 This is a video about inverse trig functions, part one. I'll start off by talking about the graph of y equals sine x. This is the graph of the sine function. This is two periods of the graph. It continues going on in the rightward direction forever and the left hand direction forever. The domain of the sine function is negative infinity to infinity and the range is negative one to one. The issue here is for a function that have an inverse, it must be one to one, meaning it must pass the horizontal line test. If I was to go upward on this graph with the horizontal line, I would touch multiple points at the same time, which means that the function cannot have an inverse. So what I have to do is I have to restrict the domain to the following, negative pi over two to pi over two. So I'm gonna cut the graph from negative pi over two to pi over two and this leaves this upward diagonal looking figure right here. That's the only thing left of the graph. So negative pi over two to pi over two is my domain for this function that have an inverse. This is what makes the function one to one. So as a result, the inverse sine function, sine inverse of X or inverse sine of X. The sine function, when the domain is restricted to negative pi over two to pi over two and the range is negative one to one, it always is. The inverse sine function, its domain then becomes negative one to one and its range then becomes negative pi over two to pi over two. The roles of the domain and the range switch places. The range of the original becomes the domain of the inverse and then the domain of the original becomes the range of the inverse. The reason why is because trig functions as we have come to learn are applied to angles and the answer is a measurement. So I plug angles into the sine function and out comes some measurement from negative one to one. Inverse trig functions, they're applied to measurements. So in the case of the inverse sine function, they're applied to negative one to one and then the result is an angle. So we're working backwards now. We're gonna be given a measurement and our goal is to find the angle which gave that measurement. That's what we're going to be doing. So let's talk exclusively about the inverse sine function. Find the value of each of the following. So inverse sine of one, what this means is, sine of what angle theta is equal to one. So where does sine at what angle the sine equal one? I'll write this out in words for this example or this part here. So I'll say where is the sine function equal to one. Where does the sine function equal to one? Now let's think about this for just a minute. I'm gonna draw my unit circle. I'm gonna focus on my key points for this part here. So remember I have one zero upper portion here. I have zero one left hand portion. I have negative one zero bottom portion. I have zero negative one. Where is sine equal to one? So remember my angle that is spit out from inverse sine has to be between negative pi over two and pi over two. That would be negative quadrant four, the quadrant one. So somewhere in this right hand side of my coordinate plane. Now remember sine is the second coordinate. Which of these ordered pairs has a second coordinate of one? And that would be pi over two. So in this example, theta is pi over two. So inverse sine of one is equal to pi over two. The original angle was 90 degrees. All right, moving on to our next example. Inverse sine of negative one half. Now remember this value that we get from this inverse sine function has to be in either quadrant one or quadrant four. So I have a choice. It's going to be quadrant four or quadrant one. However, since sine is negative, we are in which quadrant it all trig functions are positive in one, sine is negative and quadrant four. Only cosine and secant are positive. All right, so we are in four. We are in quadrant four. Keep that in mind as I go through and I say, okay, sine of what angle is equal to negative one half? So I'm going to draw a triangle here. In this triangle, I'm going to have some sort of reference angle here. And that angle will be moved around to the appropriate quadrant. So sine of theta equals negative one half. Sine is opposite over hypotenuse. Opposite's one hypotenuse is two. So one, two, which means the third side has to be square root of three. So this is a 30, 60, 90 triangle. What angle is always across from the side with measure one? Well, the angle is pi over six or 30 degrees. But we need to move this to quadrant four. We must move to quadrant four. It has to be negative quadrant four because remember we have to be between negative pi over two and pi over two. Negative pi over two to zero is quadrant four, the negative part of it. And zero to pi over two is quadrant one. So to move an angle to quadrant four from quadrant one, you can simply just make the angle negative. So if an angle is positive in quadrant one, you can move it to quadrant four just by making it negative. So my angle is actually going to be negative pi over six. Theta is negative pi over six. So the answer to this inverse trick expression is negative pi over six. That is our answer. Next, part C, inverse sine of square root of three over two. Now my issue is when I go to draw my triangle, do I know anything that has a square root of three and a two? And the answer is yes, that would be 30, 60, 90 triangle relationships here. So we'll figure out what the answer is, but remember I have to have a final angle that rests in quadrant one or quadrant four. So I have a positive value here since sine is, since sine is positive, we are in one. We are in quadrant one in this case. So I have sine of what angle, that will be a quadrant one angle is equal to square root of three over two. Let's draw our triangle. We have some mystery angle we're trying to find and sine, it's square root of three over two. Opposite is square root of three, hypotenuse is two, which means the other side would have to be one. Who is always across from square root of three? What is my mystery angle? The angle is, well, it's a 60 degree angle. It's pi over three. And this is in quadrant one already. So it can be my answer because I need angle in quadrant one. So theta is equal to pi over three. So I have pi over three or 60 degrees is my answer. So this is using the inverse sine function to find original angle measures. Now we're gonna go in and we're gonna talk about all of the other inverse trig functions. But first I want us to kind of have a throwback and a review here of how to find reference angles. For any angle theta, the reference angle gamma is found from the following. Quadrant two. So if your original angle is in quadrant two, the reference angle is 180 minus the original angle or pi minus the original angle. Quadrant