 Hi, I'm Zor. Welcome to a new Zor education. I would like to talk about something I would say interesting and creative in trigonometry which is related to trigonometric series. Now, series as we know is basically a sum of certain members. Now, trigonometric series is obviously is when these members represent certain trigonometric function of some arguments. So, just as an example again, this would be an example of trigonometric series. Well, what's interesting about these is that there is actually an interesting practical usage of these things. You see, sine and cosine functions, they represent oscillations. So, whenever certain oscillations are added together, for instance, you have some source of radio frequencies and another source of radio frequencies, they are actually somehow combined. Now, let's just forget about all these practical issues and let's concentrate on mathematics. What I would like to present today is one particular problem related to trigonometric series. And what's interesting about this problem is, it's on the border between trigonometry and algebra and geometry, if you wish. And it's using this connection between trigonometric functions and complex numbers. So, let me just present to the problem and you'll judge yourself whether I'm right, actually, to be quite excited about what kind of a problem this actually is. OK, so here is the problem. Let's say you have to summarize sine of phi plus phi plus some delta plus sine of phi plus 2 delta plus, et cetera, plus sine of phi plus, what's the last one, n minus 1 delta. So I have n members from no delta to n minus 1 delta, added to original angle phi. Now, absolutely similar can be a cosine. So it's two problems, but actually it's one problem. I'll combine them together. Now, there are different ways to solve this problem, how to find this particular sum, let's call it, as sine or as cosine. I will offer the way, the methodology, which I consider very interesting, very creative. And somewhat, well, I'm not afraid of this word beautiful in this particular case. That's one of the things in mathematics which definitely can be beautiful, but to reach them, you have to really accumulate certain amount of knowledge to be able to use it. So how can I solve this particular problem? How can I express this sum in a more concise way? Here is the way which I suggest. And again, this is not the only way, obviously there are others, but this I particularly like. Let's just recall that any complex number, let's say we have this complex number, it can be expressed geometrically as a point with coordinates x and y. Now, you remember that this distance from the origin is called the modules. Now, let's consider only complex numbers on the unit circle. So those complex numbers with modules equal to 1. Now, modules is obviously x squared plus y squared in this case. So x squared plus y squared equals to 1. That's the numbers which we are considering. Now, every point on the unit circle can be represented by its angle phi with an x-axis. And at the same time, there is a definition that the abscissa and originate of this angle phi are abscissa is cosine phi and originate is sine of phi. That's by definition of cosine and sine. So if this is an angle phi, this point belongs to a unit circle, then this particular piece is equal to cosine of this angle and this particular piece is equal to sine. Which means that x is equal to cosine and y is equal to sine. So I can always put this. So any complex number with a modulus 1, which y is basically on the unit circle around the origin, can be expressed in this way, where phi is an angle, which ranges to this point, makes with the positive direction of the x-axis. Alright, so I just reminded you this. I obviously talked about this in the corresponding lecture. Okay, that's one thing. Now, another thing is, let's say that I have two different complex numbers. This is one and this is another. Both of them are on the unit circle. Both of them have modulus 1. Now, obviously their product also has modulus 1, because the modulus of the product of complex numbers is the product of modulus. So it's 1. Now, where exactly this particular product be on this unit circle? So if this is angle phi 1, this is angle, let's say phi 2, where exactly is the product of these two numbers be? Well, let's just multiply them, right? I'm basically repeating something which I have already presented to you in the corresponding lecture about the connection between complex numbers and trigonometry. But anyway, let's just derive this again. I mean, I don't mind to derive it. So multiply these two things. So what will be the real part? This will be cosine by cosine and also sine by sine, because I square would be equal to minus 1. So the real part of the product would be cosine phi 1 times cosine phi 2 minus sine phi 1 sine phi 2. What will be the complex, the imaginary part, coefficient with imaginary parts? I times sine phi 1, but cosine phi 2, right? Sine phi 1 cosine phi 2 plus sine phi 2 cosine phi 1. Now, we immediately recognize that this is a formula for cosine of a sum of two angles and this is the formula for a sine of sum of two angles. So basically what you can say is z1 times z2 is equal to cosine of their sum plus i sum, sine of their sum. So that's all as a kind of introduction to the methodology which I would like to present you. Now, this is all basically the old knowledge which we all know about. Whenever you are multiplying two complex numbers with the modulus one, their product would be also the complex number of modulus one and the angle would be equal to the sum of angles. In some way, by the way, it resembles logarithms, if you remember. Logarithms of the product is equal to sum of logarithms. Here is actually the same thing. Angle of the product is equal to sum of angles. In some way, it's similar, if you wish. All right, so knowing this, I would like to use this particular feature for the following purpose. Let's use one complex number, b cosine phi plus i sine phi. This is one complex number which I am talking about. Another complex number, v, is cosine of delta plus i sine of delta. Two complex numbers. Now, using these two complex numbers, what will be u times v? Well, if I multiply one complex number with an angle phi by another with an angle delta, I will have a complex number with angle equal to sum of these angles. What if I will multiply again by v? This by v. Again, my angles will be added together. Obviously, if I put the third power of three, it will be three. And if I will have power of n minus one, I