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. 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 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 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 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.