 Hi, I'm Zor. Welcome to a new Zor education. I would like to present a few problems, very simple problems which are related to inverse functions, inverse trigonometric functions. Basically, I hope they will help you to better understand the concepts behind these inverse trigonometric functions. If you did not attempt to do these problems yourself first, before listening to this lecture, I do suggest to press pause and try to do it. It's on unizor.com in this chapter about inverse functions. Now, let me just address all these. I have two lectures actually. One is just a little bit simpler, and it's related to just inverse functions, and another combines with some trigonometric functions. All right. We have six different trigonometric functions. We have six different inverse trigonometric functions. The main problem with inverse functions in trigonometry is that the regular trigonometric functions are periodical and obviously, they do not have inverse like everywhere. We have to restrict the domain of the main function to the area where it's monotonic. In that area, we can talk about inverse functions. The first problem is you variate r sine of sine of 3 pi over 2. In this case, as in any other case, the problem is that we have to really consider this particular value, but we have to find the angle which has the same sign as the 3 pi over 2. But this angle is not 3 pi over 2. I mean, obviously, you can think about, okay, r sine is inverse to sine. So if you start from something, apply the function, then apply the inverse function, we should return back to this one. That's not true, and the reason is, if you will consider the graph of the sine, now this is 0 pi pi over 2, 3 pi over 2, this is minus pi over 2. So the function sine is considered as inversible only between these two points. So this is a monotonic branch or monotonic part of the graph of the function sine, where inverse is actually defined. 3 pi over 2, which is an argument to sine, is not part of this area. So what is part of this area, which has exactly the same sign as 3 pi over 2? Well, this is obviously minus 1. And the same value of minus 1 has this particular point, minus pi over 2. So sine of minus pi over 2 and sine of 3 pi over 2 are the same, they are minus 1. So basically what I'm asking here to do is, I'm asking to find the value r sine of minus 1. And I have to find this value only between these two boundaries, which designate the area where sine is inversible between minus pi over 2 and pi over 2. Now within this area, sine takes also all the possible values from minus 1 to plus 1. And the minus 1, it takes in this particular point. So the answer to this is minus pi over 2. So regardless of the fact that we started with 3 pi over 2, when we perform sine and arc sine, we have to basically shift to the point, which has exactly the same sign as 3 pi over 2. But it belongs to the area where the function arc sine is supposed to take values. So the main for the arc sine is obviously from minus 1 to 1. But the range only between minus pi over 2 to pi over 2. And that's where we have to find the value in that range defined for this function. So I have actually explained it in more details. And the rest of the problems, which are very similar, I will explain it in less details. But the idea is exactly the same. We have to find within the range of this function the value which would give the same value of trigonometric function as an argument. Next one is arc cosine of minus pi. Same thing. Let's start with a graph. This is the cosine. Cosine is an even function. It's symmetrical. So that's how I remember it. And this is pi over 2. This is pi. This is 3 pi over 2. Now, where the function is monotonic and where it's reversible between these boundaries? So this is the piece, the monotonically decreasing piece. And that's where we have to really find the resulting value between this and this. Now, what is cosine of minus pi? Well, let's just continue the graph. Now, this is minus pi over 2. This is minus pi. So it's minus 1, right? Now, within this area, this is the point where the cosine also shakes the value of minus 1. But this point belongs to the range of the arc cosine. So that's why the answer to this is pi. Same story. We started with minus pi. We found what's the value of the cosine in this place. And then we shift the whole thing within the domain where the cosine can be inverse. And the domain for the cosine where it's reversible is from 0 to pi. Next, arc tangent of tangent of minus pi, OK? The graph of the tangent is minus pi over 2, 0, pi over 2. And then it repeats itself here as well, minus pi. So tangent of minus pi is 0. But we have to find an angle within this area