 Hi, I'm Zor. Welcome to a new Zor education. I would like to spend some time and talk about graphics of trigonometric functions. Everyone of them will be addressed in separate lectures. So today I will talk about sign. The first of all, before we go into concrete problems, which I would like to basically present to you, let me summarize what you really have to understand about the function before you attempt to graph it. Well, I have a few points here, which I will just address one after another. First of all, trigonometric functions are either odd or even. Odd means they change the sign, if argument changed the signs, like sign for instance or tangent, and even function like cosine do not change. Very little if argument changes the sign. Now, next is symmetry. Well, if the function is odd, then the graph is centrally symmetrical relatively to the beginning of the coordinates. So it's like turning by 180 degree, that's central symmetry. If function is even like cosine, then it's symmetrical reflective symmetry relative to vertical axis, y-axis. Now, next it's important to know when the function equals to zero, zero points. For sign, for instance, zero pi, two pi, etc. for cosine, everything is shifted by p over two. So it's p over two, p over two plus pi, p over two plus two pi, etc. Periodicity obviously is important. Now, the periodicity is two pi for sine cosine, it's pi for tangent for tangent. Next is asymptotes. Functions like tangent and cotangent and second and cosecant, they have something in the denominator which can be equal to zero. At these points, when denominator of these functions equals to zero, the function itself has an asymptote. You have to analyze what's the sign of the asymptote on the left and on the right. What's the sign of the function on the left and on the right of the asymptote? Next is maximum and minimum. Some functions have maximum and minimum like sine. Some functions don't have anything. They're going to negative infinity to positive infinity like tangent. Some functions have only local minimums and local maximum like second and cosecant. So understanding all these issues, let's just go through examples one by one. So the first one is, you have to draw the graph of this function. Now, let's just think about it. If function y is equal to sine of x has a graph and there is a point a, b which belongs to this graph, what does it mean? It means that b is equal to sine of a. That's what it means. Point a, b, a is x-coordinate, b is y-coordinate belongs to this graph. If it satisfies this particular equation. Now, if this is true for point a, b and graph of this function, then let's take a point minus a, b and see if it will belong to this graph. Well, the answer is yes, because if I will substitute minus a to this and b to this, I will get this particular equation and we know that this particular equation is true. So for every point of a graph a, b of a graph of this function, point a, b, point minus a, b belongs to this particular graph. So if you have a graph of function sine, which looks like this, so this is 0, this is pi, this is minus pi, this is pi over 2, this is minus pi over 2. So if you have a graph like this and you have a point a, b which belongs to this graph, that I know that the point minus a, b which is here approximately, it's symmetrical relatively to the y-axis. This is y-axis, this is x-axis. So if point a, b is here, point minus a, b is symmetrical relatively to this. So for every point of this graph, for every point, there is a corresponding point of this graph which is symmetrical relatively to the y-axis, which means that the whole graph is symmetrical relatively to the y-axis. So we will just have to flip it. So it will be like this. So this is y is equal to sine of minus x. So whenever you change the sine of the attribute, and that's true not only for this function, it's true for any function. If you have any function y is equal to f of minus x, the graph is symmetrical relatively to the y-axis to the graph of y is equal to f of x. Okay. That's the problem number one. I have eight of these problems and they're all more or less the same kind of easy. Sine of 3x. Well, same consideration. Let's start with a function y is equal to sine of x, and let's consider the point a, b belongs to it. Now, if that's true, which means actually that b is equal to sine of f of a, if that is true, then point a over 3, b belongs to this graph, right? Because if you will substitute a over 3 times 3, it would be a and this would be b. So that's the true equality. So for every point a, b, the point a divided by 3b belongs to this graph. So let me start from the beginning. So this is sine 0 pi to pi minus pi yx. Now, if this is a point a, b, now where is the point a divided by 3b? Well, this particular segment is divided by 3, and same thing will be on the negative side. So the whole graph will squeeze in by the ratio of 3. So instead of pi, I will have here pi over 3, and the wave will be like this. Instead of 2pi, I will have 2pi over 3, and the wave will be like this. And same thing here. So the whole graph is squeezed in, and it becomes like the oscillation becomes more frequent, if you wish, three times more frequent oscillation. So whenever you multiply the argument by 3, you will have the whole graph horizontally squeezed towards the beginning of coordinate by the same ratio of 2. Next, sine x over divided by 3. Now, this is absolutely the same, so I don't want to repeat a lot. If point a, b belongs to the graph of sine of x, then point 