Hey, everyone. If you were asked to find the inverse cosine of one half or maybe even the sine of some angle, like, say, pi over 3, you would be able to come over to your unit circle and get an answer rather easily. But what if instead you were asked to find the sine of the inverse cosine of one half? That doesn't seem quite so simple. But here, I'm going to show you that it actually is that simple because we're going to be able to deal with each of these functions separately using what we already know. Now, when dealing with functions such as this one, these are referred to as composite functions because we have one function inside of another function. And when dealing with composite functions, we only want to deal with one function at a time. So, in order to do that, we're always going to evaluate the function that is inside of the parentheses first. Now, in doing this, we're essentially going to be solving this composite function from the inside out. So, we're going to start with that inside function. Now, with this one, the sine of the inverse cosine of one half, our inside function is the inverse cosine of one half. So that's what we're going to deal with first, just completely ignoring that outside function, the sine. So, here to find the inverse cosine of one half, remember when dealing with inverse trig functions, you can also think of this as, okay, the cosine of what angle gives me a value of one half. Now, coming over here to our unit circle, I see that for my angle of pi over 3, I end up with a cosine value of one half. So that tells me that my inside function, I get an answer of pi over 3. Now that I've dealt with that inside function, I can now deal with my outside function of sine. Now, I'm just left to find the sine of pi over 3, which we can again do relatively easily just coming over to our unit circle. Now again, for my angle of pi over 3, I have a sine value of 32. So the sine of pi over 3 is 32. And that gives me my answer to this entire expression. The sine of the inverse cosine of one half is 32. So now that we know the basics of solving these composite trig functions, let's go ahead and come down here to solve some more examples together. Now looking at this first example, I have the cosine of the inverse tangent of 0. Now remember that when working with these composite functions, we want to look at our inside function first. Now our inside function here is the inverse tangent of 0. Now when working with inverse trig functions, remember you can always think of this as the tangent of what angle will give me a value of 0? Now, coming over here to our unit circle, we can see that for our angle of 0, we also get a tangent value of 0. Then also for our angle of pi, we will end up with a tangent value of 0 as well. But remember that when working with inverse trig functions, we have to consider the interval for which they're defined. So we can only use values within that correct interval. Now for the inverse tangent, I know that my interval for my angles goes from negative pi over 2 to positive pi over 2. So coming back down to my unit circle, I only want to deal with angles in that correct interval. So this angle of pi is not within my interval from negative pi over 2 to pi over 2. So that is not my solution here. And here, my solution is 0 for that inside trig function. So here, the inverse tangent of that value of 0 is my angle of 0. Now I'm just left to deal with that outside function, the cosine. So here, I want to find the cosine of my angle 0. Now, for my cosine of 0, I can come back down to my unit circle here. The cosine of 0 is simply 1. So that gives me my final solution of 1 for the cosine of the inverse tangent of 0, and we're done here. Now let's take a look at one final example. Here, we're asked to find the inverse cosine of the sine of pi over 3. Now something that you might notice here is that my inverse trig function is now on the outside rather than on the inside as it was in example a. Now you'll see these either way. It doesn't matter which way because we're still going to solve these the same exact way by starting with our inside trig function. Now our inside function here is the sine of pi over 3. So coming over to my unit circle, I want to find the sine of pi over 3, which, looking at my angle pi over 3, I know that the sine is the square root of 3 over 2. So that gives me my solution to that inside function, the square root of 3 over 2. Now I'm just left to find the inverse cosine of that value. Now remember when working with inverse trig functions, you can also think of this as the cosine of what angle gives me a value of the square root of 3 over 2. But remember, we are also considering our interval here. So since we're working with the inverse cosine in this case, our interval for the inverse cosine goes from 0 to pi when considering what angles we want to deal with. So on our unit circle, I only want to find an angle that is between 0 and pi. So an angle between 0 and pi that gives me a cosine value of the square root of 3 over 2. I see that for pi over 6, the cosine is the square root of 3 over 2. And that is within my specified interval. So that's my final solution here, pi over 6. So the inverse cosine of the sine of pi over 3 is pi over 6, and we're done here. Now that we know how to evaluate composite trig functions, let's get a bit more practice together. Thanks for watching, and let me know if you have any questions.

