 Hi, I'm Zor. Welcome to Unisor Education. I would like to solve a few trigonometric problems, and the only condition is please try to do it yourself first. So if you just see this lecture for the first time, just pause it, go to the website Unisor.com, to the corresponding chapter in trigonometry. You have these problems presented to you with some hints even, and try to do it yourself, only then whether you will succeed or will not succeed, only then listen to this lecture. Okay, so let's do it. The first problem is you have a product of two cosines, and the problem is converted into a sum of some trigonometric function of some argument. So in short, convert product of trigonometric functions into sum of trigonometric functions. All right. Now, it's really a very, very simple problem as long as you remember that the sum, the cosine of the sum of two angles is represented as, remember this, right? Now, similarly, the difference between two angles as a cosine represented as this. So, how can I get this? Look, just add them together, and you will have cosine alpha plus beta plus cosine of alpha minus beta equals to two cosine alpha cosine beta, and sinuses will be reduced, minus and plus, right? So that's the formula, so that's how you can convert. So this is one half of cosine plus cosine of difference. That's very simple in basically a couple of lines. There is nothing to do except you really have to recall what's the cosine of sum of two angles and difference of two angles actually is. Now, next problem is about sines. Well, it's basically the same approach. So let me repeat it. I wiped it out, forgetting that I will need it for the second one. Then I get only sines. Well, obviously I have to subtract from this, this, and I will get cosine alpha minus beta minus cosine of alpha plus beta equals. Cosines will reduce, and from sine I subtract minus sine, so I will get two sines. So how to get this is one half of the difference between two cosines. So again, we can convert a product of cosines or a product of two sines into sum or difference of cosines. All right, now the third one is the mixture, sine and cosine. I have sine alpha times cosine beta. How to convert this into a product or a difference of sines and cosines? How to get rid of the multiplication basically. All right, now again, recall the formula for sine of two, of sum of two angles. It's sine alpha cosine beta plus cosine alpha sine beta, right? Now we need cosine, we need sine alpha cosine beta. We need this one. This one we don't need. So let's just recall the difference, which can be actually obtained from this, replacing b with minus b. And since cosine is an even function and sine is an odd function, you will get this minus cosine alpha sine beta. So how to get sine alpha cosine beta? Well, just add them together. You will get this reduced. And this will be two sine alpha sine beta. So this will be one half of sine alpha plus beta plus sine alpha minus beta. Now let's reverse the problems. I would like to convert sum or difference of two trigonometric functions into a product. So far, we have converted multiplication into additional suppression. Now let's do it the other way around. I will still base this type of thing on the function which we were talking about before. Cosine alpha minus beta is equal to cosine alpha cosine beta plus sine alpha sine beta, right? Remember that. And cosine alpha plus beta equals cosine alpha cosine beta minus sine alpha sine beta. So my point is that something like cosine alpha times cosine beta can be represented. Let me just do it again. 2 cosine alpha cosine beta is represented as sum of these two things. And sine alpha sine beta would be represented as a difference. So what's written here is basically the conversion between the product and the summation. Now here, we have two summations. And similarly, let me go with the sine. Sine alpha minus beta is equal to sine alpha sine beta minus, sorry, sine alpha cosine beta. Let's start with plus, plus cosine alpha sine beta. And sine of alpha minus beta equals sine alpha cosine beta minus cosine alpha sine beta, from which we can conclude that if we will add them together, we will have the product of sines, right? So sine alpha plus beta plus sine alpha minus beta is equal to 2 sine alpha cosine beta, because this will be reduced. And this is conversion. So these two are conversions of product of two cosines, or two sines. And this is the conversion of sine and cosine. Now we have two sines here, right? And here we have two sines. But here we have alpha and beta. And here we have alpha plus beta and alpha minus beta. So what if I will represent alpha as, let's say, x plus y, and beta as x minus y? If I would be able to do this, then this particular equation would look exactly like this one, where instead of alpha you have x and instead of beta you have y, right? And this sine is beta would be like this. So how can we do this? Well, that's very