For more information, please see full course syllabus of Trigonometry

For more information, please see full course syllabus of Trigonometry

### Inverse Trigonometric Functions

**Main definitions and formulas**:

- The
*arcsine*function, also known as the*inverse sine*function, is a function that, for each*x*between -1 and 1, produces an angle θ between − (π /2) and (π /2) whose sine is*x*. If arcsin*x*= θ , then sinθ =*x*. It is sometimes written sin^{− 1}*x*, but this notation is misleading because it could also be interpreted as [1/(sin*x*)], which is different from arcsin*x*. - The
*arccosine*function, also known as the*inverse cosine*function, is a function that, for each*x*between -1 and 1, produces an angle θ between 0 and π whose cosine is*x*. If arccos*x*= θ , then cosθ =*x*. It is sometimes written cos^{− 1}*x*, but this notation is misleading because it could also be interpreted as [1/(cos*x*)], which is different from arccos*x*. - The
*arctangent*function, also known as the*inverse tangent*function, is a function that, for each*x*, produces an angle θ between − (π/2) and (π/2) whose tangent is*x*. If arctan*x*= θ , then tanθ =*x*. It is sometimes written tan^{− 1}*x*, but this notation is misleading because it could also be interpreted as [1/(tan*x*)], which is different from arctan*x*.

**Example 1**:

**Example 2**:

**Example 3**:

**Example 4**:

**Example 5**:

### Inverse Trigonometric Functions

- Plot the point [(3π)/4] on the unit circle. Note: [(3π)/4] = 135
^{°} - Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of [(3π)/4] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arcsin ≤ [(π)/2]
- Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]

- Plot the point [(2π)/3] on the unit circle. Note: [(2π)/3] = 120
^{°} - Recall that tangent is [x/y] . Now find the angle that has the same tangent value of [(2π)/3] between − [(π)/2] and [(π)/2] because - [(π)/2] ≤ arctan ≤ [(π)/2]
- Draw a line diagonally across to locate a corresponding point that is between − [(π)/2] and [(π)/2]

- Plot the point − [(2π)/3] on the unit circle. Note: − [(2π)/3] = − 120
^{°} - Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of − [(2π)/3] between 0 and π because 0 ≤ arccos
- Draw a line across the x - axis to locate a corresponding point that is between 0 and π

- Plot the point [(5π)/6] on the unit circle. Note: [(5π)/6] = 150
^{°} - Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of [(5π)/6] between − [(π)/2] and [(π)/2] because - [(π)/2] ≤ arcsin ≤ [(π)/2]
- Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]

- Plot the point − [p/6] on the unit circle. Note: − [(π)/6] = − 30
^{°} - Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of − [(π)/6] between 0 and π because 0 ≤ arccos ≤ π
- Draw a line across the x - axis to locate a corresponding point that is between 0 and π

- Plot the point [(4π)/3] on the unit circle. Note: [(4π)/3] = 240
^{°} - Recall that tangent is [x/y]. Now find the angle that has the same tangent value of [(4π)/3] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arctan ≤ [(π)/2]
- Draw a line diagonally across to locate a corresponding point that is between − [(π)/2] and [(π)/2]

- Plot the point [(5π)/4] on the unit circle. Note: [(5π)/4] = 225
^{°} - Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of [(5π)/4] between − [(π)/2] and [(π)/2] because - [(π)/2] ≤ arcsin ≤ [(π)/2]
- Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]

- Plot the point [(5π)/3] on the unit circle. Note: [(5π)/3] = 300
^{°} - Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of [(5π)/3] between 0 and π because 0 ≤ arccos ≤ π
- Draw a line across the x - axis to locate a corresponding point that is between 0 and π

- Plot the point − [(7π)/6] on the unit circle. Note: − [(7π)/6] = 210
^{°} - Recall that tangent is [x/y]. Now find the angle that has the same tangent value of − [(7π)/6] between − [(π)/2] and [(π)/2] because - [(π)/2] ≤ arctan ≤ [(π)/2]
- Draw a line diagonally across to locate a corresponding point that is between − [(π)/2] and [(π)/2]

- Plot the point − [(5π)/6] on the unit circle. Note: − [(5π)/6] = 150
^{°} - Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of − [(5π)/6] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arcsin ≤ [(π)/2]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Inverse Trigonometric Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Arcsine Function 0:24
- Restrictions between -1 and 1
- Arcsine Notation
- Arccosine Function 3:07
- Restrictions between -1 and 1
- Cosine Notation
- Arctangent Function 4:30
- Between -Pi/2 and Pi/2
- Tangent Notation
- 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
- Extra Example 2: Arcsin/Arccos/Arctan Values