- 0. Fundamental Concepts of Algebra3h 29m
- 1. Equations and Inequalities3h 27m
- 2. Graphs43m
- 3. Functions & Graphs2h 17m
- 4. Polynomial Functions1h 54m
- 5. Rational Functions1h 23m
- 6. Exponential and Logarithmic Functions2h 28m
- 7. Measuring Angles39m
- 8. Trigonometric Functions on Right Triangles2h 5m
- 9. Unit Circle1h 19m
- 10. Graphing Trigonometric Functions1h 19m
- 11. Inverse Trigonometric Functions and Basic Trig Equations1h 41m
- 12. Trigonometric Identities 34m
- 13. Non-Right Triangles1h 38m
- 14. Vectors2h 25m
- 15. Polar Equations2h 5m
- 16. Parametric Equations1h 6m
- 17. Graphing Complex Numbers1h 7m
- 18. Systems of Equations and Matrices1h 6m
- 19. Conic Sections2h 36m
- 20. Sequences, Series & Induction1h 15m
- 21. Combinatorics and Probability1h 45m
- 22. Limits & Continuity1h 49m
- 23. Intro to Derivatives & Area Under the Curve2h 9m

# Evaluate Composite Trig Functions - Online Tutor, Practice Problems & Exam Prep

To evaluate composite trigonometric functions, start with the inside function, considering the appropriate intervals. For example, the sine of the inverse tangent of 3/4 can be solved by drawing a right triangle, leading to a sine value of 3/5. Similarly, for the inverse cosine of negative 5/13, identify the quadrant, draw the triangle, and apply the Pythagorean theorem to find the sine, resulting in 12/13. Understanding these principles is crucial for solving complex trigonometric expressions effectively.

### Evaluate Composite Functions - Values on Unit Circle

#### Video transcript

Evaluate the expression.

$\cos\left(\sin^{-1}1\right)$

$0$

$1$

$-1$

$\frac12$

Evaluate the expression.

$\sin^{-1}\left(\cos\frac{2\pi}{3}\right)$

$\frac{\pi}{6}$

$\frac{5\pi}{6}$

$\frac{\pi}{3}$

$-\frac{\pi}{6}$

### Example 1

#### Video transcript

Hey, everyone. In this problem, we're asked to evaluate the expression that the secant of the inverse cosine of negative square root of 3 over 2. Now this function might look a little bit complicated to you, but we can still break this down and solve this the way that we already know how. So let's start with that inside function here, the inverse cosine of negative square root of 3 over 2. Now because this is an inverse trigonometric function, remember that we can also think of this as, okay, the cosine of what angle will give me negative square root of 3 over 2? Now also remember, because this is an inverse trigonometric function, we're looking for an angle within a specified interval. Now for the inverse cosine, I only want to work with angles that are between 0 and pi. So, I am looking for an angle between 0 and pi for which my cosine is equal to negative square root of 3 over 2. Now looking at my unit circle, I know that for 5 pi over 6, that will give me a cosine of negative square root of 3 over 2. So that tells me that this inside function is 5 pi over 6. And I am now left to find the secant of that 5 pi over 6. Now from working with these trigonometric functions previously, you may remember that the secant is really just one over the cosine. So this secant of 5 pi over 6 can really be rewritten as one over the cosine of 5 pi over 6. Now we actually just saw what the cosine of 5 pi over 6 was. We know that it's negative square root of 3 over 2. So this is really one over negative square root of 3 over 2. Now whenever I have one over a fraction, I can really just flip that fraction. So this is going to give me negative two over the square root of 3. Now this is technically a correct answer, but whenever we have a radical in our denominator, we really don't want that there. So we want to rationalize this denominator, which we can do by multiplying by square root of 3 over square root of 3. Now that will give me a final answer of negative 2 square root of 3 over 3. And that's my final answer. The secant of the inverse cosine of negative square root of 3 over 2 is negative 2 square root of 3 over 3. Thanks for watching, and let me know if you have questions.

