 Hi, I'm Zor. Welcome to a new Zor education. We will talk about graphs of different functions related to cosine. Well, the function is the same, but different combination of changes in the arguments and the function, et cetera. This is a continuation of lectures about graphs of trigonomic functions. I have already covered sine, so today's cosine, and then will be the other ones. So before we go into graphs, let's just recall certain basic properties of the cosine function. OK, number one is even function, which means cosine of minus x equals to cosine of x. Secondly, it's periodical function. 2 pi is a period. What's next? Where is it equal to 0? Well, remember, the main graph of the cosine goes like this. This is 0. This is p over 2. This is pi. This is 3 pi over 2. And this is 2 pi. The period is 2 pi. And the function is equal to 0 at pi over 2 plus pi k. The function gets maximum at 0 and 2 pi in both directions. So it's basically 2 pi n. And the minimum is equal to minus 1 when x is equal to pi plus 2 pi n, where n is n. So basically, that's the properties of this particular function. Now using these properties, we will try to draw certain graphs, which I was planning. I think I have 8 of them or something like this. OK, function number 1 is cosine of minus x. OK, graph of this function, well, you remember that cosine is an even function. So basically, cosine of x and cosine of minus x are exactly the same value, which means that the graph should remain the same. On the other hand, you know that if point A, B belongs to graph of function y is equal to cosine of x, which means b is equal to cosine of a, then minus A, B belongs to the graphic cosine of minus x. So together with A, B, we have minus A, B. So the graph of this function should be symmetrical relative to the vertical y-axis to the graph of the base function, y is equal to cosine of x. But again, our function cosine of x is even, which means it's already symmetrical. So basically, if this is the graph of the function cosine, that would be exactly the same graph of the function cosine of minus x. Because together with any point A, B, we have a point minus A, B, which already belongs to the graph. So graph of this function, not only symmetrical to the graph of the base function relative to the y-axis, it's exactly the same because this graph is already symmetrical. Next, cosine of 3x. OK, based on the functions, which I was already explaining, like sine, for instance, and based on the general graphs of the functions, you know that if you multiply the argument by some factor, it basically squeezes the graph horizontally towards 0 point, by, in this case, the factor of 3y. For obvious reasons, again, if you have a function f-attacks and point A, B belongs to this graph, which means B is equal to f of A, then for the function y is equal to f of 3x, obviously, 1 third A and B point belongs to this graph. And what's the difference between these points, A, B, and 1 third B? Well, because abscissa is actually three times smaller. So for every point like this, there is a point which is three times closer to the y-axis horizontally squeezed, which belongs to this graph. So in theory, we should expect this particular graph to look exactly like it was before because it's symmetrical, right? However, it's squeezed by three times. So this was what, p over 2 in the original graph. So now it will be p over 6. And this will be minus p over 6. Then this function, this value, used to be what, pi over 2, right? No, it was pi, it was pi. So now it's pi over 3. And this is corresponding to minus pi over 3. So basically, the whole period, this is 2 pi over 3. And this is minus 2 pi over 3. So the whole period is squeezed from 2 pi to 2 pi over 3. So the graph looks the same, but it's more frequently oscillating around the abscissa. Now reverse, instead of 3, I will put 1 third. What's the difference? Basically, the original difference, instead of squeezing by three times, we have stretching by three times. So again, it's my original graph. Now let me use just two colors. My original graph from 0 to 2 pi looked like this. Now everything will be stretched by three times. Now I just used one period. So now it will be the same, basically. But this would be 6 pi. This would be 9 pi over 2. This one would be 3 pi. And this one would be 3 pi over 2. And the graph would look the same. But it's scaled different. Again, on the same scale, it would look like something like this. So where is the 0? 3 pi over 2. It would be 9 pi over 2 at somewhere here. So on the same scale, it would look like this, much more rarer oscillating around the x-axis. OK, next. Next, 3 fx cosine of x. Well, now we are multiplying the value of the function, not the value of the argument, which means stretching or squeezing is along the y-axis. So whenever I'm multiplying my value of the function by 3, well, it's just three times higher, or three times lower, depending on the sign. But the function, again, would look exactly the same. So if my original function looked like this, and this is 2 pi, this is pi, this is pi over 2, this is 3 pi over 2. Then the new function, when we use the same scale in this case, would look like three times higher, which means instead of 1, it would be 3. Instead of minus 1, it would be minus 3. So what's the graph? It goes through 0. 