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Most of us are unaware of how trigonometry factors into our daily lives. Dr. William Murray is very aware of the importance of trigonometry and he is ready to help you master it all. His in-depth course covers everything from Functions, Identities, and Complex/Polar Coordinates to Word Problems and the Law of Sines.  This course meets or exceeds all state standards and is essential to those having trouble with trigonometry in any setting. Professor Murray received his Ph.D from UC Berkeley, B.S. from Georgetown University, and has been teaching in the university setting for 10+ years.

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I. Trigonometric Functions
  Angles 39:05
   Intro 0:00 
   Degrees 0:22 
    Circle is 360 Degrees 0:48 
    Splitting a Circle 1:13 
   Radians 2:08 
    Circle is 2 Pi Radians 2:31 
    One Radian 2:52 
    Half-Circle and Right Angle 4:00 
   Converting Between Degrees and Radians 6:24 
    Formulas for Degrees and Radians 6:52 
   Coterminal, Complementary, Supplementary Angles 7:23 
    Coterminal Angles 7:30 
    Complementary Angles 9:40 
    Supplementary Angles 10:08 
   Example 1: Dividing a Circle 10:38 
   Example 2: Converting Between Degrees and Radians 11:56 
   Example 3: Quadrants and Coterminal Angles 14:18 
   Extra Example 1: Common Angle Conversions 8:02 
   Extra Example 2: Quadrants and Coterminal Angles 7:14 
  Sine and Cosine Functions 43:16
   Intro 0:00 
   Sine and Cosine 0:15 
    Unit Circle 0:22 
    Coordinates on Unit Circle 1:03 
    Right Triangles 1:52 
    Adjacent, Opposite, Hypotenuse 2:25 
    Master Right Triangle Formula: SOHCAHTOA 2:48 
   Odd Functions, Even Functions 4:40 
    Example: Odd Function 4:56 
    Example: Even Function 7:30 
   Example 1: Sine and Cosine 10:27 
   Example 2: Graphing Sine and Cosine Functions 14:39 
   Example 3: Right Triangle 21:40 
   Example 4: Odd, Even, or Neither 26:01 
   Extra Example 1: Right Triangle 4:05 
   Extra Example 2: Graphing Sine and Cosine Functions 5:23 
  Sine and Cosine Values of Special Angles 33:05
   Intro 0:00 
   45-45-90 Triangle and 30-60-90 Triangle 0:08 
    45-45-90 Triangle 0:21 
    30-60-90 Triangle 2:06 
   Mnemonic: All Students Take Calculus (ASTC) 5:21 
    Using the Unit Circle 5:59 
    New Angles 6:21 
    Other Quadrants 9:43 
    Mnemonic: All Students Take Calculus 10:13 
   Example 1: Convert, Quadrant, Sine/Cosine 13:11 
   Example 2: Convert, Quadrant, Sine/Cosine 16:48 
   Example 3: All Angles and Quadrants 20:21 
   Extra Example 1: Convert, Quadrant, Sine/Cosine 4:15 
   Extra Example 2: All Angles and Quadrants 4:03 
  Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D 52:03
   Intro 0:00 
   Amplitude and Period of a Sine Wave 0:38 
    Sine Wave Graph 0:58 
    Amplitude: Distance from Middle to Peak 1:18 
    Peak: Distance from Peak to Peak 2:41 
   Phase Shift and Vertical Shift 4:13 
    Phase Shift: Distance Shifted Horizontally 4:16 
    Vertical Shift: Distance Shifted Vertically 6:48 
   Example 1: Amplitude/Period/Phase and Vertical Shift 8:04 
   Example 2: Amplitude/Period/Phase and Vertical Shift 17:39 
   Example 3: Find Sine Wave Given Attributes 25:23 
   Extra Example 1: Amplitude/Period/Phase and Vertical Shift 7:27 
   Extra Example 2: Find Cosine Wave Given Attributes 10:27 
  Tangent and Cotangent Functions 36:04
   Intro 0:00 
   Tangent and Cotangent Definitions 0:21 
    Tangent Definition 0:25 
    Cotangent Definition 0:47 
   Master Formula: SOHCAHTOA 1:01 
    Mnemonic 1:16 
   Tangent and Cotangent Values 2:29 
    Remember Common Values of Sine and Cosine 2:46 
    90 Degrees Undefined 4:36 
   Slope and Mnemonic: ASTC 5:47 
    Uses of Tangent 5:54 
    Example: Tangent of Angle is Slope 6:09 
    Sign of Tangent in Quadrants 7:49 
   Example 1: Graph Tangent and Cotangent Functions 10:42 
   Example 2: Tangent and Cotangent of Angles 16:09 
   Example 3: Odd, Even, or Neither 18:56 
   Extra Example 1: Tangent and Cotangent of Angles 2:27 
   Extra Example 2: Tangent and Cotangent of Angles 5:02 
  Secant and Cosecant Functions 27:18
   Intro 0:00 
   Secant and Cosecant Definitions 0:17 
    Secant Definition 0:18 
    Cosecant Definition 0:33 
   Example 1: Graph Secant Function 0:48 
   Example 2: Values of Secant and Cosecant 6:49 
   Example 3: Odd, Even, or Neither 12:49 
   Extra Example 1: Graph of Cosecant Function 4:58 
   Extra Example 2: Values of Secant and Cosecant 5:19 
  Inverse Trigonometric Functions 32:58