### Evaluate Composite Functions - Special Cases

#### Video transcript

Hey, everyone. As you work through problems dealing with composite trig functions, you may come across an expression that looks something like this one: the inverse cosine of the cosine of 11 pi over 6. Now, looking at this, knowing that trig functions and their inverses undo each other, you may be tempted to cancel this inverse cosine and cosine, leaving you with just 11 pi over 6 as your final answer. But this would actually be entirely wrong. Because remember, when dealing with inverse trig functions, you have to consider the interval for which they are defined. Now, you may be worried that this will make these sorts of problems really tricky, but you don't have to worry about that because we're actually going to solve these the same exact way we would any composite trig function, evaluating them from the inside out and considering the interval along the way. So here, I'm going to walk you through exactly how to deal with these types of problems, and you'll be a pro in no time. So let's go ahead and get started.

Now, like I said, you cannot assume that your argument is your final answer here. In this first example, the inverse cosine of the cosine of 11 pi over 6, we cannot assume that our final answer is 11 pi over 6. Because of that, we're just going to solve this as we would any composite trig function, starting with that inside function, which in this case is the cosine of 11 pi over 6. Now, in finding the cosine of 11 pi over 6, I can just come right on down to my unit circle. And I see that for 11pi over 6, the cosine is 32. So for that inside trig function here, I get an answer of 32. Now that I've dealt with that inside trig function, I can deal with my outer trig function, which is the inverse cosine of that 32.

Now remember, when dealing with inverse trig functions, we can also think of this as, okay, the cosine of what angle will give me a value of 32? And also when dealing with these inverse trig functions, we have to consider their interval. Specifically, when dealing with the inverse cosine, our interval for our angles is between 0 and pi. So, looking at our unit circle, I only want to look for angles between 0 and pi for which the cosine is 32. And looking at this unit circle within this specified interval, I see that for my angle pi over 6, my cosine is 32. So that gives me my final answer here: The inverse cosine of 32 is pi over 6, and that gives me my final answer. And you see that indeed this is not equal to 11pi over 6, so we have to be super careful here.

Now, let's go ahead and move on to our next example where we are asked to find the sine of the inverse sine of 2. Now remember, we're going to solve these the way we would any other composite trig function. So when you are tempted here to cancel this sine and inverse sine, don't. Remember that we have to solve these the way we would any other composite trig function, starting from the inside. So here our inside function is the inverse sine of 2. Now remember when dealing with inverse trig functions, you can also think of this as, okay, the sine of what angle gives me a value of 2. Now, coming over to our unit circle and going around looking at all these different angles, I don't see one for which the sine is going to be 2. So this seems a bit strange. But if we consider the interval for our inverse sine function, we know that we can only take the inverse sine of values between negative one and one, and 2 is not within that interval. So if I try to take the inverse sine of 2, I can't do that because this is simply an undefined value. Now even if you were to try to type this into your calculator, the sine of the inverse sine of 2, you would end up with an error, which is something that you might not have considered if you were to have just canceled this sine and inverse sine. So remember, whenever you see functions that look like these and you're tempted to just cancel those functions out, don't. Remember, that you have to consider the interval of your inverse trig function. Thanks for watching, and I'll see you in the next one.

### Example 2

#### Video transcript

Hey, everyone. In this problem, we're asked to evaluate the expression the inverse sine of the sine of π6. Now here, you might be tempted to just cancel the inverse sine with the sine and end up with your answer of π6. But remember, we have to be extra careful when working with these composite trig functions because our inverse trig functions are only defined for a particular interval. So let's go ahead and break this down the way we would any composite trig function, starting with that inside function, the sine of π6. Now coming over here to my unit circle, the sine of π6 is equal to 1/2. So, that inside function gives me a value of 1/2, and now I'm just left to find the inverse sine of that 1/2.

Now remember when working with inverse trig functions, we can also think of this as, okay, the sine of what angle is equal to 1/2. But that angle can only be within our specified interval for the inverse sine, which happens to be from -π2 to π2. So, you only want an angle within this interval for which the sine is equal to 1/2. Now inside of this interval, where is my sine equal to 1/2? Well, the sine of π6 is equal to 1/2, and that is within my interval. So that actually gives me a final answer of π6. Now here you might be confused because I just told you that you can't cancel the inverse sine with the sine. But here, we actually could. Now, the reason that we could do that is because our angle from the beginning was within our specified interval.

