 Hi, I'm Zor. Welcome to a new Zor education. Continuing graphing different trigonometric functions today. Today's subject is cosine, sorry, cosecant, cosecant. Cosine have already been covered. So, we will talk about the original cosecant graph, and then we will have certain transformations of this graph. I have something like eight or so, yeah, eight different graphs which are basically variations of the same main original cosecant graph. Okay, let's do it. By definition, the cosecant function is one over sine of x. Now, I will probably start with just the original graph and that will be easier and then it will transform it. Now, I don't remember any graphs other than sine and cosine. Everything else I derived basically. So, let's derive this graph from the sine. So, first we will draw the sine which looks like this. These are minus pi, zero, pi, two pi, etc. Now, this is one, this is minus one. Now, if I would like to invert basically this graph, so build the graph which is one over this, well, let's just do it based on the behavior of the sine. Wherever the sine is equal to zero, the cosecant is undefined and in the immediate vicinity of this, obviously it goes to infinity. Now, so we will have asymptotes in these and this is as well. So, these are asymptotes for our graph of secant because denominator equals to zero at this point. Now, so as I was saying in immediate vicinity of these points, the graph of the cosecant should go to infinity because the denominator goes to zero. Now, here is a very important detail. Here, sine is positive and that's why the cosecant is positive. Here, sine is negative and cosecant is also therefore negative. What actually follows from here? That this infinity is positive and this infinity is negative. Same thing here, when we approach zero from both sides, from this side it will be positive, from this side it will be negative. Now, obviously, in the midpoint, in this case it's pi over two, it's the sine is equal to one, so one over one is one. So, we are basically having situation like this. So, the red ones are components of the graph of the secant. Okay, from zero to pi it's upside, it's horns up and the local minimum is pi over two and it's equal to one. And then the next from pi to two pi it's upside down and it's local maximum which is equal to minus one. And then the situation repeats itself. Positive negative, positive negative and to the left at the same time. Okay, so this is the red ones is the graph of the cosecant. Now, let's do the properties of this graph. First of all, odd even or neither? Well, obviously it's odd because sine is odd. Sine changes the sine if the argument changes the sine. So, sine of minus x is minus sine of x. Obviously the same goes to one over sine of x which is the cosecant. And as you know odd functions are symmetrical relative to the center. It's a central symmetry. And if you will turn the whole picture by 180 degrees clockwise or counterclockwise doesn't matter, the graph will turn into itself. This will turn into this, this will turn into this, etc. So, the function is odd, function has asymptotes as we were talking about. What else? It's not equal to zero anywhere at all. Doesn't have any roots. It has local minimum and maximum at pi over two plus two pi and where n is any integer, it's local minimum which is equal to one. And at three pi over two plus two pi and where n is integer. It's local maximum which is equal to minus one. Well, that's it about general properties of this particular graph. Now let's do the problems. Problem number one, cosecant of minus x. Okay, how the cosecant of minus x looks like. In lectures which were about trigonometric graphs and in general lectures about graphs, I was using exactly the same methodology. Namely, consider the function y equals to f of x. And consider the point AB belongs to the graph of this function. Which means f of A is equal to B. That's what it means, right? That the AB belongs to a graph. Well, then if you consider the function f of minus x and consider the point minus AB, obviously this point satisfies this equation, y. Well, since f of A is equal to B, if you substitute minus A instead of x, you will have minus minus A which is plus A. So it will be f of A which is equal to B according to the first equation. So for any AB which belongs to the graph of this function, the symmetrical point relative to the y-axis, which is negative apsis and the same ordinate. It's reflection relative to the y-axis. This is AB, this is minus AB, this is AB, this is minus AB. So this point belongs to the graph of this function. So the whole graph actually of the function y is equal to f of x. If you would like to convert it into a graph f of minus x, you just have to symmetrically reflect it relatively to the y-axis. Now in this particular case, I'll use a different color. If you will symmetrically reflect it, you will have this. This will go to this, this will go to this, and then continue the same way. So these are components of this particular graph. Let me wipe out this, and we don't need sign anymore. So this is the graph of the cosecant of minus x. Next problem is this. Similar consideration. If you have the graph of this function and point AB belongs to this, now which point belongs to this? Well, obviously, A divided by 3, B. So if AB belongs to this, A over 3, B belongs to graph of this function. Because if you will substitute A over 3 instead of x, you will get f of 3 times A divided by 3, which is A, which is B, as we know, so this one satisfies this particular equation. What does it mean? It means that for every point of this graph, there is a point which is 3 times closer to 0 along the horizontal x. It's squeezed, which belongs to this graph. So let me basically draw it. It's very simple. So let's use, let's say, from 0 to 2 pi. We have the graph of original function looks like this. This is original cosecant. Now, if I want to