   Intro 0:00 
   Arcsine Function 0:24 
    Restrictions between -1 and 1 0:43 
    Arcsine Notation 1:26 
   Arccosine Function 3:07 
    Restrictions between -1 and 1 3:36 
    Cosine Notation 3:53 
   Arctangent Function 4:30 
    Between -Pi/2 and Pi/2 4:44 
    Tangent Notation 5:02 
   Example 1: Domain/Range/Graph of Arcsine 5:45 
   Example 2: Arcsin/Arccos/Arctan Values 10:46 
   Example 3: Domain/Range/Graph of Arctangent 17:14 
   Extra Example 1: Domain/Range/Graph of Arccosine 4:30 
   Extra Example 2: Arcsin/Arccos/Arctan Values 5:40 
  Computations of Inverse Trigonometric Functions 31:08
   Intro 0:00 
   Inverse Trigonometric Function Domains and Ranges 0:31 
    Arcsine 0:41 
    Arccosine 1:14 
    Arctangent 1:41 
   Example 1: Arcsines of Common Values 2:44 
   Example 2: Odd, Even, or Neither 5:57 
   Example 3: Arccosines of Common Values 12:24 
   Extra Example 1: Arctangents of Common Values 5:50 
   Extra Example 2: Arcsin/Arccos/Arctan Values 8:51 
II. Trigonometric Identities
  Pythagorean Identity 19:11
   Intro 0:00 
   Pythagorean Identity 0:17 
    Pythagorean Triangle 0:27 
    Pythagorean Identity 0:45 
   Example 1: Use Pythagorean Theorem to Prove Pythagorean Identity 1:14 
   Example 2: Find Angle Given Cosine and Quadrant 4:18 
   Example 3: Verify Trigonometric Identity 8:00 
   Extra Example 1: Use Pythagorean Identity to Prove Pythagorean Theorem 3:32 
   Extra Example 2: Find Angle Given Cosine and Quadrant 3:55 
  Identity Tan(squared)x+1=Sec(squared)x 23:16
   Intro 0:00 
   Main Formulas 0:19 
    Companion to Pythagorean Identity 0:27 
    For Cotangents and Cosecants 0:52 
    How to Remember 0:58 
   Example 1: Prove the Identity 1:40 
   Example 2: Given Tan Find Sec 3:42 
   Example 3: Prove the Identity 7:45 
   Extra Example 1: Prove the Identity 2:22 
   Extra Example 2: Given Sec Find Tan 4:34 
  Addition and Subtraction Formulas 52:52
   Intro 0:00 
   Addition and Subtraction Formulas 0:09 
    How to Remember 0:48 
   Cofunction Identities 1:31 
    How to Remember Graphically 1:44 
    Where to Use Cofunction Identities 2:52 
   Example 1: Derive the Formula for cos(A-B) 3:08 
   Example 2: Use Addition and Subtraction Formulas 16:03 
   Example 3: Use Addition and Subtraction Formulas to Prove Identity 25:11 
   Extra Example 1: Use cos(A-B) and Cofunction Identities 7:54 
   Extra Example 2: Convert to Radians and use Formulas 11:32 
  Double Angle Formulas 29:05
   Intro 0:00 
   Main Formula 0:07 
    How to Remember from Addition Formula 0:18 
    Two Other Forms 1:35 
   Example 1: Find Sine and Cosine of Angle using Double Angle 3:16 
   Example 2: Prove Trigonometric Identity using Double Angle 9:37 
   Example 3: Use Addition and Subtraction Formulas 12:38 
   Extra Example 1: Find Sine and Cosine of Angle using Double Angle 6:10 
   Extra Example 2: Prove Trigonometric Identity using Double Angle 3:18 
  Half-Angle Formulas 43:55
   Intro 0:00 
   Main Formulas 0:09 
    Confusing Part 0:34 
   Example 1: Find Sine and Cosine of Angle using Half-Angle 0:54 
   Example 2: Prove Trigonometric Identity using Half-Angle 11:51 
   Example 3: Prove the Half-Angle Formula for Tangents 18:39 
   Extra Example 1: Find Sine and Cosine of Angle using Half-Angle 7:16 
   Extra Example 2: Prove Trigonometric Identity using Half-Angle 3:34 
III. Applications of Trigonometry
  Trigonometry in Right Angles 25:43
   Intro 0:00 
   Master Formula for Right Angles 0:11 
    SOHCAHTOA 0:15 
    Only for Right Triangles 1:26 
   Example 1: Find All Angles in a Triangle 2:19 
   Example 2: Find Lengths of All Sides of Triangle 7:39 
   Example 3: Find All Angles in a Triangle 11:00 
   Extra Example 1: Find All Angles in a Triangle 5:10 
   Extra Example 2: Find Lengths of All Sides of Triangle 4:18 
  Law of Sines 56:40
   Intro 0:00 
   Law of Sines Formula 0:18 
    SOHCAHTOA 0:27 
    Any Triangle 0:59 
    Graphical Representation 1:25 
    Solving Triangle Completely 2:37 
   When to Use Law of Sines 2:55 
    ASA, SAA, SSA, AAA 2:59 
    SAS, SSS for Law of Cosines 7:11 
   Example 1: How Many Triangles Satisfy Conditions, Solve Completely 8:44 
   Example 2: How Many Triangles Satisfy Conditions, Solve Completely 15:30 
   Example 3: How Many Triangles Satisfy Conditions, Solve Completely 28:32 
   Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely 8:01 
   Extra Example 2: How Many Triangles Satisfy Conditions, Solve Completely 15:11 
  Law of Cosines 49:05
   Intro 0:00 
