 In our previous video, we introduced the idea of the inverse trigonometric functions, specifically arc sine, arc cosine, and arc tangent. In this video, what I wanna do is show you how you can do calculations with the inverse trigonometric functions without necessarily grabbing our calculator. I mean, sometimes you're gonna have to, but like with sine, cosine, and tangent, there are some special angles that we memorized from our unit circle diagrams, right? And so like, you know, if you're doing the angle of pi-sixth, for example, we know that the x-coordinate there coincides with cosine, which then, in that case, we're gonna end up with root three over two, and then the y-coordinate, which represents sine, is gonna be one-half. So using special angles, we can also turn this around and do some special calculations, special ratios associated to sine inverse. So if you wanna do sine inverse of one-half, what is that equal to? Well, one thing I like to do when I think of inverse trigonometric functions, you have to think of it as an angle, right? Sine inverse of one-half means you want the angle which is put into the sine function gives you the ratio of one-half, all right? So whenever you see an inverse trigonometric function, sometimes I like to denote it as theta. Theta is sort of like my go-to variable for angles. It makes me think of an angle that way. Sine inverse of one-half is an angle. And because it's an inverse function, we could move sine inverse to the side and we try to solve the equation one-half equals sine of theta. So we have to solve this. Computing sine inverse of one-half comes down to solving one-half equals sine theta. And so if we think of that in terms of the unit circle, we're looking for those coordinates, those Y coordinates, which the ratio, that is equal to one-half, excuse me. So what angle will give you a ratio of one-half? And so we can already see this labeled on the screen right here. We're gonna see that theta is equal to pi-sixth. In other words, sine inverse of one-half equals pi-sixth. Now when it comes to the equation sine in, when it comes to the equation one-half equals sine theta, there's technically two solutions to this. There's gonna be pi-sixth and pi-sixth. But the range of sine inverse is restricted. Sine inverse here, sine inverse of X, it's always gonna be in the first or fourth quadrant. It's always gonna sit between positive pi-halves and negative pi-halves. And so if you were to compute this with your calculator, sine inverse of one-half, assuming you're in radians, it would give you the answer, pi-sixth. It would not notice the other angle, pi-sixth. We'd have to use reference angles to find that. And that's a topic for another video. Suppose we wanna compute cosine inverse of negative square root of three over two. So if we think of our unit circle, right? We're looking for negative square root of three over two. Cosine represents an X-coordinate. We're thinking of something over here. Now arc cosine will only grab something in the first or second quadrant. So we're looking for this point right here. What angle is gonna do that? So then playing around with it, we want cosine of theta. This is supposed to equal to theta, right? Cosine of theta equals negative root three over two. And so we know it's gonna be in the second quadrant. Then we have to kind of figure, okay, in the second quadrant, you have this point over here, negative root three over two, positive one half, right? Oh, that references over here to this point. This point, I guess I didn't draw this perfectly to scale, but that's fine. No big deal. So in the reference angle here, so theta hat, the reference angle, this is gonna be pi-sixth. But we're supposed to be in the second quadrant. And so that's where we get, oh, this is gonna be five pi-sixth, for which if you can check with the calculator, that's exactly what we're gonna get right here. Let's do one more of these things. If we add tangent inverse of negative one, so again, think of this as an angle. We're looking for the angle theta, such that tangent is equal to negative one. Which also remember that tangent is just sine over cosine, right? We're looking for the angle so that negative cosine is equal to sine. So we're looking for the angle in the first or fourth quadrant because when it comes to arc tangent, arc tangent here is always gonna be between negative pi halves and pi halves, not including pi halves. Equality is not allowed there because that corresponds to an asymptote. So we were looking for what, where in the first or fourth quadrant, well, if sine and cosine have different signs, one's positive and negative, that's gonna be in the fourth quadrant. But so where are they the same but different by a sine? Okay, that feels like a 45 degree angle but it's in the second quadrant. So that's gonna give us negative 45 degrees or negative pi force, which is what the answer would be on this one. So this seems a little bit, maybe a little bit clunky at first. This is something you get used to as you turn an inverse trig function and just to a trig problem, right? You can switch from inverse functions to the trig functions there, just insert the symbol theta and try to think of