 Hi, I'm Zor. Welcome to a new Zor education. Let's talk about other simple trigonometric identities. In this lecture, I would like to talk about law of cosines. It's actually about triangles which trigonometry, which actually this word seems to be related to triangles, right? It's the measurements of triangles. So this is actually about triangles, not about any angles, angles in the triangles, which basically has certain restrictions on the angles. It's greater than zero and less than 180 degrees. Now, in other cases, we were talking about cosines of any angles. But in this case, we are talking about only angles in the triangles. Now, what's interesting about this theorem is kind of a generalization of a well-known theorem in geometry, Pythagorean theorem. Pythagorean theorem is about right triangles, right? So if you have right triangle, A, B, C, C squared equals A squared plus B squared. Now, the law of cosines generalizes this particular theorem for any triangle. In particular, if triangle is the right one, then law of cosine actually states exactly the same as the Pythagorean theorem, which it should be, right? So let's talk about any triangle. Let's start with this one. The C, usually the opposite side is marked with lowercase similar letters, capital A lowercase A side, capital B lower B, and capital C lowercase C. And what I would like to prove is the following. Let's call this angle gamma. Why gamma? Well, because at A you have alpha, at B you have weight, and for C you have gamma, right? So the law of cosines states the following. C squared equals to A squared plus B squared. So we start as a Pythagorean theorem, but then there is another member, two A, D, and cosine of this angle between them. So this particular member is this generalization from right triangle to any triangle. Because if gamma is 90 degrees, so if this is the right triangle, then cosine is equal to zero, and this member disappears completely, and we will have just a Pythagorean theorem. All right? Seems to be easy, right? Okay, so how can we prove this theorem? Well, let's wipe out what we don't really need. We don't need this angle alpha, and we don't need this angle beta. We will just have these, and what I will do, I will put an altitude H from B to AC. Now, let's consider triangle BHC. Now, this is altitude, which means this is a right angle, right? Now, considering this is the right angle, I can always say that C squared equals AH squared plus BH squared. AH squared plus BH squared equals to C squared, right? This is the hypogenesis. These are two categories. Now, what do we know about each of these? Well, AH we can always replace with AC minus CH. AC minus CH squared. Now, BH we will leave as is squared equals. Now, we will switch to trigonometry. Why? Because AC we know what it is. It's B, right? Now, what is CH? CH is a catheter in this triangle, BH. It's adjacent to gamma, so it's equal to hypotenuse, which is A times cosine of this. That's what we have. What's BH? In the same triangle, BH is opposite catheter to the gamma. So, it's equal to hypotenuse times sine. That's it. I mean, that's all ingenuity which is required, because after that, it's just technical thing. Let's open the square, which is B squared minus 2AB cosine gamma plus A squared cosine squared gamma plus A squared sine squared gamma. Now, look at this. What is this? Obviously, this is equal to... Well, let me repeat these two members. B squared plus 2AB cosine gamma plus A squared factor out. And what's inside? Cosine squared plus sine squared of the angle gamma. And we all know that this is equal to 1. This is the major fundamental identity of trigonometry. So, I just leave A squared. And that's exactly... I'm sorry, it should be minus 2. Which is exactly what I need here. So, as you see, we do have a very simple proof of the law of cosine. All we needed, that's basically the creative part of this proof, is to drop a perpendicular in altitude, and then just use it if you guarantee it twice. Now, is it the end of the proof? Well, I make a pause, so you will think about this. Well, obviously if I ask, it means it's not. Now, why is it not? Well, for a very simple reason. What if our triangle looks like this? So, this is point B. This is side B. This is side C. So, this is an obtuse angle, which means that the altitude doesn't really drop in between B and C. So, we don't have exactly the same picture. You see, here is the problem. If you are using a drawing to prove certain things, your drawing should be, well, general enough. And if in the particular drawing you represent a particular relationship between sides and angles, etc., you better make sure that some other relationship also is represented in some other drawing and your proof is exactly the same or at least similar. And that's what we are going to do. So, we do have different cases when this perpendicular drops inside this particular side B-C or outside. And actually, outside also can be on both ends, right? It can be on the left from the B or it can be on the right from the C. So, that's what we have to