   Law of Cosines Formula 0:23 
    Graphical Representation 0:34 
    Relates Sides to Angles 1:00 
    Any Triangle 1:20 
    Generalization of Pythagorean Theorem 1:32 
   When to Use Law of Cosines 2:26 
    SAS, SSS 2:30 
   Heron's Formula 4:49 
    Semiperimeter S 5:11 
   Example 1: How Many Triangles Satisfy Conditions, Solve Completely 5:53 
   Example 2: How Many Triangles Satisfy Conditions, Solve Completely 15:19 
   Example 3: Find Area of a Triangle Given All Side Lengths 26:33 
   Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely 11:05 
   Extra Example 2: Length of Third Side and Area of Triangle 9:17 
  Finding the Area of a Triangle 27:37
   Intro 0:00 
   Master Right Triangle Formula and Law of Cosines 0:19 
    SOHCAHTOA 0:27 
    Law of Cosines 1:23 
   Heron's Formula 2:22 
    Semiperimeter S 2:37 
   Example 1: Area of Triangle with Two Sides and One Angle 3:12 
   Example 2: Area of Triangle with Three Sides 6:11 
   Example 3: Area of Triangle with Three Sides, No Heron's Formula 8:50 
   Extra Example 1: Area of Triangle with Two Sides and One Angle 2:54 
   Extra Example 2: Area of Triangle with Two Sides and One Angle 6:48 
  Word Problems and Applications of Trigonometry 34:25
   Intro 0:00 
   Formulas to Remember 0:11 
    SOHCAHTOA 0:15 
    Law of Sines 0:55 
    Law of Cosines 1:48 
    Heron's Formula 2:46 
   Example 1: Telephone Pole Height 4:01 
   Example 2: Bridge Length 7:48 
   Example 3: Area of Triangular Field 14:20 
   Extra Example 1: Kite Height 4:36 
   Extra Example 2: Roads to a Town 10:34 
  Vectors 46:42
   Intro 0:00 
   Vector Formulas and Concepts 0:12 
    Vectors as Arrows 0:28 
    Magnitude 0:38 
    Direction 0:50 
    Drawing Vectors 1:16 
    Uses of Vectors: Velocity, Force 1:37 
    Vector Magnitude Formula 3:15 
    Vector Direction Formula 3:28 
    Vector Components 6:27 
   Example 1: Magnitude and Direction of Vector 8:00 
   Example 2: Force to a Box on a Ramp 12:25 
   Example 3: Plane with Wind 18:30 
   Extra Example 1: Components of a Vector 2:54 
   Extra Example 2: Ship with a Current 13:13 
IV. Complex Numbers and Polar Coordinates
  Polar Coordinates 67:35
   Intro 0:00 
   Polar Coordinates vs Rectangular/Cartesian Coordinates 0:12 
    Rectangular Coordinates, Cartesian Coordinates 0:23 
    Polar Coordinates 0:59 
   Converting Between Polar and Rectangular Coordinates 2:06 
    R 2:16 
    Theta 2:48 
   Example 1: Convert Rectangular to Polar Coordinates 6:53 
   Example 2: Convert Polar to Rectangular Coordinates 17:28 
   Example 3: Graph the Polar Equation 28:00 
   Extra Example 1: Convert Polar to Rectangular Coordinates 10:01 
   Extra Example 2: Graph the Polar Equation 10:53 
  Complex Numbers 35:59
   Intro 0:00 
   Main Definition 0:07 
    Number i 0:23 
    Complex Number Form 0:33 
   Powers of Imaginary Number i 1:00 
    Repeating Pattern 1:43 
   Operations on Complex Numbers 3:30 
    Adding and Subtracting Complex Numbers 3:39 
    Multiplying Complex Numbers 4:39 
    FOIL Method 5:06 
    Conjugation 6:29 
   Dividing Complex Numbers 7:34 
    Conjugate of Denominator 7:45 
   Example 1: Solve For Complex Number z 11:02 
   Example 2: Expand and Simplify 15:34 
   Example 3: Simplify the Powers of i 17:50 
   Extra Example 1: Simplify 4:37 
   Extra Example 2: All Complex Numbers Satisfying Equation 10:00 
  Polar Form of Complex Numbers 40:43
   Intro 0:00 
   Polar Coordinates 0:49 
    Rectangular Form 0:52 
    Polar Form 1:25 
    R and Theta 1:51 
   Polar Form Conversion 2:27 
    R and Theta 2:35 
    Optimal Values 4:05 
    Euler's Formula 4:25 
   Multiplying Two Complex Numbers in Polar Form 6:10 
    Multiply r's Together and Add Exponents 6:32 
   Example 1: Convert Rectangular to Polar Form 7:17 
   Example 2: Convert Polar to Rectangular Form 13:49 
   Example 3: Multiply Two Complex Numbers 17:28 
   Extra Example 1: Convert Between Rectangular and Polar Forms 6:48 
   Extra Example 2: Simplify Expression to Polar Form 7:48 
  DeMoivre's Theorem 57:37
   Intro 0:00 
   Introduction to DeMoivre's Theorem 0:10 
    n nth Roots 3:06 
   DeMoivre's Theorem: Finding nth Roots 3:52 
    Relation to Unit Circle 6:29 
    One nth Root for Each Value of k 7:11 
   Example 1: Convert to Polar Form and Use DeMoivre's Theorem 8:24 
   Example 2: Find Complex Eighth Roots 15:27 
   Example 3: Find Complex Roots 27:49 
   Extra Example 1: Convert to Polar Form and Use DeMoivre's Theorem 7:41 
   Extra Example 2: Find Complex Fourth Roots 14:36 