So when that does happen, you actually are able to effectively cancel the inverse sine with the sine. But you always want to be extra careful when working with these because remember, that isn't always true. So always, always double-check your interval and make sure you're getting a value only within your specified interval. Thanks for watching, and let me know if you have questions.

Evaluate the expression.

$\cos\left(\cos^{-1}\left(-\sqrt3\right)\right)$

$\frac{\pi}{3}$

$\frac{2\pi}{3}$

$\pi$

Undefined

Evaluate the expression.

$\cos^{-1}\left(\cos\left(\frac{\pi}{2}\right)\right)$

$0$

$\frac{\pi}{2}$

$\pi$

Undefined

Evaluate the expression.

$\tan^{-1}\left(\tan\frac{2\pi}{3}\right)$

$-\frac{\pi}{3}$

$\frac{\pi}{3}$

$\frac{2\pi}{3}$

$\frac{5\pi}{3}$

### Evaluate Composite Functions - Values Not on Unit Circle

#### Video transcript

Hey, everyone. We've been evaluating composite trig functions using the unit circle, but what if you were given a composite trig function like this one: the sine of the inverse tangent of three fourths? Three fourths is not a value that is directly on our unit circle. So, how can we go about evaluating this? Well, you may be worried that this sort of problem is going to be really tricky, but we can actually just solve this by drawing a right triangle, and then we can use the absolute basics of right triangles that we already know in order to come to a solution. So here, I'm going to walk you through exactly how to solve these problems step by step, and soon, you'll be able to evaluate any composite trig function, whether you can use the unit circle or not. So let's go ahead and get started here.

Now here in this example, we're tasked with evaluating the composite trig function, the sine of the inverse tangent of three fourths. Now remember, for any composite trig function, we're always going to start with that inside function, which here is the inverse tangent of three fourths. So we can also think of this by remembering how we think about inverse trig functions, as the tangent of what angle theta gives me a value of three fourths. So for what angle is this true? Well, we can actually just make this angle by drawing a right triangle. Now here, my right triangle is in quadrant 1, which makes sense because we're taking the inverse tangent of a positive value. So if I draw my angle theta right here and I want the tangent of this angle theta to be equal to three fourths, I can just label my opposite side as 3, my adjacent side as 4, and the tangent of it is three fourths. Now you'll notice here that I don't actually care about the value of the angle, but just that its tangent is equal to three fourths.

Now from here, I'm just left to find the sine of this angle and evaluate that outside function. Now in order to get the sine based on SOHCAHTOA, I know that I need my opposite side and my hypotenuse. So I first need to find my hypotenuse, which I can do using the Pythagorean theorem. Or here, you might also notice that since my leg lengths are 3 and 4, that means my hypotenuse has to be 5 because this is a 3, 4, 5 triangle. Now from here, I can take that opposite side 3 divided by my hypotenuse 5 in order to get my final solution. The sine of the inverse tangent of three fourths is equal to three fifths.

Now let's recap how we got there. Here, we took our inside function, used it to draw a right triangle, and then found our missing side length, allowing us to evaluate that outside function and come to a final solution.

Now, not all these problems will be quite as straightforward. So here, I'm going to walk you through some steps that will work for any composite trig function for which you can't use the unit circle. Let's come down to this example here. Here, we're asked to evaluate the expression, the sine of the inverse cosine of negative five over thirteen. Now, you'll notice here that our argument is negative. And that's kind of what complicates this a little bit. But no worries. We're going to walk through some steps here.

Now, if you're ever unsure whether or not you should use the unit circle or draw a right triangle, if you look at your composite trig function and you see that your inside function is an inverse function, and your argument is kind of a strange number that you don't recognize as being on the unit circle, that's usually a good hint that you should draw a right triangle, because the unit circle is not going to help you here. So here, seeing that both of those things are true, let's go ahead and get started with our steps and get to drawing this right triangle.

