 Hi, I'm Zor. Welcome to a new Zor education. This lecture is about certain identities in trigonometry. They are not really related to anything specific. Well, most likely the formulas for some of two angles will be used. So, in this particular regard, they are not really very complicated. However, it's still interesting to go through these relatively simple exercises just to gain some fluency in manipulating with trigonometric functions. All right, so problem number one. Prove that. Sine of alpha plus sine of beta equals 2 sine of alpha plus beta over 2 times cosine of alpha minus beta. All right, so we need some idea about how to prove this particular identity. Well, whenever I see these type of things, it immediately comes to my mind that their sum is equal to alpha and their difference is equal to beta. Right? If you add them together, then you will have, well, b and minus b would be 0, so alpha and alpha would be 2 alpha and divided by 2, so it's alpha. So, I think it makes sense to convert this alpha into sum of these and this beta into difference between these. And let's see what happens. We will use the formula for sum and the difference between different angles. So on the left, it's equal to sine of alpha plus beta over 2 plus alpha minus beta over 2. That's alpha plus sine of alpha plus beta over 2 minus alpha minus beta over 2. Right? Equals. Now we will use the formula for sine of 2, sine of the sum of 2 angles. So it's sine and cosine plus cosine sine. Okay, so it's sine of alpha plus beta over 2 times cosine alpha minus beta divided by 2. Plus cosine of this, sine of that. Plus Now, this is the difference between two angles. So the sine of difference is the same as the sine of the sum, but instead of this plus, we will use minus. So it's a sine of alpha plus beta over 2 times cosine of alpha minus beta over 2 minus cosine alpha plus beta over 2 times sine of alpha minus beta over 2. And what do we see now? Well, here. Plus and minus. And this thing is the same as this thing, which makes them 2 sine alpha plus beta over 2 cosine alpha minus beta over 2, which is exactly the same as we wanted to prove. So what's the creative, ingenious idea, which I used in this particular case, representation of alpha as sum of these and beta as difference of these. How can you come up with this idea of just by yourself? I don't know. But looking at the right side of what's necessary to prove, I think it just kind of, it invites you to think in these particular terms. Okay, so anyway, after solving like dozens or hundreds of problems like that, obviously this particular idea of representing alpha as the sum of these and beta as a difference between them would probably come to your mind automatically, because the number of these techniques is really very limited. They're all used in many different problems. So after certain number of problems, it will come to you. Next, sine alpha plus cosine alpha equals square root of 2 cosine of pi over 4 minus alpha. Well, again, it actually invites you to think about how can I represent alpha if I want to get something like this? Well, I think what actually comes to my mind, at least, is the representation of alpha in this format. This is still alpha, right? Subtract pi over 4 and add pi over 4. So, sine of this plus cosine of this. These are equal. I don't change anything. I just add it and subtract pi over 4. But now, what I will do, I'll put parentheses around these and parentheses around these. And that's how I get alpha minus pi 4. Now, pi 4 minus alpha and alpha minus pi 4, it's actually the same thing because the cosine is an even function. If you change the sine of the argument, the function doesn't really change. So alpha minus pi over 4 is fine. And now, I can use this as a sum of two angles. One is a minus pi over 4 and another is pi over 4. And again, this square root of 2 should remind you sine of 45 degrees pi over 4, which is square root of 2 over 2. All right, so sine of alpha minus pi over 4 and a cosine of pi over 4, which is square root of 2 over 2 plus cosine of this times sine pi over 4, which is also sine of 45 degrees and cosine of 45 degrees is exactly the same square root of 2 over 2 plus. Okay, cosine of sum, cosine times cosine minus sine times sine. If this is a plus, then the cosine will have the difference between two. So it's cosine of alpha minus pi over 4 times cosine of pi over 4, which is square root of 2 over 2 minus sine pi over 4, sorry, alpha minus pi over 4 times sine of pi over 4, which is again square root of 2 over 2. And what do we see now? This plus and minus. And what we have? Cosine of alpha minus pi over 4 square root of 2 once and square root of 2 over 2 actually, here and here. So if I add them together, it would be twice as much, which is this. And the proof. Again, what's the idea behind this? Representation of alpha as a sum of these two. So just looking at this and prompted by the value of the square root of 2, because it's definitely something related to sine of pi over 4, 45 degrees, or cosine at the same time. All right, cosine alpha plus beta plus sine alpha minus beta. It's supposed to be equal to cosine alpha plus sine alpha cosine beta minus sine beta. Well, quite frankly, I don't see much ingenuity in this particular thing. All we have to do is just open up these two things and open up that thing and see if they're equal. All right, cosine of the sum is cosine times cosine minus sine and sine. Sine of difference is sine times cosine minus cosine alpha sine beta. So that's what we have on the left. Now, on the right, what we have? Cosine times cosine cosine minus sine cosine alpha sine beta sine cosine sine minus sine. Now let's compare. Okay, this is the same as this. Fine. This is the same as this. Okay. Minus cosine and sine. This is the same as this. And this is the same as this. So basically, there is no ingenuity here. Just use the formula, open the parenthesis, and you'll get it. That's simple. Okay, the last one, sine of alpha plus beta sine of alpha minus beta. That's a problem. Should be equal to sine square alpha minus sine square beta. Well, again, I don't see any other way. Just do the sum of the angles and see what happens. Sine alpha cosine beta plus cosine alpha sine beta times this, which is sine alpha cosine beta minus cosine alpha sine beta equals. All right, let's open it up. Sine square cosine square. Okay, sine square alpha cosine square beta. Now, this times this sine alpha cosine alpha sine beta cosine. Okay, so it's minus sine alpha cosine alpha sine beta cosine beta. Now, this one times this one plus sine alpha cosine and sine and cosine. Again, all four. Sine alpha cosine alpha sine beta cosine beta. And this times this minus cosine square sine alpha sine square beta. Okay, now this is reduced. Now, how different this is from this? Well, what we can do, we can probably transfer cosine of beta as one minus, I mean cosine square beta as one minus sine square beta. So, let's see what happens. I don't know, but it seems to be reasonable. One minus sine square beta minus cosine square alpha sine square beta equals. So, what happens here? Sine square alpha minus sine square alpha times sine square beta minus cosine square alpha sine square beta equals sine square alpha minus. I can factor out sine square beta. What will be inside? Sine square plus cosine square minus and minus. So, I have only one minus. That's why I have two pluses. But this is equal to one sine square alpha plus cosine square. So, I have sine square alpha minus sine square beta whatever we have to prove. Okay, easy. All right. Now, there are really very, very interesting and very large quantities of problems like this. I cannot possibly solve all of them. I will try to put some more on the website and try to discuss them. Probably, I will try to go into more difficult ones because these are relatively simple and trivial. In any case, again, they serve the purpose. After going through these simple things, I do recommend you, by the way, to do it yourself a couple of times. Okay, next will be a little bit more difficult. That's it for today. Thanks very much and good luck.