0 remains 0. But 1 becomes 3, and minus 1 becomes minus 3. So the graph would look like this. And the repetition based on the period. Period is exactly the same, 2 pi. OK, next. y is equal to cosine of pi over 2 minus x. OK, before grabbing this particular function, let me do a little manipulation. Now, obviously, since cosine is an even function, I can change the sign of the argument without changing the value of the function. Now, we know, again, from the general theory of graphs for the functions, if I add or subtract something from the argument, it shifts the graph left or right, depending on the sign. So in this particular case, for instance, if I have a point AB, which belongs to the function cosine of x, then a plus p over 2 b belongs to cosine of x minus pi over 2. So instead of x, you substitute a plus p over 2 minus p over 2. It would be a, and cosine of a is b, as we know. So this particular point belongs, which means that the point, which is pi over 2 to the right from this point, belongs to the new graph. So now, we basically have a shift. So if we used to have something like this, and this is 2 pi, and this is pi, this is pi over 2, this is 3 pi over 2. Now, yes, and obviously, this goes minus p over 2. Now, everything together with a point AB, I have a point A plus pi over 2 b, which belongs to the graph. For every point here, I have a point here. So the whole graph actually is moving that way. So this point, when it's 0, it's at p over 2 in the old graph. In the new one, it will be pi, right? So at pi over 2, it would be 1, and then it goes down to pi, and then it's minus 1 here, and 0 here, and continuation. So that's the graph of y is equal to cosine of pi over 2 minus x. Now, if you remember the previous lecture about sine, this red one is exactly the graph of the sine, if you remember. So basically, you kind of feel that maybe, just maybe, cosine of pi over 2 minus x is equal to sine of x. Well, it is indeed true, and I will talk about this in a separate lecture, the relationship between different functions, trigonometric functions. Right now, let's just carve it as a hunch. Yes, it is true, but let's not get any deeper into this. But basically, this red line is the graph of this function. That's what I wanted to show. What's next? More complicated. y is equal minus 1 third cosine minus 3x minus 3 pi over 2. Well, again, I would first convert it into something simpler, which is what? Minus 1 third cosine. First of all, I can change the sine under the cosine without changing the way. So it would be 3x plus 3 pi over 2. Now, it would be even better if I will factor out 3, and I will have 3 times x plus pi over 2. OK, that I can actually draw. So my plan is the following. First, I will start with a cosine of x. That's obvious. Then the next one would be cosine of x plus pi over 2. That's easy. That's almost the same as the previous problem, but shift will be to the left. Because this is positive, so the shift will be to the left. Excuse me. Then I will multiply my argument by 3, which means it will be squeezed three times by factor of 3 to the horizontal squeezed towards 0. And finally, multiply by minus 1 third. It means my, in this case, squeezing would be again also 3 times, but also reverse the sine of the graph, which means it will be symmetrically reflected over the x-axis. So this is the plan. Number one, I will skip the cosine of x. You'll know how it looks. And let me immediately switch to cosine of x plus p over 2. So the graph will be shifted. So instead of this, it will be this. So this is minus p over 2. This is 0. This is pi over 2. This is pi. And this is 3 pi over 2. So I shifted the graph from 0 to 2 pi to the left by pi over 2 and from minus pi over 2 to 3 pi over 2. It's also the length of 2 pi. So the period is preserved. And then it repeats itself. OK, that's the first one. And let me mark it. This is y is equal to cosine of x plus pi over 2. Next, we will multiply the argument by 3, which means we should squeeze it down three times. So it was 3 pi over 2. Now this point will basically shift to the pi over 2. This pi would be to pi divided by 3. So let me take another color. So this is something like pi over 3, right? So the graph would reach its minimum at pi over 3. We'll go here. Then 0, it would be at pi over 6. And this is pi over 3. That's where it would be 0. This point would shift here. Then at pi over 2, it would be 1. So this is this part. And this part would be minus pi over 6. It goes this. So again, it's periodical. But the period now instead of 2 pi is 2 pi over 3. That's how it's supposed to be, right? From minus pi over 6, it would be pi over 3 plus pi over 6 is equal to 4 pi over 6, which is 2 pi over 3. Exactly. So it's 3 times less than 2 pi. So from minus pi over 6 to pi over 3 is exactly 2 pi over 3 lengths. So that's the new period after which it repeats itself. And finally, I have to multiply it by minus 1 third, which means it will be shorter three times. So it's not up to the 1 or minus 1. It would be up to the 1 third and minus 1 third. But we have to change the sign. So this is minus 1 third. This is 1 third. So changing the sign would be, correspondingly from this, down 1 third and change the sign so it starts from here. Then it goes this way. This 1 third of this, again, changing the sign here and here. So that's the period. And then it repeats itself, obviously. So that's