Now, of course, we're going to start by using our inside function. This is true of any composite trig function. So in step 1, we're going to take that inside trig function, and we are going to use our interval to determine what quadrant we will eventually draw our right triangle in. So here, my inside function is the inverse cosine. And for the inverse cosine, I know that my values have to be between negative one and one and my angles between 0 and pi. Now, looking at my coordinate system based on that angle interval, I know that my angle has to be in quadrant 1 or quadrant 2 because this is my interval from 0 to pi.

Now here's where we want to consider the sign of our argument. And here, we're taking the inverse cosine of a negative value. Our argument is negative here. Now based on our intervals, we're either in quadrant 1 or quadrant 2, but cosine values can only be negative in quadrant 2. So that's where my triangle is going to go, and we've completed step number 1. Quadrant 2 is where we'll draw our triangle, moving on to step 2.

So we're actually going to draw this triangle, and then we're going to use our argument to label our angle and our two sides. So here in quadrant 2, I'm going to go ahead and draw my right triangle. Now, the orientation of your right triangle might change just depending on what quadrant you're in. But you're always going to draw it with one side length against the x-axis so that your angle can be measured from that x-axis. So here, this is where my angle theta is. And looking at my argument negative five over thirteen, this tells me that the cosine of my angle that I labeled has to be equal to negative five over thirteen to satisfy this inside function. So using SOHCAHTOA, I know that means my adjacent side has to be negative 5 and my hypotenuse has to be 13. Now, it might be kind of weird here that I labeled this side length as being a negative value. But if we just think about where we're located on our coordinate system, my x values are negative here in quadrant 2, so it makes sense that this side length is negative. So always consider your location on the graph when you're looking at the signs of your values.

So now we've completed step number 2. We can move on to step number 3 and use the Pythagorean theorem to find that missing third side. Now, here, our missing side is a leg length, so we're either solving for a or b. It doesn't matter which one. So I'm going to go ahead and set up my Pythagorean theorem, a2+b2=c2. And I'm going to be solving for a here. So I have a2. I'm going to plug in that other side length of negative 5, square that, and that is equal to my hypotenuse, which is 132.

Now, from here, we can just do some algebra. a2+(-5)2 = 25, and that's equal to 169. Now subtracting 25 from both sides cancels on that left side, leaving me with a2=144, having subtracted that 25 from 169. Now, from here, my last step is to take the square root, giving me my side length of 12, which I can go ahead and label up here on my triangle that missing side length of 12.

Now, we've completed step number 3, and we can move on to our final step where we're going to use our triangle to evaluate that outside function. So moving on to that outside function, which here is the sine. So here we want to find the sine of our angle theta that we have labeled on our triangle here. And I know that using SOHCAHTOA, that means I want to take my opposite side, divide it by my hypotenuse, and I have all the information I need. So my opposite side here is 12, and then my hypotenuse is 13. So that gives me my final answer of 12 over 13. So that tells me the sine of the inverse cosine of negative 5 over 13 is 12 over 13.

Now, you may notice here that our argument started out as negative. We were taking the sine of the inverse cosine of a negative value, but then we ended up with a final answer that's positive. But again, think about where we're located on the graph. In quadrant 2, my sine values are going to be positive, so this makes perfect sense. So now, you should be able to evaluate any composite trig function, whether you can use the unit circle or not. Let's continue practicin