### Trigonometry Online Course

I. Trigonometric Functions | ||
---|---|---|

Angles | 39:05 | |

Sine and Cosine Functions | 43:16 | |

Sine and Cosine Values of Special Angles | 33:05 | |

Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D | 52:03 | |

Tangent and Cotangent Functions | 36:04 | |

Secant and Cosecant Functions | 27:18 | |

Inverse Trigonometric Functions | 32:58 | |

Computations of Inverse Trigonometric Functions | 31:08 | |

II. Trigonometric Identities | ||

Pythagorean Identity | 19:11 | |

Identity Tan(squared)x+1=Sec(squared)x | 23:16 | |

Addition and Subtraction Formulas | 52:52 | |

Double Angle Formulas | 29:05 | |

Half-Angle Formulas | 43:55 | |

III. Applications of Trigonometry | ||

Trigonometry in Right Angles | 25:43 | |

Law of Sines | 56:40 | |

Law of Cosines | 49:05 | |

Finding the Area of a Triangle | 27:37 | |

Word Problems and Applications of Trigonometry | 34:25 | |

Vectors | 46:42 | |

IV. Complex Numbers and Polar Coordinates | ||

Polar Coordinates | 1:07:35 | |

Complex Numbers | 35:59 | |

Polar Form of Complex Numbers | 40:43 | |

DeMoivre's Theorem | 57:37 |

### Transcription: Inverse Trigonometric Functions

*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 Uâ€™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

*Iâ€™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 â€“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 â€“pi/3.*0093

*Arcsin(sin4pi/3), an angle that has the same sin as 4pi/3 is â€“pi/3.*0110

*Make that negative sign a little clear, we have solved the first one.*0125

*Next one is arcos(cos â€“pi/4), well there is â€“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 â€“pi/4.*0158

*Cos is the (x) value we are keeping the (x) value constant and Iâ€™m moving up to find an angle in the first two quadrants, there it is at pi/4.*0165

*Arcos(cos â€“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 â€“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 â€“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, Iâ€™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 â€“pi/3, pi/4 and pi/6, those are angles that has the same sin, cos, and tan as the ones we are given.*0323

*That was our first lecture on inverse trigonometric functions, these are the trigonometry lectures for www.educator.com.*0334

*Hi, these are the trigonometry lectures for educator.com, and we're here to learn about the inverse trigonometric functions.*0000

*We've already learned about sine, cosine and tangent, and today we're going to learn about the inverse sine, inverse cosine, and inverse tangent, probably better known as arc sine, arc cosine, and arc tangent.*0008

*The arc sine function is also known as the inverse sine function, basically, it's kind of the opposite of the sine function.*0022

*The idea is that you're given a value of x, and you want to find an angle whose sine is that value of x.*0030

*In order to make this work, sines only occur between -1 and 1, so you have to be given a value of x between -1 and 1.*0041

*You're going to try to give an angle between -π/2 and π/2.*0050

*You'll try to give an angle between -π/2 and π/2 that has the given value as a sine.*0060

*If arcsin(x)=θ, what that really means is that the sin(θ)=x.*0078

*Now, there's some unfortunate notation in mathematics which is that arcsin is sometimes written as sin to the negative 1 of x.*0087

*This is very unfortunate because people talk about, for example, sin ^{2}x means (sin(x))^{2}.*0095

*You might think that sin ^{-1}(x), would be (sin(x))^{-1}, which would be 1/sin(x).*0106

*Now, that's not what it means, the sine inverse of x doesn't mean 1/sin(x), it means arcsin(x).*0119

*This notation really is very ambiguous because this inverse sine notation could mean arcsin(x) or it could mean 1/sin(x), and so this notation is ambiguous because arcsin(x) and 1/sin(x) are not the same.*0129

*The safest thing to do is to not use this notation sin ^{-1} at all.*0151

*Let me just say, avoid this notation completely because it is ambiguous, it could be interpreted to mean these two different things that are not equal to each other.*0159

*Instead, it's probably safer to use the notation arcsin(x), which definitely means inverse sin(x), and cannot be confused.*0175

*The arccos(x), the arc cosine function is sort of the opposite of the cosine function.*0187

