 Hi, I'm Zor. Welcome to Unisor Education. I would like to present a couple of problems which I have already discussed in the previous lecture about trigonometric series. Problems are exactly the same. However, I would like to offer an alternative solution. And this alternative solution, I would say, is quite artificial. So it's like if you know this solution, it's kind of easy to come up with it. But if you don't know the solution, the whole idea, well, not necessarily comes as the first thing in your mind. However, knowing the answer to these problems which I have derived in the previous lecture, I can actually think about some other solution and that's exactly what I'm going to present right now. So, the first problem was about some of cosines and the second problem was about sum of sines. So let's start with a cosine. So we have a problem of summation of this particular series. So we have n numbers in the series. The base is n times 5 and then on each step it's added delta 2 delta 3 delta up to n minus 1 delta. Now, our problem is to come up with a formula which will be in a concise form reflecting the result of this summation. Now, using the correspondence between trigonometry and complex numbers, I have derived with a formula which after certain transformations, which I also put in the notes in the previous lecture, looks like this. Cosine 5 plus n minus 1 delta over 2 sin n delta over 2 divided by sin delta over 2. So that's the result. And again, knowing the result, I can come up with actually something really alternative to whatever the approach I took. I still like the approach related to complex numbers and geometrical progression better, which is in the previous lecture. However, it's interesting that it can be derived differently and the idea about how it can be derived comes from this particular formula. Now, what does it mean? It means that the sine of delta over 2 times this should be equal to this product of these cosine and sine. And I also remember, and again, that was addressed in the previous lecture, something like if you have cosine of u minus v minus cosine of u plus v, it's equal to what? Cosine cosine plus sine sine. This is also cosine cosine and minus sine sine. So cosine will reduce itself and what I will have left is 2 sine sine. Now similarly, if I will add them together, I will have cosine cosine plus sine sine, cosine cosine minus sine sine, so sines will be reduced and I will have as a result this, cosine u cosine v. Now, if I want to have sine and cosine mixed together, I should really do with the sine. Sine u plus v is sine cosine plus cosine sine, right? Now, if I will add sine u minus v, what I will have? Now, this is sine cosine and plus cosine v and sine v and this is sine cosine minus. So what I will have is 2 sine u cosine v. So basically these are transformations which I used in the previous lecture and they are kind of trivial actually. These are trivial identities. And here is what I have, sine. If I will multiply it by all of these, somehow sine and cosine must be played against each other. So let's see how it happens. My general member of this particular sequence is cosine of phi plus sum n multiplied by delta. And I am multiplying this by sine of delta over 2. Now, what will happen? Now, I will use something like this, right? So my n plus delta would be equal to, this is a cosine, so this is v, and delta over 2 would be u, right? So I will have, well, this is 2, so I will have to have one half of this, right here. One half of sine of their sum, sine of u plus v, this plus this. So it's phi plus n delta plus delta over 2, which is sine of phi plus, how about this? 2n plus 1 over 2 delta. Is that right? n delta plus delta over 2, n delta plus delta, correct. And the cosine, no, sine plus sum plus sine of u minus v. Now, what's u minus v? It's this minus this. Now, if I will change the sine and put this minus this, I will have minus, right? That's easier for me. Sine of this minus this. Now, in this case, this is phi plus 2n minus 1 over 2 delta. Correct? Now, I am supposed to have u minus v with a plus sine, but if I will have v minus u, sine is an odd function than this minus would be, right? And in this case, I do want v minus u, because this is kind of greater than this. So this is my formula for the common member of this series, where lowercase n is changing from 0 to uppercase n minus 1. Now, this is actually the key to the whole thing, because if you will consider this and you will really use this common member and rewrite your series substituting for each member of this, we have multiplied by this sine of delta over 2, right? So whenever we multiply by this and then multiply by the common member, we will have this difference, and let's just do it from the very beginning. So 1 half over. Now, this is member with n is equal to 0, and so in my parentheses, I will have sine of n is equal to 0, and sine of n is phi plus delta over 2. This is minus sine of phi minus delta over 2. That's n equals to 0. Now, n is equal to 1 plus, well, actually I'll give this square bracket. This is over the entire thing. With n is equal to 2, I have plus sine n is equal to 2, sorry, n is equal to 1. n is equal to 1, so it's delta plus half delta, so it's 3 delta over 2, 3 delta over 2 minus, and this for n is equal to 1 sine of delta over 2, phi plus delta over 2 plus. Now, for n is equal to 2 now, sine of phi plus. If n is equal to 2, I have 5 seconds delta. Minus sine of phi plus n is equal to 2, so it's 3 seconds delta plus delta. Now, do you see what happens here? This is what happens. This is reduced to this, this is reduced to this, and basically every member would be reduced with the one which is step over it. Now, what will be remaining is the second member from the first, which is this, and the last member for lowercase n equals to n minus 1, the bigger one. So, the remaining one, if lowercase n is equal to n minus 1, it would be sine of phi plus 2n minus 1 over 2 delta. That's what it is. This will also remain. If I'm not mistaken, right? Again, if lowercase n is equal to capital N minus 1, then this would be 2n minus 2 plus 1, which is minus 1, divided by 2 delta. And the one which has a minus will be reduced with one of the previous ones. So, the only thing which remains in this case is one