 Hi, I'm Zor. Welcome to a new Zor education. Today we will graph the cotangent function and all the related to this particular issue items we will consider. So cotangent, you know the definition of the cotangent is cosine. That's all one. That's cosine divided by sine. And I do have a few different problems related to graph of this particular function. Alright, so let's start. Function is odd or even. Well, function is definitely odd because cosine is even. Sine is odd, which means sine changes the sine if argument changes the sine and cosine doesn't. So the ratio obviously changes the sine. Now, is there any symmetry? Well, cosine is an even function and it's symmetrical relative to the vertical y-axis. Sine is odd and therefore it's centrally symmetrical. So the result would be obviously the centrally symmetrical graph. Now, does it have zero points? Yes, when cosine is equal to zero, which is pi over two plus pi plus two pi, etc. Yes, asymptotes exist when sine is equal to zero, which is zero pi, two pi, etc. And finally, maximum and minimum, no, they do not have, this function does not have maximum and minimum because it goes to either minus infinity or plus infinity when the sine goes closer to zero right near the asymptote. So let's draw the graphs. The first one is y is equal to cotangent of minus x. Now, in the previous lectures, which were dedicated to other trigonometric functions and in the general lecture about functions and their graphs, I was talking about the graph of this function. Basically, we can generalize these two functions. If the point a, b, let's say, belongs to this graph, then point minus a, b belongs to this graph. Because obviously, if you substitute minus a into the x, you would have same as if you substitute a into this function. So these two points are symmetrical relative to the y-axis because the ordinances are the same, b, but the abscissa is changing the sine. So basically, the graphs are symmetrical relative to the y-axis. For every point of the graph of this graph, there is a point symmetrical relative to the y-axis in this graph. So if I know the graph of the cotangent, I will know the graph of the cotangent of minus x just by symmetrically reflecting it relatively to the y-axis. So let me start from the cotangent. Now, the asymptotes are obviously where sine is equal to zero, which is zero pi to pi. So these are asymptotes. Right near zero, this denominator is almost zero, right? Just a little bit on the positive side. And cosine is equal to one at point x equals to zero. So if I divide one into a very small positive number, I will have a very large positive number. So this how it goes. Then at point pi over two, when the cosine is equal to zero and sine is equal to one, the cotangent is obviously zero, and then it continues to minus infinity as the sine approaches to pi because the cosine turns negative after the pi over two. And then everything goes symmetrical by a period. So now if I want to, let me continue this periodicity to the left as well. So if I want to reflect this graph relative to the y-axis, what will I have? Well, obviously I will have this. So these red curves represent graph of y is equal to cotangent of minus x. Good. Now, let's wipe out this so I don't have to draw again the cotangent. And we will go to the next problem. And the next problem is cotangent of 3x. Okay. Now, again, from the previous lectures about other trigonometric functions and from the general lecture about the graph of the functions per se, when you multiply an argument by, let's say, 3 in this particular case, what happens with the graph? Well, the answer is the graph is squeezed in horizontally towards the vertical y-axis by the factor of 3. So if you multiply by 3, graph is squeezing by 3 times. y, very simple explanation. If you have two functions, y is equal to f at x, and there is a point a, b which belongs to this particular graph, and then you have f of 3x, then obviously 1 third a, b belongs to this graph. Well, if you substitute 1 third a into the x, that multiplied by 3 would be a. So an f of a, as we know, is b because a, b belongs. So that would be the true equality. So if point a, b belongs to the original graph, point 1 third a, b, which is the same b, the same order, the same y-axis coordinate. But the x coordinate is squeezed by 3 times. So the whole graph was squeezing, which means that these asymptotes would not be at 0 pi, 2 pi, or minus pi and minus 2 pi. They would be corresponding with 3 times narrower, which means I will have here asymptote as at pi over 3. And instead of 2 pi, I will have 2 pi over 3, which is somewhere here. And the graph would be exactly like this in shape, but squeezed in. So let me just do it this way. So it's