*You're given a value of x, and you have to find an angle whose cosine is that value of x.*0194

*Again, the value of x you must be given would have to be between -1 and 1, because those are the only values that come up as answers for cosine.*0201

*What you try to do is produce an angle between 0 and π, so there's 0, π/2, π.*0215

*You try to produce an angle between 0 and π that has that value as its cosine.*0229

*Just like we have with sine, there's the problem of this misleading notation cos ^{-1}(x), it could be interpreted as 1/cos(x) or it could be interpreted as arccos(x).*0233

*The best thing to do is to avoid using this notation completely, cos ^{-1}(x) is just misleading, it could be interpreted either way.*0247

*Try not to use it at all, instead stick to the notation arccos(x).*0263

*Finally, arc tangent is known as the inverse tangent function.*0270

*You're given a value of x, and you want to find an angle whose tangent is x.*0275

*For arc tangent, we're going to try to find the angles between -π/2 and π/2 because that covers all the possible tangents we could get.*0282

*If arctan(x)=θ, that really means that tan(θ)=x.*0297

*Just like with sine and cosine, we have this potentially misleading notation, tan ^{-1}(x), could be interpreted to mean arctan(x) or 1/tan(x).*0302

*Those are both reasonable interpretations but they mean two different things.*0314

*Again, let's try to avoid this notation completely because it could mean two completely different things.*0317

*We'll just try not to use that at all when we're talking about inverse tangents we'll say arctan instead of tan ^{-1}.*0337

*Let's get started with some examples.*0345

*First off we have to identify the domain and range of the arc sine function, and then graph the function.*0347

*Let's start out with the graph of sin(x), because arcsin is really meant to be an inverse to sin(x).*0354

*I'll start with the graph of sin(x) here.*0362

*Remember that when you're trying to find inverse functions, you take the function and you reflect it across the line y=x.*0372

*In order for something to be a function, it has to pass the vertical line test.*0388

*If you draw a vertical line, you shouldn't cross the graph twice.*0393

*Since we're reflecting across the line y=x, that kind of switches the x's and y's, so we wanted something that will pass the horizontal line test.*0397

*Let me draw this in red.*0408

*We don't want to be able to draw a horizontal line and cross the graph twice.*0411

*As you see, when I've drawn these horizontal lines, we crossed the graph in lots of places.*0415

*We have a problem when we wanted to find the inverse sine function.*0421

*The way we solve that is by not using all of the sine graph, we just cut off a piece of the sine graph that will pass the horizontal line test.*0425

*I'm going to cut off a piece of the sine graph from -π/2 to π/2.*0439

*If you just look at this portion of the sine graph, you see that it passes the horizontal line test.*0448

*That means we can take the inverse just at that part of the sine function.*0459

*Let me draw what that looks like when we reflect it.*0465

*We're just taking this thick piece here, and I'm going to reflect that across that line y=x.*0472

*Now, the notations that I had on the y and x axis are going to switch, it goes from now -1 to 1 on the x-axis, and -π/2 to π/2 on the y-axis.*0489

*I can't keep going with this, I can't draw the rest of the graph because if I do draw any more, I'm going to get something that fails the vertical line test, it won't be a function anymore.*0510

*This is the entire arcsin function.*0521

*The domain here, it's all the numbers that you can plug into the arcsin, that's all values of x with -1 less than or equal to x, less than or equal to +1.*0532

*We can't plug any other values of x into arcsin, and that's really because in the other direction the only values that come out of the sine function are between -1 and 1.*0560

*The only values that you can plug into the arcsin function are between -1 and 1.*0571

*The range, the numbers that come out of the arcsin, all values of x between -π/2 and π/2.*0580

*Let me write that as y because when we think of those are the values coming out of the arcsin function.*0596

*Those are all the y values you see here and you see that the smallest y-value is -π/2 and the biggest value we see is π/2.*0605

*Arcsin takes in a number between -1 or 1, gives you a number between -π/2 and π/2 and its graph looks like a sort of chopped of piece of the sine graph.*0615

*Remember, the reason we had to chop it off is to get a piece of the sine graph that would satisfy the horizontal line test, so that the arcsin graph satisfies the vertical line test and really is a function.*0629

*In our second example here, we have to find the arcsin of the sin(2π/3) and then some similar values for our cosine and our tangent.*0647

*The first thing to do here is really to figure out where we are on the unit circle.*0663

