chart, the angle, (cos), (sin) and then it would be an easy matter to find the (sec) and (cosec).*0099

*Let us start with 5pi/6, that is over here and that is a 30, 60, 90 triangle so we know the values.*0115

*The (cos) is (root 3)/2, (sin) is ½, but then we have to worry about whether they are positive or negative.*0123

*The (x) coordinate is negative so I will make the (cos) negative.*0130

*The (sec) now is just 1/cos, so that is 2/square root 3, but if we rationalized that, that is 2 root 3/3 and of course it is negative.*0135

*The (cosec) is 1/sin so 1/(1/2) is just 2.*0148

*Now the next one is 240 degrees, if you like to translate into radians, that is 4pi/3, that is this angle down here.*0156

*We see a 30, 60, 90 triangle, this time the (cos) is the short side that is ½, the (sin) is the long side root 3/2.*0176

*But it is in the third quadrant so they are both negative.*0187

*The (sec) here is 1/cos, so that is -2, (cosec) 1/sin, 2/root 3 rationalizes to (2 root 3)/3 and that is negative.*0192

*Finally, 7pi/4 that is this angle over here.*0207

*It helps to start out with the (cos) and (sin), that is a 45 degrees triangle so we know they are both root 2/2.*0219

*We just have to figure out which one is positive and which one is negative.*0226

*The (y) coordinate is negative there so the (sin) must be negative.*0230

*The (sec) is 1/(cos), 2 /root 2 is just root 2, (cosec)=1/(sin) is 2/root 2 so that is root 2 again but it is negative.*0235

*I really emphasize to my students that you do not really need to memorize the (sec) and (cosec) of the common values if you really have the (sin) and (cos) memorized well.*0250

*I think it is more important to memorize the (sin) and (cos) very well and the way you get those is by knowing those common values for the 30, 60, 90 triangles and for the 45, 45, 90 triangles.*0264

*You use those to figure out the (sin) and (cos), you figure out which quadrant you are in, which tells you whether they are positive or negative.*0278

*Now you know (sin) and (cos), whether they are positive or negative.*0287

*For (sec) and (cosec), all you have to remember is that sec(theta)=1/cos(theta).*0291

*If you can figure out the (cos), you can figure out (sec).*0300

*Similarly for the cosec(theta) is 1/sin(theta), again if you figure out the sin(theta) it is a simple matter to figure out the cosec(theta).*0303

*I hope all those worked out well for you, this is part of the trigonometry lectures series for www.educator.com.*0315

*Hi welcome back to the trigonometry lectures on educator.com*0000

*Today, we're going to learn about the last trigonometric functions.*0004

*We've learned about the sine and cosine, and tangent and cotangent.*0008

*Today we're going to learn the secant and cosecant function.*0011

*Let's start with their definitions.*0017

*The secant function is just defined by the sec(θ)=1/cos(θ).*0019

*That only works when the cos(θ) is non-zero.*0025

*If the (cosθ)=0, we just say the secant is undefined.*0027

*The cosecant, which people shorten to csc, is just 1/sin(θ).*0033

*If the sin(θ)=0, we just say that the cosecant is undefined.*0040

*Let's start with the first example right away.*0048

*We have to draw a graph of the secant function.*0051

*In particular, we have to label all zeros, max's, mins, and asymptotes and figure what the period of the secant function is.*0055

*Remember now that the secant function, sec(θ), by definition is 1/cos(θ).*0064

*A really good place to start here when you're trying to understand sec(θ) is with the graph of cos(θ).*0073

*Let me start with the graph of cos(θ).*0081

*There's π, there's 2π, and I'm going to extend this out a bit, 3π, and -π.*0086

*Now, remember that the cosine function starts at 1, goes down to 0 at π/2, so there's π/2, goes down to -1 at π, comes back to 0 at 3π/2, back up to 1 at 2π.*0098

*The period of cosine is 2π, so it's repeating itself after 2π.*0130

*I'm not drawing secant yet, I'm drawing cos(x), y=cos(x).*0140

*In black here, I've got, we'll call it cos(θ).*0146

*Now, I'm going to draw the secant function in red.*0153

*That means we're doing 1/cos(x), or 1/cos(θ).*0159

*In particular, 1/1 is 1.*0165

*Then as the cosine goes down to 0, secant is 1/cos, so it goes up to infinity there, and does the same thing on the other side, so secant looks like that.*0170

*When the cosine is negative, when the cosine is -1, secant is -1.*0201

*When the cosine goes to 0, secant blows up but since cosine is negative here, secant goes to negative infinity when the cosine is negative.*0217

*Then when cosine is positive again, secant is positive again, going up to positive infinity on both ends there.*0235

*That's what the secant function looks like.*0247

*While the cosine goes between -1 and 1, secant is just the reciprocal of that, so it goes 1 up to negative infinity, and from -1 down to negative infinity.*0249

*Now, I've got the secant graph in red.*0260

*Let me label all the things that we've been asked to label.*0267

*First of all, zeros, while the sec(θ) has no zeros because it never crosses the x-axis, so there are no zeros to label.*0270

*Max's, all local max's of the secant function, well here's one down here at (π,-1) is a local max.*0282

*Minimum at (0,1) is a local min, (2π,1) is a local min, and so on, as a local min and a local max, every π units.*0295

*We've got the max's and the mins.*0314

*The asymptotes are the places where the secant blows up to infinity or drops down to negative infinity and so that's an asymptote at -π/2, and at π/2.*0318

