Inverse Trigonometric Functions - Online Tutor, Practice Problems & Exam Prep

Inverse Cosine

Hey, everyone. You may remember that just like a square root undoes something being squared, an inverse trig function, like the inverse cosine, undoes its corresponding trig function. But that's pretty much where we left off when dealing with inverse trig functions in the context of right triangles. We now know much more about trig functions in general, and we also know about the unit circle. So, we can dive much deeper into our inverse trig functions using the unit circle.

And you'll now be asked to evaluate expressions like the inverse cosine of one half. I'm going to show you exactly how to get there. Here, I will specifically break down everything you need to know about the inverse cosine function. So let's go ahead and get started. When working with a basic cosine function, we start with an angle and we take the cosine to end up with a value.

Now with the inverse cosine, because we're undoing the cosine function, we instead start with that value and then take the inverse cosine to end back up with an angle. It's important to note that when working with our cosine function, we start with an angle and we end up with a value. Whereas with the inverse cosine, we instead start with that value and then end up back at an angle. So, when evaluating an expression like the inverse cosine of one half, we want to find the angle that has the corresponding cosine value which we can find on our unit circle. So let's come down here to this example and go ahead and evaluate this expression.

Here, we're asked to find the inverse cosine of one half. Another way to think about this, and this will work for any inverse trig function, is as "the cosine of what angle theta will give me a value of one half?" I want to find the angle that gives me that corresponding cosine value. So, coming over here to my unit circle, I know that in quadrant 1, for my angle pi over 3, this has a corresponding cosine value of one half.

π / 3

This pi over 3 represents an angle that gives me my corresponding value. Now, on my unit circle, as I continue around, I also know that in quadrant 4, all of my cosine values are positive here as well. So for my angle of 5pi over 3, that also has a cosine value of positive one half. So 5pi over 3 represents another angle that gives me my corresponding value. But both of these angles are not the solution to my original expression.

5 π / 3

Only one of them is. So how do we determine which one of these is my solution? Well, we need to dive a bit deeper into our inverse cosine function. We're going to look at the graph of our cosine function.

Remember, to get our inverse function, we can simply take our original function and reflect it over the line y equals x. But because my cosine is not a one-to-one function, if I were to reflect this entire cosine graph over that line y equals x, I would end up with something that is not a function at all, and it would not pass the vertical line test. Because of that, we actually only want to reflect part of our cosine graph to get our inverse cosine. And that specific part is between 0 and pi. So if I'd reflect just this part over the line y equals x, I end up with the graph of inverse cosine.

The most important aspect, however, is not how these graphs look, but rather the specific values for which my inverse cosine function is defined. Now remember, I said when working with the inverse cosine function, we put values into it, and we get angles back out. The values that we put into our inverse cosine function must be between negative one and one, and the angles that we get back out must be between 0 and pi. So when faced with evaluating an inverse cosine expression, our solution can only be angles within the interval 0 to pi. So, with that in mind, let's come back up to our example here and determine what our final solution actually is.

For the inverse cosine of one half, looking at my unit circle here, I only want to consider angles between 0 and pi, my specified interval here. So those bottom two quadrants are not going to have any of my solutions. That's just not going to work. I only want angles in that top half. So 5 pi over 3 is right down here in quadrant 4, but this cannot be a solution because it is not within my specified interval.

So that leaves me with my final answer of pi over 3, which is up here in quadrant 1 between 0 and pi. So my final answer here is that the inverse cosine of one half is pi over 3. When evaluating inverse cosine expressions, we are still going to look for angles with our corresponding values on the unit circle, but only within our specified interval from 0 to pi. So now that we know how to evaluate inverse cosine expressions, let's get some more practice. Thanks for watching, and I'll see you in the next one.

Example 1

Hey everyone. In this problem, we're asked to evaluate the expression the inverse cosine of negative 32. Now whenever working with an inverse trig function, remember that we can also think of this as, "Okay, the cosine of what angle is equal to negative 32?" And we want to find the angle for which that is true. Now when working with the inverse cosine, we know that our values, our angles, can only be between 0 and π.

But we can actually get even more specific here because we know that all of our cosine values are going to be positive in quadrant 1, and all of our cosine values in quadrant 2 are going to be negative. So whenever we're taking the inverse cosine of a positive number, we know that our solution has to be in quadrant 1. And whenever we're taking the inverse cosine of a negative number, just like we are here, we know that our solution has to be in quadrant 2. So here we already know that our angle has to be in the second quadrant. So for which one of these angles is the cosine equal to negative 32?

Well, I know that the cosine of 5π over 6 is equal to just that so that represents my solution. My angle here is 5π over 6 and we are done here. Thanks for watching, and I'll see you in the next one.

Evaluate the expression.

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

• A. $0$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. $\frac{3\pi }{2}$

Evaluate the expression.

