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.