*We have an asymptote at π/2, and again at 3π/2, and again at 5π/2.*0333

*Basically, every π units, we have an asymptote where the secant function blows up to infinity or drops down to negative infinity.*0350

*Finally, what is the period of secant function.*0362

*The secant function is really dependent on the cosine function, and the cosine function repeats itself once every 2π.*0367

*The period of cosine is 2π.*0374

*The period of secant is also 2π.*0376

*You can see that from the graph, it starts repeating itself after a multiple of 2π.*0384

*That's what the graph of the secant function looks like.*0394

*It kind of has this hues and this upside down hues really dependent on the cosine function because the secant is just 1/cos(θ).*0399

*For the second example, we have to figure out some common values of secant and cosecant at the angles in the first quadrant, 0, π/6, π/4, π/3, and π/2.*0410

*Now, these angles are probably so common that you really should have memorized the sine and cosine.*0426

*I'm going to start by writing down the sine and cosine of these values.*0433

*I'm going to write them down both in degrees and radians because it's very important to be able to identify these common values either way.*0440

*I'll write down degrees, radians, I'll write down the cosine and the sine, and then the secant and cosecant of each one.*0449

*I'll make a nice chart here.*0465

*The values given were 0, π/6, π/4, π/3, and π/2, in terms of degrees, that's 0, 30, 45, 60, and 90.*0467

*Now, you really should have probably memorized the cosine and sine of these already.*0486

*You probably shouldn't even have to check the unit circle.*0493

*But if you need to, go ahead and draw yourself a unit circle.*0496

*Then draw out those triangles in the first quadrant, and you'll be able to figure out the cosine and sine very quickly as long as you remember the values of the 30-60-90 triangles and the 45-45-90 triangles.*0500

*In particular, the cosine and sine of 0, are 1 and 0.*0510

*For 30-degree angle, cosine is root 3 over 2, sine is 1/2.*0517

*For 45-degree angle, they're both root 2 over 2.*0524

*For 60-degree angle, they're just the opposite of what they were for 30, 1/2 and root 3 over 2.*0529

*For 90, they're just the opposite of what they were for 0, 0 and 1.*0537

*Now, the secant is just the reciprocal of cosine, it's just 1/cos.*0542

*I'll just take 1/1, 2 divided by root 3, if you rationalize that, you get 2 root 3 over 3, 2 divided by root 2, is just root 2, the reciprocal of 1/2 is 2, and the reciprocal of 0 is undefined.*0547

*Those are the secants of those common values.*0568

*The cosecant is 1/sin, while 1/0 is undefined, 1/(1/2) is 2.*0574

*The reciprocal of root 2 over 2 is 2 divided by root 2, which again is root 2.*0585

*Then 2 divided by root 3 rationalizes into 2 root 3 over 3, and then 1/1 is just 1.*0592

*I've been really trying to drill you on memorizing the values of sine and cosine.*0601

*I don't think it's really worth memorizing the values of secant and cosecant, they don't come up as often as sine and cosine.*0606

*The key thing to remember is that sec(θ) is just 1/cos(θ), and csc(θ) is just 1/sin(θ).*0615

*As long as you really have memorized the values of sine and cosine, you can always work out the values of secant and cosecant.*0630

*I don't think you need to memorize these values of secant and cosecant.*0639

*It helps if you practice them but it's not really worth memorizing them as long as you know your sine and cosine really well.*0643

*You can always work out secant and cosecant.*0649

*Last thing this example asked is, which other quadrants the secant and cosecant are positive?*0652

*Let's go back and remember our little mnemonic here, All Students Take Calculus.*0662

*That's in quadrant 1, quadrant 2, quadrant 3, and quadrant 4.*0671

*That tells us which of the common functions are positive in which quadrant.*0679

*In the first quadrant, they're all positive, in the second quadrant, only sine, in the third quadrant, only tangent, and in the fourth quadrant, only cosine.*0684

*Let's figure out what that means for secant and cosecant in each case.*0697

*On the first quadrant, they're both positive, both sine and cosine are positive, so secant and cosecant are both positive.*0712

*In the second quadrant, sine is positive, which means that secant is positive, but cosine is negative so secant is negative.*0720

*In the third quadrant, tangent is the only thing that's positive, sine and cosine are both negative, so secant and cosecant are both negative.*0733

*Finally, in the fourth quadrant, cosine is positive so secant is positive, sine is negative so cosecant is negative.*0739

*Now, we have a little chart that tells us which quadrant secant and cosecant are positive and negative in.*0748

*Again, I don't think you really need to memorize this as long as you remember very well where sine and cosine are positive, you can always work out where secant and cosecant are positive and negative.*0756

*For our next example, we're asked to find whether the secant and cosecant functions are odd, even or neither.*0770

*Let's remember what the definition of odd and even are.*0779

*Odd is where f(-x)=-f(x), and that also, by looking at the graph, you can identify odd functions, they have rotational symmetry around the origin.*0784

*Even functions, f(-x)=f(x), and they have mirror symmetry across the y-axis.*0806

*Now, let's look at sec(x), sec(x), actually we have to look at sec(-x) to check whether it's odd or even.*0830

*So, sec(-x), secant remember is 1/cos, 1/cos(-x), cosine is an even function, so this is just 1/cos(x), which is sec(x) again, sec(x) is even.*0840

