 Hi, I'm Zor. Welcome to a new Zor education. We will continue solving different problems related to trigonometric series. There is nothing really new about these problems. They are based on the same principles which the previous problems were built upon. It's just to train your thought to go into this direction when dealing with these problems. One more comment which probably is necessary. The result is a question, well, why do we need this? I did answer this question many times, so let me repeat it again. I doubt very much that in your practical experience you will need to solve problems like this. They are just the brain exercises. All the different things which we can throw into the mind of a student which would basically force him or her to creatively approach whatever problem is in front of him. That's good. That's the purpose of this website. Again, do not consider these problems as anything practical. I just can't even imagine if there are any practical problems related to this. Maybe, but I just don't know. It's only for training your mind to be creative, analytical, logical, etc. That's the purpose and you should really approach all these problems, most of the problems in this course as just the training exercise for your mind. All right, now let's do the... Again, I have two problems for today and I will try to be accurate and not make a mistake. So, the first one is sin x times sin 2x plus sin 2x times sin 3x plus, etc. plus sin nx sin n plus 1x. So, as we see, we have this particular series of n members and each member is a sin of some number multiplied by x times sin of the next number multiplied by x. Now, we have to convert this series which is very long actually into a compact formula where obviously n and x and some trigonometric functions will be the arguments. All right, now, we did use this particular technique before and we will use it again. We will convert the product of two sines into a sum of something. Now, sum of what? Obviously trigonometric functions. Now, since I never remember any formulas except the very basic ones, I do remember that in the formula for a cosine of a sum of two angles, there are two members, one is the product of cosines and the other is the product of sines. You just remind it. So, cosine of u plus v equals cosine u cosine v minus sin u sin v. So, that's product of sines and obviously cosine of u minus v is equal to a sum of these two. How can I get only the product of sines? Well, I have to subtract. From this, I have to subtract this. Cosines will be reduced and I will have only the product of two sines. So, sin times sin minus sin times sin will be two. Sin u sin v is equal to, from this, we subtract that, right? So, it's cosine u minus v minus cosine of u plus v. So, that's the formula. Now, if I will use this formula for u and v equal to x and 2x and then 2x and 3x and then, etcetera, nx and n plus 1x, well, let's see what happens. Alright. So, x is u, 2x is v, right? So, I have, well, first of all, this is 2, so I have to divide by 2. So, it will be one half. One half and let's open the square bracket. Now, cosine of u minus v, u minus v. So, x minus 2x, that's minus x, but the cosine is an even function. So, the cosine of a negative angle is the same as cosine of the corresponding positive one. So, I will just put cosine of x minus cosine of their sum, u plus v. Now, what's the u plus v? It's x plus 2x, right? So, that's 3x. Good. Now, next one is, we will use 2x and 3x. 2x and 3x. The difference is, again, minus x, but cosine is even function, so I can put cosine of x minus cosine. 2 and 3 is 5, 5x. So, et cetera, plus. And the last one would be cosine of the difference, which is, again, minus x, but the cosine is even function, and minus cosine of 2n plus 1x. So, that's what our sum has become. Now, is it better? Well, it's obviously better because these guys are the same. The number of these guys is obviously the number of members in this sum, which is n. So, I can put immediately 1, 2, 1 half, n times cosine x minus 1 half of 3x, 5x, et cetera. They are all numbers. So, cosine 3x plus cosine 5x plus, et cetera, plus cosine 2n plus 1x. Also, n numbers in this sequence. Now, this is something which we have already done in the previous lecture. We're summarizing a more general, more general series. It was cosine x plus cosine x plus d plus cosine x plus 2g, et cetera. Remember this? Now, in this case, instead of x, we have 3x, and instead of d, we have 2x, right? So, it would be basically the same thing. So, if I remember the formula for this, I can just substitute into this formula. But obviously, I don't. So, I will repeat the derivation for this particular formula. So, let me completely omit this particular member from all further calculations. And I will just concentrate on this sum, all right? Now, what can I wipe out? I can wipe out, I guess, this one. So, we have to calculate the sum of these. Now, if you remember, again, in the previous lecture, it was quite obvious, it's very helpful in this case to multiply all the members of this trigonometric series by the sign of an angle which is half of the difference between consecutive members. So, the difference is always 2x, right? 