$\cos^{-1}\left(0\right)$

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $\frac{3\pi }{2}$

Evaluate the expression.

$\cos^{-1}\left(-\frac{\sqrt2}{2}\right)$

• A. $\frac{\pi }{4}$
• B. $\frac{3\pi }{4}$
• C. $\frac{5\pi }{4}$
• D. $\frac{7\pi }{4}$

Inverse Sine

Hey, everyone. We just saw that for our inverse cosine function, we were able to flip our cosine graph but only over a specific interval. Well, you'll also be faced with evaluating inverse sine expressions. And that may seem like it's going to be annoying or a bit difficult, but it's actually the exact same principle. So we're going to do the exact same thing here that we did with our inverse cosine function, and we'll see that our graph and our intervals will just work out a little bit differently, but I'm going to walk you through all of that here.

So let's go ahead and jump right in. Now here we have our sine graph, and remember that to get the graph of our inverse sine function, we need to flip this over the line y=x. But because this function is not 1 to 1, I only want to flip part of it. The specific part I want to flip is between -π2 and positive π2. So flipping that, reflecting it over the line y=x, I end up with a graph that looks something like this for my inverse sine function.

But remember, the graph is not the most important part, but rather the values for which this function is defined. Now remember, when working with any inverse trig functions, we put values into them, and we get angles back out. The values that we put into our inverse sine function have to be between negative one and positive one. And the angles that we get back out have to be between -π2 and π2. So now with this interval in mind, we can go ahead and evaluate inverse sine expressions with no problem.

So let's come down here to our first example, which asks us to evaluate the expression, the inverse sine of one half. Now remember, when working with inverse trig functions, we want to find angles that have corresponding values on the unit circle. So we can also think of this expression as asking us, Okay, the sine of what angle θ gives me a value of one-half. But remember, we must consider our interval here. So remember, that for our inverse sine functions, the value that we put in has to be between negative one and positive one, which one-half definitely is.

And the angle that we get back out for our answer has to be between -π2 and π2. So let's come down to our unit circle here. Now, on our unit circle, we want to find an angle θ for which our sine value is one half. And looking at my unit circle here, I see that for my angle of π6, I end up with a sine value of one-half. So this represents the solution to my expression, π6.

Now, you may be tempted here to come over to 5π6 and think, okay, isn't the sine of 5π6 also equal to one-half? And it is, but the problem is that this is not within our specified interval from -π2 to π2. So I actually don't care about that half of my unit circle at all when dealing with the inverse sine. I only want to consider angles on this half of my unit circle between -π2 and π2. So this solution, π6, is my final solution here and is my only solution to the expression inverse sine of one-half.

Let's come to our next example here, which is the inverse sine of negative root 2 over 2. Now remember, we can also think of this as, okay, the sine of what angle θ gives me a value of negative square root 2 over 2. And coming over here to my unit circle, I see that down here in quadrant 4, for negative π4, I know that my sine value is going to be negative square root 2 over 2. But something else that you might have noticed here is that all of these values in quadrant 4 are negative. And on a typical unit circle, this negative π4 would actually be 7π4.

So what’s going on here? Well, the problem is that 7π4 is not within my specified interval of -π2 to π2. So whenever dealing with the inverse sine function, we consider this quadrant 4 as being measured clockwise from 0 rather than counterclockwise so that all of our values are within that interval, -π2 to π2. So here, my solution is negative π4 within that specified interval. So the inverse sine of negative root 2 over 2 is negative π4.

So now that we know how to evaluate inverse sine expressions, let's get a bit more practice together. Thanks for watching, and I'll see you in the next one.

Example 2

Hey everyone. In this problem, we're asked to evaluate the expression the inverse sine of negative one. Now, remember when working with inverse trig functions, we can also think of this as "the sine of what angle is equal to negative one," and we're looking for the angle for which that is true. Now when working with the inverse sine specifically, we know that our solution can only be angles between negative pi over 2 and positive pi over 2, but we can get even more specific here because all of my sine values in this first quadrant, quadrant 1, from 0 to pi over 2, are going to be positive. Whereas in the 4th quadrant, from negative pi over 2 to 0, all of these sine values are going to be negative.

So if I'm taking the inverse sine of a positive value, I know that my solution has to be in quadrant 1. Whereas if I'm taking the inverse sine of a negative value, like we are here, the inverse sine of negative one, we know that our solution has to be in quadrant 4 between negative pi over 2 and 0. So here, since we're taking the inverse sine of negative one, I already know that my solution has to be down here in this 4th quadrant. So, where is the sine equal to negative one? Well, for negative pi over 2, I know that my sine, or my y-value, is equal to negative one.

So here my angle, and my solution to this expression is negative pi over 2, and we are done here. Thanks for watching and I'll see you in the next one.

Evaluate the expression.