### Example 3

#### Video transcript

Hey everyone. In this problem, we're asked to evaluate the expression the sine of the inverse tangent of 2 thirds. Now, looking at this composite trig function, my inside function is an inverse, and that value is not something I recognize as being on the unit circle, so I know that I need to solve this by using a right triangle. So let's go ahead and jump into our steps here. Now, step number 1 is to use our interval to identify our quadrant. Now we're, of course, working with our inside function first here. So our inside function is the inverse tangent. And the inverse tangent has an angle interval from negative pi over 2 to positive pi over 2. So I know that my right triangle either needs to be in quadrant 1 or quadrant 2 to match that interval. Now here we want to consider the sign of our argument. Now our argument here is a positive value of 2 thirds. And I know that within this interval, my tangent values can only be positive in quadrant 1, so that's where my triangle is going to go. Now we've completed step number 1. We can move on to step number 2. And now we want to go ahead and actually draw that right triangle in quadrant 1 with that side length down against the x-axis. And then I'm going to label my angle theta as that inside angle right there. Now looking at my argument, I have 2 thirds, and my inside function tells me that the tangent of that angle that I just labeled has to be equal to that 2 thirds argument. So based on SOHCAHTOA, I know that my opposite side has to be 2 and my adjacent side has to be 3. Now, we've completed step number 2. Moving on to step number 3, we want to use the Pythagorean theorem in order to find that missing third side. Now here our missing side length is our hypotenuse. So setting up our Pythagorean theorem here, a2+b2=c2. I'm solving for c here because my missing side is the hypotenuse. So here I take 22+32, those two side lengths that I already know. That's equal to c squared. Now 22 is 4, and 32 is 9. That's equal to c squared. Adding those two values together, I get that 13 is equal to c squared. Now my final step here in getting this missing side length is taking the square root of both sides. Now root 13 cannot be simplified anymore, so that's just what my side length is. Root 13 is equal to my hypotenuse, that missing side length. So I can go ahead and label that up here on my triangle. And we have completed step number 3. Now moving on to our final step, we are going to use our right triangle in order to evaluate that outside function. Now here, my outside function is the sine. So I am tasked with finding the sine of my angle theta, which I know I need my opposite side and my hypotenuse based on SOHCAHTOA. So I have all the information I need here. Looking at my triangle, I'm going to take that opposite side of 2 and divide it by my hypotenuse, which is the square root of 13. Now this is technically a correct answer. But whenever we have a radical in the denominator, we want to go ahead and rationalize that denominator, which we can do here by multiplying this by root 13 over root 13. So that gets our radical, out of that denominator, leaving me with 2 root 13 on the top and just 13 on the bottom. So my final answer here is that the sine of the inverse tangent of 2 thirds is equal to 2 root 13 over 13. Thanks for watching, and let me know if you have questions.

Evaluate the expression.

$\tan\left(\cos^{-1}\frac{12}{13}\right)$

$\frac{12}{13}$

$\frac{12}{5}$

$\frac{5}{13}$

$\frac{5}{12}$

Evaluate the expression.

$\sin\left(\tan^{-1}\frac{15}{8}\right)$

$\frac{8}{15}$

$\frac{15}{17}$

$\frac{8}{17}$

$\frac{15}{289}$

Evaluate the expression.

$\cos\left(\sin^{-1}\left(-\frac{7}{25}\right)\right)$

$\frac{7}{24}$

$\frac{8}{25}$

$\frac{24}{25}$

$-\frac{24}{25}$

### Example 4

#### Video transcript

Everyone, in this problem, we're asked to evaluate the expression, the tangent of the inverse sine of xx2+4. Now, this problem might look really overwhelming because there are multiple variables. There's a square root. There's a bunch of stuff going on. But we can still attack this the same way we would any of these composite trig functions that are definitely not on our unit circle. And xx2+4, definitely not on there. So let's go ahead and get started with step number 1.

Now for step 1, we're going to take that inside function, and we're going to use our interval to determine our quadrant. Now here, looking at this inside function, I am faced with finding the inverse sine. And now when working with the inverse sine, I know that I am working with angles from from negative π/2 to positive π/2. So my triangle either has to be in quadrant 4 or quadrant 1. Now looking at the sine of our argument here, our argument is xx2+4, this is a positive value, so that means that my triangle has to go in quadrant 1, or my sine values are positive. So step 1 is done. We can move on to step number 2, which is to draw our triangle and use our argument in order to label our angle theta and our two sides.

So up here in quadrant 1, let's go ahead and draw our right triangle. And with my right triangle, I'm going to go ahead and label my angle theta right here. And with this argument, this tells me that the sine of my angle theta is equal to xx2+4. Now this seems strange, but hang with me here. Using SOHCAHTOA, this tells me that my opposite side has to be x, and my hypotenuse is the square root of x2+4. So we're still working through this the same way. We've completed step number 2. We can move on to step number 3, where we're going to use the pythagorean theorem to find that missing third side.

