0. Functions
Inverse Trigonometric Functions
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everyone. You may remember that just like a square root undoes something being squared, an inverse trig function, like, say, the inverse cosine undoes its corresponding trig function. But that's pretty much where we last left off when dealing with inverse trig functions in the context of right triangles. But 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 1 half. Now I'm going to show you exactly how to get there. And here, I'm going to specifically break down everything that you need to know about the inverse cosine function. So let's go ahead and get started. Now 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. So 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, say, the inverse cosine of 1 half, we want to find the angle that has the corresponding cosine value, which we can do on our unit circle. So let's come down here to this example here and go ahead and evaluate this expression. Here, we're asked to find the inverse cosine of 1 half. Now, another way that you can think of this, and this will work for any inverse trig function, is for the inverse cosine of onetwo. I can also think of this as, okay, the cosine of what angle theta will give me a value of onetwo. And 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 1 half. So 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 onetwo. 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. Only one of them is. So how do we determine which one of these is actually my solution? Well, in order to do that, we actually need to dive a bit deeper into our inverse cosine function. So we're actually going to take a look at the graph of cosine. So here we have the graph of our cosine function. And 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. So because of that, we actually only wanna reflect part of our cosine graph in order 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, x, I end up with the graph of inverse cosine. But the way that these graphs look is not the most important part, 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. Now the values that we put into our inverse cosine function have to 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. So for the inverse cosine of 1 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 1 half is pi over 3. So 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.
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