 Hi, I'm Zor. Welcome to Unisor Education. Well, let's continue solving problems. Today's problems will be about trigonometry and geometry with trigonometry. Two problems. Now, these problems, this lecture actually is part of the course called Mass Plus and Problems. It's presented at Unisor.com. I do suggest you to watch this lecture from the website, because together with the lecture we have a note, basically, for this lecture, which is basically like a textbook. So you can read in a very concise manner. I might miss something during the lecture or make a mistake, and I think that the notes are verified that they seem to be definitely better. But at the same time, I might say something during the lecture, which I just didn't mention in the notes, and that would be useful information as well. So you need both. You need the lecture with live video and the notes as basically a textbook for every lecture. Okay, now the website Unisor.com contains prerequisite course called Mass for Teens. It's definitely, whatever is in that course is definitely used in Mass Plus and Problems, because I'm presenting problems based on theoretical material, which is presented in the main course, main theoretical course, Mass for Teens. So I do suggest you to at least glance through that course. Now, all the courses on the Unisor.com are totally free. There are no advertisements, so it's just for your consumption of knowledge. All for free and free for all. Okay, so let's get back to problems. My first problem is purely trigonometric. So I need a sum of sine square of x plus k times pi over n, where k is from 0 to n minus 1. So I have n components in this sum, and I would like to find this sum. Why precisely this particular one? Because the next problem, which is geometrical problem with trigonometry, will use it. So let me first concentrate on trigonometric part, and then we will go through the second problem. Okay, so how can I approach it? Well, as usually, when I present the problem, it makes sense for you to pause the video and think about this problem yourself. Well, as a hint, I might actually tell you that first I will convert sine square into cosine of a double angle, and then I will use the results of the previous lecture where I calculated the sum of cosines. So this is my plan. Okay, so now let's just proceed. First of all, how to convert sine square into cosine of double angle? Well, again, if you don't remember the formula, I usually don't remember. I do remember what is the cosine of 2 phi. That's a cosine square of phi minus sine square of phi. Now, one of the formulas which I do remember is the sine square plus cosine square equals to 1 for every angle. So instead of cosine, I can put 1 minus sine square phi and minus another sine square phi, which can be basically written as 2 sine square, right? From which sine square of phi is equal to this 1 minus cosine of 2 phi divided by 2, right? So let me just write this formula here. Sine square phi is equal to 1 half of 1 minus cosine of 2 phi, okay? All right, now, so let me just replace this with that. So I will have sum of n components. So 1 half I can move here. 1 minus cosine of double angle, which is 2x plus k 2 pi over n. Now, there are n components in this sum from 0 to n minus 1. So it's n times, I repeat, 1. So let me just put it, this is n over 2 minus 1 half sum of cosine 2x plus k times 2 pi divided by n. Okay? That's the same sum, right? I just opened this parenthesis. It's n times 1, which is n and 1 half. And this 1 half for sum of cosines. And this is exactly what we were actually dealing with in the previous lecture. So let me just remind you the results of the previous lecture. Sum of cosines of x plus ky, where k is changing from 0 to n. That was n in the previous lecture, but that doesn't remember. We will use it anyway. Equals to sin of x plus 2n plus 1 times y over 2 minus sin of x minus y over 2 divided by sin of y over 2. Now, this formula was derived in the previous lecture with all the details. So I'll just use it. And again, if you don't recall how it was done, just take a look at the previous lecture. It's on the same Unisor.com, you go to mass plus. Problems, trigonometry, and this will be trigonometry 0,3. Because this one is trigonometry 0,4. All right, so I just borrowed that formula. It was derived completely with all the details in the previous lecture. In two ways, actually, derived. So I'll just use it. Now, what do I have to change in this formula? The only thing I have to change is, you see, this is from k0 to k equals to n. In our case, it's to n minus 1. So if I would change it to minus 1, I have to change here to minus 1. If n is changed to n minus 1, it would be 2 times n minus 2 plus 1, so it's minus 1. So that's the only difference which I will have here. So that's exactly the same formula, just the number of members, number of terms. Instead of n plus 1, it was from 0 to n, it's n plus 1. So I put n numbers here. All right, fine. So how can I use this formula to calculate this? To calculate this, actually. This is something which I would like to calculate. Well, they look more or less the same, but what should they change? Well, instead of x, it's 2x, fine. And instead of y, 2 pi over n, right? All right, so instead of x, put 2x. That's simple. I just put 2 here. Since the formula is exactly the same, 2 here and 2 here. And instead of y, I will put 2 pi divided by n. Okay. 2 pi divided by n, y over 2 would be pi divided by n. This y over 2 would be pi divided by n and pi divided by n here. All right, so let's check what's in the numerator of this. It would be sine of 2x plus 2n times pi divided by n would be 2 pi minus pi over n. Minus sine of 2x minus pi over n. Wow, that's what's interesting. What's the difference between them? This is the difference. But sine is a periodic function and 2 pi is a period, which means that the value of this is exactly the same as the value of this. And the whole thing actually becomes 0 and the answer is n over 2. Simple. Again, if you don't remember the formula itself, I don't remember, nobody remembers this formula. But I do remember how we derived it and I do encourage you to go to the previous lecture and take a look at the notes. There are two different ways to derive this formula and I suggest you to basically familiarize yourself with it. I think without it, it would not really be very interesting quite frankly. If I just used the ready formula without basically specification how I did it, that would not be interesting. So we have derived the answer. And the answer is n over 2. n over 2. Good. Now this is the first problem. And now we will