the graph from minus pi over 6 to pi over 2 pi over 3. And well, actually, I prolonged it. It was further than pi over 3. And so this is the period. It's 2 pi over 3. And the amplitude, minimal maximum from minus 1 third to 1 third. OK, that's it. Cosine of 2x plus cosine of minus x, minus x, not x, minus x. Which is the same, actually, since cosine is an even function. Now, here we just have to add two graphs together. So let's add them together. The first one is 2x, which means it's squeezed by a factor of 2 from the original. So instead of 0 to 2 pi, it would be from 0 to pi. Now, this function is the same as cosine of x, as I was saying. So that has an original period of 2 pi. So this is 1 minus 1. It goes, now, this is pi over 2, 3 pi over 2. That's my cosine of minus x. And cosine of x is squeezed by a factor of 2. So this would be pi over 4, pi with the integer pi over. So the minus 1 would be here. So the function would go here, like this. Then 1 should be here, which means it should be here. So and the period would be pi, obviously. But since I have to really draw it on the whole period, and the whole period is 2 pi, so let me throw on to the second period in this case. And the second period would be 3 pi over 4, 0 would be, yeah. How is it going? This is pi, OK. Now, 0.0 would be 3 pi over 4. So it would be something like this. And at 2 pi, it's equal to one more loop, like this. So that line is cosine of 2x. And solid line is cosine of minus x. Now, let's add them together. At 0.0, 1 and 1 would be, obviously, 2. So we have this point. Now, at 0.5 over 4, this one is equal to 0, so I will have exactly the same as here. So the graph would go this way. At pi over 2, this one is equal to 0, and this one is equal to minus 1. So graph goes this way. Then continues to go down. Well, actually it's, yeah, it would be double here. And it will eventually shoot to 0, because this is plus 1 and this is minus 1. So it will reach 0 at pi. And then it would reach minus. So it will be something like that. And then both of them go up, up, up, up to 2, something like this. I believe that would be a more or less closed picture of what happens. So it's kind of a double wave, but these waves are bigger and these waves are smaller. That's when you are combining oscillation with different frequencies. You will have a complicated picture of your mechanical device or whatever, electronic device, which will oscillate with different amplitude in a different period of time. And I showed this particular mechanical model of a pendulum if you have one pendulum. And then at the end of it, there is another pendulum of a different length. Since it's different lengths, this is 2 and this is 1, let's say. It more or less corresponds to this picture. So if this is where there are joints, then the final point would make a very interesting kind of oscillations which resemble this particular line. So that's what happened when you are adding things with different frequencies, with different periods. OK, and the last function I have when I'm adding frequencies which are of the same period but are shifted one relative to another. Cosine of x plus cosine of x plus p over 2. So let's examine this. The last graph. Again, this is cosine. This is 2 pi. This is pi minus 1, 0, 1, y, x. That's this one. Now this one is different by shifting to the left. So let's shift it to the left by pi over 2. Now this is pi over 2. This is minus pi over 2. So it would look like this. This is 2 pi over 2. And let's bring them to the same period. Well, period is the same, 2 pi. So all I have to do is to analyze the graphs on one particular period, let's say from 0 to 2 pi. All right, so what will that be? Well, immediately we can see that in this case I will have 0.2. No, I'm sorry. This will still be 1 because I'm adding 1 and 0. So it's still 1. Now here it's still minus 1. Now in between, 1 is positive and going down. Another is negative and going up. So basically the function will behave like this. Now here is also addition. But now we're adding two negative things. So it would be a little bit down. And then at this particular point, it should be minus 1, right? So it goes something like this. Then at this point, it would be at 3 pi over 2, it would be 1 because we are adding 1 and nothing. So it would be something like this. And then we will increase it just a little bit and then go to 1 at this point. So this is how the graph would look. So from 1 it goes down to minus 1, then a little bit more down so the minimum will be less than minus 1. And then the maximum will be a little bit greater than 1. And then it goes back to 1. And then we repeat the same thing. So that's what happens when you are adding two oscillations with shifted argument. Well, that's it. I hope you will review these graphs again. What's interesting about sine and cosine bicycles, basically, these trigonometric functions, they describe oscillations. It's a very physical process. So all these trigonometric properties, they have not just appeared in the same air in abstract maths. They are all having a very practical application in real physics, mechanics, in real practical life. So it's quite interesting actually that purely mathematical theory actually gets such a huge application in engineering and electronics, et cetera. Well, that's it for today. Thank you very much and good luck.