So let's go ahead and set our pythagorean theorem up here. We have a^{2} + b^{2} = c^{2}, and we're going to be solving for one of our leg lengths, so either a or b. Here, I'm going to solve for a. So I'm going to keep a as variable, a^{2}, and then plus my b squared. B here is x. I'm squaring that value. And that's equal to c^{2}. Now c here is this whole expression, the square root of x2+4, and all of that gets squared. Now from here, let's simplify this algebraically. Now here on my left side, I can't really do anything. I'm just left with a^{2} + x2. But on that right side, if I square this, I'm going to get rid of that square root. So I'm just left with x2+4. Now from here, if I'm solving for a, I want to go ahead and isolate a by subtracting x2 from both sides. Now when I do that, it cancels on that left side, but it also cancels on that right side. So I'm simply left with a^{2} = 4. This looks like a much more friendly expression. So here, we're just left to take the square root of both sides. We're left with a equals the square root of 4, which is 2. So our missing side length here is 2. Now we've completed step number 3. We can move on to our final step, which is to use our triangle in order to evaluate that outside function.

Now, here, our outside function is the tangent. And in order to find the tangent of our angle theta, using SOHCAHTOA, we're going to take our opposite side. We're going to divide it by our adjacent side. We have all of that information. Here, our opposite side is x. Our adjacent side is 2. So my final answer here is x2. The tangent of the inverse sine of xx2+4 is equal to x2. And we're completely done here. Thanks for watching, and I'll see you in the next one.

### Here’s what students ask on this topic:

How do you evaluate the sine of the inverse cosine of 1/2?

To evaluate the sine of the inverse cosine of 1/2, start with the inside function, which is the inverse cosine of 1/2. This asks for the angle whose cosine is 1/2. From the unit circle, we know that cos(π/3) = 1/2. So, the inverse cosine of 1/2 is π/3. Next, evaluate the sine of π/3. From the unit circle, sin(π/3) = √3/2. Therefore, the sine of the inverse cosine of 1/2 is √3/2.

What is the cosine of the inverse tangent of 0?

To find the cosine of the inverse tangent of 0, start with the inside function, which is the inverse tangent of 0. This asks for the angle whose tangent is 0. From the unit circle, tan(0) = 0. So, the inverse tangent of 0 is 0. Next, evaluate the cosine of 0. From the unit circle, cos(0) = 1. Therefore, the cosine of the inverse tangent of 0 is 1.

How do you solve the inverse cosine of the sine of π/3?

To solve the inverse cosine of the sine of π/3, start with the inside function, which is the sine of π/3. From the unit circle, sin(π/3) = √3/2. Next, evaluate the inverse cosine of √3/2. This asks for the angle whose cosine is √3/2. From the unit circle, cos(π/6) = √3/2. Therefore, the inverse cosine of the sine of π/3 is π/6.

How do you evaluate the sine of the inverse tangent of 3/4?

To evaluate the sine of the inverse tangent of 3/4, start with the inside function, which is the inverse tangent of 3/4. This asks for the angle whose tangent is 3/4. Draw a right triangle where the opposite side is 3 and the adjacent side is 4. Using the Pythagorean theorem, the hypotenuse is √(3² + 4²) = 5. Now, evaluate the sine of this angle, which is the opposite side over the hypotenuse: sin(θ) = 3/5. Therefore, the sine of the inverse tangent of 3/4 is 3/5.

What is the sine of the inverse cosine of -5/13?

To find the sine of the inverse cosine of -5/13, start with the inside function, which is the inverse cosine of -5/13. This asks for the angle whose cosine is -5/13. Draw a right triangle in the second quadrant where the adjacent side is -5 and the hypotenuse is 13. Using the Pythagorean theorem, the opposite side is √(13² - 5²) = 12. Now, evaluate the sine of this angle, which is the opposite side over the hypotenuse: sin(θ) = 12/13. Therefore, the sine of the inverse cosine of -5/13 is 12/13.