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 angle0047

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 radiance0192

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 there0174

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 degrees0050

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 3/30113

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 x3, y=x3 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=x2 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=x3, but it's not the same as the graph of y=x3.0609

One big difference is that tan(x) has asymptotes at π/2 and -π/2, and of course y=x3 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 20863

-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, (x2-x1)2+(y2-y1)2, you add those together and you take the square root of the whole thing.0365

That's the distance formula.0379

Here, the x2 and the x1 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 d2=(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, d2=(cos(a)-1)2+(sin2(a-b)).0566

What I'm going to do is look at these two different expressions here for d2.0590

Well, they're both describing the same d2, 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+(sin2(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=a2-2ab+b2.0659

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 cos2(a)-2cos(a)cos(b)+cos2(b)+sin2(a)-2sin(a)sin(b)+sin2(b)=cos2(a-b)-2×1×cos(a-b)+sin2(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, cos2(a) and sin2(a), that gives me 1-2cos(a)cos(b).0735

Now I have a cos2(b) and a sin2(b), so that's another 1-2sin(a)sin(b), is equal to, now look, cos2(a-b) and sin2(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)-12, I got cos2(a-b)-2cos(a-b), then there should be +12, 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 sin2(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, sin2+cos2=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 later2004

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 X2.0156

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 X2) / (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 cos2(x)-sin2(x).0082

Now, there's two other ways that you often see this formula written as 2cos2(x)-1, and 1-2sin2(x).0096

Those might look different but actually you can figure them out very quickly, or check them very quickly, because 2cos2(x)-1 is 2cos2(x) minus, now remember 1 is the same as sin2(x)+cos2(x).0106

If you work with that a little bit, you have 2cos2-cos2.0126

That's just a single cos2(x)-sin2(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 sin2+cos2, 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, sin2+cos2=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 cos2(x)-sin2(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 cos2(π/3)-sin2(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)=2cos2(x)-1.0357

It was also supposed to be equal to 1-2sin2(x).0370

We're going to check out each one of those.0376

Cos(2π/3), using those other formulas, is equal to 2cos2(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, 1-2sin2(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 22 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+tan2(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+tan2, that's 1+sin2(x)/cos2(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 cos2(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 cos2, that gives me cos(x) in the numerator.0665

In the bottom, I have cos2 times 1+sin2(x) over cos2, that gives me 1×cos2 is cos2(x), plus the cos2(x) cancels with the denominator sin2(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, cos2+sin2(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 cos2(x).0736

Then I start noticing some formulas that I recognized, 2sin(x)×cos(x) is a double angle formula, and cos2(x)+sin2(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 cos2(x)-sin2(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 cos2(x), and I'll divide the bottom by cos2(x).0866

We're dividing top and bottom by cos2(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 cos2, we get 2sin(x)/cos(x).0882

In the bottom, we're dividing everything by cos2(x), we get 1-sin2(x)/cos2(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-sin2(x)/cos2(x) is tan2(x).0918

