 Hi, I'm Zor. Welcome to a new Zor education. Continuing building different graphs related to trigonometric functions, today's term is for secant. Secant is by definition, as you know, one over cosine of x. That's secant. Now, before addressing any individual problem which I have here, let's talk about secant itself very briefly. Now, considering this definition, and considering that cosine looks like this, where this is pi over 2 minus pi over 2, 0, y, x, pi, minus pi. One over cosine of x obviously has asymptotes where the cosine is equal to 0, which is pi over 2 plus pi plus 2 pi, etc., or minus pi minus 2 pi, etc. So these are asymptotes for then how it looks. Well, let's consider from 0, we move let's say towards the increasing argument x. At point 0, cosine is equal to 1, and obviously, inverse would be 1 as well. So this is a starting point. Our secant will grow because cosine goes down to 0 being positive, and this is denominator, which means one over cosine would go to plus infinity in this particular case. So let me just draw it this way. It goes asymptotically goes to this vertical line through pi over 2, and same thing goes here as well. Now, in this interval, everything really is similar except it's negative. So from minus 1, inverse to minus 1 is minus 1 obviously, and then it goes again since cosine goes to 0 being negative, one over cosine goes to infinity but again negative infinity. So it goes this way, and same thing here, and in this case on the positive side, so that's basically the graph of the second. Second is the black ones. Now, let's talk about properties. First of all, cosine is an even function. It doesn't change the value if you change an argument from positive to negative, and obviously, it stands exactly the same way for an inverted cosine, which is secant function is even. It has exactly the same value for positive as well as negative argument of the same absolute value, and that's why the graph is symmetrical relative to the y-axis. Now, does it have zero points? No, secant never equals to zero. It goes from either 1 to the positive infinity or from minus 1 to the negative infinity. So the range of this function is a combination of two intervals, this and this. Now that I'm using square bracket here and parenthesis here, meaning the end is not included because obviously the end is infinity. So it's a symbolical representation that obviously goes to infinity, obviously it's never reaching it. Same thing here. All right, so that's the range of the function. Asymptotes, yes. Asymptotes are where the cosine is equal to zero, which is pi over 2 plus pi times any integer number, positive or negative. Maximum and minimum? Well, there is no absolute maximum and minimum, but there is a local one. On each interval, there is either minimum on this interval, which is one, or maximum on this interval, which is minus one. Now let's talk about graphs. So the first example which I have is y is equal to second of minus x. Now, second is an even function, and as we were just talking about, which means second of minus x is exactly the same as second of x, right? So this is the same as second of x, and the graph is obviously the same. And as you know, the graph is something like this, where these are asymptotes. And this is minus pi over 2, and this is pi over 2. And this is 3 pi over 2. So that's the graph. That's simple. Now, how about second of 3x? Well, second of 3x, as I have explained in many previous lectures and in a general lecture about graphs of the functions, whenever you multiply an argument by 3, it means that the graph is squeezing towards the y-axis, which is, in this case, this one. So along the x towards y, towards 0, y, the simple explanation is if you have a function and point AB belongs to its graph, it means basically that B is equal to f of A. And then you have a function of 3x, then obviously 1 third AB would belong to this particular graph. Because if you substitute 1 third A as an argument, you will get f of A, and you know that f of A is B. So that would be B. So for each point AB of the original graph y is equal to f of x. There is a point with coordinates 1 third AB to the function which has a 3 multiplied by an argument. So this point 1 third AB is different from this one by its x coordinates 1 third, which means it's squeezed in 3 times. So if my asymptotes are p over 2 minus p over 2 and p over 2 and 3 over 2, then my new asymptotes would be, instead of p over 2, I would put p over 6 and p over 6 here and p over 2 here. So everything is divided by 3. So that's basically the graph. It's just changing the scale if you wish. So it looks exactly the same. Just asymptotes are spread apart, not at pi interval from pi over 2 to 3 pi over 2 to 5 pi over 2, et cetera. But it's spread at pi over 3 in 3 times less interval. So the asymptotes are at intervals of pi over 3. From minus pi over 6 to pi over 6, from pi over 6 to pi over 2, et cetera. But the general shape remains exactly the same. Next. Next is, instead of multiplying by 3, I will multiply by 1 third or divide by 3 if you wish. Well, obviously, the transformation of the graph is similar, but