 Hi, I'm Zor. Welcome to a user education. I would like to present a few problems related to triple angle, sine of triple angle, cosine, et cetera. Now, obviously, I will bring the triple angle into a sum of double angle and the single angle, and then double angle itself should be broken down, et cetera. So we do need formulas for sine and cosine of sum of angles. To tell the truth, I don't remember most of these formulas. I do remember actually only one, which is sine of sum of two angles is sine of one, cosine another, plus cosine of the first, and sine of another. I do remember this formula. Now, quite frankly, I do remember the formula for cosine, but let's assume I don't remember it. Am I, can I derive the formula for cosine from the formula for sine? Well, maybe it's not very, very easy, but at least I can show some method which you can use yourself to recall that formula. The method which I can suggest is the following. Now, we know that sine square plus cosine square are equal to 1, right? So which means that cosine square of alpha plus beta equals 1 minus sine square of alpha plus beta equals, let's square this. So 1 minus square of this, which is sine square alpha cosine square beta plus 2 times this, and there is a minus sign, so it's minus 2 sine alpha cosine beta cosine alpha sine beta. And minus square of this guy, cosine square alpha sine square beta equals. Now, in this case, I will replace cosine square of beta with 1 minus sine square beta. So what I will have is sine square alpha 1 minus sine square beta. Now, this member leaves as is sine alpha cosine beta cosine alpha sine beta minus cosine square alpha. And this one, I also replace as 1 minus cosine square beta. Now, why did I do it? You will see in a second. So if I multiply sine square by 1, I will have sine square, cosine square by 1, I will have also cosine square, which added to this cosine square, and both are with the sine minus, will give me minus 1, right? Sine square alpha and cosine square alpha times 1 in both cases gives me 1, and the sine is minus, so it's minus 1. And there is a 1 here, and that will be reduced. So what's remaining is minus and minus, it's a plus sine square alpha sine square beta. Now, this is minus 2 sine sine alpha cosine beta cosine alpha sine beta. And minus and minus, it's plus cosine square alpha cosine square beta. And what is this? Well, you obviously recognize this as sine alpha sine beta minus cosine alpha cosine beta square. Now, basically, that's it. The only problem is, this is the square and this is the square. The real formula is, I remember, well, first of all, it doesn't really matter whether it's sine minus cosine or cosine minus sine, because it's square, right? So question is, which one of them is the real one? So we definitely know that cosine of alpha plus beta equals to either sine minus cosine or cosine minus sine. So how to determine which sine to choose? Because this is the correct formula. Well, you can actually take a couple of known angles, like, for instance, cosine is equal to, I mean, alpha is equal to, let's say, pi over 2, alpha is equal pi over 2, and beta is equal to, let's say, 0. Then the sum will be pi over 2, and cosine of pi over 2, it's 0. Now, 0 is not good. Let's do it differently. Let's do it also pi over 2. Oh, wait a minute. What if I take just 0? Yeah, 0 would be better. 0 and 0, it would be even faster. Yeah. Then this is 0, so it's cosine of 0 is 1. And this is 0, and this is 0. And this is also 1, and this is sine of 0 is equal to 0. So 1, yeah, so this is 1. Now, if you will take sine minus cosine, in this case, you will have minus 1 on the right, which is definitely wrong. So this substitution of alpha and beta equals to 0 will give you the proper sine of these members. So cosine, cosine with a plus, and sine, sine will be minus. Now, why did I do it? Because I don't want you to remember everything by heart. And then there are certain things we should have to remember. But other things, it's much better to find the way to derive it from whatever you already know. Don't overload your memory. Try to develop your analytical abilities to compensate for, because you cannot remember everything. However, you can actually, using very few techniques, derive everything we should need. So that was just an illustration. Now, so let me just put again for sine of alpha plus beta equals sine alpha cosine beta plus cosine alpha sine beta, which means that sine of 2 alpha, which is alpha plus alpha. So beta is equal to alpha, so I have two members like this. So it's 2 sine alpha cosine alpha. I will use this to derive the formula for a triple angle. That's my purpose today. So this is sine of alpha plus 2 alpha, right? That's the triple alpha. Now, I'll use this formula. So in this case, beta is equal to 2 alpha. So let me just put it for cosine. Yeah, I'll put it here. Cosine of alpha plus beta, cosine alpha cosine beta minus sine alpha sine beta. We just derived it. So cosine of 2 alpha equals cosine square alpha minus sine square alpha, right? If alpha is equal to that. So I will use these two things. And now I will take care of this game. So sine of 2 of sum of two angles, sine first alpha, cosine of 2 alpha. And cosine of 2 alpha is this plus, so it's sine by cosine. Now we need cosine of alpha times sine of 2 alpha, which is 2 sine alpha cosine alpha. Is that right? Sine alpha times cosine 2 alpha, which is this plus cosine alpha and sine of 2 alpha, which is this. Yeah, seems right. So equals, let's open the parenthesis. And I will replace cosine square with 1 minus sine square, because it would like everything to be with sine only. So I will have 1 minus 2 