3ab belongs to this graph of this, right? Because it's 3a divided by 3 with the a, and this is b, and this is the true thing. So for every point a, b which belongs to this graph, which you know, this graph has a point which is 3 times further. So it's not, in the previous problem, we had a squeezing. In this problem, we have a stretching, the whole graph. So if this is 0pi 2pi, minus pi, minus 2pi, then the next will be, this is 3pi. So I'm squeezing, sorry, I'm stretching three times. So this point, this particular piece of the curve will stretch, and this point from pi will go to 3pi, so it will be like this. And continue. So next point will be 6pi over there. Same thing can I think, it goes like this, minus 3pi. So the graph is stretched, and the frequency of oscillation becomes obviously small. Now whatever I was just talking about is true for any function, not only sine or cosine. So basically if you know the properties of graphs, how the graph is transformed, if you are manipulating in some way attributes, like you divide the argument by something or multiply or add something. So if you know this, this should be no problem. Now 3 sine of x. OK, now I will be brief. If point AB belongs to the sine of x, then point A3B belongs to this sine. Why? Well, obviously, if you substitute A here, we'll have sine A, which is equal to B, because AB belongs to the sine, and multiply by 3 would be 3B. So what does it mean? If point AB belongs to this one, let me start again. So where is the 3AB? Well, obviously, it's three times higher. So everything goes three times higher. It would be something like this. So it's stretching, but vertically in this case. Previous problems were squeezing or stretching horizontally. In this case, this particular factor, it is greater than 1. It's stretching. If it's smaller than 1, it's squeezing. So the maximum was 1, but in this case, maximum would be 3, obviously, when x is equal to pi over 2. As you see, all these problems are simple, and all they require is knowledge of how the graph is transformed when argument is transformed as well. Next, sine pi over 2 minus x. Let me transform it a little bit. First, I will do sine of minus x. And they know what it is. Then I will do sine of minus x minus pi over 2. And I know how to do that. And this is exactly what I need, right? So let's start from this. If I change the sign of the argument, you remember the graph would be symmetrical relatively too. So instead of this, I will get this. So that's my first transformation, from blue to red, from sine x to sine of minus x. Now, if I subtract some constant from the argument, what happens with the function? Well, if I subtract positive constant, then the whole graph moves to the right. Again, that's the property of the graphs, which I'm sure you know what it is. So right now, my graph is moved to the right by pi over 2. So my next would be shift by pi over 2. This is, by the way, pi over 2. So the whole thing minus pi over 2, pi over 2. So the whole thing actually moves here. So I will have this way. So that's my new graph. And this is what I need. Now, does it look familiar? Yes, it's the graph of a cosine. Because sine of pi over 2 minus x is actually a cosine. And we will talk about this in a separate lecture. So I was using things from the general graph theory that if you add some constant to the argument, then the graph is moving left and right, depending on the sine of this particular constant is your adding. If it's negative, it's to the right. If it's positive, it's to the left. Next, minus 1 third sine of 3 minus 3x minus 3 pi over 2. OK, let's, again, do something simple. First, I will start with this. Then I will change it to this. Then I will change it to this, or minus 3,000. And then I will change it to this. These are transformations which I would like actually to perform. Now, I don't really need this. Let's just do two steps in one shot. That would be easy, right? So first, I start with my sine. Then I multiply it by minus 3, which means I have to, minus means I'm reflecting relatively to the y. And then I'm squeezing by three times. So this was pi. This was pi over 2. Now this is pi over 6. And this is pi over 3, right? So I'm inverting this. I'm reflecting. OK, and that would be something like this. The wave would be like this. Same thing here. Pi over 6, pi over 3, like this. That's the first transformation. Reflecting and squeezing by three times. OK, next, I have to shift it by minus p over 2. So at p over 2, it would be where? Let me just think about it. So it's something like this, right? This is 2 pi over 3. So the graph would be like this. OK, now, if I add pi over 2, which means I'm shifting everything to the left by pi over 2, which is this particular piece. So it would be something like this, the brown one. So the red one would be shifted by pi over 2, which is this particular piece. It's not on the maximum, obviously. But anyway, that's basically the description of the graph. So you take this one and shift it to the left by pi over 2. That's it. So far, everything which I was just doing was related to general properties of the graphs and their transformation. Next is slightly different. Next is sine of 2x plus sine of minus x. That's interesting. So what is this? We have to basically add two functions. So if this is 2 pi pi pi over 2, now sine x would be this way, but we are reflecting to the negative sine minus pi minus pi over 2 minus 2 pi. So the graph would be this. That's sine of minus x. Now sine of x would be the regular sine