really consider. That's what's covering all the cases. So, let's do it. What happens in this particular case? Well, let's do exactly the same thing. This is still a hypotenuse. So, C-square is still equals to B-H-square plus, I'm sorry, this is letter A, plus A-H-square. Now, well, B-H is still an opposite casualties in the triangle B-H-C, so it's still A times hypotenuse, A times sine of gamma. But A-H right now is different if H-C minus B plus H-C minus A-C-square, which is A-square sine-square gamma plus H-C is A times cosine minus A-B is B-square. Now, what difference between this and the one which is before? Well, actually, not much. The sine was here in reverse, but this is the square. So, we basically go to the same result. A-square sine-square gamma plus A-square cosine-square gamma minus 2AB cosine gamma plus B-square, which is exactly the same thing because this is equal to A-square, A-square plus B-square minus 2AB times cosine. So, this is a simple case as we see. It's exactly the same basically as before. But we did have to consider it because the drawing is different and our relationship between sides is different. And let's talk about another one right here. So, we have this type of triangle. This is A, this is C, this is B, and this is H-B-C-A. Now, in this case, let's do it again. C-square equals from AHB, it's BH. Now, where is the gamma? Gamma is here. This is my gamma. So, C-square is equal to AH-square plus BH-square. So far, exactly the same as before. Equals. Now, AH-square. Now, it's B plus something. Or AAC, rather, plus CH plus BH-square. Equals. Now, AC is B. CH is from this triangle BCH. This is the hypotenuse. This is an adjacent angle, adjacent side to the angle, but not to the angle gamma, but to the angle 180 degree minus gamma, or pi minus gamma. So, we now have A cosine pi minus gamma. That's what CH is. CH is hypotenuse A times cosine of this angle, which is pi minus gamma. That's what we have. Now, BH is still A times sine of pi minus gamma-square. So, instead of the angle gamma, now we are using angle which is equal to pi minus gamma. And that actually presents just, you know, a slight difficulty. But now, let's consider this. Okay. We do know the properties of the functions sine and cosine. So, let me just put it there. I don't want to do it from the memory. Now, the sine looks like this. This is 0. This is pi over 2. This is pi. Now, if you have this angle, this is the sine. Now, the graph is symmetrical relatively to the line through pi over 2. So, if you go with the same segment, you go from the pi to the left, you will get exactly the same. That's on the graph. Now, very similarly, if you consider the unit circle, sine is ordinate, right? So, this is one angle, and this is pi minus this angle, right? And the ordinate is exactly the same. So, this angle and this angle have the same ordinate. Now, sine of pi minus gamma is exactly the same as sine of gamma. What about cosine? Now, cosine is a system. So, on the unit circle, it looks very good. So, this is by absolute value the same as this, but they have opposite signs. So, my statement is that cosine of pi minus gamma is minus cosine of gamma. It looks like this. Cosine goes through zero on this way. So, this is pi over 2, this is pi. So, look, if you have an angle here, this is a cosine. If you have an angle, the same segment you put here, you have exactly the same height between negative side. So, either through the unit circle or through the graph, you can always say that the cosine of pi minus gamma is minus cosine of gamma. So, this is equal to pi b minus a cosine gamma. And as you see, we have come to exactly the same equality, but slightly differently. We still have this minus here, because if you will open all the parentheses and put together b square minus 2ab cosine gamma plus a square cosine square gamma plus a square sine square gamma. And this is obviously equal to a square. You will get exactly the same. A square plus b square minus 2ab cosine. We just came to this slightly differently, and our drawing in this particular case is dramatically different from the one which we had before. And we have to consider the angle hundreds of a degrees or pi minus gamma instead of gamma. So, my point is that, first of all, this is the law of cosine. I have proven it in all the cases. And number two, which is also very important, if you are using the drawing, you have to be very careful to basically cover all the situations which can come. Well, that's it for the law of cosine. I do recommend you to go to unison.com and go through the notes for this lecture as well. This is the trigonometer part. And then I encourage parents and supervisors to also participate in this website. If you register and then you register your students as students which you supervise, then you can enroll these students to different courses. You can ask them to take exams because exams are available only for registered students. And basically you can make a judgment on how the whole studying is going. That's it for today. Thank you very much and good luck.