3x, 5x, 7x, et cetera. So, the half of this distance is x. Now, why am I doing it? Well, basically, because of this thing, if I will be able to represent this as something minus angle, and this is something plus angle, then they will probably reduce each other. I mean, if you remember, that was actually the point. So, if the difference between consecutive arguments is, in this case, 2x, I will multiply everything by sign of half of the distance, which is sine of x, and see what happens. All right. What happens is, so I do this cosine x without any changes. So, I divide by sine x, and I multiply everything by sine of x. Cosine 3x sine x plus cosine 5x sine x plus cosine of 2m plus 1x sine x. Now, why did I do it? Because a product of sine by cosine, I can also represent in a similar fashion. So, let me just represent it first. We don't need this for this. Now, same as the difference between cosines, give you the product sine by sine. The product of sine by cosine can be obtained from the difference of the sine. So, let's remember the formula for sines. Sine u plus v equals to sine u cosine v plus cosine u sine v. And sine of u minus v equals sine u cosine v minus cosine u sine v. So, how can I get cosine u and sine v, which we need here? I have to subtract like this. So, what I will have is 2 cosine u sine v, because this will be reduced, and this will be doubled, equals to sine u plus v minus sine u minus v. Right? So, that's the formula which I will use. Now, u is 3x, v is x. So, and there is a 2 here. So, I have to substitute one half of sine of their sum, which is 4x minus sine of their difference, which is 2x plus. Now, this sine of their sum, which is 6x minus sine of their difference, which is 4x. Next will be 8 minus 6. You see, this is reduced with this. Next will be 8 sine of 8x minus sine of 6x, and this will be reduced, etc. So, it's always the first one would be, the first from the previous will be reduced to the second from the next one. So, the very last one would be sine of their sum, which is 2n plus 2 of x minus sine of their difference, which is 2nx, which is also will be reduced. So, what's left is this and this. So, it's n over 2 cosine x minus 1 over 2 sine x. Then another 2, so I will have to put 4 here. The only thing in the parenthesis in square brackets would be sine of 2n plus 2 minus sine 2x. This is basically the end of it. I mean, if you want, you can convert the difference between sines into a product, so the formula might actually look a little better or not better. I don't know, it just doesn't really matter. But in any case, this is one of the final formulas which can actually be stopped at and no further reduction is necessary because you don't need all these dot things which signify there are certain numbers, certain terms, members of the sequence which we didn't really write. So, this is a compact formula which represents this particular series, which we can again, we can a little bit change as well, but doesn't really matter anymore. We have the compact formula without any kind of unnecessary dots. All right, that's basically it for this particular problem. As you see, I did use a couple of formulas here, but it's not really something which I remember. I derived it right on the fly, and the only thing which I do remember, and quite frankly, I do remember it quite well, is sine of the sum and sine of the difference between two angles and cosine of the sum and cosine of the difference. This actually is something which you probably have to remember, and it's not really a big deal, only like two formulas, you don't really need the second one because the second one can be derived from the first one. Like for instance, if instead of v you put minus v, cosine of minus v would be the same as cosine of v, but the sine of minus v would be minus sine of v, right? So, this second formula is derived from the first one because sine is an odd function and cosine is an even function. The same thing here. If you know the sum, the cosine for the sum, and that's the minus sign here, you have to really remember that for a cosine, this plus corresponds to this plus, for a cosine, this plus corresponds to minus. But how to derive it from minus v? Well, if you substitute minus v, the cosine part remains the same because the cosine is even function, and the sine would reverse, and that's why you have, instead of minus, you have plus. So, only two formulas, basically. Sine of sum of two angles and cosine of sum of two angles. Everything else is derivable. And derivable quite easily as you see, I did it right on the fly, basically. All right, that's it with this particular problem. Let's wipe it out and go to the next one. Second and the last problem for this lecture is the following. Sine qx plus sine cube x plus d plus sine cube x plus 2d, etc. plus sine cube n minus 1, sorry, sine cube x plus n minus 1d. So, from x plus 0d to x plus n minus 1d, n-members, and we