closer asymptotically closing to the left asymptote on the top and to the right asymptote on the bottom. Same thing here. Starts here, goes in the middle and goes here. And obviously repeated after every p divided by 3 period, because the period of dysfunction is now p over 3, not pi. Again, p and pi I'm mixing together. One is Greek and now there is English. So, okay, that's it basically. The graph is squeezed in. Next, now I will have to wipe it out because it's too much. Now here and here we will have different numbers. We will have one-third and here we will have also one-third. What does it mean? Well, it means that instead of one-third a, I should put here 3a. Because now if I substitute 3a into the x, then I will have basically the equality. Which means if point a b belongs to original graph, point 3a b, which is three times stretched further from the middle point, positive are going to the right, negative are going to the left. Everything is stretched horizontally from the central point, from the y axis, left and right. So the graph, again, it's very easy to put it together. If you have original graph, which is pi, 2pi, 3pi, 4pi. So, original graph is like this. Now my new graph would be stretched three times from the point zero. This is my zero. This is my y axis and this is my x axis. Now, period used to be pi, now the period will be 3pi. So this would be my first asymptote and at 6pi would be my next asymptote. Because this 2pi point would stretch three times to 6pi, right? So that would be my next asymptote. So the graph would look exactly like this as far as the shape is concerned. But stretched three times, which means, again, on the left it's closer to the plus infinity and on the right, so it's somewhere here. It's like this. So that would be my graph. And obviously repeated with a period equal to 3pi, because this is the period, 3pi repeated left and right. Alright, fine. That's done. And next is y is equal to 3 cotangent of x. So now I'm multiplying the function, the argument. Well, again, from the previous lectures about trigonometric functions and from general functions and their graphs, you know that this is a vertical stretch. Well, stretchers, please, depending on this particular coefficient. Three means stretch. So whenever, again, let me get back to my general functions. If you have general function f at x and the point AB belongs to this particular graph, which means B is equal to f at A, then if you have the graph this, then obviously A3B belongs to this graph. Why? Well, that's substitute. A would be f at A. f at A we know equals to B. So it would be 3B. And y is equal to 3B. So for each function, for each point AB which belongs to this graph, the point which has coordinate A3B, which is three times larger the ordinate, the y coordinate, belongs to this graph. So everything seems to be like stretching in a vertical direction. From zero up, it would be stretching up. From zero down, it would be stretching down because positive numbers are multiplied by three and negative are multiplied by the same three. So it goes further from zero. And now about the graph. So let me start again. I will do it only on one particular period because the period is exactly the same. So my original graph seems to be like this. Now my new graph would be stretched three times, which means every ordinate is tripled. So it's something like this. Whatever it was here, it's three times more here. So that's what means stretching three times. Obviously, if this coefficient is less than one, let's say it's one-half, it would be squeezing. But anyway, every ordinate is multiplied by this particular factor. Okay, next. The tangent of pi over 2 minus x. Okay. The way to approach problems like this, the easiest way is the following. From the general function theory, you know that if you have a graph of function and a point in this graph, then you have a graph of this function, well, I shouldn't use the same letter a, plus some constant c, would be a minus cb, right? Because if you substitute a minus c instead of x, you would have a minus c plus c, which is a, and you know that f of a is b, so this would be the quality. And this means what? This means that the point which is c, item c units to the left from the original a, b. So the whole graph is shifted by c to the left for positive c. Obviously, if c is negative, then it's to the right. So we will use this, but it's not exactly like what we have here, right? What we have here is minus x here, and here we have x with plus. But now we know that the function, below the graph of the function, cot engines of minus x, it's already been done. It's symmetrical relative to the y-axis to the original. So I will convert this into cot engine of minus x