 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue talking about some of Engels and trigonometric functions of these sums. Now, the previous lecture was about the same thing, basically. But I geometrically proved the formula. Well, obviously, geometrical approach assumes that Engels are. In this particular case, it was right triangle, so the Engels were acute. In theory, the formula actually is valid for any Engels. So what I'm doing right now is I'm trying to prove this formula for basically any Engels. And obviously, it might actually involve a couple of special cases. But still, the approach is very, very general. And that's what I'm going to present to you. So first, what I would like to do is I will derive the formula for a cosine of the difference between two Engels. Cosine of alpha minus beta is cosine alpha times cosine beta plus sine alpha sine beta. This is the same formula, by the way, which I have derived in the previous lecture for acute Engels. But now, I will not use this restriction. So alpha and beta can be basically any Engels. Positive, negative, greater than pi, greater than 7 pi, or whatever you want to do. All right, now, with this generality in mind, let's go back to the definition of trigonometric functions with the general definition, which is using the unit circle. So here is our unit circle. This is the positive direction of the x-axis. And let's consider two Engels. Now, I will put them here. But actually, the proof doesn't really depend on. So this is alpha, and this is beta. These are two Engels. Now, this is A. This is B. And obviously, the angle A or B is equal to alpha minus beta. Now, from the definition of sine and cosine, what are coordinates of these coordinates A and B? Well, we know that A is on the unit circle, which means ordinate is sine, and abscissa is cosine. So A has coordinates cosine, alpha, sine, alpha. Now, B has correspondingly cosine, beta, and sine, beta. Now, let's recall the Cartesian coordinates in this plane. And knowing coordinates of these two points, I can use the formula for a distance between these points. Now, the square of a distance is square of a difference between abscissa and plus square of a difference between the coordinates. If you don't remember it, it's trivially derived from the Pythagorean theorem, or you can refer to the corresponding lecture in the mass concepts, where I describe different systems of coordinates, including Cartesian, and including the distance between two points in the Cartesian coordinate system. So the distance between A and B, which is basically the length of this segment, AB, equals square, the square of a distance, is equal to square of a difference between abscissa plus square of a difference between ordnance equals. OK, let's open up. Cosine square plus minus 2 cosine alpha cosine beta plus cosine square beta plus sine square alpha minus 2 sine alpha sine beta plus sine square beta equals. Cosine square plus sine square, we know this is a fundamental identity of the three denominator is 1. Cosine plus sine beta is also 1. And this, so I will just replace it with 2. And this is also 2, so I will factor out 2, and I will have 1 minus cosine alpha cosine beta minus sine alpha sine beta. Let's just remember this formula for square of a distance between these two points. And by the way, if you notice, this piece looks very much like this one, right? OK, now let's derive the same square of a distance between A and B differently. I can suggest two ways of doing this. I will present the simpler one right now here in the lecture. And the second one is presented in the notes. I'll just mention how I do it. But basically, all the details I would like to not to waste any time, because it's really simple. So I will use the law of cosine of cosines for a triangle OAB. Now, the law of cosine, cosines was discussed in the topic with simple trigonometric advantages. So for any triangle, it works. So there is one little difference, which I have to really consider here, because we are talking about general angles, alpha and beta. So in case this angle beta is large so much, that it's larger than angle alpha by more than the pi. So OV would be, for instance, somewhere here. Then triangle OAB would be this one, not this one. However, I will actually address this issue after I will prove it in a relatively more simple case, because it results in exactly the same formula. And it's based on the fact that the cosine is an even function. And the cosine of pi minus angle are also sines and cosines, maybe with different sines, but they're squared anyway, so everything will be fine. So let's just forget about this case, and let's assume that the difference between alpha and beta is less than pi, so I can consider this triangle. Now, the law of cosines for this particular triangle states the following. AB squared is equal to OA squared plus OB squared minus 2AOAB cosine of the angle between them, which is beta minus alpha equals. OA is 1, OB is 1, so it's 1 plus 1. It's 2 minus 2. Again, this is 1, and this is 1, so this is cosine of alpha minus beta. Or if you wish, I can factor out 2, and I will get 2 times 1 minus. Well, look, exactly the same square of a distance between A and B is expressed in this way or in this way. So what follows? Well, obviously, the cosine of alpha minus beta is equal to cosine alpha cosine beta plus cosine alpha sine beta, exactly what we wanted to prove. Now, let's go back now with this case, if this is angle beta. Oops. OK, this is beta. Well, if this is beta, then I can always consider 2 pi minus beta, which is this angle in this triangle. Now, if I replace beta with 2 pi minus beta, actually nothing happens. Number one, 2 pi is a period, right? Now, cosine is an even function, so plus or minus doesn't really change. So cosine beta is basically the same as cosine of 2 pi minus beta. And the same thing would be exactly with any other function which participates in this. And in case the sign does change the sign, but there are two changes of the sign will be exactly the same. So just trust me, if you don't just try to do it yourself, but you will have exactly the same result. I just don't want to go into these small details. The proof is exactly the same. So that's what we've got. We've got cosine for alpha minus beta equals to this sign. Well, this is basically the end of the big story. Everything else is technicality. For instance, how can we obtain for the sum of two angles? Well, the formula for the sum of two angles, if you remember, is for acute angles, which I have proven, is basically the same except this sine cosine alpha plus beta equals cosine