*Csc(x), well csc(-x), cosecant is 1/sin, so that's 1/sin(-x), but sine's an odd function, so this is 1/-sin(x), which is -csc(x), so cosecant is odd.*0865

*That was just a matter of checking the definitions of odd and even, plugging -x into secant, cosecant, and seeing what we came up with.*0895

*We can also figure it out from the graphs if we remember what those look like.*0904

*Secant, remember ...*0911

*Let me draw a cosine.*0914

*Secant is 1/cos, so that was the one that look like this, that's sec(x) in red there.*0918

*If you look, it has mirror symmetry across the y-axis, which checks that sec(x) is really an even function.*0928

*Mirror symmetry across the y-axis.*0948

*Csc(x), let me draw a quick graph of csc(x).*0955

*Remember, that's based on the graph of sin(x), so start by drawing a graph of sin(x).*0960

*Now, we'll fill in a graph of csc(x), it has asymptotes wherever sine is 0, so that's now our graph of csc(x).*0968

*Clearly, that does not have mirror symmetry across the y-axis, but it does have rotational symmetry around the origin.*0981

*If you spun that around 180 degrees, it would look the same, so it does have rotational symmetry.*0992

*That confirms that cosecant is an odd function.*1005

*We'll try some more examples later.*1017

*You should try working them out yourself, then we'll work them out together.*1019

*We are talking about the trigonometric identities in particular to the Pythagorean identities for (tan) and (cot) of (theta), and (sec) and (cosec).*0000

*Here we are being asked to prove the identity (cot)*^{2} +1 = (cosec)^{2}.0012

*The trick there is to remember the original Pythagorean identity.*0020

*Let me write that down to start with, sin*^{2}x + cos^{2}x = 1, that is the original Pythagorean identity.0024

*That should be very familiar to you because you use that all the time in trigonometry.*0035

*We will start with that and I know I’m trying to find (cot) and (cosec) in this somehow.*0043

*What I’m going to do is divide both sides by sin*^{2}x.0049

*Divide everything here by sin*^{2}x and that gives me 1.0063

*Now (cos)*^{2}/(sin)^{2} that is (cos)/(sin)^{2}, that is (cot)^{2}x.0072

*1/(sin)*^{2} that is 1/(sin), that is (cosec) by definition.0081

*There you have it, that is a pretty quick one.*0090

*We start with the original Pythagorean identity and we just divide both sides by (sin)*^{2}x to make it look exactly like what we are asked for.0093

*I can rearrange terms here and I can get (cot)*^{2}x + 1 is equal to (cosec)^{2}x.0104

*This just comes back to knowing the original Pythagorean identity, if you remember that original Pythagorean identity.*0119

*And if you remember the definitions of sec, tan, cosec, and cot, then you can pretty quickly derive the new one cot*^{2} + 1 = cosec^{2}.0125

*You might not even really need to memorize that one as long as you know the others vey well because you can work it out quickly.*0136

*Hi we are given (sec) of an angle here, 13/12 and we are given the angle in terms of degrees is between 270 and 360 and we are asked to find the tan(theta).*0000

*This is pretty clearly asking us to use the Pythagorean identity 10 *^{2}(theta) +1= sec^{2}(theta).0012

*Sec*^{2}(theta) is we plug in 13/12, that is 13/12^{2} and that is 169/144.0023

*10*^{2}(theta) if we subtract 1 from both sides is 169/144 – 1 which is 169-144/144 which is 25/144.0038

*Tan(theta) if we take the square root of both sides is + or – the square root of 25/144, which is + or -, those are both perfect squares so it comes out neatly 5/12.*0067

*Now the question is, whether it is the positive one or the negative one and we need to figure that out for the problem.*0088

*There is extra information given in the problem that we have not use yet.*0094

*We have not use the fact that (theta) is between 270 and 360 degrees .*0098

*Let me draw our (theta) would be approximately, here is my unit circle, in degrees that is 0, 90, 180, 270, and 360.*0104

*(theta) is between 270 and 360, (theta) is down there somewhere, so (theta) is an angle around there.*0123

*If you remember all students take calculus, that tells you which of these basic functions are positive and negative in each quadrants.*0133

*In the first quadrant they are all positive, in the second quadrant only (sin) is, third quadrant only (tan) is, fourth quadrant only (cos) is.*0145

*Our (theta) is in the fourth quadrant, let me write that as (theta) is in quadrant 4.*0156

*Tan(theta) only cos is positive there, tan is not, so tan(theta) is negative.*0177

*When we are choosing between these positive and negative square root here, we know we have to pick the negative one, -5/12.*0188

*There are really two steps to figuring out how to do this problem, one is to remember the Pythagorean identity tan*^{2}(theta) + 1=sec^{2}(theta).0210

*Then we take the value of sec(theta) that we are given, we plug it in, we work it down through and we try to solve for tan(theta) and we end up taking the square root.*0221

*We get plus or minus in there and we do not know what we want if positive or negative.*0232

*The second step there was to use the information about what quadrant (theta) is in and once we know that (theta) is in quadrant 4, we know that its tan it has to be negative.*0237

*That is from the all student take calculus rule and tan(theta) is negative, we know we need to take the negative square root.*0253

*That is how we get tan(theta) is equal to -5/12.*0260

*That is our lesson on the Pythagorean identity tan*^{2}(theta) + 1= sec^{2}(theta).0265