from here to here. This is the range of the function tangent, which is reversible. So we have to find within these boundaries the value of the angle, which is equal to 0, same as tangent of minus pi over 2. So we have to find this. So what's the angle tangent of which is equal to 0, but which belongs to this area? Obviously, it's 0. So the problems are simple, as you see. All you have to do is you have to understand where exactly each trigonometric function is inversely. So which interval, where the function is monotonic and takes all the possible values, should be taken as the interval where we have to look for an angle which we need. Next, cotangent of cotangent 3 pi over 2, OK, 3 pi over 2. The graph of the cotangent is this is pi over 2, this is pi. Now here we have the next one. So that's the cotangent graph. Now, we are interested in arc cotangent from the cotangent of 3 pi over 2. So 3 pi over 2 cotangent is 0. So this is 0. So question where exactly we should consider the function cotangent to make it inversible. Well, traditionally it's from 0 to pi. So this is the branch where it is inversible. And in this particular branch, the value of 0 is where the angle is equal to pi over 2. So that's why the answer to this is pi over 2. By the way, for those who do not really remember how to construct the graphs of whatever the functions, trigonometric functions I was just using, go to the corresponding lectures. Every lecture for every function has this graph. Next arc secant of secant of minus 2 pi. All right. So let's draw the graph. Now arc secant is 1 over cosine. So the cosine is equal to 0 at minus pi over 2 and pi over 2. So these would be the asymptotes. Now cosine goes this way, so secant goes this way. Now from pi over 2 to 3 pi over 2, it would be the negative. And here I will have minus pi minus 3 pi over 2, and another asymptote, it would be like this. And then next is minus 2 pi. That's what we need. So the function would be like this. So it looks like our function is equal to secant of minus 2 pi. It looks like it's 1. Now where exactly the function secant is monotonic and therefore inversible? Well, traditionally it's this one from 0 to pi. It's this branch and this branch. So the combination of these two branches is a piece of the graph where it's monotonic. And the function is defined from 0 to pi except the point pi over 2 where there is an asymptote. Now the question is, we need 1, right? So this is secant of minus pi is 1. So within these boundaries, we have to find where exactly the function is equal to 1. And this only point 1, 0. That's where it's 1. So the answer to this is 0. I spent more time writing the board than explaining the material. All right, the last one, arc cosecant. Arc cosecant of minus 3 pi over 2. Now cosecant is 1 over sine. So the asymptotes are where the sine is equal to 0. So it's 0 pi minus pi and minus 2 pi. So these are asymptotes. Now the graph looks like, so the sine is from 0 to pi. The sine is this way, so 1 over sine would be this way. And similarly would be here and similarly would be here. So minus 3 pi over 2, the function is equal to 1. So my question is, where exactly the sine is, or actually cosecant, cosecant is monotonic, where we can invert it. Well, this is from minus pi over 2 to pi over 2. And the monotonic branch is this monotonically. So within these boundaries from minus pi over 2 to pi over 2, the function cosecant is monotonic and therefore inversible. So question is, where exactly it's equal to 1? In this interval from minus pi over 2 to pi over 2? Well, obviously it's this point. So it's pi over 2 the result of this. So the angle where the arc, the angle, the cosecant of which is the same as cosecant of minus 3 pi over 2 is pi over 2. That's it. That completes this particular lecture. These are very simple exercises in number one, graphing, by the way, that's an important thing. Number two, you have to understand which part of the graph you should use for inversing the trigonometric function. Usually it's either from minus pi over 2 to pi over 2 or from 0 to pi, either or. Wherever it's monotonic. Like sine is from minus pi over 2 to pi over 2. It's monotonic or increasing. Cosine is from 0 to pi, where it's monotonic, monotonically decreasing, et cetera. So either this or that. Within these intervals, we should look for the answer to all problems where we have to really evaluate arc something, arc sine arc, cosine arc, tangent, et cetera. Now, next lecture would be more or less the same. I'll just slightly change the arguments. So basically, again, that's it for today. Thank you very much. And try again to do it yourself. That probably would be very helpful. Thanks a lot.