Our formula, our double angle formula for tangent is tan(2x)=2tan(x)/(1-tan2(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-tan2(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 cos2(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 sin2 + cos2 = 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 sin2 + cos2 = 1.0336

Let us work those out, sin2(pi/8) + cos2(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 root2, so they cancelled each other we get ((2 – root 2)/denominator 22 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 sin2(pi/8) + cos2(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 sin2+cos2=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 a2-2ab+b2, 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 sin2+cos2=1.0602

Sin2(15)+cos2(15), sin2(15) was square root of 2 minus root 3/2.0613

We're going to square that.0627

Plus cos2(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 sin2(15)+cos2(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 a2-b2.0781

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 cos2(1/2 x)+2sin(1/2 x)×cos(1/2 x)+ sin2(1/2 x).0848

We can invoke this difference of squares formula.0873

We get cos2(1/2 x)-sin2(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)=cos2(θ)-sin2(θ).0901

Now, look at what we have here.0922

There's several good things that are going to happen.0923

First of all, cos2 and sin2, 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 cos2 of something minus sin2 of something, and I know that cos2 of something minus sin2 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, sin2+cos2 gives you 1.1046

Secondly, 2 sine of something cosine of something, that's the double angle formula for sine.1051

Then, cos2 of something minus sin2 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)=a2-b2.1282

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-cos2(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-cos2(x), that's something that should set off some warning bells in your brain.1389

Certainly doesn't mind, because I remember that sin2+cos2=1, the Pythagorean identity.1398

If you move that around, if you see 1-cos2, that's equal to sin2.1407

So, 1-cos2(x)=sin2(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

x2 + 32 = 72, so x2 + 9 = 49.0096

X2 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 1.32 + 2.72 and that should be approximately equal to 1.32 + 2.72.0168

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

h2=32+42, 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 62+7.22 gives me 87.8, which is approximately equal to 9.32.0623

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

h2=22+52, 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, c2=(a2) +(( b2 - (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 162= (302+342) – 2(30)(34) x cos(C).0154

(162)=256, (302=900)+(342=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, (c2)=(a2+b2)-((2ab cos (C)).0328

I will work through and I will solve for (C), that is (302)=(162+342)-((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

c2 is (342)=a2is (162)+b2 is now (302)- 2(16)(30) cos(C).0491

342 is 1156, 162 is 256, 302 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 162 + 302 = 342.0558

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 (c2)=(a2+b2) – ((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, (c2=(82 is 64), (162 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 c2=a2+b2-2abcos(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 c2=a2+b2, 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, c2=a2+b2.0115

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 c2=a2+b2-2abcos(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 c2=a2+b2-2abcos(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, a2 would be 9, b2 would be 16 because 42 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, c2 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 c2=a2+b2-2abcos(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 102 equals, a and b is 52+72-2×5×7×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 c2=a2+b2-2abcos(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 c2, that's 49 is equal to a2, that's 100, plus b2 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

c2 is 25, equals 49, 72, plus b2 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, c2 = a2 + b2 - ((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), c2 = 25 that is 52 + 62 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 c2 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, c2=a2+b2-2abcos(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, c2=a2+b2.0124

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 c2=a2+b2-2abcos(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, 142=72+92-2×7×9×cos(C).0625

I'll do a little algebra on that.0646

142 is 196, 72 is 49, 92 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 c2=a2+b2-2abcos(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, c2=a2+b2-2abcos(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 c2, I don't know that yet, a2 is 1602, plus b2 is 2802, 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

1602=25600, 2802=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 c2, 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 South0150

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 (x2 + y2).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 (x2 + y2), 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 x2+y2.0197

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 x2+y2, which is the square root of -32+42.0518

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 x2 + 52 = 132.0044

That is x2 + 25 =169, x2 = 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 neither0113

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 x3.0299

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 x3,0310

(-1)3 is -1, so that's just -x3, and that's negative of the original f(x).0326

For x3, 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)=x3, 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 x3, x to the third, because 3 is an odd number and x3 is an odd function.0417

Something, kind of, has that property that x3 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 x2.0456

Well, f(-x) is equal to, you plugin -x into the function, so (-x)2 well that's just the same as x2, 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 x2, 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 x2.0494

If you graph those, let me graph x2 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 either0590

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, h2 = 32 + 42, 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 adjacent1381

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 x2+y2.0149

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=x2+y2.0160

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 x2+y2.0465

Let me remind you of those formulas up here because we're going to use them quite a bit, x2+y2.0472

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 42+32.0628

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 22+52, 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 x2+y2.0995

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 r2=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 x2+y2, r2=x2+y2.2468

With this nice new version of the equation r2=2rsin(θ), I can write that as x2+y2.2480

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 x2+y2-2y=0.2503

I like to simplify that y2-2y.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 y2-22.2535