*These are the trigonometry lectures for www.educator.com.*0271

*OK, welcome back to the trigonometry lectures on educator.com, and today we're talking about the identity, tan*^{2}(x)+1=sec^{2}(x).0000

*You really want to think about this as a kind of companion identity to the main Pythagorean identity, which is that sin*^{2}(x)+cos^{2}(x)=1.0010

*Hopefully, you've really memorized this identity, sin*^{2}(x)+cos^{2}(x)=1, that's one that comes up everywhere in trigonometry.0022

*Then sort of the companion Pythagorean identity to that for secants and tangents is tan*^{2}(x)+1=sec^{2}(x).0034

*There's another related identity, essentially just the same for cotangents and cosecants, is that cot*^{2}(x)+1=csc^{2}(x).0045

*You may wonder how you'll remember this identities.*0058

*For example, how do you remember whether it's tan*^{2}+1=sec^{2}, or maybe it's the other way around, sec^{2}+1=tan^{2}.0062

*Well, really the answer to that lies in the exercises that we're about to do.*0072

*We'll show you how to derive those identities from the original one sin*^{2}+cos^{2}=1.0077

*As long as you can remember sin*^{2}+cos^{2}=1, you'll learn how you can use that to figure out the others and you really won't have to work hard to remember this new Pythagorean identities.0085

*Let's jump right into that.*0099

*The first example here is to prove the identity tan*^{2}(x)+1=sec^{2}(x).0102

*The trick there is to remember the original Pythagorean identity, which is cos*^{2}(x)+sin^{2}(x)=1, that's the original Pythagorean identity.0109

*That's one that you really should memorize and remember throughout all your work with trigonometry.*0127

*What you do to manipulate this into the new identity, is you just divide both sides by cos*^{2}(x).0136

*On the left, we get cos*^{2}(x)/cos^{2}(x) plus, let me write it as (sin(x)/cos(x)^{2}=1/cos^{2}(x).0152

*Then of course, cos*^{2}(x)/cos^{2}(x) is just 1, sin/cos, remember that's the definition of tangent, so this is tan^{2}(x), 1/cos, that's the definition of sec(x), so sec^{2}(x).0173

*You can just rearrange this into tan*^{2}(x)+1=sec^{2}(x).0191

*That shows that this new identity is just a straight consequence of the original Pythagorean identity, so really if you can remember that Pythagorean identity, you can pretty quickly work out this new Pythagorean identity for tangents and cotangents.*0203

*Let's try using the Pythagorean identity for tangents and cotangents.*0224

*In this problem, we're given the tan(θ)=-4.21, and θ is between π/2 and π, we want to find secθ.*0230

*Remembering the Pythagorean identity, tan*^{2}(θ)+1=sec^{2}(θ).0238

*If we plug in what we're given here, tan*^{2}(θ), that's (-4.21^{2})+1=sec^{2}(θ).0246

*Well, 4.21*^{2}, that's something I'm going to workout on my calculator, is 17.7241.0259

*So, 17.7241+1=sec*^{2}(θ), that's 18.7241=sec^{2}(θ).0274

*If I take the square root of both sides, I get sec(θ) is equal to plus or minus the square root of 18.7241.*0290

*Again, I'm going to do that square root on the calculator, and I get approximately 4.33.*0303

*The question here is whether we want the positive or negative square root, and that's where we use the other piece of information given in the problem.*0323

*Θ is in the second quadrant here, θ is between π/2 and π, so θ is somewhere over here.*0332

*Now, sec(θ) remember, is 1/cos(θ).*0348

*In the second quadrant, if you remember All Students Take Calculus.*0353

*In the second quadrant, sine is positive, but cosine is not, cosine is negative.*0358

*That means, sec(θ) is negative.*0365

*θ is in quadrant 2, cos(θ) is negative, sec(θ), because that's 1/(θ), is negative.*0370

*So, sec(θ), we want to take the negative square root there, and we get an approximate value of -4.33.*0394

*That negative is very important.*0411

*The key to that problem is first of all invoking this Pythagorean identity, tan*^{2}+1=sec^{2}.0415

*That's very important to remember.*0423

*We plug in the value that we're given and then we work it through and we'll try to solve for sec(θ).*0426

*We get this plus or minus at the end because we don't know if we want the positive square root or the negative square root.*0431

*That's where we use the fact that θ is in the second quadrant.*0437

*Since θ is in the second quadrant, cos(θ) must be negative, sec(θ), remember, is just 1/cos(θ), must also be negative.*0448

*That's how we know we want the negative square root there, so we get -4.33.*0458

*In our next example, we're given trigonometric identity and we're asked to prove it, (csc(θ)-cot(θ))/(sec(θ)-1) = cot(θ).*0467

*I'm going to start with the left-hand side of this trigonometric identity and I'm going to manipulate it.*0480

*I'm going to try and manipulate it into the right-hand side.*0490

*The left-hand side which is (csc(θ)-cot(θ))/(sec(θ)-1).*0496

*Now, I'm going to do something clever here and it's based on this old principle of whenever you have an (a-b) in the denominator, try multiplying by the conjugate, (a+b)/(a+b).*0511

*If it's the other way around, if you have (a+b), you multiply by the conjugate (a-b)/(a-b).*0524

*The reason you do that is that it makes the denominator (a*^{2}-b^{2}).0530

*We're taking advantage of that old formula from algebra, the difference of squares formula, and a lot of times the (a*^{2}-b^{2}) is something that it'll simplify into something nice.0538

