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.