instead of squeezing, you have stretching. Because again, if function f of x has a point a b, then f of 1 third x has a point 3 a b. So this means that b is equal to f of a. And if you will substitute this into this, 3a into the x, you will have again a. And you know that f of a is b. So this particular point belongs to this graph, which means the y-coordinate is the same, but x-coordinate is three times further. Now, when I'm talking about further, I'm talking about stretching from the 0 point to the right and to the left. So again, my original asymptotes were pi over 2 and 3 pi over 2. Now, my new asymptotes would be spread three times further from each other, which means that this is supposed to be 3 pi over 2. This is supposed to be 9 pi over 2. And this is supposed to be minus 3 pi over 2. So the difference between them from minus 3 pi over 2 to 3 pi over 2 is 6 pi over 2, which is 6 pi over 2, right? Which is 3 pi, right? So used to be pi, now it's 3 pi, the distance between different asymptotes. And from 3 pi over 2 to 9 pi over 2, it's also 6 pi over 2, which is 3 pi. So these asymptotes are rarer, three times rarer. So that's the graph again. It's the same shape, just different scaling along the x-coordinates. OK. Next is, again, very much like it, but stretching would be instead of horizontal, it would be vertical. So the function is 3 secant of x. So now we are multiplying the function by 3, not the argument. Now what happens? Well, we used to have minus pi over 2, pi over 2, 3 pi over 2, minus 1, 1. Now what happens if every value of the function I will multiply by 3? Well, obviously the function will be stretched vertically. Now in this particular case, if I will use this example, what does it mean? It means that if I have a function 3 f of x, then a 3b would be a point of this function. Indeed, if I will substitute a into this, I will have f of a. But f of a is equal to b according to this. So the times 3 would be 3b, exactly. So what I'm saying is that if point a b belongs to this graph, which is b is equal to f of a, then this point belongs to this graph. Because again, 3b is equal to 3 times f of a, which is b. So it's 3b. Now what does it mean? It means that the graph is stretching vertically from the x-axis up and down. So this would not be 1. This would be 3. And this would not be minus 1. This would be minus 3. And the whole graph would go vertically a little steeper. But what's important is the asymptotes will remain exactly where they are. So the graph would be steeper. It will start from 3, and it will go three times faster up to the infinity, plus infinity or minus infinity. But it will be still within the same asymptotic boundaries. Next, secant of pi over 2 minus x. OK. Now, let me convert it slightly. Again, based on the fact that the secant is even function, which means it doesn't change the variable if I change the sign of the argument, I would rather use it as a secant of x minus pi over 2. So I inverted the sign. Now, what is this? Let's go back to this general function, general properties of the graphs. If I have minus c, what happens to the graph? Obviously, a plus c comma b point would belong to this graph. Why? If a b belongs to this, which means b is equal to f of a, that's given. Now, this point, a pi by cb, if you will substitute it instead of x, a plus c, and minus c would be f of a. And that's equal to b, exactly. So for each point of this graph, there is a point which is shifted to the right by c. Well, right if c is positive, if c is negative, it's to the left. So in this particular case, minus c, this is minus pi over 2, so everything is shifted to the right by pi over 2. What does it mean? It means that the graph actually remains as it used to be before, which is this is 1 and minus 1. But everything is shifted to the right by pi over 2. So this asymptote, which used to be at minus pi over 2, now will be at 0. So let me just draw another one. So now I will have a 0 asymptote. And then instead of pi over 2, it will be pi. And then 2 pi, et cetera. So these are new asymptotes. But other than this, from 1 and from minus 1, other than this, asymptotes, they direct. Basically, they restrict these lands of this. But other than that, behavior is exactly as it was before. So only shift to the right by pi over 2. So the asymptotes are changing, pulling the whole graph to the right. Now, a combination of all the above, just to make sure that everything is in place, as far as we understand it. So the combination is y is equal to minus 1 third secant of minus 3x minus 3 pi over 2. OK. How to deal with this? Well, we can deal with this in steps. We know how the graph is transformed on each elementary transformation of argument or function, et cetera. So let's just transform it in this way. First, I would draw secant of x. Then I would draw secant of 3x. Then I will draw secant of 3x plus pi over 2. Then I will draw secant of minus 3x plus pi over 2. And finally, I will multiply the whole graph by minus 1 third. Well, I can do it first by 1 third. And then minus 1 third. So a chain of these transformation will lead us to the final graph. But let's just