 Hi, I'm Zor. Welcome to Unizor Education. Let me continue talking about trigonometric aspects of geometry. So, I'm going to solve some geometric problems using the trigonometry. Well, in this particular case, I'm not attempting to solve like any problem. However, it probably can be categorized as a nice illustration of geometric property using the trigonometry. I will be talking about the lengths of circumference of the circle and the area. We know the formulas. So, the circumference is 2 pi r where r is the radius and the area is pi r square. Now, I would like to approach these particular formulas from trigonometric aspects which we know. Again, I'm not really pretending that I'm going to prove or solve something. It's more of an illustration. Now, in one of the previous lectures, the one which was about trigonometric inequalities, I have derived this inequality between signs and tangents and the arguments in regions. Now, this formula was actually based on certain geometric properties. That's why it wouldn't be correct to derive geometric properties from the formula. However, as an illustration of this particular fact, which just proves that there is no internal contradiction in the theory, I would like, based on this particular inequality as given, I would like to derive these formulas for the circumference and the area of the circle. Now, to do that, I have to basically define what is the circumference and what is the area. Now, these aspects were actually touched in the geometry lectures of this particular course. I do recommend you to review lengths and area in the geometry course. Also, we will need some information from the limits theory. These aspects are also presented in the algebra course in one of the lectures. It's called basically sequences and limits. Please familiarize yourself with these particular topics before we go any further. Again, what I am trying to accomplish here is to derive these formulas using this particular trigonometric inequality. Now, let's start with, let's say, the circumference. By the way, this particular trigonometric inequality implies very simply the following. Now, for x equals to zero, obviously, these are all equal, so I can say that x is greater than zero and less than pi over two. I mean, that was the condition of the original, and it's enough, actually, for us to concentrate on these particular angles. Now, it's very easy to derive from here. If we divide by x, I will have sine of x over x less than one from this one, right? Now, if I divide this by x, I will have one less than, and instead of tangent, I will put sine of x over cosine of x and x, right? Over one over cosine of x, sine x over x, right? So, what do I have here? If I multiply by cosine both sides, I will have cosine of x less than sine of x over x. And combined with this, I have this, right? Now, obviously, as x becomes smaller and smaller, this thing becomes closer and closer to cosine of zero, which is one. Now, this is one as a constant, so these two are narrowing and narrowing the interval where sine of x over x can actually be, and it's narrowing to one, which means that the limit of that thing is equal to one whenever x is going to zero, right? So, let's just have this on a side, and I will put it here, that limit of sine of x over x is equal to one as x goes to zero. Alright, so we'll keep it in mind. Now, let's talk about circumference. There is no direct definition of what is the circumference of a circle. It's more of a definition using the limit theory. So, if you have a circle, and let's say you inscribe some polygon into this circle, and then you increase the number of sides of this polygon in such a way that the maximum side, like in this particular case, is diminishing all the time down to zero. So, if you have this process, and for instance, how do we can accomplish this process? You can always define a new polygon by dividing each side in two and replacing old polygon with new one, like this. This would cause every side to be smaller and smaller and smaller. Now, as this process continues infinitely, so the sides of the polygons are getting smaller and smaller down to zero, then we can say that the limit of this process is the circumference. Now, obviously, you have to prove the existence of this limit and the uniqueness of this limit, regardless of how you arrange this process of inscribing polygons into a circle. So, these are very interesting aspects of the whole theory of the circumference or the length of any curve, basically. And it's beyond the scope of this course. So, I will just state that if this process is arranged in such a way that the length of the biggest segment is going down to zero, then the result limit, its unique limit regardless of the process, and this limit is called the circumference. So, I will use this particular definition of the circumference. And here is what I'm going to do right now. So, let's say I would like to know what's the limit of my circumference. So, I would like to basically derive this formula using these trigonometric inequalities. So, let's say I have some kind of a polygon inscribed. In this case, it's a square. And now, in this particular case, the number of sides is, let's say, n. Now, on this particular drawing, n is equal to 4. Now, and let's consider this to be a regular polygon with n sides inscribed into a circle. Now, regular polygon is the one which has all the sides equal and all the angles are equal. So, in this case, all the angles are at 90 degrees and all the sides are equal because it's a square. Now, what exactly is this particular segment AC in this case? Well, what is this angle? Now, if my entire angle is 2 pi, right, 2 pi regions, and again, why the entire angle is 2 pi regions? Because we have defined the region as an angle supported by the arc which has the lengths of r. So, basically, I'm implying that if I use the entire circle which is 2 pi, it's obviously 2 pi, the circumference should be 2 pi regions because