minus y over 2. Minus x minus and minus would be plus. So that's the same thing. So if I will start from cot engine of minus x, then I can use this particular principle. So let me start from the cot engine of minus x, and we already know what that actually is. So original, now this is pi, and this is minus pi, and we know that the original cot engine is like this. So symmetrical, I was already saying, is like this. Oops, that's wrong too much. Something like this. So that's original, that's cot engine of minus x. Now I have to subtract pi over 2 from the argument, and from this principle it goes, well in this case since this is negative, it goes to the right by pi over 2. So whatever asymptotes were, and they were at 0 pi or minus pi, the asymptotes will shift together with the graph. So the new asymptotes would be at pi over 2, which is in the middle of this. And minus pi over 2, and 3 pi over 2. But the graph basically will remain the same in between these asymptotes. So it would be like this. So my new graph is black, something like this. So it goes from the black one. So it's minus pi over 2, this is pi over 2, and this is 3 pi over 2. So that's where new asymptotes will be, and the black represents this function, the graph of this function. Okay, done. A little bit more complicated. Y is equal to minus 1 third cot engine of minus 3x minus 3 pi over 2. Okay, what should we do with this? Well, again, let's just transform this into something a little bit more palatable, which is minus 1 third cot engine. Well, it's not working. Minus 3x plus pi over 2. Right? So what are the steps? Well, steps are like this. First, we will do cot engine of x. That's original, right? From this, we will do cot engine of minus 3x. Now, what does this mean? It means that we are using the minus, which means we are reflecting relative to the y-axis, as we know, but we also are multiplying argument by 3, which means the graph would be squeezed by 3 times. And again, we already had this. This is just a combination of whatever it was before. Next would be cot engine of minus 3x plus pi over 2, which means we will shift the whole graph by minus pi over 2, which means pi over 2 to the left. And finally, we will multiply it by 1 third, which means we will squeeze it vertically 3 times. So that's the plan, right? So I think the best thing, if you are working with functions like tangent or cot engine, to work around asymptotes. So you know that basically the cot engine is like this relative to the asymptotes, right? Now, if you multiply it by minus 3, it would be symmetrical relative to the y-axis, which means instead of this, you will have this, and this would be minus pi, and this is 0, this is pi, and obviously repeat it. Now, since you are multiplying by factor... Now, this is a cot engine of minus x, and now we have to multiply argument by 3, which means we will squeeze it. So instead of pi, this actually would be pi over 3, and this will be pi over 3, and this is 0, because it's squeezed in. Instead of 0 pi, it would be 0 pi over 3. Now, we will shift it by minus pi over 2, which means pi over 2 to the left. So 0 would become minus pi over 2, and minus pi over 3, if we shift it to the left, pi over 3 plus pi over 2 would be what? It's 2, 5, 5 over 6, right? So it's somewhere right here. Minus 5 pi over 6. So this would be our second, and the graph would be between them. So this is, right now, from minus pi over 5 pi over 6 to minus pi over 2, is our periodicity. And obviously we can repeat these particular asymptotes every pi over 3 times. So the next one would be minus pi over 2 minus pi over 3, that's pi over 6, right? So that would be my next. And plus pi over 2, that's pi over 2 minus pi over 6, it's pi over 3. Next is pi over 3. So this is my next. So anyway, you understand how to basically calculate this particular period. Oh no, not pi over, it's pi over 3. It's not pi over 3, it's pi over pi over 6, yes, pi over 6 plus pi over 6, because the period is pi over 3, right? So from minus pi over 5 pi over 6 to minus pi over 2, from minus pi over 2 to minus pi over 6, minus pi over 6 to plus pi over 6, etc. So that's how it looks. This is y-axis in the middle, and this is the graph. So key points are minus pi over 6, minus pi over 2, minus pi over 6, etc., with a period of pi over 3. And the only thing we have to add right now is multiply graph by one-third, which means it will be flatter, if you wish. So instead of this, it would be, let me get another color, it would be this. It's flatter. Closer to the x-axis. So that's the graph. What's the lesson of this thing? We can build the graph step-by-step, transforming our original graph into whatever we need, using the methodology which we know. I mean, multiply