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.