easy, actually, because from this you can derive that x is equal to, add them together will be alpha plus beta will be 2x. So it's alpha plus beta divided by 2. And y is equal to alpha minus beta, right? Let me wipe out this. We don't need it. And let's wipe out this. We don't need it. So I only have these conversions, OK? So if I will do this particular replacement, so let's assign x is equal to this and y is equal to this, then I can write that this is equal to sine of x plus y plus sine of x minus y, right? Beta is x minus y and alpha is x plus y, where x and y are these. Now, this looks exactly like this, correct? So it's equal to 2 sine x cosine y. So instead of alpha I use x, instead of beta I use y. And I get this formula, basically it's this one. And now let's return back to our original alpha and beta. So it could be 2 sine of alpha plus beta over 2 and cosine of y, which is alpha minus beta. So that's how we did it. That's how we converted addition into multiplication. So we know both, actually, how to convert multiplication into addition and how to convert addition to multiplication. Now, another problem on addition with cosines. And we will do exactly the same trick. Cosine alpha plus cosine beta. So we have something like this, which we can use. But this is alpha minus beta and alpha plus beta. So we will do exactly the same thing. So we will do it as cosine of alpha is x plus y plus cosine x minus y, where x and y are these. Now, this looks exactly like this. Instead of x is alpha, instead of y is beta. So I can write it as 2 cosine x times cosine y. And now I will go back to the original alpha and beta. So it's 2 cosine alpha plus beta over 2 and cosine alpha minus beta. So again, we converted summation with a product. Addition with multiplication. All right, so we can convert sum into product and product into sum. That's very important. And the importance of this will actually be much better visible in the next lecture when I will talk about series. But in this lecture, I still have a small final problem which would help to basically understand why it's sometimes important to have these formulas to convert product into sum and sum into product. Here is the problem. You have to somehow simplify a sum of these things. Actually, I could have written up to n, basically, components. I have chosen to do it only with four components in this lecture, and I will do the real series in the next lecture. OK, so how can we simplify this? If you have a certain number of sine functions with arguments which make up actually a arithmetic progression, you see, x2, x3, x4, x, et cetera, well, here is the way how these problems can be addressed using these elementary techniques which we have already learned today. You see, the difference between them is always constant. It's x in this particular case, right? So what if I will multiply it by sine of half of the difference between them? So s times sine of x over 2 is equal to sine of x over 2 times sine x plus sine x over 2 times sine 2x plus sine x over 2 sine 3x plus sine x over 2 sine 4x equals. Now let's convert the product into sum, actually into a difference. And you will see what exactly I mean. Let's wipe out this. And let's recall our formula for sine alpha times sine beta is equal. So it's sine times sine. So what do we do? We have cosine of difference. That would be cosine cosine plus sine sine. Now if I will add cosine of their sum, it will be cosine cosine minus sine sine. No, but I need not addition, I need subtraction. So cosine cosine and cosine cosine will be reduced plus sine sine minus and again minus. So it will be plus sine sine will be 2. So I have 2 here. Now why did they do it? Why did they use x over 2 here? Well, look at this. This is alpha and this is beta. All right? So sine times sine, sine times sine, so let's put 1 half outside, will give me cosine of a difference which is cosine of x minus x over 2 minus cosine of x plus x over 2. Right? That's the first one. Plus 1 half cosine, now this is alpha and this is beta. So I have 2x minus x over 2 minus cosine 2x plus x over 2 plus. Now you understand I can just write it without much commenting, 3x plus x over 2, OK? And the last one is cosine of 4x plus x over 2 minus cosine of 4x minus x over 2. Now, why did I choose x over 2? It's exactly half of the distance between the arguments between x and 2x between 2x and 3x, et cetera. Here is why. x plus x over 2 and 2x minus x over 2. They are the same value because x2 is actually in the middle between x and 2x. So this is 1 half x and this is 1 half x. And this is minus, this is plus. So they are reduced. Same thing here. Minus 2 and 1 half and plus 2 and 1 half. These are reduced. 