squeeze it towards y, I can access by the factor of 3. It means that these asymptotes would be closer 3 times. So instead of pi, I will have pi over 3, and instead of 2 pi, I will have 2 pi over 3 asymptotes. And the period will be, instead of 2 pi, obviously the period will be 2 pi over 3. And the function would look like this. Now, this piece is squeezed by the factor of 3 of this component of the original graph, and this piece is squeezed by the factor of 3 of this particular component. And the periodicity is 2 pi over 3, which means then it repeats left and right, the same thing. This, this, this, et cetera. And the interval in between the asymptotes is pi over 3. Next one is quite analogous with a very small difference. The factor is not greater than 1. The factor is smaller than 1, 1 third. Now, what does it change? It obviously change the quantitative characteristics, but not the type of deformation which is being done with the graph. Because obviously you see that if AB belongs to this graph, then 3AB belongs to this graph. 3A substituted into x would get f of 1 third of 3A, which is A, and we know it's B, so this equality is satisfied. Now, what does it mean that for each point of this, like this, this point belongs to this graph? It means the graph is stretched by 3 times. Horizontally, abscissa is increased by 3 times. Well, actually both. Left and right, as you understand. Negative go to the left, positive to the right. So again, if your original graph is, this is 2pi, I'm using a smaller scale now, and your original graph is this, then your new graph will have a period, not 2pi, but 6pi, 3 times greater. And this asymptote at x equals to pi would be here at x equals to 3pi, and instead of 2pi, we will get 6pi somewhere here. That would be another asymptote. And the graph would be stretched 3 times to the right and to the left. So it would be something like this, and something like this. This component is stretched 3 times this, component of the original graph, and this component is stretched 3 times this component of the original graph. And again, everything is repeated to the left and to the right. That's it for this exercise. Well, while I'm writing this couple of words, you see, I'm paying a lot of attention in these lectures to graphs of the functions, because graph explains you how the function behaves much better than formula-rendering else. Next. Now we will multiply the function itself by the factor of 3. y is equal to 3 cosecant of x. So, before we were multiplying argument by some factor, 3 or 1 third, so it's squeezing or stretching. If we multiply the value of the function, then obviously it's stretching or squeezing depending on this particular factor vertically. Again, same logic. If for the function f attacks, a, b belongs to its graph, then for the function 3 f attacks, a, 3b belongs to this graph, right? If you will substitute a and 3b, on the left you will have 3b, on the right you will have 3 f at a, but f at a is equal to b, so 3 f at a would be 3b, so it's equal to b. Now, what does it mean? Obsessor is the same, but originate is stretched 3 times. So, that's vertical. So, how the graph looks like. Pi 2 pi 1 minus 1. Okay. This is original cosecant, right? Now, if I multiply it by 3, well, the behavior actually remains the same. It's just the numbers are different. Instead of minimum equal to 1, it would be equal to 3. And the whole curve would be 3 times cheaper, right? So, it goes like here. And same thing here. From minus 3, it goes this way. So, the periodicity remains exactly the same, but the local minimum and local maximum are 3 and minus 3. But asymptotes are exactly in the same place, and this behavior of going to infinity plus infinity or minus infinity is retained. Interesting detail. If you multiply an argument by a number 3, then the graph is squeezed by 3 times. If you multiply the function by 3, the function is stretched 3 times. So, the horizontal is kind of opposite to vertical transformation. If horizontally, the greater the factor, the more you squeeze it, and the smaller the factor, the more you stretch it. In this case, for the vertical transformation, the greater this factor is, the greater you're stretching. Well, that's life. I don't know how to explain it in philosophical or anything like this, because that's how it looks like. Okay, so that's it for this particular example. Next. Next is cosecant of pi over 2 minus 6. Okay, now, before I start this, let me do another example of this type. Let's say we have a function and it's graph, and point AB belongs to the graph. Then, if you have a function X minus C, which point belongs to this particular graph? Well, obviously, A plus C comma B, right? Because if you substitute A plus C into X, you will get F of A, and F of A, you know it's equal to B, so that will be an equality. So, what does it mean? Well, if point AB belongs to this graph, let's assume for a second that C is positive. Then, the point which is by C unit shifted to the right belongs to this graph, which means the whole graph shifted to the right if C is positive. And obviously, if C is negative, it's to the left, right? So, let's just remember this. Now, what I will do next is using the fact that the function is odd, I can rewrite it as minus secant of X minus P over 2, right? I changed the sign of the argument, so I changed the sign of the function. That's the definition of the odd function. Sine is an odd function, so one over sine, which is cosecant, is also an odd function. I changed the sign of the argument, and that's why I changed the sign of the function. And now, let's grab this one. Now, what is the graph of this guy? That's the graph of the original secant shifted to the right by pi over 2. So, that's what I will draw right now. That will be my first step. So, this is 0 pi 2 pi minus pi. And my original is from 0 to pi. I have these things, right? But now, since the whole graph is