sine square, right? So it's sine alpha minus 2 sine cube alpha, right? Plus. Now this is cosine and cosine is cosine square, which I will replace with 1 minus sine square. Times this will be 2 sine alpha minus 4. Am I right? So minus 2 sine cube of equals 3 sine alpha minus 4 sine cube alpha. So that's the formula. That's the sine of 3 alpha. Next, the cosine. And now I remember all these sine and cosine of some of the two angles. So I will just use them from memory. All right, cosine of 3 alpha. Again, it's cosine of alpha plus 2 alpha equals cosine of a sum of two angles. It's cosine times cosine minus sine times sine. Now multiplied by cosine of 2 alpha, which is cosine square alpha minus sine square alpha minus sine by sine. Sine alpha times 2 sine alpha cosine alpha. Because this is the sine of 12. Equals. Now I would like everything to be cosines. For sine of 3 and triple angle I used sine. And for cosine I will use only cosines. So this sine will be 1 minus cosine square. And in this sine square will be 1 minus cosine square. So it will be 2 cosine square minus 1 times cosine. 2 cosine cube alpha minus cosine of here minus 2. Now sine square again 1 minus cosine square. So it's 2 cosine alpha plus 2 cosine cube alpha. Because this is minus, this is plus. This is minus. So 1 minus cosine square, yeah, seems to be right. Which gives me 4 cosine cube alpha minus 3 cosine alpha. That's the formula for cosine of a triple angle. Next. Next is tangent. Well, here I can go two different ways. Number one, I can use the definition of tangent as sine over cosine. Number two, I can actually deal with sum of two angles expressed in the tangent form. Well, let me do first like sine over cosine and then we will see how the time goes. So tangent of 3 alpha equals 2 sine of 3 alpha over cosine of 3 alpha equals. Now sine of 3 alpha is 3 sine alpha minus 4 sine cube alpha. That's what we have just determined, right? Now cosine is 4 cosine cube alpha minus 3 cosine alpha. Okay. Well, let's just think what can we do about it. Is it possible to reduce it to tangent only? Interesting. Actually, yes. Here it is. Here is what I suggest. Sine alpha open parenthesis 3 minus 4 sine square alpha. Right? That's on the top. Sine alpha 4 cosine square minus 3. So right now, what I can always write sine over cosine, I can replace with the tangent, right? That's easier. Next, I do recall one very interesting identity. 1 plus tangent square alpha equals 2 1 plus sine square alpha over cosine square alpha equals. If I will use common denominator, I will have cosine square plus cosine square on the top, which is 1. Alright? So cosine square is 1 over 1 plus tangent square from this, right? I inverted both. I can use it now instead of sine square, I can put 1 minus cosine square. And here I have cosine square. And here is my expression of cosine with a tangent. Alright, so instead of this, I will use this. Now, instead of this, I can use 4 cosine square minus 1, right? 4 cosine square minus 1, right? 4 cosine square minus 1 equals. So instead of cosine square, I will use this one. So I have... Well, let me just try it from here. 4 cosine square is equal to 4, right? And now what I will have is 4 divided by 1 plus tangent square alpha minus 1. And here I have 4 over 1 plus tangent square alpha minus 3 equals. So here I will have tangent alpha times 4 minus 1 minus tangent. So it's 3 minus tangent square alpha divided by... Now, this denominator will be the same as here, so they will reduce each other. So I will not write it down. So 4 minus 3 times this. So it's 1 minus 3 tangent square alpha. Yeah, that's the formula. So this is the tangent of 3 alpha. That's how it is represented in terms of tangent of alpha. Now, so I did this. Now, I can do exactly the same thing by using the formula for tangent of some of the two angles, which will give me exactly the same result. You can actually do it yourself as an exercise. And finally, the cotangent. Well, cotangent is basically very easy to derive from the tangent, right? Since my... Let me repeat it again. The tangent of 3 alpha is equal to 3 tangent alpha minus tangent cube alpha. I multiplied tangent by the parentheses. And here I had 1 minus 3 tangent square alpha. Now, the cotangent of 3 alpha is 1 over tangent of 3 alpha, right? So I have to invert it. So let me invert it. On the top, I will have 1 minus 3. And instead of tangent square, I will put 1 over cotangent square, right? And on the bottom, I have 3 1 over cotangent alpha. No. I will have 3, yes, 1 over cotangent alpha minus 3 1 over cotangent cube alpha, right? So I replace tangent with 1 over cotangent. And all I have to do is get everything to the common denominator. No, I think I'm wrong here. This is 1. 3 minus 1, right? 1 minus 3, right? Okay, so let's multiply by cotangent cube both, top and bottom. What I will have is cotangent cube alpha minus 3 cotangent alpha, right? Cube, and this is the second thing we have. And here, I will have 3 cotangent square alpha minus 1. So that's the formula for cotangent. Okay, that's it. That's the whole business about triple angles. So I expressed the triple of angle, the sine in terms of sine, cosine of triple angle in terms of cosine of a single angle, tangent and cotangent. Well, quite frankly, there is nothing here but pure technicality. There is nothing like very creative or inventive, but it's a good exercise and it gives you a certain fluency in dealing with trigonometric functions. So it has its own purpose, obviously, besides the purely technical aspect of this. So that's it for today. Thank you very much and I'll try to be more creative next time. Thanks very much. Good luck.