I have to divide the whole circumference into this to get the number of regions to get the 2 pi, right? So, I'm kind of trying to prove something which I already know, but again, I'm saying it's not really a proof, it's illustration. So, let's just relate to this as an illustration that the theory doesn't have any internal contradictions. Alright, so, now, the angle AOB, the angle AOB is 1 nth of the entire circle, right? Since this is my n-sided regular polygon, then this particular angle AOB is 1 nth because it's supported by one side. Now, the angle AOC, therefore, is equal to half of this, right? Because these are two radiuses, this is the altitude of the triangle AOB. So, it's pi over nth. So, this is pi over nth. This is nth. Now, OA is radius, so AC is equal to R times sine of pi over nth. And therefore, AB, which is the side, is equal to 2R sine pi over nth. And therefore, the perimeter of the entire n-sided polygon is equal to n times side, which is 2R sine pi over nth. So, that's my perimeter. What I was saying was that if I inscribe regular polygons one after another with increasing number of sides, let's say I divide each side into, like, from four sides, from square I go to eight sides, then to 16 sides, etc., that will reduce this length of each side. So, what's the limit of this? What's the limit of 2Rn sine of pi over n as n goes to infinity? Well, we can say it's slightly different with 2R. That's the multiplier, and the limit doesn't really depend on it. Then limit, then I will have sine of pi over n divided by pi over n. Now, as n equals to infinity. Now, I have this pi. It's an extra pi. So, I have to multiply by pi to neutralize it, right? So, sine of pi over n divided by pi over n and multiplied by pi. That would be exactly n sine of pi over n, right? n goes to the numerator and pi will be reduced. So, that's exactly what it is. Now, this is exactly the same as limit of sine x divided by x as x goes to zero where x is pi over n, right? If n goes to infinity, pi over n goes to zero. So, I can just say it this way, right? Which means that the entire thing has 2 pi R as a limit. Because this is 1. And that's what I actually wanted to derive. This same formula, which again, it's not like I don't know it. I know from the definition, basically, of the radian and some geometric properties. But still, it's interesting to illustrate the connection between trigonometry and geometry, which we already know. Now, the second part will be about the area. And it will be very, very similar, actually. So, let's talk about area. Again, let's inscribe a square. It doesn't really matter what kind of... Okay. So, again, the area of a circle would be a limit of the area of polygons inscribed into a circle and the process would be exactly the same. That the number of sides goes to infinity and the lengths of each side of the polygon goes to zero. So, let's just assume that we start with this particular thing and this is, again, n is equal to 4. It's four-sided polygon. And we are talking about regular polygons because it doesn't really matter which polygons we are inscribing. There is a theory which shows that it doesn't really matter what's the kind of polygons you're inscribing as long as the lengths of the biggest side goes to zero. All right. Now, so we have to approximate the area of this sector with area of this triangle because as the number of sides is increasing, the triangle A O B, triangle A O B would be closer and closer to a sector A O B. And that's why the area of the rectangle, so what we have to calculate right now is the area of the square based on the n and then goes to infinity and we will see how it goes. Now, the same considerations. So, this is R. This is, as we already know, P over n. So, AC is equal to R sine of Pi over n, right? Now, OC, which is an altitude, is R cosine Pi over n. Okay? Now, so AC is half of the base. Now, OC is the altitude in this triangle and therefore the area is equal to the product of these two things, right? So, the area is R square sine Pi over n cosine Pi over n. Or if you wish, it's 1 half R square sine 2 Pi n over n, right? Remember, sine of double angle is 2 sine of single angle times cosine of single angle. So, the angle is Pi over n. So, 2 Pi over n, if I will take the sine, it will be 2 sine Pi over n and cosine Pi over n. And I have 1 half and that's why it goes this way. So, the area of the entire polygon would be n times greater, right? Now I have only one side. Now I have all n sides. So, the area would be n times R square times 1 half times sine of 2 Pi over n. And since I would like, actually, instead of n going to infinity, which you have some kind of x going to zero, where x probably should be 2 Pi over n in this particular case, right? So, this would be equal to 1 half R square. Now sine 2 Pi over n divided by 2 Pi over n. And I have to neutralize Pi and 2. So, I have to multiply by 2 Pi, right? And now, what's the limit of that thing? Now, this is the variable which depends on this x, which goes to zero, which means the limit is 1. 1 half R square 2 and Pi. So, the limit is Pi R square, which is the area of the circle. So, again, my purpose was to illustrate that there are no contradiction between the geometry and trigonometry. And obviously, there should be none. But I think it's a nice exercise of how to basically breach these two things together. None of this is actually a proof of the circumference length or the area of the circle. But it's just an illustration or explanation, but they're very wanted to treat it, of the validity of all these formulas. They're all connected somewhere. And it's probably quite an interesting phenomenon. You learn something and then you learn something else. And all of a sudden, you see that somewhere there is a connection between these things. And the more you know, the more connections you will see. So, this is just one of the connections which I wanted to illustrate. Okay, thanks very much and good luck.