shifted by pi over 2, then this asymptote becomes actually this asymptote. And this asymptote. And this becomes this asymptote. And this becomes this asymptote. So, that's where my asymptotes of this graph actually are. And this is pi over 2. This is 3 pi over 2. This is minus pi over 2. Okay? Now, everything else remains the same. So, my graph, which used to be here from 0 to pi, now it will be from pi over 2 to 3 pi over 2. And then everything else will be, as we know. So, that's the cosecant of x minus pi over 2. Now, I have to multiply it by minus 1. Now, what does it mean that I'm multiplying the graph by minus 1? Well, again, let me resort to my favorite example. That's what happens. It belongs to the graph of the function y is equal to f of x, which means b is equal to f of a. Then, for the graph of f of y is equal to minus f of x, obviously the function, the point a minus b belongs to this graph. Because if you substitute a, you will get minus f of a. Now, f of a is b, so minus f of a is minus b. So, minus b would be here, and that's an equality. Which means that the graph is actually converted from every point which used to be a b becomes a minus b, which means the same abscissa, but opposite origin. It's a reflection relative to the x-axis in this case. So, let me just use a different color in this case. So, I'll use black now. So, it will be reflection of this, and reflection of this, and the reflection of this, etc. So, the black ones represent this particular function. And local minimum and local maximum remain the same. It's one and minus one. Okay, that's it. Next example is, as they say, all together now. So, this is a combination of everything which we have already done. You just have to step by step approach this particular graph, transforming from original form into the form which represents this particular graph. So, let me just modify slightly. First, I will start with original one, right? Then, I will do cosecant of 3x. Then, I will do cosecant of minus 3x. Then, I will do cosecant of minus 3x plus pi over 2, right? Minus 3x minus 3 pi over 2. And finally, well, not finally, two more steps. One-third of minus 3x plus pi over 2, too. And then, minus one-third cosecant of minus 3x plus pi over 2. Now, if I will be able to do with my graph these steps one after another, I will convert and transform original cosecant x into whatever I need. Now, I'm not going to do it right now because I did the same thing with any other function, but let me just explain what happens. In this case, original graph is squeezed by the ratio of 3 horizontally towards the vertical y-axis. Now, what happens with this? Whenever I change the argument from plus to minus, I'm reflecting, as we saw in the very beginning, the first problem, I'm reflecting the whole graph relative to the y-axis. Done that. Now, I want to shift my graph by pi over 2, in this case to the left because there is a plus sign here. Remember, minus positive goes to the right, plus positive goes to the left. Then, I will squeeze the graph vertically by the ratio of 3. And finally, I will reflect the graph relative to the x-axis, vertically, transform it, turn around. And that would be the end of this story. What's interesting, by the way, for this, and for two next problems, which are a little bit more complicated, is to use some scientific calculator or some web page which allows you to enter the formula and it will plot the function. So, enter this formula and see if you will have something which resembles what you would do following these steps. What's important is, again, to represent this original function as a combination of these steps, these transformations. Do it on the paper and then try to verify this with some computer program which is available. Okay, so that's it with this particular thing. I would rather spend a little bit more time for the next two problems which are a little bit more interesting, I would say. So, this is a known fact because all it is is factorization, vertically or horizontally, reflection, vertically or horizontally, and shift. That's it. No more. That's it. Okay, now let's do something slightly more interesting, cosecant 2x plus cosecant minus x. Well, there is no recipe, basically, but to draw one graph and on the same co-ordinate plane, draw another graph and then try to add them together. Basically, that's what I'm going to do. Now, the periodicity of this graph is 2pi and asymptotes are at every pi, right? So, you have pi and you have 2pi. Now, also, I will need pi over 2 and 3pi over 2. Now, one graph will have these asymptotes. This is my first graph, the original one. Now, first what I want to do is to draw this one. What's the difference? Well, we know it's squeezed by the factor of 2, which means let me continue. So, when I squeeze everything by the factor of 2, these asymptotes become correspondingly, this one becomes pi over 2 and the whole curve would look like this. This one from pi to 2pi will be from pi over 2 to pi. It's squeezed by 2, so it's here. Now, this guy would be from 2pi to 3pi, so it would be from pi to 3pi over 2. So, that would be this. And then the next one would be here. So, this is the graph of the function secant of 2x. Now, on the same scale, on the same coordinate plane, we will draw the cosecant of minus x. Now, we already did that. It's the cosecant reflected on the left and the left to the right of the original one, right? It's just a mirror image relative to the y-axis. So, let me just do it this way. It used to be like this and here. Now, when I overlap it here, I will have here. Basically, that's all I need from 0 to 2pi because everything else is repeated by the periodicity, right? So, now I have to add these two graphs together. Well, let's do it. From 0, well, obviously, asymptotes would be on every pi over 2 boundary because one of the components is equal to infinity, which means both together will be equal to