*That's what's going to happen in here.*0549

*We're going to multiply, since we have (sec(θ)-1 in the denominator, by sec(θ)+1.*0552

*What that turns into in the denominator is this (a*^{2}-b^{2}) form, so sec^{2}(θ)-1.0565

*The numerator really doesn't have anything very good happening yet, (csc(θ)-cot(θ))×sec(θ)+1, nothing very good happening in the numerator yet.*0577

*In the denominator, we're going to take advantage of the fact that tan*^{2}(θ)+1=sec^{2}(θ), that's the Pythagorean identity that we're studying today.0593

*If you move that around, you get sec*^{2}(θ)-1=tan^{2}(θ).0607

*That converts the denominator into tan*^{2}(θ), so that's very useful.0614

*In the numerator, we have csc(θ)-cot(θ) and sec(θ)+1.*0623

*I think now I'm going to translate everything into sines and cosines because those will be easier to understand than cosecants and cotangents.*0628

*So, cosecant, remember is 1/sin(θ), minus cotangent is cos(θ)/sin(θ), sec(θ)+1, well, sec(θ) is 1/cos(θ), and tan(θ), if we translate into sines over cosines, that's sin*^{2}(θ)/cos^{2}(θ).0637

*Now, I've got fractions divided by fractions, I think the easiest thing to do here is to bring the denominator, flip it up and multiply it by the numerator.*0670

*We get cos*^{2}(θ)/sin^{2}(θ), that's flipping up the denominator, and then the old numerator is, if I combined the first two terms of the first one, it's (1-cos(θ))/sin(θ).0681

*In the second one, we have 1/cos(θ)+1.*0706

*Now, I think what I'm going to do is I'm going to look at this cosine squared, write it as cos(θ)×cos(θ).*0715

*Then pull one of those cosines over and multiply it by this term, and try to simplify things a little bit that way.*0726

*That will leave me with just one cosine left, still have sin*^{2} in the denominator, (1-cos(θ)/sin(θ)) times, now, cos(θ)×(1/cos(θ) is 1, plus 1×cos(θ).0736

*Now, look, we have (1-cos(θ))×(1+cos(θ)).*0767

*We're going to use this pattern again, the (a-b)×(a+b)=a*^{2}-b^{2}.0772

*If cos(θ)/sin*^{2}(θ), now multiply (1-cos)×(1+cos) gives us (1-cos^{2}(θ)) and sin(θ) in the denominator.0780

*Now, we're going to use the other Pythagorean identity, the original one which are right over here, sin*^{2}(θ)+cos^{2}(θ)=1, if you switched that around 1-cos^{2}(θ)=sin^{2}(θ).0798

*I'll plug that in, cos(θ)/sin(θ), 1-cos*^{2}(θ), now translates into sin^{2}(θ) all divided by sin(θ).0819

*I forgot my squared there, that's from the line above.*0836

*Now, we have some cancellations, sin*^{2}(θ) cancel, we get cos(θ)/sin(θ), but that's cot(θ).0840

*By definition of cotangent, cotangent is defined to be cos/sin.*0852

*That's exactly equal to the right-hand side of the identity.*0858

*When you're proving this trigonometric identities, you pick one side and you multiply it as much as you can.*0864

*You try to manipulate it into the other side.*0870

*It does takes some trial and error.*0874

*I worked this problem out ahead of time, I tried a couple of different things and I finally found a way that gets us through it relatively quickly.*0876

*It's not like you'll always know exactly which path to follow, it takes a little bit of experimentation.*0882

*There are some patterns that you see over and over again, and the two patterns that we really invoke strongly in this one are these algebra pattern where (a-b)×(a+b)=a*^{2}-b^{2}.0888

*Essentially, whenever you see an (a-b) or an (a+b), think about multiplying it by the conjugate, and then taking advantage of that pattern.*0904

*That's what really got us started on the first step here.*0913

*We had a sec(θ)-1, and I thought, okay, let's multiply that by sec(θ)+1, that gives us sec*^{2}-1.0917

*The second pattern that we use there was to invoke these Pythagorean identities, tan*^{2}+1=sec^{2}, and sin^{2}+cos^{2}=1.0927

*We invoked that one here, converting sec*^{2}-1 into tan^{2}.0942

*Then later on, when he had another (a-b)×(a+b), it converted into 1-cos*^{2} and that in turn, by the Pythagorean identity converted into sin^{2}.0947

*It's just a lot of manipulation but you can let your manipulation be kind of guided by these common algebraic identities and these common Pythagorean identities.*0962

*Then you just try to keep manipulating until you get to the other side of the equation.*0973

*We'll try some more examples later on.*0978

*We are trying some more examples of inverse trigonometric functions, now we are asked to identify the domain and range of the arcos function and graph the function.*0000

*Let me start by graphing the cos function itself.*0012

*There is pi/2, pi/ -pi/2, 3pi/2, 2pi.*0027

*What I just graphed there is just cos(theta), I did not graph the arcos function yet and the key thing here is we are going to flip this function around the y=x line.*0039

*We are going to flip it around the line y=x and we want it to be a function after we flip it.*0054

*Right now if we flip it the way it is then it would not be a function because it would not pass the vertical line test.*0059

*We need to cut off just a little piece of the cos function and flip that little piece around and that would give us the arcos function.*0067