summarize how it would look. Secant, we know what it is. Secant of 3x, we also actually did. It's squeezing of the graph horizontally along the x-axis towards 0 by a factor of 3. Then I add positive number, or subtract negative, if you wish, which means I'm moving to the left, the whole graph by pi over 2. Then I change the sign of the argument. But secant is even function, so it doesn't change the graph. So from here to here, there is no difference. Then I multiply graph by 1 third and we already did something like this. It means I'm squeezing the graph vertically by the factor of 3 towards 0 along the y-axis. And finally, I multiply it by minus 1. When you multiply the graph by minus 1, whatever was positive becomes symmetrically negative. Whatever negative becomes symmetrically positive. So graph is just turned this way. All right, I can do it relatively quickly, I hope, without much detail. So if you have, right now, we have from minus pi over 2 to pi over 2, pi 3 pi over 2, 2 pi. OK, so let's do this. And this is 0, obviously. So we start with, OK, this is minus 1, this is 1, and this is minus 1. So we start with this picture, that secant. Times 3 for x, that squeezes everything by a factor of 3, which means this becomes 2 pi, OK, let me just write it this way. This becomes 2 pi over 3, this becomes over 2, this becomes pi over 3, this becomes pi over 6, and this becomes minus pi over 6. So that's after this. Now we should move the whole graph to the left by pi over 2. So 0 was here. Now when we move it here, this would be 0. So the new graph would be like here. 0, and then you have pi over 6 to the right, and pi over 6 to the left. And these are new, wait a moment. Did I shift it? No, I'm sorry, I think I mixed that a little bit. Yes, that's not supposed to be 0. I started from this, this is, I didn't shift it yet. This is already a shifted variant. OK, let me start in the beginning. We should start with 0 and minus pi over 2 and plus pi over 2. And these are asymptotes, so it goes like this. Now this is pi and this is 3 pi over 2 and this is 2 pi. So the next is this. OK, now when I squeeze it, that's pi over 3. This is pi over 6. This is pi over 6. This is pi over 2. And this is 2 pi over 3. That's how I squeeze it by a factor of 3. Now when I move it to the left by pi over 2, so this point goes here. OK, so my 0 becomes an asymptote. OK, my 0 becomes an asymptote. So on the left and on the right, I will have this. And this is pi over 3 because asymptotes are now every pi over 3. Used to have asymptotes every pi, but now since we multiply by 3, every asymptote is the distance of pi over 3. So that's how it will be after this particular transformation. Now after this, nothing's changed because function is even. So if I change the argument, nothing changed. And all I have to do is multiply by 1 third, which means whatever used to be 1 now will become 1 third. And this was minus 1, so now it's minus 1 third. Because everything is squeezed in this way. Obviously, I didn't do it in a scale. My minus is definitely smaller than my plus. So this is my minus 1 third, so the graph is minus. So that's how the graph looks like. It's period is instead of 2 pi, the period is 2 pi over 3. The asymptotes are after each pi over 3. The minimum, local minimum is 1 third. And local maximum is minus 1 third. And these are asymptotes, so positive and negative, et cetera. And then the whole thing repeats, obviously. So you have something here and here. So that's the graph. Well, it might be a little more complicated. But in theory, the most important is to represent the complicated problem as a consecutive steps, each of them is not really complicated when you are familiar with this. OK, we have a couple of more problems related to graphs. This one is y is equal to secant of 2x plus secant of minus x. OK, so let's just add two graphs together. Now, we know that secant of x or secant of minus x is exactly the same thing. And we have pi over 2. Let's make it wider. Pi over 2 minus and pi over 2 plus. That would be pi. That would be pi over 4, 3 pi over 4, minus pi over 4, minus 3 pi over 4, and minus pi. So asymptotes of this guy are, as usually, from minus pi over 2 to plus pi over 2. And this is one. So that's this guy, this one. How about this one? We already talked about that the whole graph would be squeezed along the x-axis horizontally towards y-axis by a factor of 2. So let's just draw the second one. Now, it means that in this case, the asymptotes would be also squeezed in from pi over 2 to pi over 4. So these are asymptotes. And next asymptote was 3 pi over 2, right? 