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 -12=1.2550

Add that to both sides.2555

The point of that is that now we have a perfect square.2558

We have y2-2y+1, that's (y-1)2=1.2562

This is actually an equation I recognize.2571

Remember that if you have x2+y2 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 r2, on the right, we get 2rsin(θ).2755

The point of that is that the rsin(θ) converts into y, and the x2+y2 is what you get from r2.2760

That of course comes from this, r2=x2+y2.2770

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 x2 - y(i)2 but i2 is -1 so this is really x2 + y2.0036

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 32 + 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), 32 + 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/(52) that is 23 + i/(52).0144

Now remember that (a + b)2)is a2 + 2ab + b2.0166

232 is 529 + 2ab 2(23)(i)that is 46i + i2 that is -1, then 52 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 x2 + y2 formula, that is why I get 32 + 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 = a2 + 2ab + b2.0262

That is how I get to this line remembering that i2 is -1 and then I simplified it down to get the answer.0270

Finally we have to find all complex numbers satisfying z2= 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 x2 + 2xy(i) + yi2 but that is the same as -y2 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 x2 -y2=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 x2 - y2=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 x2 - (2/x)2 = 3.0121

Now I narrowed it down to one equation and I will try to solve for that on variable x2 - 4/x2 = 3.0148

We got to multiply both sides here by x2 to clear my denominator.0160

I get( x4 - 4) =3(x) 2, I will move the 3(x) 2 over so I get x4 - 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 x2 and x4.0191

I’m going to make a little substitution, I’m going to let w=x2 and the point is that is we have not done yet x4=w2 - 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 x2=4 or x2=-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 x2= -1 since (x) is a real number.0265

That x2=-1 really does not make sense and it must be x2=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 x2=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, z2 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 22 + ((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 a2 + 2ab + b2.0443

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=x2, 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 x2= something or x2= 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 x2=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 i2=-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 i2 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 i0, just like anything else to the 0, we define to be 1, we say 10 is 1.0060

The first power is i, i2=-1, i3=i2×i, remember i2=-1, so i3=-i, i4 remember is i2×i2=-1×-1, i4=1.0072

Since we've come back to 1, if we multiply on one more power of i, i5 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 i2.0311

If you simplify that, you get x×a, the ybi2, remember i2 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 x2-(yi)2.0401

But x2, remember i2 is -1, this turns into x2+y2.0421

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 x2+y2 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 a2+b2.0485

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)=a2+b2.0500

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 -ybi2.0517

If we simplify that and collect the real terms, remember i2 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)/(a2+b2), that's a real number, then ((ya-xb)/(a2+b2))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 a2+b2 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 12-(2i)2 but since i2=-1, this is just +22.0798

Remember that's because i2=-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 -14i2, but i2 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+22.0905

The reason it's plus is because we have an i2 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+2i3.0935

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=a2+b2+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 +8i2.1003

The 8i2 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, i1=i, i2=-1, i3=1, after that they start repeating.1084

i3=-i, i4=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

i19, 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 i3 because we went down.1168

Essentially what we did was we went down by 4 at a time, and i3 I remember from my pattern over here is -i.1171

i33, if you're counting by multiples of 4, you have 4, 8, 12, 16, 20, 24, 28, 32, 33 is one more, that's i1.1183

It cycles every 4 powers of i.1208

This i1 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 i3, then you remember i3=-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 x2 of x2 + y2.0048

(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), 2e5pi/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 esin(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 (x2 + y2).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 6e5pi/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(x2 + y2) 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 epi/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 27, now pi 4(i).0123

e to 1 power raised to another power, you just multiply the exponents, that just turns into e7th(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 21/2 to the 7th power and then you could write that as 27 ½.0165

27 ½ is the same as 2, 7 ½ is the same as 3½.0178

That is the same as 2 cubed x 2, law of exponents there and 2 cubed is 8, 21 ½ is root 2 there.0186

Root (27 ) is 8 (root 2), so this whole thing turns into 8 (root 2) x e27 pi/4(i).0199

I like to expand out e7 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 epi/4(i) and we have to raise that up to the 7th power.0385

We got (root 2) 7 and epi/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 21/2 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 e7pi/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.0085

That represents the polar coordinates of the same point.0093

reiθ, sometimes people write it as reθi.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.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 x2+y2.0147

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 as x+yi.0265

I'll write that as iy.0277

Polar form is re, 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=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=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, r1×e1, it's got an r and a θ, and r2×e2.0375

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 r1×r2.0390

Remember the laws of exponents xa×xb=xa+b.0399

Here, we have x or e1×e2, you add the exponents, iθ1+iθ1, just gives you i(θ1)+θ2).0409

You add the exponents there.0426

You end up just multiplying the r's and adding the angles θ12 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 x2+y2.0454

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 x2, negative root 3 squared is just 3, +y2 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 r2e, 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 x2, 6 root 2 squared is 36, times 2 is 72, +y2 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=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.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 x2+y2.1071

Let me write these formulas generically, x2+y2, 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 12 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=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, i2, but i2=-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 x2+y2.1484

