 Hi, I'm Zor. Welcome to a new Zor education. Today's lecture would be the last one in series about different trigonometric functions as far as their very basic properties are concerned. It's about co-second. That's how it's, sometimes it's actually co-second. But I will use this particular abbreviation. I'll like three letters in this case. So, about co-second. Now, the previous lectures where I was explaining basic functions of sine, cosine, etc., I was using primarily the unit circle to demonstrate how the function behaves if we are increasing angle from zero to pi over two to pi to three pi over two to two pi, etc. In this particular case, I'll just use a slightly different approach. I will use graphs and the definition of a co-second, and the definition is one over sine. So, you remember, second is one over cosine and co-second is one over sine. So, instead of analyzing this particular function as basically one over ordinate of a point in the unit circle, I will just start from the graph of sine and I'll divide one by this particular graph. We know how to manipulate graphs and that will give me the basic properties as much as the other approach with unit circle. Why I decided to do it? Well, just to be different from the previous lectures and just to make sure that you understand that there are different approaches to the same problems and you can basically use one or another. And plus it's a good exercise in the graph transformation. So, let's start with graph of the function sine. Now, we know it's from minus one to one. At zero, it's zero. Then, at pi over two, it reaches its maximum. At pi, it goes to zero. At three pi over two, it goes to minus one. And at two pi, it ends up at zero again. So, that's how the sine behaves. Now, let's try to have one over sine. So, this is y is equal to sine of x. Now, here, I will have also one and minus one. And I'll try to be on the same scale as in this particular case. All right, so let's think how I can divide one by this particular graph. Well, first of all, if I'm dividing something by something, I should really worry about denominator being equal to zero. Because in this particular case, obviously the inverted graph should go to infinity, positive or negative, when the sine goes to zero. So, sine goes to zero at point zero pi and two pi, etc. Every pi is zero, which means we have asymptotes in these cases. In case of x is equal to zero, in case x is equal to pi, in case x is equal to two pi. And if you wish to continue this, we can do it to a negative territory. And asymptote is here as well. So, we know that around these points, the graph goes to infinity, plus or minus. We just have to analyze which one. Okay, now, from zero to pi, not including the edges of this interval, the graph is positive, which means one over sine will also be positive. So, in this particular area from zero to pi, the graph will be above the x axis. Now, at point pi over two, it's one. One over one is still one, so we still have this point. This point of sine represents this point of one over sine. After which, if we go to the left or to the right from this point, sine is decreasing to zero, which means one over sine will be increasing to infinity. And it's a positive infinity, so the graph will go this way. By the way, if my denominator is decreasing, it's obvious that the fraction actually is increasing, right? Why is it obvious, by the way? Well, for positive numbers, it's very easy. If you have a greater than b, this is an inequality considering a and b are positive numbers. b is smaller, a is greater. I would like to prove that one over a is less than one over b. How can I prove it? Well, considering this is an inequality, we can divide inequality by a positive number, and it will do the same inequality. So, let's divide it by a. I will have a over a greater than b over a, or one greater than b over a. Now, let's divide it by b, also positive number, and it will retain. So, one over b greater than b over a b, which is one over b greater than one over a. So, we started a greater b and ended with one over a is less than one over b, right? One over b is greater than one over a. So, that's why when the denominator is going down, being positive, the fraction goes up to infinity because the denominator goes to zero. Well, this is just a simple exercise, let's begin it. I just don't want you to take basically for granted anything which I'm saying without the proof. You have to always look for the proof. Now, let's continue, left and right. Now, in this particular case, let me continue this, it would be like this, right? So, in this area, in this interval from pi to 2pi, not including the edges, now, in the middle, I have minus one. Now, one over minus one will be minus one. Now, the whole graph is negative, which means one over will be negative as well, it will be below the x-axis and it will go to infinity, in this case minus infinity since we are talking about negative numbers as from the middle point we go up, so it goes this way. And obviously the whole thing is very similar, that will be the graph. So, this is a behavior of the function y equals second, no, cos second, sorry, cos second of x, which is one over sin of x. So, purely graphically, we just divided one over the graph and we got this one. So, we can conclude basically everything about behavior of this function, cos second. Now, what else can we say, asymptotes of 0pi, 2pi, 3pi, minus pi, etc., so every pi we have an asymptote or generally speaking if x is equal to pi n, where n is any integer number, this function is undefined and it has an asymptote. What else is important? Now, in the middle of each interval it has either one or minus one, so if you have something like if x is equal pi over 2 plus 2 pi n, y is equal to 1 and if x is equal to minus 2, minus plus 2pn, y is equal to minus 1, right? So, pi over 2 and it will be pi over 2 plus 2pn, etc., now if minus pi over 2, this is minus pi over 2, then plus 2pn, etc., it will be minus 1. So, these are local maximums and local minimums of the function. Okay, what else is important? Cos second of x plus pi is equal to, let's just think about it. From the graphical standpoint, if I add a pi, for instance in this case, add pi to p over 2, I will get 3pi over 2, so it changes the sign. So, it looks like function changes the sign. Now, why is it obvious? Well, because it's 1 over sign and we know that sign actually also changes the sign if we add pi, so it's always this. Now, if you consider the sign as a function, it's an odd function, right? Sign is an odd function, it's a symmetrical relative to this point and it changes the sign if argument changes the sign, right? So, we were talking about this. Well, obviously, this is transferred to cos second. So, cos second of minus x is equal to minus cos second of x because of the same property because cos second is 1 over sign. Finally, x minus pi, okay. So, cos second of x minus pi is equal to, well, let's think about it. It's a periodic function, so you can always add to pi. It will be the same as cos second of x minus pi plus 2pi, right? Since 2pi is a period, which is cos second of x plus pi, which is, we just talk about this, it's minus cos second of x. Now, the cos second of pi minus x, since cos second is an odd function, would be equal to cos second of x and always pi minus x. That would be always equal to cos second of x. Why? Because this is negative to this and the function is odd, so this should be, this should be negative to this. And in addition, if you remember, the sign of pi minus x is the same as sin x. And obviously, 1 over sign would be the same. So, that's these properties. Basically, that's it, no more interesting material. So, what's important about this particular graph is that, first of all, we can say that the domain of the function is real numbers except those which are multiple of pi, so 0, pi, 2pi, minus pi, minus 2pi, et cetera. These are all points where the function is not defined. It has asymptotes. Now, what about the range? Well, you remember, the range of the sign is from minus 1 to 1. The range of 1 over sign would be either greater than 1 or less than minus 1. The interval from minus 1 to 1 is not part of the range of this function, okay? It doesn't take these values within these. Starting from 1 up and from minus 1 down, yes it does. So, that's the range. Basically, that's it. I will spend probably some time talking about different various dysfunction takes for different angles, but basic properties are just like that. That's it for this particular lecture. That's it about explanation of what Cosecond actually is all about. Thanks very much.