will have cosine phi plus n minus one delta plus i sine phi plus n minus one delta. Looks familiar, right? The real part of these numbers is equal to my sum of cosines. And the imaginary part equals to the sum of sines. So, let's just sum them up together. So, sum u and all these guys. So, what I will have. Cosine, cosine of phi plus delta, phi plus 2 delta, et cetera, phi plus n minus one delta. So, this would be s cosine plus i sine s sine. That's what I will have on the right side. What will I have on the left side? What is u plus uv plus uv square, et cetera, plus u to the power of n minus one? Well, obviously, it's geometric progression. And we know how to sum geometric progression, right? Well, for those who don't remember the formula, and that actually includes myself, let's write it. So, if you have s equals to u plus uv plus uv square plus, et cetera, plus uv to the power of n minus one, it's capital N. Then I multiply this sum by v. So, v times s equals to u times v would be uv. Uv times v would be uv square. And the last one would be, whenever I multiply this by v, I will have uv to the power of n. Now, if I will subtract them, these numbers will all be reduced because I'm subtracting. The only thing which will be remaining is this and this, from which I can derive that s is equal to uv to the power of n minus one divided by, if I subtract this, it would be ds times minus s, it's v minus one s, right? So, I divide it by v minus one. So, this is my formula. And that would be on the left. I don't need this anymore. All I need is this formula where u is a cosine phi plus i sine and v is cosine delta, right? So, my result would be cosine phi plus i sine phi. This is u, cosine phi delta, sorry, delta. That's v plus i sine delta to the power of n minus one divided by cosine delta plus i sine delta minus one. That's how we use this formula. And this is, as I was just saying, it's s cosine plus i s sine. So, almost done except, I mean I can probably multiply something without any problems except I have to raise to the power of n, right? But that's exactly the same as we did before because what is this? Well, we are multiplying one complex number with an angle delta by itself and times, right? So, whenever we multiply it once, if I do cosine delta plus i sine delta, if I will multiply it by itself once, what will I have? Well, whenever I multiply two different complex numbers, which both have modules one and angles such and such, angles are added together, right? So, cosine two delta plus i sine two delta. Now, if I will multiply it again by the same number, I will have three, right? I will multiply this by another cosine plus i sine. It would be three. So, if I multiply it n times, obviously I will have, that's what cosine delta plus i sine delta to the power of n is. So, that's how I can put it here and my formula would be n delta n delta. Is that the end of it? Well, not quite because I have derived a complex number and this is the representation. If I will be able to represent this, which is a complex number in the form, like a plus bi, this, then I can say that a is s cosine and b is s sine, right? Now, this is a problem, but it's a technical problem. There is nothing creative about this. All I have to do is really do the multiplication and division and I know how to do multiplication and division in complex numbers. Just as an example, I mean, if you really want to know what, for instance, you have to do if you have to multiply this by this. Well, you all know that the real part would be a c and bi times di would be b di square, which is minus bd. So, that would be the new real part and imaginary coefficient would be bc plus ad. So, that's how you multiply and you get separately real part and the imaginary part. Now, we want to divide something. Well, that's easy too. If you want to divide a plus bi divided by c plus di, all you have to do is multiply by, how is it called? Conjugate or something like this, c plus id, oh, I'm sorry, it's minus. What will be in the denominator c square plus d square, right? i times i, it's i square, which is minus one and minus. So, it will be c square plus d square, that's the real part. Imaginary part will be cd and minus cd, it will be zero, right? So, here I will have a square plus c square. Now, on the top, I will have the real multiplication, which is ac plus bd, right? bd i square and minus right. That's the real part and imaginary part will be bc minus ad, something like this. Now, but this is real, so I can actually divide it separately. So, it's ac plus bd divided by a square plus c square. So, that's my real part, which is this and imaginary part will be this, which is this. Because now, if I have two complex numbers, both are represented in a normalized form, the real part is equal to real and the complex is equal to this. So, I would like to finish here, I don't want to really do all this transformation, multiplication, division, etc., on the board. But what I have done, I put in the notes for this lecture on unizor.com, which is on the right from the video, all these calculations. So, if anybody is really interested and I do suggest you to go through these calculations, no matter how tedious they are. I mean, they are kind of tedious because the formulas will be big, etc., and not very exciting. But I think it's a good practice to go through these calculations and derive the formula in its final form. So, that's one way of doing this thing. I might actually suggest some other ways in some other examples of doing exactly the same thing, but without using the complex numbers and the geometric sequence, etc. I think I wanted to start with this because that's what really opens up completely the whole mathematics as a field where everything is somehow interrelated. Here we have trigonometry and complex numbers and algebra of geometric series combined together, which is a rare occasion, and that's why I think it's very important to understand. The more you know, the more you see connections between different things, and that's not only the math, it's just in the world as well. You know this and you know that, and it's not really separate facts. Somehow everything is related in this world. That's philosophy. So anyway, I think it was an interesting lecture. Again, I do suggest you to go through calculations yourself and derive from this the final formula. And again, if you would like to go with my calculations on the website, on the user.com, and everything is in the notes for this lecture. Thanks very much. I hope it was interesting and good luck.