3 and 1 half with a minus. I should have actually say minus first. I'm sorry, plus second. So 3 and 1 half plus and 4 minus 1 half is also 3 and 1 half with a plus. They're reduced. So everything will be reduced. By the way, if I will have 5x, 6x, whatever x, it will be exactly the same. And only the last member and the first member will be retained. So the answer to this is basically 1 half of cosine. Let me write it here. This is all reduced, right? So what I have left is this minus this 1 half of cosine of x over 2 minus cosine of 5x over 2. So that's what s times sine of x over 2. Now, if I want to convert it into even more compact way, I have to convert this difference into a product of two functions. And we already, again, addressed that particular issue, how to convert cosine minus cosine. So if you convert it into a product and then considering you have this multiplier, so if you have s is equal to you have sine of x over 2 in the denominator. And in the numerator, you have this one. So this is 2. This is cosine of x over 2 minus cosine 5x over 2. And it would be nice if I can convert it into a product. So I will have only multiplication and division. So I have to recall, again, how to convert this into product. OK, let's go back to the previous problem. Remember, we had to convert cosine alpha. Well, in this case, it's minus. So minus cosine beta. How to convert it? Remember, we were representing alpha s. Let's use different letters. Let's say u plus v and beta is u minus v, where u is equal to alpha plus beta over 2, and v is equal to alpha minus beta over 2. Then this difference is equal to cosine of v u plus v, and minus cosine of u minus v equals, this is cosine cosine minus sine sine. This is cosine cosine plus sine sine. So we'll have 2 minus, first of all, it will be minus 2 sine u sine, right, which is minus 2 sine alpha plus beta over 2 by sine of alpha minus beta over 2. So that's the result of cosine minus cosine, all right? Now, what's alpha, what's beta? Let's just think about this. Well, this is alpha. This is beta. So alpha plus beta over 2 is equal to x over 2 and 5x over 2. That's 6x over 2, over 2 is 3x over 2. Alpha minus beta over 2 is equal to minus 4 divided by 2 is 2. So it's minus x, correct? So in the denominator, I can write s is equal to minus 2 sine 3x over 2 and sine of minus x. Now sine is odd function, so I can put minus here and minus here and divide by 2 sine x over 2. 2 is reduced. And this is my final answer if I did not make any mistakes. So that's the final answer. That's the sum of sine x plus sine 2x plus sine 3x plus sine 4x. Just as a precaution, I mean, when you derive formula like this, it's good to test it in some way. I mean, maybe I made a mistake. If I did, I'm not going to repeat it because it's just too long. You will do it yourself. But if I didn't make a mistake, that's what it is. But let's just check it. For x is equal to 0, obviously on the right, I have 0. But on the left, I shouldn't really have 0 because I cannot divide by 0. So let's do something else. What other nice angle? Let's say 30 degrees. 30 degrees sine. Now, let's put 60 because we have over 2. OK, so sine of 60 is what? Square root of 3 over 2 sine of 2x. That's 120. Sine is an ordinate. So 120 should have the same as 60. So it's also square root of 3. 3 is 180. Sine of 180 is 0. And 4 is 240. 240 is 360 minus 120. So it's minus 120. So it's minus square root of 3 over 2. So the result seems to be this. OK, let's check this one. x over 2 is 30 degrees. So I have 1 half sine of 30 degrees is 1 half. Sine of sine x is square root of 3 over 2. And 3x is 45. Am I right? No, no, no. 180 divided by 2. 180 divided by 2, it's 90. It's 1. OK, so it looks like it's square root of 3 over 4, actually. I have over 3. Well, I did make a mistake most likely. In any case, probably I forgot some multiply 2 in one of two cases. Well, sorry about this. In any case, the approach which I have shown to you is very important. Whenever you have a series like this, it's one of the ways to calculate the result of this summation is to multiply by, in this case, sine of the argument, which is right half of the distance between these two. Because then this plus this half is the same as this minus half. And it will reduce each other. OK? All right, so that's it for this particular lecture. Try to do it more accurately than me. Right now, repeat all these calculations which I did by yourself. Make sure you did everything accurately. It's very important. And don't forget that you can actually use Unisor.com as an educational tool which contains not only the lectures like this one, but you also can be enrolled by your supervisor or your parent into the course. You can take exam. That's very important. And you can see the results of this exam. And your supervisor or your parent can see the results of the exam. And arrange this type of gradual movement along the course from one topic to another. Something would be completed. Something would still be necessary to repeat. You can repeat it as many times as you want. You can take as many times your exam as you want. So basically it's a good educational tool which I do recommend you to use. Thanks very much and good luck.