*What we are going to do is we are going to cut off the piece from here at 0 to pi.*0077

*We are going to cut off this piece, we are going to cut here to make arcos function.*0091

*The point of doing it like that is if we cut it off that way, we will get something that will pass the vertical line test after we flip it.*0111

*Let me draw that now, let me draw what we will get when we flip it.*0120

*If we just take that little piece that we cut off, it started at 0,1.*0125

*Let me start this one at 1,0, and went down to pi/2,0.*0130

*Run it up to 0 pi/2 and then it went to pi -1.*0141

*There is pi on the y axis now and -1 on the x axis and that is our graph of arcos(x).*0151

*The domain is all the numbers that you can plug into that.*0165

*It is all (x) would be negative -1, less than or equal to (x), less than or equal to 1.*0177

*The way you know that is it is all the numbers that can come out of the cos function.*0182

*Cos always gives you numbers between -1 and 1, those are the numbers that you can plug into arcos.*0188

*Let me emphasize here that the end points -1 and 1 are included, those are less than and equal to sin.*0196

*Back when we studied the arctan function, we have to leave the end points out, -pi/2 and pi/2 for arc tan.*0209

*We left those points out because the function never actually got to those points.*0217

*Here we include -1 and 1 because the function does get to those points.*0225

*The range is all (y) with 0 less than (y), less than or equal to pi.*0231

*0 less than or equal to (y), less than or equal to pi.*0245

*It is all the (y) values that you could get from the arcos function which in turn corresponds to all the values that go into this little piece of the cos function.*0249

*It is all about U’s from 0 to pi.*0262

*If we identified the domain and range and we have got ourselves a graph, we are done with that example.*0266

*Ok one more example of finding the values of arcsin, arcos, and arctan.*0000

*I’m going to start by drawing a unit circle.*0008

*Let me graph the angle that we are going to start with, 4pi/3, there is 0, pi/2, 3pi/2 and 2pi.*0032

*4pi/3 is down here, there is 4pi/3, it is between pi and 3pi/2.*0050

*Sin(4pi/3) is the (y) coordinate, the arcsin is always between –pi/2 and pi/2.*0061

*-pi/2 is at the same place as 3pi/2, we want to find arcsin(sin 4pi/3), we want to find an angle in that range that has the same sin as 4pi/3.*0077

*Sin is the (y) coordinate, let me keep the (y) coordinate fixed and I will go over and find the point over here, that is –pi/3.*0093

*Arcsin(sin4pi/3), an angle that has the same sin as 4pi/3 is –pi/3.*0110

*Make that negative sign a little clear, we have solved the first one.*0125

*Next one is arcos(cos –pi/4), well there is –pi/4 down there.*0131

*We want to find an angle that has the same cos but arcos remember is always between 0 and pi, it is always in the first 2 quadrants.*0144

*We want to find an angle in the first two quadrants that has the same cos as –pi/4.*0158

*Cos is the (x) value we are keeping the (x) value constant and I’m moving up to find an angle in the first two quadrants, there it is at pi/4.*0165

*Arcos(cos –pi/4) is pi/4.*0181

*Finally we have arctan(tan 7pi/6), 7pi/6 is just bigger than pi, there it is down there 7pi/6.*0196

*We want to find an angle that has the same tan as that.*0209

*Remember with tan, if we reflect across the origin we will get an angle that has the same sin and cos except they switched from being negative to positive.*0213

*But they both switched which mean tan(sin/cos) will actually end up the same.*0233

*This angle is pi/6 will have the same tan as 7pi/6, that is good because arctan is forced to be between –pi/2 and pi/2.*0240

*We got this angle pi/6 that has the same tan as what we are looking for 7pi/6.*0259

*The key to solving that problem is to take each one of these angles and find them on the unit circle.*0280

*Figure out what the range of our function is we are looking for, arctan between –pi/2 and pi/2, arcsin in the same range, arcos little different between 0 and pi.*0287

*Then we try to find an angle within that range that has the same sin, cos, or tan, as the angle that we started with.*0302

*That is why I drawn this little dotted lines here on the unit circle, I’m trying to find angles with same cos or same sin, or same tan as what we started with.*0313

*That what gave us our answers –pi/3, pi/4 and pi/6, those are angles that has the same sin, cos, and tan as the ones we are given.*0323

*That was our first lecture on inverse trigonometric functions, these are the trigonometry lectures for www.educator.com.*0334

*Hi, these are the trigonometry lectures for educator.com, and we're here to learn about the inverse trigonometric functions.*0000

*We've already learned about sine, cosine and tangent, and today we're going to learn about the inverse sine, inverse cosine, and inverse tangent, probably better known as arc sine, arc cosine, and arc tangent.*0008

*The arc sine function is also known as the inverse sine function, basically, it's kind of the opposite of the sine function.*0022

*The idea is that you're given a value of x, and you want to find an angle whose sine is that value of x.*0030

*In order to make this work, sines only occur between -1 and 1, so you have to be given a value of x between -1 and 1.*0041

*You're going to try to give an angle between -π/2 and π/2.*0050

*You'll try to give an angle between -π/2 and π/2 that has the given value as a sine.*0060

*If arcsin(x)=θ, what that really means is that the sin(θ)=x.*0078

*Now, there's some unfortunate notation in mathematics which is that arcsin is sometimes written as sin to the negative 1 of x.*0087