*There's 0, π/2, π, 3π/2 and 2π.*0675

*Let's figure out where each of this angle is on the unit circle.*0685

*2π/3 is over here, there's 2π/3.*0691

*Remember, arcsin is always between -π/2 and π/2.*0698

*Arcsin is between -π/2 and π/2, so there's -π/2 down here.*0712

*We want to find an angle between -π/2 and π/2.*0717

*On the right-hand side that has the same sine as 2π/3.*0724

*Well, sine is the y-value so we want something that has the same y-value as 2π/3.*0729

*That angle right there has the same y-coordinate as 2π/3, that's π/3.*0741

*From the graph, we see that π/3 has the same sine as sin(2π/3).*0750

*So, arcsin(2π/3) is π/3.*0771

*It's an angle whose sine is the sin(2π/3).*0780

*Let's try the next one, -5π/6.*0785

*-π is over there, so 5π/6 in the negative direction puts you over there, there's -5π/6.*0792

*We want an angle whose cosine is the same as the cosine(-5π/6).*0809

*Cosine is the x-value, so we want an angle that has the same x-value as the one we just found.*0818

*There it is right there, 5π/6.*0830

*That's between 0 and π, which is the range for the arccos function, it's always between 0 an π.*0838

*That means that arccos of the cos(-5π/6) is 5π/6, that's an angle between 0 and π that has the right cosine.*0850

*Finally, we want to find the arctan of tan(3π/4).*0875

*3π/4 is over there.*0882

*There's 3π/4.*0890

*We want to find an angle that has the same tangent as that one.*0893

*Remember, arctan is forced to be between π/2 and -π/2.*0898

*We want to find an angle between π/2 and -π/2 that has the same tangent as 3π/4.*0909

*If we go straight across the origin there, we started with an angle whose cosine is negative and sine is positive.*0920

*This angle right here has a positive cosine and negative sine, so it'll end up having the same tangent, and that angle is -π/4.*0934

*That's an angle inside the range we want that has the same tangent.*0946

*Arctan of tan(3π/4) is -π/4.*0952

*Those are really quite tricky.*0967

*What we're being asked to do here in each case is we're given an angle, for example 2π/3.*0969

*We want to find arcsin of sin(2π/3).*0977

*We look at the sine of 2π/3, and then we want to find another angle that has that sine but it has to be in the specified range for arcsin.*0981

*We're really trying to find what's an angle between -π/2 and π/2 whose sine is the same as sin(2π/3).*0993

*That's why we did this reversal to find the angle π/3 that has the same sine as the sin(2π/3).*1003

*That was kind of the same process of all three of them, finding angles that have the same cosine as -5π/6, but is in the specified range, finding an angle that has the same tangent as 3π/4 but is in the specified range.*1017

*Our third example here, we're asked to identify the domain and range of the arctan function, and graph the function.*1037

*Remember when we graphed arcsin, we started out with a graph of sine, so let me start out here with a graph of tangent.*1044

*I'll graph it in blue.*1054

*Has asymptotes of π/2, and then it starts repeating itself.*1057

*It has period π.*1065

*This is 0, π/2, -π/2, and that's π, 3π/2.*1084

*What I graphed in blue there is tan(θ).*1096

*Now, I want to take this graph and I want to flip it around the line y=x.*1102

*Let me graph the line y=x.*1107

*I want to get something that it will be a function, it has to pass the vertical line test after I flip it.*1112

*That means it has to pass the horizontal line test before I flip it.*1120

*Just like with sine, the tangent function fails the horizontal line test badly.*1124

*You can draw a horizontal lines and intersect it in lots of places.*1132

*Just like with sine, we're going to cut off part of the tangent function and use that to form the arctan function.*1137

*We're going to cut off just one of these curves.*1143

*I'll cut it off these asymptotes at -π/2 and π/2.*1151

*I'll just take this part of the tangent function and I'll flip it around to make the arctan function.*1156

*I'll do this in red.*1171

*Arctan(x).*1179

*I'm going to flip this around just that one main branch of the tangent function.*1181

*Since the tangent function had vertical asymptotes, the arctan function is going to have horizontal asymptotes at -π/2 and π/2.*1192

*Those are my asymptotes right there at -π/2 and π/2.*1210

*We're also asked to identify the domain and range.*1216

*The domain means what numbers can we plug in to arctan(x).*1219

*Domain is all values of x because we can plug any number in on the x-axis here and we'll get an arctan value.*1228