alpha cosine beta minus sine alpha sine beta. Now, how can I obtain this from this? Well, let's change beta to minus beta in this particular case. So what happens? Instead of alpha plus beta, I will put minus minus beta. Now I have minus, right? According to this formula, it's a cosine alpha cosine minus beta plus sine alpha sine minus beta. Now, cosine minus beta is equal to cosine beta because the beta is even function. Now sine of minus beta is minus sine of beta because this is an alpha function. So minus goes here, and then I have this. So I have this formula. Now, if beta is very, very big, it's so big that it's more than a pi greater than alpha. That's the case when I was trying to use with this red radius. Well, let's just do it this way. If I will change beta to, if I will change, let's say cosine of alpha and beta is very big. So we can have this beta as pi plus, let's use a different letter, pi plus gamma. Now, so if alpha is greater than beta by more than pi, then I can always subtract pi from the beta and have gamma smaller. Now, what happens with this, according to this formula? Well, let's just think about it. It's cosine alpha times cosine pi plus gamma is, well, let me just put it as is, plus sine alpha times sine pi plus alpha. Now, pi plus alpha is changing the sine. So it's basically minus cosine alpha cosine gamma minus sine alpha sine gamma, sorry. Because cosine of pi plus gamma is equal to cosine gamma. And sine of pi plus gamma is minus sine gamma. So again, we can, using this manipulation, we can always change the formula from whatever we have proven into anything for any other angle. So basically, oh yes, one more thing, sine, how can I obtain it for sine? Sine is also very simple, because sine of alpha minus beta is equal to cosine of pi over 2 minus alpha minus beta. Remember this fundamental, sine of n is equal to cosine of pi over 2 minus the center. Well, let's just find out what it is. Now, I know about this formula, which is what? Cosine of pi over 2 minus alpha plus beta. And we will put parenthesis like here, right? And then I will use this one. So this is what? Cosine times cosine. Cosine pi over 2 minus alpha cosine beta minus sine sine sine pi over 2 minus alpha sine beta equals. Now, cosine pi over 2 minus alpha is, again, this is this fundamental identity between sine and cosine. So this is sine of alpha, cosine beta minus. And this is also the same fundamental. Sine is changing to cosine and sine beta. So this is the formula for a difference between two angles. Now, what's the formula for sine of two angles? For sum of two angles, sir, sum of two angles is. Well, let's just substitute minus b instead of b in this formula. Well, cosine is even function, so it doesn't change. Minus cosine alpha. And sine of minus b, sine is odd function, so it changes to sine, so it will be plus here. Sine beta. So this is the formula for sine. So again, there are a couple of cases when alpha minus beta is equal to pi or 0 or 2 pi, when I cannot really construct this triangle. Because if the difference between angles is pi, then it will be just a straight line. One is going this, one is going that. Well, but again, if you will just substitute instead of beta, if you will substitute, let's say, alpha plus pi n, where n is any integer number. So that would be a general approach. So if you substitute instead of beta, you will put it's equal alpha plus pi n, where n is any integer number. Now, what happens with difference? Well, on the left, I will have cosine of pi n. Now, what's the cosine of pi n? Well, 0 is 1, so pi is minus 1. So it's plus or minus 1, depending on whether n is odd. So it's 1 or minus 1. This is for n is equal to even, and this is for n is equal to odd. Now, how about this guy? Now, cosine alpha times cosine of alpha plus pi n plus sine of alpha times sine of alpha plus pi n. Now, for n even, this would be the same as cosine alpha. So for n is equal to 2k, and this would be the same as sine alpha. So I will have cosine square plus sine square, and the result is 1. Now, for n equals to some odd number, this would be with a minus sine minus cosine alpha. So cosine times cosine times minus cosine would be minus cosine square. And this would be minus sine alpha. So it will be minus sine square. So it will be minus cosine square times minus sine square, which is minus 1. It will be minus 1, which is exactly the same. So as you see, if alpha and beta are different by pi n, then this formula is still valid. Because on the left, I will have always 1 or minus 1, depending on even or odd. And on the right, I will have 1 or minus 1, depending on whether n is even or odd. So I think that actually concludes all the cases. It's very important, actually. If you want to really rigorously prove something, you do have to consider all the different cases. And if you are using some geometry, like you're drawing a triangle, you have to make sure that angles basically are fit for being part of a triangle. OK, so that's it. Thank you very much. I basically end this with, yes, one little remark. There is another way, instead of using the law of cosine, there is another way to prove this using the following thing, which I, in details, put in the notes for this lecture. OK, back to this picture. This is P, A, B, the difference between A and B. I have derived as a law of cosine. Instead, I can do the following. I can do this. This is the bisector of the angle, which means m. Which means angle A for m is equal to A minus, alpha minus beta over 2. Now, mA is equal to, from this right triangle, it's OA, which is 1, times sine of alpha minus beta over 2. So AB, which is 2 mA, is 2 of these sines. AB squared would be equal to 4 sine squared of alpha minus beta over 2. And now, what I will do is I will use a formula, again, which we have derived in the fundamental identities, about a double angle. So if you remember, I have derived the formula, that cosine of 2 phi is equal to, what is it? Never remember it, but let me just derive it very quickly. It's cosine squared phi minus sine squared phi, or 1 minus 2 sine squared phi. So from this, I can get the sine squared and substitute it here. So you will have a formula for cosine of alpha minus beta. And you will have exactly the same formula for cosine of alpha minus beta. All right. So the details of this are in the notes on unisor.com. And well, that's it for this particular lecture. Now we know how to calculate the trigonometric functions from sum or difference between n goals. Well, we still need to go through many, many different problems. And that's in the future lectures. Thanks very much, and goodbye.