go to a geometrical problem. Where we will use the results of this one. Here is an interesting geometrical problem. Let's consider you have a circle and a regular n-sided polygon inscribed into this circle. I choose triangle. It can be a square. It can be five-sided polygon, etc. It doesn't really matter. So we have this. And let's call these vertices a 1, a 2, a 3. Now this is the center. Oh. And obviously assume that the radius is r. Now, here is the interesting problem. Take a look at any point on the circle and connect it with all vertices of the polygon. So this is p. What I am saying is that sum of p a i square where i from 1 to n. So I assume it is n-sided regular polygon. In my case it is just 3. It doesn't really depend on the position of the point p. So wherever it is, sum of squares and distances to all the vertices of the polygon is exactly the same. And I will calculate exactly how, what exactly this particular value is. Again, what's interesting about this, regardless of position of point p, it can be anywhere on the circle including on the vertex itself, on any vertex and anywhere in between. Because it's on the circle. Alright, so how can we approach it? Well, let's just calculate trigonometrically a particular side. Let's say p a 3. So I will have a perpendicular to the chord p a 3 and I will have this angle. Well, let's have this angle and call it alpha. Alpha i. By alpha i, that's angle between o p and o a i. Okay, so angle p o a i is alpha i. So we have this angle would be p o a 1 would be alpha 1, p o a 2 would be alpha 2, p o a 3 would be alpha 3, etc. Now what's important about alpha, let's say this is alpha 1 and this is alpha 2. p o a 2 is alpha 2 and p o a 1 is alpha 1. Well, obviously alpha 2 is equal to alpha 1 plus 2 pi divided by n. Why? Because this is a regular polygon which means every angle from a 1, a 1, 0, a 2 or a 2, 0, a 3 or a 3, 0, a 1, they are all equal to 2 pi which is the whole circle divided by n number of side. So this is equal to alpha 1 plus, let's use the index k if you don't remember, plus k minus 1 2 pi over n. So for k is equal to 1 I have 0, so I have only alpha 1. For k is equal to 2, alpha 2 is equal to alpha 1 plus 2 pi over n like this one, etc. So every next angle p o next a k's will be equal to a 1 plus k minus 1 times 2 pi over n. Okay, that's simple. Now, if I do have this angle, let me just draw it again. This is p, this is a k, this is o. So p a k is double, let's call it q k. It's double p q p q. So it's obvious that the perpendicular to the chord is dividing this chord into equal parts and divides angle into equal parts. So this angle is alpha k's. So p q k is equal to r times, this is r, this is half of the angle, so it's a sine times sine alpha k divided by 2. So this is alpha k, this is alpha k divided by 2, this is p q k. So p a k is equal to 2 p q k, right? p a k is double p q k. So p a k is equal to 2 r sine and a k is equal to alpha 1 plus k minus 1 2 pi over n, right? And, therefore, we are interested in summing this thing from k equals 1 to n, right? Well, let me change slightly the variable k. Instead of k, I will use i minus 1. So this is equal to sigma a from 0 to n minus 1. Now, we need square, right? So it would be sigma square. So it would be 4 r square sine square of alpha first plus i 2 pi over n. k minus 1, oh, I'm sorry, k plus 1. So if k is equal to 1, i is equal to 0. If k is equal to n, i is equal to n minus 1. And this would be i instead of k minus 1. So this is the sum which we have to calculate. Sum of the squares of distances from p to o a k's, okay? And look at this. This is basically the same thing. This is index k, this is index i, doesn't matter. X is a 1, alpha 1, sorry. This is exactly the same index. So it's all the same. So the answer is 4 r square times sum of these signs, which is n over 2, which is 2 pi, not pi, 2 r square, 2 r square n, 2 pi again. I used 2 pi r. Okay, it's 2 r square times n, where n is the number of sides in our regular polygon. And as you see, it's independent on the position of this point p. It depends only on the radius and number of sides in the polygon. It's quite remarkable to tell the truth. I mean, I would not expect it. If somebody tells me that sum of the squares of these distances is constant, it doesn't really depend on the position, well, I would be skeptical about this. Nevertheless, calculations show exactly this. Okay, so that was my second problem. Now, I would like to finish it basically with very important thing. It's the whole purpose of studying mathematics in school. I'm not talking about college where there are some specifications and some directions towards some profession or industry or whatever. No, in high school, it's general knowledge basically. You will not use it as much, quite frankly, in your practical life. I mean, the problems like this will never probably be in your practical life. However, why do we do it? We do it exactly for the same purpose you go to gym for your muscles. Solving problems is the gym for your brain. It's very important to be creative because you really have to be creative. It develops your creativity when you solve the problem which nobody told you how to solve. It's not like you're given a recipe and you just basically do exactly what this particular recipe tells you and you come up with a result. There is no recipe. I mean, whenever you're facing something like this, you have to think about, okay, how do I approach it? Well, sine square kind of complicated. Let's convert it into cosine. Whenever we have to summarize the cosines, we used to do it before and that's exactly the purpose because the more problems you solve, the wider will be your repertoire of different approaches, different tools how to solve new problems. So that's why it's very important to first present the theory which I did in Math 14's course and then some, well, obviously some maybe easier problems and illustrative problems were in that course but problems which do require a certain level of creativity I do present in this course and that's why I suggest you to go diligently from one problem to another and try to solve them yourself. I mean, if you don't solve it, it's fine. Then read the notes or which to lecture. But the more you watch, the more you read, the wider will be your horizons and the better will be your chest of tools which you can use to solve different problems and that's what will prepare you to solve the practical problems whenever you will be working somewhere in whatever industry you will be. Okay, that's it. Thank you very much and again suggest you to read the notes for this lecture and good luck.