3 pi over 2 and minus 3 pi over 2. So everything is shifted. So we squeeze this one and let's squeeze this one. It would be 3 pi over 4, right? So that would be this one, another asymptote. And 3 pi over 4, OK? Now, what does it mean as far as our graph is concerned? Well, this piece would be here between these two guys. And this guy would squeeze in between these two. And this piece would also squeeze in between these two. So we have to add brown to red. OK, let's do it. In this particular area in between these two things, obviously we start with 2, because 1 and 1, we will add together would be 2. And then 1, the graph goes a very limited way. And another goes to infinity, which means we will have going to infinity relative to the same two asymptotes. Obviously, asymptotes would be the most frequent one. OK, this is this piece. Now, let's move on here. In this case, the brown one is very limited, but the red one goes to infinity. So on this side, the red one is limited, and the brown one goes to minus infinity, which means that the graph should be something like this. It's plus infinity here, and then gradually going down to minus infinity. If you add this red to this brown, in this case, the red prevails with a plus infinity. In this case, brown prevails with minus infinity. So the graph would go like this. And obviously, similarly, it will go the same way here, because it's an even function. So this red piece and these two are the sum of these two graphs. And then the story actually more or less repeats itself, because in this particular case, we have these two asymptotes. In both cases, both are minus infinity, which means that their sum would be like this. It's limited, somewhere limited in the middle, but goes to minus infinity, because either one or another component of this sum goes to minus infinity. And similarly here, well, et cetera. Then we will just spread it, and it will repeat itself more or less. Well, that's how it looks like. This is the sum of two sequence with different arguments, different multipliers, actually, because they have different asymptotes. But now the next problem is in some way similar, but in one particular way, it's different. And let me tell you which way. OK, one of them is x, and another is x plus pi over 2. So in this case, we have basically similar graphs, but shifted horizontally one relative to another, not squeezed, but shifted. So that's the difference. All right, let's see what happens in this case. So one would be from minus pi over 2 to pi over 2 like this, then to pi, and then to 3 pi over 2, then to 2 pi. So that's another asymptote, and that's another one. Here you have minus pi, minus 3 pi over 2, and minus 2 pi. That's another asymptote, and that's another piece. So that's how one graph looks like. Another graph is shifted to the left by minus pi over 2. So whatever was 0 here is shifted here. So the asymptote, which was pi over 2, will become 0. So basically, we have new asymptotes at these boundaries. And the new graph, which is shifted by pi over 2, would be like this. We shifted this, let's shift this one, and let's shift this one. So now we have to add dotted line and solid line. Well, let's do it. And again, as you understand, every asymptote will retain its asymptotic quality, because either one or another argument goes to infinity around it, and it's not defined. All right, so let's start from 0, let's say. When we go to 0 from, let's say, left, the dotted one goes to plus infinity, and this one is basically limited. And on the right from the 0, dotted line goes to minus infinity. So if I have to draw their sum, it will go to plus infinity here and minus infinity there. Let's go to this point. At this point, the dotted line is limited, but this one is plus infinity. So I have to round here and go up. Here, this one is plus infinity. This is limited, which means I have to go to plus infinity. OK, so these are my brown lines, which represent this graph and this particular piece. Next, let's consider this particular thing. Well, obviously, it's minus infinity and minus infinity, and this is something limited, so it goes like this. Very much close to this one. And on this one, on this side, this is minus infinity, and this is plus infinity, so it should go from here to here. And then similarly to this, it will be here. The function is symmetrical at this point. And then it will continue, basically, based on periodicity. So the complicated graph looks like this. And obviously, it's not defined in any point where you have an asymptote, either one or another. Well, these are exercises in graphing different functions, and I think it's very useful just to understand how the function behaves. There is one more lecture for cosecant, which would be next. And that would be the end of the graphs for trigonometric functions of this type. These are just properties of trigonometric functions by themselves, everyone separately. Then we might actually combine sine and cosine, tangent and cotangent, what happens with them. So it's all related to your understanding how the function behaves. OK, that's it. I do suggest you to reexamine the notes for this particular lecture at unizord.com and try again to draw these curves. And you know what might actually be helpful if you want? After you draw it yourself in this schematic fashion, you can use probably one of the scientific calculators which is capable of graphing functions, or you can find it on the computer some tool which can do exactly the same. And you can verify if the function like this does have the graph which looks like whatever you have schematically put on your paper. That's it. Thank you very much and good luck.