squeezed by 2, which means it would be this, something like this. And now I have to add these two together. Well, let's do it in key points. Key points are obviously 0 pi over 2 pi, 3 pi over 2, et cetera. Now the periodicity is obvious because the periodicity is 2 pi. So we can basically forget about the negative side. So let's concentrate on the positive side from 0 to 2 pi. So the blue one is sine of 2x. It goes up and down, up and down, up twice. So we have two waves. And this one goes down and then up and then down again. That's one wave. So where are zero points? Well, zero points are obviously this. And this and this. That's good. Now here the blue one is going up. The black one, which is sine of minus x, goes down. So which one is faster? Well, let's think about it. Since the blue one is squeezed the regular sine, it's steeper, which means it goes up faster. So it goes up faster. And it reaches this maximum at point pi over 4, obviously. This is equal to 1 at pi over 4. Now this function at pi over 4 is equal to square root of 2 over 2. So the difference would be some positive thing. Then we go all the way down very fast on the blue one. And the black one goes to pi over 2. It goes to minus 1. So in this particular case, we have minus 1. The sum of these two, right? So basically, this is how it would look. Next, at this point they are equal. So their difference would be equal to zero. Now here, again, we have zero here. So we will have some kind of a loop here. And then we are subtracting, we are adding both of them together. But now they are added together because both of them are on the positive side. So it would be something, and maximum would be something like this. And then it will go down to this because the blue one is equal to zero. Somewhere along the line, we will get zero. Now this is steeper, so it would be negative, and it goes like this. So what we have right now is also a waving function. But it waves much more in a much more complex way than each one separately. Now it's something, the behavior of the function which is sum of signs with different factors on the argument is basically also an oscillation. But physical meaning of that thing is consider you have some kind of a pendulum. Now pendulum is changing its position also based on some trigonometric function, sign or cosine depending on what angle you are using. But then what if you not only change, not only you move the pendulum itself, but also the point where it is hanging off. And it's also changing in oscillation kind of a way. It's basically this. So you have this thing going left and right, and this is also flexible so it goes left and right. Now the question is, what's the movement of the ending point? Well, this is the physical meaning of that particular thing. So it would be a waving kind of a position changing, waving function, but the waving will be much more complex. So it will seem like jerking actually. But basically it's not jerking. It's small oscillation over applied on the bigger ones. So that's basically the meaning of that thing. And that's the law which you can use to basically to grab the position of this point or investigate its position. And so what's important here, we have two different factors. So one oscillation is more frequent than another. And that's how you have this kind of difficult to describe behavior. By the way, if you will take a look at this particular construction, mechanical construction, you will be surprised how complex the movement of that particular point actually is. Now, and the last one is sine of x plus sine of x plus pi over 2. Now, what's the specificity of this particular thing? Well, the frequency of the oscillation is exactly the same. It's just one that's shifted, in this case, to the left. So let me try to draw the graph. So one graph would be the normal x. So it's like this. Another is shifted by pi over 2 to the left. So pi over 2 to the left would do this. Now we have to add them together. Well, in this case, it's 0. It's 1. So this is the sum. And in this case, it's also one graph is 0. Another is 1. And somewhere in between, this would be a little bit greater. Then it goes down to minus 1, even further down, because this is adding together absolute values to a negative thing. Then it goes to 1 here, because it's 0 and 1 and 1 here. So somewhere in between, I shouldn't really go in this way. It's smooth. So this is basically the sum of these two graphs. And then, by periodicity, it's actually repeating itself like this. This is a smooth thing. And here you have something like this. So in this particular case, the frequency is the same. We're just shifting them together. So it looks more like a regular graph of a sign, as far as number of maximum and minimum. But it's just shifted and stretched a little, whatever. So that's the transformation. We just add it together, these two graphs. Well, as you see, it's not really difficult if you do remember the function's graphs and how the graph are changing with small manipulation of the function itself, like changing of the argument by some linear factor or adding a constant or changing the value of the function, like multiplying by some second. So I do suggest you to repeat again the properties of the function graphs in general. And then all these problems would actually be very easy. Well, that's it for this particular lecture. I will repeat something like this for all trigonometric functions, cosine, tangent, cotangent, second, and cos second. But that's the future lectures. Thank you very much.