*-infinity less than x less than infinity.*1243

*The range is all x with, okay, let me write this as y because these are values on the y-axis.*1248

*All y with -π/2 less than y less than π/2.*1264

*Let me emphasize here that π/2 and -π/2 are not included, π/2 themselves are not in the range.*1269

*These inequalities, they're not less than or equal to they're strictly less than.*1291

*The reason for that is that the arctan function never quite gets to -π/2 or π/2, it goes down close -π/2 and it goes up close to π/2 but it never quite gets there.*1297

*The reason for that, if you kind of look back at the tangent function, was that these asymptotes never quite get to -π/2 or π/2.*1312

*You can't take the tangent of -π/2 or π/2, because you're trying to divide by zero back in the tangent function.*1323

*To understand this problem really, you have to remember what the graph of tan(θ) looks like.*1333

*Then you take its one main branch and you flip it over, and you get the graph of arctan(x), and the domain and range just come back to any things in the domain.*1339

*The range can go from π/2 to -π/2, but you don't include those end points.*1354

*The asymptotes are the horizontal lines at -π/2 and π/2.*1362

1 answer

Last reply by: Dr. William Murray

Thu Jul 21, 2016 2:27 PM

Post by Julian Xiao on July 18 at 01:55:47 PM

I have a question about Example 3 of the domain and range of arctangent. The domain is -inf. < x < inf. and the range is -pi/2 < y < pi/2, but can't you also include infinity and pi/2 and -infinity and -pi/2? Because technically the function intersects pi/2 at infinity, and also intersects -pi/2 at -infinity.

1 answer

Last reply by: Dr. William Murray

Tue Aug 5, 2014 11:56 AM

Post by edder villegas on July 19, 2014

a don't see exercises to practice

1 answer

Last reply by: Dr. William Murray

Tue Dec 17, 2013 9:28 PM

Post by Daniel Bristow on December 16, 2013

Hi Dr. Murray.

I was wondering if you could better summarize how to "flip" a line, (ex, when you drew that line at 20:06). I understand perfectly how to flip a coordinate on the unit circle, but for my less critical mind, is there some way of knowing how to flip a (ex. tangent) of a line?

Thank you.

2 answers

Last reply by: Dr. William Murray

Wed Aug 14, 2013 11:57 AM

Post by Charles Zhou on July 31, 2013

For example 2, arcsin(sin2pi/3), can I calculate the sin2pi/3 first, and then use arcsin(root3/2) to get the answer?

1 answer

Last reply by: Dr. William Murray

Wed May 22, 2013 3:44 PM

Post by Timothy Davis on May 20, 2013

Why are arcsecant, arc cosecant, and arctan not covered?

1 answer

Last reply by: Dr. William Murray

Thu Nov 29, 2012 11:17 AM

Post by Norman Cervantes on November 27, 2012

I would strongly recommend doing the EXTRA examples for new beginners, if you haven't done so already.

-MAGGIE, the same thing has happened to me when my laptop loses connection all of a sudden while it is still loading the lesson, refreshing the page wont solve the problem, go back to the educator menu and try again.

1 answer

Last reply by: Dr. William Murray

Sun Jan 20, 2013 6:30 PM

Post by Dr. William Murray on October 26, 2012

Hi Maggie,

It's not doing that on my computer. Is anyone else having this problem? Maggie, please try it again, and if it happens repeatedly, we can contact the tech administrators for Educator. They're generally very good about fixing problems quickly.

Thanks for being part of the course!

Will Murray

1 answer

Last reply by: Dr. William Murray

Sun Jan 20, 2013 6:29 PM

Post by Maggie Henry on October 24, 2012

I want to know if I am the only one who is having the issue of the lecture skipping back to the very beginning after a couple slides have already been shown?

3 answers

Last reply by: Dr. William Murray

Sun Jan 20, 2013 6:33 PM

Post by Michael Trow on September 27, 2012

useless explanation of how to get arctan x

1 answer

Last reply by: Dr. William Murray

Sun Jan 20, 2013 6:38 PM

Post by Cain Blaha on November 28, 2011

how do you use this function for angles that do not have pi in them, or that arent in the unit circle?

3 answers

Last reply by: Dr. William Murray

Wed May 22, 2013 3:35 PM

Post by Elina Bugar on August 15, 2011

what do u mean by

-1<x<1

and

-pie/2 < y < pie/2