three. The reference angle gamma is equal to theta minus 180. Quadrant four, the reference angle is equal to 360 minus theta. Therefore, I can do the same thing. If I have an angle that is in quadrant one, I can move it back to its original quadrant using these formulas because that's what I'm doing here is I'm getting quadrant one angles and then I have to either remove them to quadrant four, I'll have to move them possibly to quadrant two here in a second as you'll see. So for instance, if I look at the quadrant two reference angle formula, and this is the one we will be using the most, I have gamma is equal to 180 minus theta. You add theta to both sides, you get theta plus gamma is equal to 180. Then you take away gamma from both sides and then you'll get theta equals 180 minus gamma. And that is exactly how we get this first formula in the bottom part of the slide here. Quadrant two, the original angle is equal to 180 minus the reference angle or pi minus the reference angle. All of the formulas on the bottom half here can be found by rearranging the formulas on the top half. So what this shows us are the inverse trig functions of the original six trig functions we originally studied. As you can see, sine function negative one to one, range is negative pi over two to pi over two. So this would be quadrant one and negative quadrant four. Cosine function that domains negative one to one but the range is zero to pi now. So we take the cosine function, we cut it from zero to pi. That's what makes it one to one. That's why the range of the inverse cosine function is zero to pi. So that would be quadrant one and quadrant two. Inverse tan, the domains negative infinity to infinity, the range is negative pi over two to pi over two. So this would be quadrant one and then you have negative quadrant four. Inverse cosecant, once again, you have quadrant one. You have negative quadrant four. That's where your answers can be. So the only places. And then inverse secant, that would be quadrant one and two. And then there's inverse cotangent, which the answer must also be in quadrant one and quadrant two. So you have two options here. You have three trig functions which their inverse gives you angles in quadrants one and two. Then you have three inverse trig functions which give you angles in quadrants one and quadrant four, the negative quadrant four. So let's work some examples. My first example is gonna be inverse tan of square root of three over three. My issue is this. I do not have a right triangle relationship that uses square root of three and three. The reason why this looks unfamiliar is because somebody took the original measurement that was in here and rationalized it. So what this is is saying tan of what angle is square root of three over three. I wanna undo the rationalizing here. So multiply by square root of three over itself. This gives me tan of theta is equal to three over three square root of three, which is one over square root of three. This looks more familiar. I know a relationship that has one and square root of three and it's a 30, 60, 90 triangle relationship. So let me go ahead and make my triangle with my mystery angle here. Tan of this angle is opposite as one adjacent to square root of three, which means hypotenuse has to be two. So looking here, who's always across from one? Well, that would be 30 degrees. That would be pi over six is our angle. Now the question is inverse tan, our answers have to either be in quadrant one or negative quadrant four. How do I figure this out? Well, look at the sign of the original value. We had square root of three over three. First, to reiterate, your final answer must be in the interval negative pi over two to pi over two. It's quadrant one or negative quadrant four. Now, question is which one is it? Since tan of theta is positive, since it's positive, theta must be in quadrant one. So whatever our mystery angle is, this question mark is just saying the reference angle. The actual angle theta, the answer to this inverse trig expression is going to be quadrant one. So since pi over six is already in one, so I'll write already in one, we don't have to do any moving around. Our answer, inverse tan of square root of three over three is pi over six. Next, inverse cosine, I have negative square root of two over two. When we're talking about cosine, inverse cosine, our final answer must be in the interval zero to pi. Zero to pi. It has to be in the interval zero to pi. That's quadrant one or quadrant two. I have cosine theta equals negative square root of two over two. Since cosine is negative, so since cosine is negative, our angle, our answer, theta must be in quadrant two. Must be in quadrant two. Now, the issue is I have the square root of two over two and I don't know any triangles with the relationship that have square root of two and two in them. Once again, we're gonna have to go through and undo the rationalizing. Multiply by square root of two over itself. That gives me two over two square roots of two. That gives me one over square root of two. So now I have cosine theta equals negative one over square root of two, which this enables me now to make my triangle. So I make my triangle. I have my mystery angle, my quadrant one reference angle, I guess is what you could call it. Dealing with cosine adjacent over hypotenuse. Adjacent is one, hypotenuse is square root of two, which means this other side has to be one. This is a 45, 45, 90 triangle. So our angle, our question mark here is pi over four. But I have to move this angle to quadrant two. That's where my angle needs to be. So my original angle theta, using the formula from the previous slides is pi minus my reference angle, pi minus pi over four. This would be four pi over four minus pi over four, which is three pi over four. So the answer here, our theta, our angle, the inverse cosine function's value for this example would be three pi over four. That is a quadrant two angle. Inverse tan of zero. Anytime you see zero or in here, you should think about the key points. That's what we wanna think about here. I'm gonna turn this into tan of what angle is zero over one, zero over one. We need to have a ratio, so that way I can know what cosine is and sin is. My angle, my final answer, final answer must be in the interval negative pi over two, two pi over two. So now I'm gonna draw my unit circle. Looks a little bit nicer than the first one we did on this lecture, label these key points and which of these points has sin of zero and cosine of one. Remember cosine is the first of the ordered pairs and sin is the second. So one zero is what I'm looking for. And that puts me right here at zero. So my angle here, theta is zero. So that is how to evaluate inverse trig expressions.