by something, an argument or multiply function by something, or add to argument, etc. So we know these simple manipulations, and from these simple manipulations, we combine them together to get to the real graph. Okay. We have two more problems. These problems are related to addition of the graphs. So y is equal to cotangent of 2x plus cotangent of minus x. All right. So let's think about it. The periodicity is pi, obviously, because the periodicity of cotangent is pi. Now, this is, as you know, a squeezed cotangent function, squeezed horizontally towards the y-axis. So its period is actually pi over 2. So what we have to consider is pi over 2 points on the surface. So its pi over 2, pi 3 pi over 2, 2 pi minus pi over 2, minus pi, etc. 0, y, x. Now, these are asymptotes for cotangent of minus x. These are on the pi boundary. And it's cotangent of minus x, which means it goes this way. Now, now we have to put cotangent of 2x. Well, 2x has a period of pi over 2, which means these would be asymptotes. On every pi over 2 point. So the asymptotes would be here. So the graph would be here. It's not easy, etc. Now we have to add them together. Well, obviously, we have to consider these points. These are key points. So what happens at points, well, 0, pi over 2, pi, etc. Well, obviously these are asymptotes. You can deny that. Now, how the function behaves in between. Well, that's not such an easy question, obviously. But let's try. Well, let's approach from left to right. We approach 0, both graphs are going to plus infinity. So, obviously, here the function, which is sum of these. Oh, wait a moment. I think I made a mistake. These are cosine of 2x, not x, which means the direction would be different. My mistake. The direction would be this way. Now let's sum them together. This is not easy. Because as we approach 0 from the left, one function goes to plus infinity, another goes to minus infinity. Now, which one is bigger? Which infinity is stronger? Well, let's just think about it. This one is squeezed in function cotangent of x. As we squeeze it in, it seems to be steeper, right? So, this one is steeper than this one, which means the ultimate result would be minus infinity. And in this case, when one of the functions is 0, another is plus infinity. Obviously, it would be result of the plus infinity. And it will cross somewhere in between. That's what I believe will be in this particular case. Now, in this particular case, again, this is as steeper than this, because this one is a squeezed in. So, in this case, infinity will be on this side, and eventually it goes to the negative infinity, because one of them is minus infinity and this one is 0, right? So, this one goes this way. Well, basically that's it, because after that, it repeats itself. After that, it's a period. So, it will be something like this. So, it will resemble actually the cotangent of 2x more or less. I mean, obviously, it will cross the x-axis in different positions, et cetera. But as far as the shape is concerned, as far as asymptotes are concerned, it will behave more or less like this function. That's how it looks. I think I'm right. It's not well-corrected. And one more last task. It's also on summation of two graphs. One is cotangent of x, and another is x plus pi over 2. So, if before we had more or less the same points where the function has asymptotes, now the asymptotes will be shifted, right? One of them will be from 0 to pi. That's the cotangent of x. And another would be shifted by pi over 2 to the left, which means it would be from pi over 2 to minus pi over 2. So, well, let's put pi. So, this is my one function. I'll use it like this. And another function is like this. So, what will be on this particular case? Now, I will continue here and here. So, that's how it will go. So, it's one curve which belongs to one graph and another belongs to another, and then again one and again another. How we should add them up? Well, obviously, it will be an asymptote on every pi over 2 boundary, right? So, let's do it from 0 to the right, let's say. One function is 0, another is plus infinity, and when we move to pi over 2, it will be here. So, it's something like this would be the combination. Same thing here. This is 0, this is plus infinity. So, it goes this way, and then to minus infinity. And same thing here. And same thing here. So, this is a function which seems to be periodic on a period of pi over 2, and within each period again it looks more or less like a cot engine. It's just slightly different in shape, but in principle it goes along the same type of curve. Okay, basically that's all I wanted to show today. Thank you very much. And we have two more lectures about the graphs, secant and cosecant. Good luck.