 Hi, I'm Zor. Welcome to a new Zor education. This is the final lecture in this mini-series which define different inverse trigonometric functions. In this case, it's arc cosecant. Now, as usually, we define inverse function based on the original function. So for arc cosecant, we have to examine the cosecant first. So what is the cosecant? Again, it's not really a direct definition. We usually define cosecant of x as 1 over sin of x. So what I would like to start is with sin, then invert it to get the cosecant, and then go to inverse function to cosecant to get its properties. And there are not many properties. It's like are there even if any domain and range and how it behaves, the graph, etc. Alright, so 1 by 1. Start with a sin. Sin is easy. Sin is this. Something like this. Minus 1. This is pi. This is pi over 2. This is minus pi over 2. Minus pi, 2 pi, 3 pi over 2. Okay, we start with a sin. Now, let's turn it around. 1 over sin. Well, obviously those points where the sin is equal to 0, we will have asymptotes. Alright, so, okay, these are asymptotes. Now, how the sin looks like? Well, let's consider this interval from one asymptote to another. In the middle, it's equal to 1. So 1 over sin will also be equal to 1, right? So from this point, if you go left and right, sin goes down to 0 while still being positive. So 1 over sin would go to infinity remaining positive. In this case, we have a similar situation, but with a negative sin. So at 3 pi over 2, it's minus 1. So 1 over sin will also be minus 1. And then, left and right from this point, sin goes down, which means 1 over sin goes up to infinity, but in this case, it's a negative infinity. And same thing here. And obviously it's over repeated many times. So let me wipe out the sin. So I will have only the cosecant. We've got the function cosecant. Now, let's talk about inverse function. We don't need this anymore. We have the graph of the cosecant. And let's talk about inverging this function. Now, inverse function, first of all, let's define what's the values of x and what's the domain and the range of this function. Now, in this particular case, x can be anything except 0, pi, 2, pi, et cetera. So let's put it this way. x not equal to pi n, where n is integer number. Now, y is 1n above or minus 1n below. So absolute value of y is greater or equal to 1. So these are our domain and the range. Now, we know that whenever we are switching to inverse function, whatever used to be domain becomes a range, whatever used to be range becomes domain. However, what we have to do is we have to be able to find for the value of the function, the value of the argument. And in this case, we cannot do it. If you take the value of the function, let's say this, minus whatever, 1 and a half, minus 1.5. Now, you draw this horizontal line and wherever we intersect our graph, so all these arguments are arguments where the function is equal to minus 1 and a half. So we cannot identify a single argument which can be put into correspondence to a single value of the function. So there is no inverse function. Now, what we want to do is we have to define and we would like to define inverse function. So what should we do? Well, the typical answer, as with many other trigonometric functions, is reduce the domain. Reduce the domain of this function cosecant to an area where it does have the inverse function. And the easiest way is to find the interval where the function is monotonous. Now, traditionally, this interval is from minus pi over 2 to pi over 2, this one. So if I will put boundaries like this, then my graph, whatever is left of it, would be this. So only these two branches, this one and this one, around zero, it's monotonously decreasing in this case from minus 1 to minus infinity, and it's monotonously increasing here from plus infinity to 1. So the function is monotonous in this particular interval. There are some other intervals where the function is monotonous, and some take me wrong, but this is traditionally taken as the definition for the domain of the cosecant if we would like to inverse the function. Why? Because it's closer to the zero, and the function with reduced domain is odd, as you see, because sine is odd function. If you change the sine, if you change the sine of the argument. All right, so on this particular interval, we can define our inverse function. So instead of this, we reduce the domain to minus pi over 2 to pi over 2 without the midpoint zero where the function is not defined because it has an asymptote. But the y is still in the same range from one up or from minus one down. So we did not really change the range of the original function. We just changed the domain, reduced it to the interval where the function is monotonous. Now, since we know that the function is monotonous, we can create its graph. So this is arc cosine of x, sorry, arc cosecant of x. Okay, so you know that the graph of this and this function are supposed to be symmetrical relatively to the bisector of the main angle. So if this graph is, let me maybe repeat it. So this is pi over 2 and this is minus pi over 2. So the function is like this. Now this is one, this is minus one. Okay, now inverse function is symmetrical. Now pi over 2 is slightly greater. So this point is to the right. And this is point to the left. Okay, now let's use a different color. Now instead of vertical asymptote, we will have a horizontal asymptote, right? Because asymptote is reflected from y-axis is reflected to x-axis, right? Now, whatever used to be here would become here. So the function would have 1 and pi over 2 and it would work like this. And here also I would have minus 1, minus pi over 2 and the function would go this way. So the red one is arc cosecant. The black one is cosecant. So the graphs are symmetrical relatively to the bisector. The domain of this function is what used to be the range for the cosecant function. So in this case, x, absolute value of x greater than 1, that's the domain. Greater than 1 equal or greater or equal or less than minus pi. These two areas. Now the range would be from minus pi over 2 to pi over 2 except 0. So y belongs from minus pi over 2 to pi over 2 except y is not equal to 0. That's the range. And also notice that the function is odd. Why is this function odd? Obviously because the arc cosecant is odd because the cosecant is odd. The cosecant is odd because it's definition 1 over sine and sine is odd. So obviously whenever you change the sine of the argument from this to this, the value of the function will also change to an opposite. It's very easy to prove. Actually the proof is in the nodes. It's just one line of proof I don't want to waste time on this. Basically that's it. These are all the properties. And this lecture concludes this mini series where I just wanted to define all the inverse trigonometric functions. Well, they're all defined and now we can do some problem solving, right? Okay, examine once more the nodes to this lecture and if you want to refresh your memory, go through the nodes for all the inverse trigonometric functions. They're very short. And in some cases I just don't put the proof that the function is odd or even on the board because it's a really very simple thing and they put it in nodes only. And be prepared for different more complex things about trigonometry. And that would be the subject of many lectures together. Thanks very much and good luck.