infinity, some infinity, plus or minus. Now, let's approach 0 from the right. This goes to plus infinity. This goes to minus infinity. Well, that's a problem because we don't know how to add up two infinities of different signs. Which infinity is stronger or which curve is steeper, if you wish. But let me give you this clue about steepness. You remember how we formed this particular curve, right? It was original form, squeezed towards the y-axis by the factor of 2, right? Now, whenever we squeeze it, it becomes slightly smaller, right? If it used to be like this and we squeeze, it would be like this, right? It would be closer and not only closer to the y-axis, but also for each value of x, the new curve is below the old curve. So it looks like this curve by absolute value will be greater than this curve and it's negative. So it looks like the result would be more on this side. Now, at the same time, when we move to p over 2, this guy goes to plus infinity and this one is around minus 1, so the whole sum would be to plus infinity. So the curve looks like it would be like this. That's my understanding. Now, what happens next? Well, here, from pi over 2 to pi, either one or another components, or both actually, in this case only one, and in this case both goes to minus infinity, which means that the whole graph would go around these two asymptotes to minus infinity. So if we will add them together, it will be something like this. Now, very similarly, in this case, the graph will go like this, because both are relatively limited here and then go to plus infinity on both sides. Next, we have a very similar situation here. Now, when we approach 3 pi over 2 from the right, this one is limited and this one is minus infinity, so it's definitely this way. But then, we have the same story when this curve is actually greater by absolute value than this one, so the graph will go up. And this looks like the graph on the period from 0 to 2 pi, and then everything is repeated. Okay? Seems to be like the right way. This is definitely the case when you can actually check this particular graph with scientific calculator on or on some web page which allows you to graph the functions and check if this looks more or less like. I mean, obviously, this is an approximation. I don't really put exact numbers, but you will get exact graph and it looks, at least from the general shape, it should look like this. If I didn't make a mistake, which I hope I didn't. All right. So that was an interesting exercise in how to add graphs. And especially interesting was consideration about infinities. When one graph goes to plus infinity and another to minus infinity, the question is which one is faster? Which one is greater by absolute value? And that infinity takes precedence, actually. Now, the last one is like this. X and this is X plus pi over 2. Well, basically, we will do exactly the same. We will add two graphs together. Now, how to do it? Well, let's just start from this graph. This is kind of obvious. And I will also use only the 0 to 2 pi because it's a period, obviously, a period for the function, so I will draw only this. So we have an asymptote here and we have an asymptote here. So this is 1. This is minus 1. The graph goes, one goes this way and another goes this way. That's cosecant of X. Now, what's the cosecant of X plus pi over 2? Well, we have to shift the graph to the left because it's plus sign, so it's to the left, by pi over 2, which means that new graph will have asymptotes at pi over 2 and 3 pi over 2. So we shift it to the left by pi over 2, so this piece will take this place. Let me use a different color. And this piece will take this and this piece will take this. But we don't care about less than 0 greater than 2 pi. So let's concentrate from 0 to 2 pi. We don't really need this one. All right, so let's add two pieces together. Now, on this interval, everything is easy because the function, one function goes to plus infinity and another goes to plus infinity. So basically, their sum would be like this. It would be to plus infinity to both sides with some midpoint. Here, interestingly enough, we have infinity only on one side, negative here and positive here. So whenever we are approaching pi over 2 from the right, this thing goes to minus infinity. So the sum with this, which is around 1, would be approaching minus infinity. And on this side, when we are approaching pi from the left, the graph will go to positive infinity. So it should be something like this. Now, here, the situation is similar to here, but reversed. So we have approaching pi from the right, we go to the minus infinity, approaching 3 pi over 2 from the left, we go also to the minus infinity. So we will have something like this. And finally, here, we will have, this is plus infinity, this is around minus 1, so the result would be here and here you have to minus infinity. So that's the graph of the function on this particular interval from 0 to 2 pi. Well, and then it repeats itself. Interesting. And again, that's very good to check on the calculator. All right. This lecture actually completes my graphic exercises for different trigonometric functions. And if you notice, I'm using only one function at a time. I had a lecture about sine, I had a lecture about cosine, tangent, cosecant in this case. I did not mix functions together, which is another very interesting component and maybe we will address it in some other lectures. So meantime, I do recommend you very strongly to go through the graphs of all trigonometric functions. It's six lectures, this one is the number six actually. And in each case, try to solve all the problems presented as the problem, draw your own graphs, and check it with some scientific calculator or on the web page whether you are correct or not. I think it will be a very good exercise to basically explain to you how the function behaves. That's it. Thank you very much and good luck.