*This is very unfortunate because people talk about, for example, sin*^{2}x means (sin(x))^{2}.0095

*You might think that sin*^{-1}(x), would be (sin(x))^{-1}, which would be 1/sin(x).0106

*Now, that's not what it means, the sine inverse of x doesn't mean 1/sin(x), it means arcsin(x).*0119

*This notation really is very ambiguous because this inverse sine notation could mean arcsin(x) or it could mean 1/sin(x), and so this notation is ambiguous because arcsin(x) and 1/sin(x) are not the same.*0129

*The safest thing to do is to not use this notation sin*^{-1} at all.0151

*Let me just say, avoid this notation completely because it is ambiguous, it could be interpreted to mean these two different things that are not equal to each other.*0159

*Instead, it's probably safer to use the notation arcsin(x), which definitely means inverse sin(x), and cannot be confused.*0175

*The arccos(x), the arc cosine function is sort of the opposite of the cosine function.*0187

*You're given a value of x, and you have to find an angle whose cosine is that value of x.*0194

*Again, the value of x you must be given would have to be between -1 and 1, because those are the only values that come up as answers for cosine.*0201

*What you try to do is produce an angle between 0 and π, so there's 0, π/2, π.*0215

*You try to produce an angle between 0 and π that has that value as its cosine.*0229

*Just like we have with sine, there's the problem of this misleading notation cos*^{-1}(x), it could be interpreted as 1/cos(x) or it could be interpreted as arccos(x).0233

*The best thing to do is to avoid using this notation completely, cos*^{-1}(x) is just misleading, it could be interpreted either way.0247

*Try not to use it at all, instead stick to the notation arccos(x).*0263

*Finally, arc tangent is known as the inverse tangent function.*0270

*You're given a value of x, and you want to find an angle whose tangent is x.*0275

*For arc tangent, we're going to try to find the angles between -π/2 and π/2 because that covers all the possible tangents we could get.*0282

*If arctan(x)=θ, that really means that tan(θ)=x.*0297

*Just like with sine and cosine, we have this potentially misleading notation, tan*^{-1}(x), could be interpreted to mean arctan(x) or 1/tan(x).0302

*Those are both reasonable interpretations but they mean two different things.*0314

*Again, let's try to avoid this notation completely because it could mean two completely different things.*0317

*We'll just try not to use that at all when we're talking about inverse tangents we'll say arctan instead of tan*^{-1}.0337

*Let's get started with some examples.*0345

*First off we have to identify the domain and range of the arc sine function, and then graph the function.*0347

*Let's start out with the graph of sin(x), because arcsin is really meant to be an inverse to sin(x).*0354

*I'll start with the graph of sin(x) here.*0362

*Remember that when you're trying to find inverse functions, you take the function and you reflect it across the line y=x.*0372

*In order for something to be a function, it has to pass the vertical line test.*0388

*If you draw a vertical line, you shouldn't cross the graph twice.*0393

*Since we're reflecting across the line y=x, that kind of switches the x's and y's, so we wanted something that will pass the horizontal line test.*0397

*Let me draw this in red.*0408

*We don't want to be able to draw a horizontal line and cross the graph twice.*0411

*As you see, when I've drawn these horizontal lines, we crossed the graph in lots of places.*0415

*We have a problem when we wanted to find the inverse sine function.*0421

*The way we solve that is by not using all of the sine graph, we just cut off a piece of the sine graph that will pass the horizontal line test.*0425

*I'm going to cut off a piece of the sine graph from -π/2 to π/2.*0439

*If you just look at this portion of the sine graph, you see that it passes the horizontal line test.*0448

*That means we can take the inverse just at that part of the sine function.*0459

*Let me draw what that looks like when we reflect it.*0465

*We're just taking this thick piece here, and I'm going to reflect that across that line y=x.*0472

*Now, the notations that I had on the y and x axis are going to switch, it goes from now -1 to 1 on the x-axis, and -π/2 to π/2 on the y-axis.*0489

*I can't keep going with this, I can't draw the rest of the graph because if I do draw any more, I'm going to get something that fails the vertical line test, it won't be a function anymore.*0510

*This is the entire arcsin function.*0521

*The domain here, it's all the numbers that you can plug into the arcsin, that's all values of x with -1 less than or equal to x, less than or equal to +1.*0532

*We can't plug any other values of x into arcsin, and that's really because in the other direction the only values that come out of the sine function are between -1 and 1.*0560

*The only values that you can plug into the arcsin function are between -1 and 1.*0571

*The range, the numbers that come out of the arcsin, all values of x between -π/2 and π/2.*0580

*Let me write that as y because when we think of those are the values coming out of the arcsin function.*0596

*Those are all the y values you see here and you see that the smallest y-value is -π/2 and the biggest value we see is π/2.*0605

*Arcsin takes in a number between -1 or 1, gives you a number between -π/2 and π/2 and its graph looks like a sort of chopped of piece of the sine graph.*0615

*Remember, the reason we had to chop it off is to get a piece of the sine graph that would satisfy the horizontal line test, so that the arcsin graph satisfies the vertical line test and really is a function.*0629

*In our second example here, we have to find the arcsin of the sin(2π/3) and then some similar values for our cosine and our tangent.*0647

*The first thing to do here is really to figure out where we are on the unit circle.*0663