$\sin^{-1}1$

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $-\frac{\pi }{2}$

Evaluate the expression.

$\sin^{-1}\left(\frac{\sqrt3}{2}\right)$

• A. $\frac{\pi }{6}$
• B. $\frac{3\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $-\frac{\pi }{3}$

Evaluate the expression.

$\sin^{-1}\left(\frac{\sqrt2}{2}\right)$

• A. $\frac{\pi }{4}$
• B. $-\frac{\pi }{4}$
• C. $\frac{3\pi }{4}$
• D. $\frac{7\pi }{4}$

Inverse Tangent

Hey, everyone. We just saw that for our inverse sine and cosine functions, we only worked with values and angles within a specified interval. So, of course, we have to do the same thing when working with our inverse tangent function. So we can work through this the exact same way. And here I'm going to show you what the interval is when dealing with the inverse tangent function, and we'll evaluate some expressions together.

So let's go ahead and just jump right into this. Now here I have the graph of my tangent function, and in order to get the graph of my inverse tangent function, I, of course, need to reflect this over the line y equals x. But this graph is not 1 to 1, as you might expect, so we only want to reflect part of it. And that specific part is between negative π/2 and positive π/2. So reflecting this over the line y equals x, I end up with a graph that looks something like this for my inverse tangent function.

But remember, the graph is not the most important part, but the intervals are. So the values that we put into our inverse tangent function are actually not anything between a negative infinity and infinity. But the angles that we get back out are restricted. They must be in between a negative π/2 and positive π/2. So with this interval in mind, let's go ahead and just evaluate some expressions.

Now the first expression that we're tasked with evaluating here is the inverse tangent of the square root of 3. Now remember, for any inverse trig functions, we can also think of this as, Okay, the tangent of what angle theta gives me a value of root 3. Now looking at our unit circle here, the specific interval that I am working with is between negative π/2 and positive π/2. So I really just don't care what's going on on this half of my graph at all. I am only looking for an angle for which the tangent is the square root of 3 on this half of my graph.

And it's actually the same exact interval we already saw when working with the inverse sine. So let's go ahead and find our angle for which the tangent is the square root of 3. Now looking in this first quadrant here, I see that for π/3 my tangent is root 3. So that tells me that my answer here is π/3. That is within my specified interval, and I don't have to worry about anything else.

Here is my answer. Now we've completed example a. We can move on to example b, which asks us to find the inverse tangent of negative one. Now, again, we can think of this as, okay, the tangent of what angle theta gives me a value of negative one, and we want to find that angle within our specified interval. So in my interval from negative π/2 to positive π/2, where is my tangent negative one?

Well, I know that for my angle negative π/4, my tangent is going to end up being negative one. So this gives me my final answer. The inverse tangent of negative one is negative π/4, and we're done here. Now as you continue to work through problems, you may have to evaluate some inverse trig functions using a calculator, either because you're explicitly asked to or because there's no other way to evaluate the function. But in order to do this, all you have to do is press the second button and then whatever your corresponding trig function is.

So in order to find the inverse sine, you would press 2nd and then sine, inverse cosine, 2nd and then cosine, or inverse tangent, 2nd and then tangent. And then you'll just type your value in and come to an answer. Now when doing this, you'll typically be in radian mode unless otherwise specified. Now that we know how to evaluate any inverse trig function, sine, cosine, or tangent, let's continue practicing together. Thanks for watching and I'll see you in the next one.

Example 3

Hey everyone. In this problem, we're asked to evaluate the expression the inverse tangent of negative square root of 3. Now when working with inverse trigonometric functions, we know that we can think of this as asking, "The tangent of what angle is equal to negative square root of 3?" and we want to find the angle for which that is true. Now looking at our unit circle, we know that when working with the inverse tangent, our value has to be between negative π/2 and π/2, which is the exact same interval that we dealt with when working with the inverse sine. Again here, we could get even more specific because all of my tangent values in quadrant 1 are going to be positive, whereas all of my tangent values in quadrant 4 are going to be negative.

So, if I'm taking the inverse tangent of a positive value, I know my solution has to be an angle in quadrant 1. Whereas if I'm taking the inverse tangent of a negative value, my answer has to be an angle in quadrant 4. Now here, since I'm taking the inverse tangent of the negative square root of 3, I am looking for an angle in this bottom quadrant, quadrant 4. So, for which of these angles is the tangent negative square root of 3? Well, for negative π/3, since this is a reference angle to π/3, I know that the tangent of this value is going to be negative square root of 3.

So that represents my solution: negative π/3. The inverse tangent of negative root 3 is negative π/3, and we are good to go here. Thanks so much for watching, and I'll see you in the next one.

Evaluate the expression.

$\tan^{-1}0$

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $-\frac{\pi }{2}$

Evaluate the expression.

$\tan^{-1}1$

• A. $0$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $-\frac{\pi }{4}$

Evaluate the expression.