it out, like when is tangent equal to negative one? When is sine equal to one-half? That's what an inverse calculation is doing here. What angle makes sine equal to one-half? What angle makes cosine equal to negative root three over two? What angle makes tangent equal to negative one? Those are special angles we know and therefore we could compute these without a calculator. On the other hand, if you had to compute something like one is arc tangent of two, that one's a lot harder because we're like, when is tangent theta equal to two? You know, that's a question we don't know just memorizing the unit circle. That's something we would have to do some type of numerical calculation on. I also, taking this principle, I wanna do a few more examples with this, but this time let's say that we have sine of sine inverse of one-fifth. In some regard, this is a super easy one but I want you to try to simplify this for one step before we just jump to the answer. Whenever you see an inverse trig function, I do want you to think of it as an angle, right? So we're trying to compute sine of theta for some angle. The angle is given by sine inverse of one-fifth. But sine inverse is equal, sine inverse of one-fifth is equal to theta, right? This tells me that sine theta is equal to one-fifth. Oh, okay, I see what I'm trying to do here. Sine theta is supposed to equal one-fifth. The answer is gonna be one-fifth, right? And so when you actually go back and analyze this, it kinda makes sense with the right perspective. Sine inverse of one-fifth, this is the angle which makes sine equal to one-fifth. Well, if I then evaluate sine, excuse me, if I evaluate the sine ratio of the angle that produces the sine ratio of one-fifth, well, that's gonna be one-fifth, okay? These are inverse functions, so they're gonna cancel each other out, all right? It's like adding two and then subtracting two. It's like putting your shoe on and taking your shoe off. The net effect is as if nothing happened. Now, before we get too zealous with that, it's like, oh, okay, sine inverse is sine of three-pi-fourths. You might be tempted to say this is three-pi-fourths because they just cancel each other out. That's actually not the answer here. It's like, what's going on here? Well, the issue comes down to sine is not actually a one-to-one function. Sine is actually infinity to one. We restricted its domain to define sine inverse right here. And so because of that, when we look at a sine wave and we think of the principal branch right here, we go from pi halves on the right to negative pi halves on the left. And what this does is we restrict the domain of sine. That means the range of sine inverse is gonna be restricted as well. Sine inverse of any ratio X will always be between pi halves and negative pi halves. And so if you spat out three-pi-fourths here, three-pi-fourths does not sit inside of here. That's a problem. Your calculator would not say that. And that's because really we're looking for the reference angle of three-pi-fourths when we do this. So sine inverse of sine of three-pi-fourths is the reference angle, which is gonna be pi-fourths. Let me give you another explanation with a different example. Take the function f of X equals X squared and g of X equals the square root of X. Well, we often would like to say that X squared and square root of X are inverse functions of each other. And that's because if we do f of g of X, this looks like the square root, excuse me, the wrong way, this is gonna be the square root of X quantity squared. This simplifies always to be X. So that's true. But on the other direction, if you do f of X, f of g of X, excuse me, excuse me, g of f of X, this is gonna turn out to be X squared times, take the square root of that. This does not equal X. This actually equals the absolute value of X. Case in point, if I take the square root of negative two squared, that's gonna give you the square root of positive four, because the double negative is the positive and the square root of four is then two, right? Which is the absolute value of negative two. When you compose the square root with X squared, in this direction, you actually get the absolute value of X. For which, if you restrict that to just be greater or equal to zero, then the absolute value in X are actually the exact same thing. The absolute value doesn't do anything to positive. That's the same phenomenon that's happening right here. That in this direction, if you take sine of sine inverse of X, this will always equal X. And that's true for cosine with its inverse tangent with its inverse, no big deal. But when we go the other way around, if you take sine inverse of sine of theta, their composition is not exactly theta. It's always gonna be the reference angle of theta. Which, if theta is in the first or fourth quadrant, that if it's between negative pi halves and pi halves, that doesn't change anything. But when you get something like three pi fourths, which is outside of the range, negative pi halves to pi halves, then you have to take the reference angle inside of those regions. So that's the only hiccup you wanna look out for when you're doing these type of calculations with inverse trigonometric functions.