 Hi, I'm Zor. Welcome to Unizor Education. I would like to introduce you to conditional trigonometric identities. They are conditional because they're not always true. Sometimes they are, sometimes they're not, and the condition basically specifies when they are true. Well, obviously there are some unconditional trigonometric identities. For instance, sine of 2 phi equals to 2 sine of phi cosine phi. Or cosine of alpha plus beta is equal to cosine alpha cosine beta minus sine alpha sine beta. So as you see, I do remember some trigonometric identities by heart, although I claim that they don't. But these are unconditional, no matter what angles are, these are true. Now, what is a conditional? Well, here is an example. Let's consider that you have three angles, alpha, beta, and gamma, such that they are angles of some triangle. So they are in triangle ABC. I'm not saying anything about what kind of triangle it is. It's just a triangle, and these are three angles of this triangle. Then the following is true. Sine 2 alpha plus sine 2 beta plus sine 2 gamma equals 4 sine alpha sine beta sine gamma. That's what I have to prove. Well, obviously this is a condition, which means that's what's given, and this is something which we have to prove. That's a theory. All right, so what can we say if three angles are angles of a triangle? Well, the most obvious thing is that their sum is equal to 180 degrees, or pi radians, right? That's the property of the triangle. And obviously we'll use this particular condition, which is given to us, and I will try to use, using this condition, I will try to prove this formula. Now, how can I do it? Well, the simplest and brute force method. Well, I shouldn't say simplest. It's the one which basically doesn't require any logical thinking. You express gamma in terms of alpha plus beta substitute here and do some transformation until you will get some kind of unconditional identity. If you do, that's the proof. Well, considering all your transformations are supposed to be obviously equivalent. All right, so let's do it. Now, from this, I will say that gamma is equal to pi minus alpha plus beta. Two gamma, I have two gammas here. It's two pi minus two alpha plus beta, and I will use this. Now, I will also use the property of the sign. You remember if you have a unit circle where the sign is an ordinate of a point, which represents an angle, so if pi minus the same angle, you will have the point which has exactly the same ordinate as this one. So a sign of pi minus some angle would be the same as a sign of the angle itself. So that's why when I will use sine of gamma, instead of sine of pi minus alpha plus beta, I can use sine of alpha plus beta. That's exactly the same thing. Now, as far as the two gamma is concerned here, now, when sine is concerned, two pi actually is a period of sine, right? So I can safely just disregard it, whether there is a two pi or there is no two pi. Now, this minus, sine is an odd function, right? So basically, I can bring it outside sine of minus two alpha plus beta. I can say this is a minus sine of two alpha plus beta. So that's what I will use. All right, let's go to the left part. And I will obviously express double angles in terms of single angles. So sine of two alpha is two sine alpha cosine alpha plus two sine beta cosine beta plus. Sine of two gamma is sine of this, which is sine of minus two. So I can put minus here. So it's minus sine of two alpha plus beta, which is two sine alpha plus beta cosine alpha plus beta. That's the left part equals. So I'll probably have to open this one up and see what happens. Minus two times. All right, opening up this. Sine alpha cosine beta plus cosine alpha sine beta. Opening up this, cosine alpha cosine beta minus sine alpha sine beta equals. Now we have to multiply them. Two sine alpha cosine alpha, two sine beta cosine beta minus. This times this sine alpha cosine alpha and cosine square. So it's two sine alpha cosine alpha cosine square beta plus or rather minus because the minus is here. This times this, so it's again minus two cosine square alpha. Sine cosine beta. Now we will multiply by this member. Now this minus and this minus, everything will be with a plus. So it's two sine alpha and sine alpha is the sine square alpha cosine and sine beta. Cosine beta sine beta. Now this times that, we have sine beta square and plus two sine alpha cosine alpha and sine square beta equals. Okay, now let's simplify it. This member sine cosine alpha. This member sine cosine alpha. And this member sine cosine alpha. Let's group them together and factor out two sine alpha cosine alpha. What do we have left? From this we have one. From this we have minus cosine square beta. And from this we have plus sine square beta. Right? Three other members, this and this, I will factor out sine beta cosine beta. So again plus two sine beta cosine beta. From this member I have one. Okay? From this member I have minus cosine square alpha. And from this member I have plus sine square alpha equals. Now look at this. One minus cosine square is a sine square, right? So this is a sine square and another sine square. So I have two sine squares. So two again goes with these two. So I have four sine alpha cosine alpha sine square beta. Very similar, one minus cosine square is a sine