$\tan^{-1}\left(-\frac{\sqrt3}{3}\right)$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $-\frac{\pi }{6}$
• D. $-\frac{\pi }{3}$

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.

$\tan^{-1}\left(5\right)$

• A. $-5$
• B. $-3.38$
• C. $1.37$
• D. $78.69$

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.

$\sin^{-1}\left(-\frac13\right)$

• A. 0.34
• B. – 0.34
• C. – 0.33
• D. – 19.47

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.

$\cos^{-1}\left(\frac14\right)$

• A. 75.52
• B. 1.82
• C. 1.32
• D. 0.97

Evaluate Composite Functions - Values on Unit Circle

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 in order to get a solution for this entire expression.

Now, here I'm going to walk you through exactly how to do that. So let's go ahead and get started. 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 like 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, 32. 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 32? 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 32. I see that for pi over 6, the cosine is 32. 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.

Evaluate the expression.

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

• A. $0$
• B. $1$
• C. $-1$
• D. $\frac12$

Evaluate the expression.

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

• A. $\frac{\pi}{6}$
• B. $\frac{5\pi}{6}$
• C. $\frac{\pi}{3}$
• D. $-\frac{\pi}{6}$

Example 1

Hey, everyone. In this problem, we're asked to evaluate the expression that the secant of the inverse cosine of negative root 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 root 3 over 2. Now because this is an inverse trig function, remember that we can also think of this as, okay, the cosine of what angle will give me negative root 3 over 2?

Now also remember, because this is an inverse trig 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 root 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 root 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 trig 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 root 3 over 2. So this is really 1 - 3 2 .

Now whenever I have one over a fraction, I can really just flip that fraction. So this is going to give me -2 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 actually want to rationalize this denominator, which we can do by multiplying by 3 3 . Now that will give me a final answer of - 2 √ 3 3 .

And that's my final answer. The secant of the inverse cosine of negative root 3 over 2 is negative 2 root 3 over 3. Thanks for watching. And let me know if you have questions.

Evaluate Composite Functions - Special Cases

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 \( \frac{11\pi}{6} \). Now looking at this, knowing that trig functions and their inverse undo each other, you may be tempted here to cancel this inverse cosine and cosine, leaving you with just \( \frac{11\pi}{6} \) as your final answer. But this would actually be entirely wrong. Because remember, when dealing with inverse functions, you have to consider the interval for which they're defined.

Now, you may be worried that this is going to make these sort 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.

So, in this first example, the inverse cosine of the cosine of \( \frac{11\pi}{6} \), we cannot assume that our final answer is \( \frac{11\pi}{6} \). And 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 \( \frac{11\pi}{6} \). Now, in finding the cosine of \( \frac{11\pi}{6} \), I can just come right on down to my unit circle, and I see that for \( \frac{11\pi}{6} \), the cosine is \( \sqrt{\frac{3}{2}} \). So, for that inside trig function here, I get an answer of \( \sqrt{\frac{3}{2}} \). Now that I've dealt with that inside trig function, I can deal with my outside trig function, which is the inverse cosine of that \( \sqrt{\frac{3}{2}} \).

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 \( \sqrt{\frac{3}{2}} \)? But remember, also when dealing with these inverse trig functions, we have to consider their interval. So specifically, when dealing with the inverse cosine, our interval for our angles is between 0 and \( \pi \). So looking at our unit circle within this specified interval, I see that for my angle \( \frac{\pi}{6} \), my cosine is \( \sqrt{\frac{3}{2}} \).

So that gives me my final answer here. The inverse cosine of \( \sqrt{\frac{3}{2}} \) is \( \frac{\pi}{6} \). And that gives me my final answer. And you see that, indeed, this is not equal to \( \frac{11\pi}{6} \), so we have to be super careful here. Now let's go ahead and move on to our next example where we're 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're 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 of 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, I 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

Hey, everyone. In this problem, we're asked to evaluate the expression the inverse sine of the sine of pi over 6. Now, here, you might be tempted to just cancel the inverse sine with the sine and end up with your answer of pi over 6. 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 pi over 6.

Now coming over here to my unit circle, the sine of pi over 6 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 a negative pi over 2 to a positive pi over 2. 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 pi over 6 is equal to 1/2, and that is within my interval. So that actually gives me a final answer of pi over 6. 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.

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

• A. $\frac{\pi}{3}$
• B. $\frac{2\pi}{3}$
• C. $\pi$
• D. Undefined

Evaluate the expression.

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

• A. $0$
• B. $\frac{\pi}{2}$
• C. $\pi$
• D. Undefined

Evaluate the expression.

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

• A. $-\frac{\pi}{3}$
• B. $\frac{\pi}{3}$
• C. $\frac{2\pi}{3}$
• D. $\frac{5\pi}{3}$