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 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 z5.0004

Let us first convert that into polar form, remember my equations x2 + y2, 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= x2 + y2.0045

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 4e7pi/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 z5 .0151

We can do that by z n=rn x cos(n)( theta) + (i)sin(n)(theta).0159

That is how DeMoivre’s theorem applies to this one z5 = r2, that is 45 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 45 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 4e7pi/4i.0415

To raise it up to the nth power, I’m going to use DeMoivre’s theorem that says zn is rn, 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, x2 + y2 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 -22 is 4, 2 (root 3) 2 is 22 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)=rei(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 z1/n is equal to r1/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 r1/n, let me figure that out.0540

R1/n =4, so ¼, there is a clever thing we can do to the exponents here I know that 4 is 22 to the ¼ so that is 22/4, that is 21/2 which is root 2.0547

What we are doing here is we are finding r1/n x cos(alpha) + (i) sin (alpha).0566

But the r1/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)=e4pi/3 x (i).0719

Now I used DeMoivre’s theorem, DeMoivre’s theorem says you look at z1/n is r1/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 r1/n, what I worked out over here is r 1/n.0811

(r) was 4 and (n) was also 4, it is 41/4, y4 is 2 squared.0818

That gives you a quick way to figure out to simplify down 41/4.0825

Simplify that into square root of 2, so I multiply r1/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.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=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 zn, we write that as re, to the nth power.0073

If you think about that, we can distribute this nth power onto the r and onto the en.0080

rn, that just gives you rn.0091

en, that's an exponent raised with an exponent.0099

You multiply the exponents.0106

That's where I get iθ×n here, so ei×n×θ.0107

If you expand that, e=cos(θ)+isin(θ), einθ 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 rn×cos(nθ)+isin(nθ).0137

Another way to start out with that is to expand e into cos(θ)+isin(θ).0145

The form we're going to be using most often is this form right here zn=rn×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, zn=rn×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 8th 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 e1/n.0236

That's ren.0240

Remember, e 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

zn=rn×eniθ, which was rn×cos(nθ), you multiply the angle by n,plus isin(θ).0266

With 1/n, replacing the n by 1/n, we get r1/ncos(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 r1/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 z7.0505

Remember to convert it into polar form.0518

You do r equals the square root of x2+y2, 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 x2+y2, x2=3, y2=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, 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

z7, the whole point is we're going to use the polar form to find z7, so this is 2e(5π/6)i7.0655

That's 27.0671

I have each one exponent raised to an exponent.0675

I just want to multiply those two exponents e7×5, 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, 27 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 is cos(θ)+isin(θ).0755

You can also use x=rcos(θ), y=rsin(θ).0766

I prefer the e 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.0870

I've got my polar form.0872

To raise it up to the 7th power, De Moivre's theorem says if you use polar form, then you just put 27, 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 8th roots of the number 16.0928

In order to find all complex 8th 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 8th roots.0952

Let me remind you that because we're asked for 8th 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 x2+y2.0981

θ=arctan(y/x), plus π if the x happens to be negative.0987

My r is the square root of 162+02, that's just 16.1001

θ, my y=0, arctan(0)=0.1007

My z=16e0i.1016

I could have worked that out, certainly 16e0=16 just by itself because e0=1.1027

That's nice to check our work.1034

z1/8.1036

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 r1/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 z1/8, r=161/8.1096

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 161/8.1229

Let me see if I can do something with that.1233

161/8, I know that 16=24, that's 24/8, 21/2 is square root of 2.1235

I'm going to multiply the square root of 2, that's 161/8 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 w8=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=16e0i.1585

Then I use De Moivre's theorem which says that you get complex nth roots by doing r1/n, that's where the 161/8 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 161/8.1648

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 x2+y2, 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 x2+y2, 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, 1e.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 r1/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 r1/n.1887

r1/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

r1/n×cos(α)+isin(α) is kind of anticlimatic because we already figured out that r1/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.2057

I go to De Moivre's theroem which says, r1/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 r1/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 wr3=-1.2106

If you worked out wr3 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 x2 + 12 = 22, which is 4.0194

So, x2 = 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 sin2(x) + cos2(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 sin2(x) + cos2(x) = 1, if I plug those in I will get b/c2 + a/c2= 1. 0112

Now I’m just going to do a little bit of algebraic manipulation b2/c2 + a2/c2 is equal to 1.0127

I’m going to multiply both sides there by c2 and that will clear my denominators on the left I got b2 + a 2 and on the right I have c2.0137

If I switch around the two terms here a2 + b2 is equal to c2.0151

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 a2 + b2=c2.0177

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 sin2(theta) + cos2(theta) = 1.0040

I will plug in sin(theta ) that is -5/132 + cos2(theta) =10051

-5/13 when you square, the negative goes away so we get 25/169 + cos2(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 cos2(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 cos2(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, sin2(theta) + cos2(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 sin2x + cos2x = 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 a2 + b2 = c2.0024

