 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue solving problems for inverse trigonometric functions. This is the second lecture dedicated to these problems. Problems are very simple. I do suggest you to do it yourself first, and only then listen to the lecture. Well, basically, this lecture will be very much like the previous one. I'll just change slightly the arguments in the functions. But again, the purpose is to evaluate the value of the inverse trigonometric function. All right. Without further ado, let's do arc cosine of 3,5 over 2. What we need right now is two functions actually. One function is sine, another function is cosine, and then we'll have to inverse the cosine. So first, let's evaluate this. So 3 pi over 2 function sine is this one. So it's 0 pi 2 pi. So 3 pi over 2 is where the function is equal to minus 1. So this is minus 1. That's done. Now, what's the arc cosine of this? Well, let's think about the graph of the arc cosine. First, graph of the cosine is this one, from 0 to pi over 2 pi, 3 pi over 2, etc. So we have to concentrate on the area where the cosine is inversible and this is the area where the cosine is monotonic. The one which is chosen is from 0 to pi, which is this piece. So the question is, within this area where the cosine equals to minus 1, obviously it's pi. So the angle cosine of which is equal to minus 1, within this boundaries is pi. So this is equal to pi. Arx sine of cosine of minus pi. So again, it's two steps first, evaluating cosine of minus pi. Now the cosine was this. This is 0 pi over 2 pi. So minus pi, this is minus pi, minus pi over 2. So the area of the cosine of minus pi is minus 1. So I have to find arc sine of minus 1. Now let's talk about what's the arc sine look like. The graph of the sine is this one. The area where the sine is monotonic is this one, from minus pi over 2 to pi over 2. So only within these boundaries, we have to look for an angle sine of which is equal to minus 1. So what is this angle? Obviously it's minus pi over 2. That's the answer. Next, arc cotangent of minus pi. OK, tangent of minus pi. Well, tangent is, I don't really have to do it with a graph. I think I can do it mentally. Tangent is sine over cosine. Now sine of minus pi is equal to 0. So tangent is equal to 0. Cosine is something else, minus 1 or something. So it's 0. Now, so where is the arc cotangent of 0? So which angle has a cotangent equal to 0? But not any angle. We need an angle from the range of the function arc cotangent. So let's just draw the cotangent first. And I do remember it looks like this. That's the area of where it's monotonic. So from 0 to pi is the range of the angles. We have to really evaluate our function arc cotangent. So which one of those where the cotangent is equal to 0? The cotangent of this should be equal to 0. Well, obviously it's this point. It's pi over 2. So that's the answer. So I deliberately did not draw the other branches of function cotangent because they are outside of the area where the inverse function is defined, versus defined only from 0 to pi. Next arc tangent of cotangent of 3 pi over 2. OK, now I just draw the cotangent. I'll just repeat it. OK, that's how it was from 0 to pi. But now we need another branch of the cotangent because we are looking for 3 pi over 2, which is here. So this is equal to 0. Now I have to find an angle tangent of which is equal to 0. Tangent, because this is arc tangent. So I have to find an angle tangent of which is equal to 0. But this angle should belong to the area where the tangent is inversible. Now tangent is inversible. I'll use the same graph, actually. Tangent is inversible from minus pi to pi. This is the tangent branch where we are considering arc tangent in that inverse function. So where exactly is equal to 0? Well, obviously it's this point where the angle is equal to 0. So the answer to this is 0. OK, two more. Arc secant of secant of minus 2 pi. Arc secant of, sorry, arc cosecant of secant of minus 2 pi. All right. Now secant is 1 over cosine, right? 2 pi is a period of the cosine. So the cosine of minus 2 pi is the same as the cosine of 0, right? I just add 2 pi, the period, to the argument. And cosine of 0, we do remember, which it is, it's 1. So this is 1. Now let's talk about cosecant. Cosecant is 1 over sine. So let's draw the graph. Now sine is equal to 0 at 0 pi 2 pi minus pi. So 1 over sine, which is the cosecant, would be obviously this. Where exactly the function is monotonic, the cosecant? Well, we have traditional put it from minus pi over 2 to pi over 2. And these two branches, this one and this one, constitute the piece of the cosecant where it's inversible. So we have to find an angle from minus pi over 2 to pi over 2, cosecant of which is equal to 1. And obviously it's this one, this point, this is 1. So it's pi over 2. Result is pi over 2. That's the angle. That's the angle. The cosecant of which is equal to 1, which is the secant of minus 2 pi. And the last problem in this series, secant of cosecant, arc secant of cosecant of minus 3 pi over 2. So let's start with the cosecant. Cosecant of minus 3 pi over 2. Again, let's just think about it. Cosecant is 1 over sine. Period is 2 pi, obviously. So if I would add 2 pi to this argument, nothing's changed. I add 2 pi to minus 3 pi over 2, I will have pi over 2. So it's the same as cosecant of pi over 2. Now, 1 over sine of pi over 2. Now sine of pi over 2 is equal to 1, right? Remember, 1. So 1 over would be 1 as well. So this is 1. Now let's talk about secant. Secant is 1 over cosine, right? So let's do the cosine. Cosine is this 0 pi over 2 pi. So this is the area of monotonicity, monotonic behavior of the cosine. So we have to find our angle from 0 to pi. Now the graph of the secant would be this. So that's where we have to really look for the value of 1. So very exactly within these boundaries there is an angle secant of which is equal to 1. Well, obviously it's this point, right? So it's angle is equal to 0, and the secant is 1. So the answer to this is 0. Well, that's it for this particular set of problems. I do suggest that you go to unison.com to notes for this lecture. That's where you have all the problems and answers actually as well. And try to do it yourself using basically the same technique. Well, other than that, thank you very much, and good luck.