square. And again sine square is two sine squares. Two goes with these two and I have four sine beta cosine beta sine square alpha equals. Now what else can I factor out? Four sine alpha. I can factor out and sine beta sine alpha. And sine beta. Sine alpha and sine beta. I have sine, I have cosine alpha and sine beta. And from this I have sine alpha sine beta sine beta. I have sine alpha cosine beta. And by the way, obviously you recognize the formula for sine of alpha plus beta. It's sine alpha cosine beta plus cosine alpha cosine beta. The sine beta. So it's equal to four sine alpha sine beta and sine alpha plus beta. Now let's go to the right side of the formula. Now obviously gamma is pi minus alpha plus beta. So sine of alpha, we were already talking about this, is equal to sine of alpha plus beta. So on the right side I have sine alpha, four sine alpha sine beta and sine of alpha plus beta, which is exactly the same as this one. And the proof. Okay. And I have another example. I have two examples. If you see, this is not difficult from the logical standpoint, but it is kind of tedious and requires a certain level of accuracy from purely technical standpoint. Well, in trigonometry problems, actually the situation is pretty typical. Some things related to technicalities rather than like creative flight of your mind, something like this. But anyway, it's all useful things to do anyway. So plus it will help you to remember certain formulas. Helps. All right. Condition. Alpha equals beta plus gamma. That's a condition. That's given. Now what we have to prove is sine square alpha plus sine square beta plus sine square gamma equals two, one minus cosine alpha cosine beta cosine gamma. Well, and the only thing I can suggest is to do exactly the same thing. Use this substitute to this and this open parentheses, of course, and see if you will get the same result. So let's start from the left side. So alpha is equal to beta plus gamma, which means sine of alpha is sine beta cosine gamma plus cosine beta sine gamma square, right? Plus sine square beta plus sine square gamma equals. We have to open this. So it's something plus something square. So it's square of the first one, which is sine square beta cosine square gamma. Plus double product to sine beta cosine beta sine gamma cosine gamma. That's what it is. Sine cosine sine cosine. Okay. And square of this guy. Cosine square beta and sine square gamma plus these two equals. All right. What can we do about this? Well, again, we have to group something. Sine square beta and sine square beta sine cosine gamma would be sine square. All right. Let's group. First, let's group this and this. It would be sine square beta times one plus cosine square gamma. That's one. Then, obviously, we can do this and this sine square gamma times one plus cosine square beta and plus two sine beta cosine beta sine gamma cosine gamma. Well, we might actually stop here and start simplifying that thing if I will not find anything better than that. Let me just think about it. Well, let's just stop here and see what happens if I will substitute my alpha here. So on the right, I will have 2 minus 2 times cosine, which is cosine alpha cosine beta minus sine alpha sine beta times cosine beta times cosine gamma equals 2 minus this times this. So it's 2. Oh, I'm sorry. It's not alpha beta. It's beta gamma. I don't have any alphas anymore. Beta gamma beta gamma. So cosine cosine cosine and cosine. So it's cosine square beta cosine square gamma. This is with the minus sign and this is with the plus sign because this minus and this minus. So I will have 2 sine beta cosine beta sine gamma cosine gamma. So what do we have? This looks like this. So all I have to do is to simplify this into this. Okay, let's think about what can be done about it. Probably the best would be to replace cosine square with a sine square here and there because then I will have only sines. So the cosine square is, well, or actually probably sine with a cosine. That would be better. So I will replace this with 1 minus cosine square beta and this I will replace with 1 minus cosine square gamma. So what do I have? 1 minus cosine square beta and 1 plus cosine square beta. That would be gamma. That would be 1 times 1. That's 1 minus cosine square beta minus cosine plus, sorry, cosine square gamma and minus cosine square beta cosine square gamma. That's what I have from this. Now from this I will have again 1 times 1 minus cosine square gamma plus cosine square beta and minus product cosine square beta cosine square. So now let's compare this and this. This is 2. This is 1 plus 1. That's the same. Cosine square beta here plus and minus nullify each other. Cosine square gamma nullify each other. And all I have left is 2 products which is exactly this. Well, that basically proves the identity. Okay, so these are two examples. As I was saying, they're a little tedious. They do require certain accuracy. Well, which is, you know, I consider it's a good exercise anyway. So thanks very much for listening to me and be prepared to go through the whole course of trigonometry filled with certain level of tedious kind of problems. I think it has its own purpose and the purpose is to develop certain level of fluency in manipulating these formulas. So thanks very much again. Good luck.