You probably heard that already.0040

The new fact for trigonometry class is that sin2x + cos2x = 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 a2 + b2 = c2.0074

We're going to try to prove the Pythagorean identity sin2x + cos2x = 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 12.0186

That's the length of the hypotenuse.0205

If we just do a little semantic cleaning up here, 12, of course, is just 1, cosine x squared, the common notation for that is cos2x + sin2x = 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, sin2θ + cos2θ = 1.0284

I'm going to fill in the one that I know, cosθ, cosθ is 0.47.0295

This is 0.472 = 1 + sin2θ.0302

Now, 0.47, that's not something I can easily find the square of, so I'll do that on my calculator.0310

0.472 = 0.2209, so that's +0.2209, sin2θ +0.2209 = 1, sin2θ = 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 sin2θ + cos2θ = 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 - b2.0554

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 (a2 + b2) formula, we get (1 - cos2θ).0570

Now, let's remember the Pythagorean identity.0586

Pythagorean identity says sin2θ + cos2θ = 1.0590

That means 1 - cos2θ = sin2θ.0596

We can substitute that in into our work here, sinθ×(1 + cosθ).0603

The denominator, by the Pythagorean identity, turns into sin2θ.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 - cos2θ) converts into sin2θ.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 function1154

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, hypotenuse2 is 52 + 122, which is 25 + 144 which is 169.0022

That is exactly 132, 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 32+42, 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-axis1588

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.com0000

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, sin2x + cos2x = 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 sin2x.0049

Divide everything here by sin2x and that gives me 1.0063

Now (cos)2/(sin)2 that is (cos)/(sin)2, that is (cot)2x.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)2x to make it look exactly like what we are asked for.0093

I can rearrange terms here and I can get (cot)2x + 1 is equal to (cosec)2x.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 cot2 + 1 = cosec2.0125

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= sec2(theta).0012

Sec2(theta) is we plug in 13/12, that is 13/122 and that is 169/144.0023

102(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 tan2(theta) + 1=sec2(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 tan2(theta) + 1= sec2(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, tan2(x)+1=sec2(x).0000

You really want to think about this as a kind of companion identity to the main Pythagorean identity, which is that sin2(x)+cos2(x)=1.0010

Hopefully, you've really memorized this identity, sin2(x)+cos2(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 tan2(x)+1=sec2(x).0034

There's another related identity, essentially just the same for cotangents and cosecants, is that cot2(x)+1=csc2(x).0045

You may wonder how you'll remember this identities.0058

For example, how do you remember whether it's tan2+1=sec2, or maybe it's the other way around, sec2+1=tan2.0062

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 sin2+cos2=1.0077

As long as you can remember sin2+cos2=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 tan2(x)+1=sec2(x).0102

The trick there is to remember the original Pythagorean identity, which is cos2(x)+sin2(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 cos2(x).0136

On the left, we get cos2(x)/cos2(x) plus, let me write it as (sin(x)/cos(x)2=1/cos2(x).0152

Then of course, cos2(x)/cos2(x) is just 1, sin/cos, remember that's the definition of tangent, so this is tan2(x), 1/cos, that's the definition of sec(x), so sec2(x).0173

You can just rearrange this into tan2(x)+1=sec2(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, tan2(θ)+1=sec2(θ).0238

If we plug in what we're given here, tan2(θ), that's (-4.212)+1=sec2(θ).0246

Well, 4.212, that's something I'm going to workout on my calculator, is 17.7241.0259

So, 17.7241+1=sec2(θ), that's 18.7241=sec2(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, tan2+1=sec2.0415

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 (a2-b2).0530

We're taking advantage of that old formula from algebra, the difference of squares formula, and a lot of times the (a2-b2) 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 (a2-b2) form, so sec2(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 tan2(θ)+1=sec2(θ), that's the Pythagorean identity that we're studying today.0593

If you move that around, you get sec2(θ)-1=tan2(θ).0607

That converts the denominator into tan2(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 sin2(θ)/cos2(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 cos2(θ)/sin2(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 sin2 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)=a2-b2.0772

If cos(θ)/sin2(θ), now multiply (1-cos)×(1+cos) gives us (1-cos2(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, sin2(θ)+cos2(θ)=1, if you switched that around 1-cos2(θ)=sin2(θ).0798

I'll plug that in, cos(θ)/sin(θ), 1-cos2(θ), now translates into sin2(θ) all divided by sin(θ).0819

I forgot my squared there, that's from the line above.0836

Now, we have some cancellations, sin2(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)=a2-b2.0888

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 sec2-1.0917

The second pattern that we use there was to invoke these Pythagorean identities, tan2+1=sec2, and sin2+cos2=1.0927

We invoked that one here, converting sec2-1 into tan2.0942

Then later on, when he had another (a-b)×(a+b), it converted into 1-cos2 and that in turn, by the Pythagorean identity converted into sin2.0947

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, sin2x 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