*There's 0, π/2, π, 3π/2 and 2π.*0675

*Let's figure out where each of this angle is on the unit circle.*0685

*2π/3 is over here, there's 2π/3.*0691

*Remember, arcsin is always between -π/2 and π/2.*0698

*Arcsin is between -π/2 and π/2, so there's -π/2 down here.*0712

*We want to find an angle between -π/2 and π/2.*0717

*On the right-hand side that has the same sine as 2π/3.*0724

*Well, sine is the y-value so we want something that has the same y-value as 2π/3.*0729

*That angle right there has the same y-coordinate as 2π/3, that's π/3.*0741

*From the graph, we see that π/3 has the same sine as sin(2π/3).*0750

*So, arcsin(2π/3) is π/3.*0771

*It's an angle whose sine is the sin(2π/3).*0780

*Let's try the next one, -5π/6.*0785

*-π is over there, so 5π/6 in the negative direction puts you over there, there's -5π/6.*0792

*We want an angle whose cosine is the same as the cosine(-5π/6).*0809

*Cosine is the x-value, so we want an angle that has the same x-value as the one we just found.*0818

*There it is right there, 5π/6.*0830

*That's between 0 and π, which is the range for the arccos function, it's always between 0 an π.*0838

*That means that arccos of the cos(-5π/6) is 5π/6, that's an angle between 0 and π that has the right cosine.*0850

*Finally, we want to find the arctan of tan(3π/4).*0875

*3π/4 is over there.*0882

*There's 3π/4.*0890

*We want to find an angle that has the same tangent as that one.*0893

*Remember, arctan is forced to be between π/2 and -π/2.*0898

*We want to find an angle between π/2 and -π/2 that has the same tangent as 3π/4.*0909

*If we go straight across the origin there, we started with an angle whose cosine is negative and sine is positive.*0920

*This angle right here has a positive cosine and negative sine, so it'll end up having the same tangent, and that angle is -π/4.*0934

*That's an angle inside the range we want that has the same tangent.*0946

*Arctan of tan(3π/4) is -π/4.*0952

*Those are really quite tricky.*0967

*What we're being asked to do here in each case is we're given an angle, for example 2π/3.*0969

*We want to find arcsin of sin(2π/3).*0977

*We look at the sine of 2π/3, and then we want to find another angle that has that sine but it has to be in the specified range for arcsin.*0981

*We're really trying to find what's an angle between -π/2 and π/2 whose sine is the same as sin(2π/3).*0993

*That's why we did this reversal to find the angle π/3 that has the same sine as the sin(2π/3).*1003

*That was kind of the same process of all three of them, finding angles that have the same cosine as -5π/6, but is in the specified range, finding an angle that has the same tangent as 3π/4 but is in the specified range.*1017

*Our third example here, we're asked to identify the domain and range of the arctan function, and graph the function.*1037

*Remember when we graphed arcsin, we started out with a graph of sine, so let me start out here with a graph of tangent.*1044

*I'll graph it in blue.*1054

*Has asymptotes of π/2, and then it starts repeating itself.*1057

*It has period π.*1065

*This is 0, π/2, -π/2, and that's π, 3π/2.*1084

*What I graphed in blue there is tan(θ).*1096

*Now, I want to take this graph and I want to flip it around the line y=x.*1102

*Let me graph the line y=x.*1107

*I want to get something that it will be a function, it has to pass the vertical line test after I flip it.*1112

*That means it has to pass the horizontal line test before I flip it.*1120

*Just like with sine, the tangent function fails the horizontal line test badly.*1124

*You can draw a horizontal lines and intersect it in lots of places.*1132

*Just like with sine, we're going to cut off part of the tangent function and use that to form the arctan function.*1137

*We're going to cut off just one of these curves.*1143

*I'll cut it off these asymptotes at -π/2 and π/2.*1151

*I'll just take this part of the tangent function and I'll flip it around to make the arctan function.*1156

*I'll do this in red.*1171

*Arctan(x).*1179

*I'm going to flip this around just that one main branch of the tangent function.*1181

*Since the tangent function had vertical asymptotes, the arctan function is going to have horizontal asymptotes at -π/2 and π/2.*1192

*Those are my asymptotes right there at -π/2 and π/2.*1210

*We're also asked to identify the domain and range.*1216

*The domain means what numbers can we plug in to arctan(x).*1219

*Domain is all values of x because we can plug any number in on the x-axis here and we'll get an arctan value.*1228

*-infinity less than x less than infinity.*1243

*The range is all x with, okay, let me write this as y because these are values on the y-axis.*1248

*All y with -π/2 less than y less than π/2.*1264

*Let me emphasize here that π/2 and -π/2 are not included, π/2 themselves are not in the range.*1269

*These inequalities, they're not less than or equal to they're strictly less than.*1291

*The reason for that is that the arctan function never quite gets to -π/2 or π/2, it goes down close -π/2 and it goes up close to π/2 but it never quite gets there.*1297

*The reason for that, if you kind of look back at the tangent function, was that these asymptotes never quite get to -π/2 or π/2.*1312

*You can't take the tangent of -π/2 or π/2, because you're trying to divide by zero back in the tangent function.*1323

*To understand this problem really, you have to remember what the graph of tan(θ) looks like.*1333

*Then you take its one main branch and you flip it over, and you get the graph of arctan(x), and the domain and range just come back to any things in the domain.*1339

*The range can go from π/2 to -π/2, but you don't include those end points.*1354

*The asymptotes are the horizontal lines at -π/2 and π/2.*1362