have to calculate this sum. Well, here is something which you might remember or might not. I, in one of the lectures, I think one of the first lectures about trigonometric identities, I derived the formula for sine of triple angle. And that formula actually contained sine, in the third degree, sine cube. So, I would like to use this formula to express sine cube as a sine of a triple angle and something else. But I do need the formula for sine of the triple angle, which again I don't remember, but we can very easily derive it, right? So, sine of 3 phi equals, well, that's obviously sine of phi plus 2 phi equals sine phi cosine 2 phi plus cosine phi sine 2 phi equals. Now, cosine 2 phi, or cosine of phi plus phi is cosine cosine minus sine sine. But angle is the same phi plus phi, so it's cosine square and sine square. So, sine phi cosine square phi minus sine square phi plus cosine phi. Now, sine of 2 phi is sine cosine plus cosine sine. But again, since angles are the same, phi and phi, I can just put 2 sine phi cosine phi equals. I would like everything to be in sines, which means this cosine I will replace with 1 minus sine square phi, if you don't mind. And here you see cosine and cosine will be cosine square. I also replace it with 1 minus sine square phi. So, what do I have? It's sine phi times 1 minus 2 sine square phi plus 2 sine phi times cosine square, which is 1 minus sine square equals. All right. Now, let's open all these brackets. Sine phi minus 2 sine cube. Sine phi minus 2 sine cube phi plus 2 sine phi minus 2 sine cube phi equals. Sine and 2 sine, that's 3 sine phi minus 4 minus 2 minus 2 sine cube phi. Now, why is it important? Well, because I can express sine cube in terms of sine phi and sine 3 phi, right? So, sine cube phi is equal to 1 fourth of 3 sine phi minus sine 3 phi. All right? If sine of 3 phi is equal to this, then 4 sine cube is equal to 3 sine minus sine of 3 phi and then divide by 4. Right. So, what it means is that every member I can replace with two members and basically completely separate this particular sequence into two different sequences. One with single angles and another with triple angles, right? So, this sum is equal to, I don't really need anything here. That is already done. So, that's my formula for sine cube. So, it equals 2. 1 fourth. Now, 3 sine x minus sine of 3x. Right? Actually, I don't need to close this. Let's just continue. 3 sine of x plus b minus sine of 3x plus 3d, etc. And the last one would be 3 sine of n minus 1. X minus sine of 3n minus 1x. As you see, I can separate these guys from these guys and have basically two different things. One is 3 quarters. This is 3 and this is 4, right? Of sine x plus sine x plus d plus sine x plus 2d plus, etc. plus sine of n minus 1, sorry, x plus n minus 1d. Oh, yeah, I have made a mistake here. It's d and this is plus x and same here. This is d and this is x, right? That's better. So, one is these guys plus and another is actually minus, not plus, minus, minus 1 fourths. This is 1 fourths. And I have here sine of 3x plus sine of 3x plus 3d plus sine of 3x plus 3n minus 1d. So, we have two sums. Each one of them is actually the same as the one which we considered during one of the first lectures dedicated to Ceres. And again, you can use exactly the same approach as in the previous problem. The difference between these two is d. So, you have to multiply everything by sine or cosine of d over 2 and each product can be represented as a difference between two different angles and everything would be basically reducible except the first and the last one. And similarly with this case, instead of actually have 3x here and the difference is 3g, not g, which means you have to multiply by sine or cosine of 3g over 2, right? And again, the whole thing will be reduced as soon as you convert the product of trigonometric functions into the sum or difference actually. So, I don't want to do these calculations because it's obvious now. We are reducing the problem which we don't know how to solve to two problems which we do know. We already did this particular thing a couple of times so that's why I don't want to waste time. But I do refer you to either corresponding lecture or basically the notes on unison.com to this lecture which do contain basically the whole sequence of formulas, etc. I do recommend, however, you do it yourself first and then look at the results, check it with the answer which is on the side, etc. Well, that's it for today. I do recommend you to use full functionality of unison.com to basically convert your studying into a real educational process. And to this you need some supervisor or responsibilities either you yourself can be your own supervisor under a different username logged into the system or your parent or your teacher or somebody else they will enroll you as a student into the course and then you can take exams and you will see the score on these exams, etc. Well, that's it for today and good luck to you. Don't forget, try to do everything yourself and only then check the answers. Thanks very much.