half, the last one, which is sine of phi plus 2n minus 1 over 2 delta, minus the second part of the first member, which is sine of phi minus delta over 2. That's the result of the summation of the cosines. Now, so if I multiply sine of delta over 2 by all this, we get this. Now, but my formula was this. Is this the same thing as this? Let's check it out. Let's check it out. So, again, if you have sine u times cosine v. Remember what it is? Now, this is sine of u plus v. It will be sine cosine plus cosine v. Then plus sine of u minus v. Am I right? Is that the right thing? And that would be too obvious, right? So, again, sine cosine, which is this, and this is sine cosine. And then the second member would be cosine u sinv and this is minus cosine v, which is reduced. So, that's what we have. Okay. So, in our case, n would be equal to nw over 2. v would be equal to 5 plus n minus 1 delta over 2. Right? So, let's check it out. What happens? u plus v equals to 5 plus n and n minus 1. So, it's 2n minus 1 over 2 delta. u minus v is equal to minus, do it again, 5 minus n delta over 2 and minus delta over 2. That's my v. And u is equal to n delta over 2. Oh, plus, sorry, plus. So, if I will subtract from uv, I will have minus 5 minus delta over 2. Okay. Seems to be working fine because this is minus 5 minus delta over 2. So, the sine of this would be equal to... So, v minus u is equal to 5 minus delta over 2. I changed the science, right? And that's what actually this is because I can always put, instead of plus sine, I can put minus sine v minus. Right? That's the same thing. And that's exactly what I have here. Minus sine of and v minus u is equal to this. And this is equal to u plus v. So, this, the numerator, is equal to this. And 2 will be reduced for... multiply by 1 hub would be exactly like this. So, this is exactly the same formula. Now, what does it mean? It means that if I am smart enough to solve my problem, the original problem as follows. Let's multiply the whole sum by... So, this is s. Let's multiply s by sine of delta over 2. If I'm smart, if I know that somehow, magically, it just occurs to me. Let's multiply by sine of delta over 2. What happens? Then I will have sine and cosine, sine and cosine, sine and cosine, I never remember. Then I will use my formula to convert sine times cosine into a difference between two sines. And they will reduce each other and all of a sudden I will get the final result very simply because only the last member and the first member actually remains. Now, it's similar to summing the geometric progression. If you remember, if I have a geometric progression like this, how can I sum it up without remembering the formula, of course? Well, I multiply it by d. In this case, the number is multiplied by v. So this will be this, this will be this, etc. And then I subtract. And as soon as I subtract, I will have this. Instead of this complicated thing, I have only u, v to the n, the very last one minus the first one, which is very similar to whatever we did. So if we're smart enough to multiply it by something, which then would be reduced and would result in reduction of all the intermediate members, only the first and the last one will be remaining. And from here I will just resolve it. And that's the formula, actually, which some people do remember, but not me. All right, so this is how you can solve this original problem if you know the results, if you kind of... Well, quite frankly, if you haven't experienced with solving some other problems similar to this one, that's where it is. Because now let me just replace this problem with a different one, and let's see if I will have similar results. Well, instead of cosines, use sines. That's the second problem, which also was resolved in the course of the lecture when I was using complex numbers. What if I have the sines? Well, again, let's be smart and let's multiply sum by sin of delta over 2. So what will I have? I will use a little bit more, like, mathematical notation. So instead of doing this, I will have s is equal to sigma, which means sum. This is the operation of summation. The general member, the common member of this series is this, where n from 0 to capital N minus 1. All right? So if I will multiply by sin of delta over 2, I will have sigma from N is equal to 0 to N minus 1, sin of delta over 2 times sin of 5 plus N delta. That's what I will have, right? So I multiply sin of delta over 2 by an entire sum, and using the distributive law, I can always replace it with multiplication by each member and then have a summation. So multiplication and summation are reversible in this particular case, obviously, for distributive law. All right, fine. Now, how can I deal with this one? Well, again, remember, cosine u plus v minus cosine of... Well, actually, let me start with u minus v, because u minus v will result in a plus. Minus cosine of u plus v is equal to 2 sin u sin v, right? Again, this is cosine cosine plus sin sin. This is cosine cosine minus sin sin. Now, cosine will be reduced, and this minus cosine and this minus will be plus, and that's why I have 2. Now, so this is basically what I have, right? So let's consider that u is equal to 5 plus N delta, sorry, and v is delta over 2, which means that my sum would be sigma N from 0 to N minus 1. So this is u, this is v. So I will have cosine of u minus v, which is 5 plus 2 N minus 1 over 2 delta. Am I right? It's 5 plus N delta minus delta over 2. Right, that's what it is. Minus cosine of 5 plus 2 N plus 1 over 2 delta, right? So that's what we have. And now, let's think about it. With N is equal to 0, I will have cosine equals... Now we can convert back into the explicit sum. Phi N is equal to 0, so it's minus delta over 2, minus cosine phi plus N is equal to 0 delta over 2. Next one. When N is equal to 1, I will have cosine of 5 plus delta over 2, minus cosine of 5 plus 3 seconds delta. That's first two members. Now, the next one for N is equal to 2. That would be 4, 3, 2, cosine of 5 plus 3 seconds of delta, minus cosine of 5 plus 5 seconds of delta, plus, etc. Now, the same thing exactly happens. This is reduced with this, this with this, this with this, etc., etc. And the only thing which would be remaining is the one with a negative sign of the last pair, which is for lower case N equals to N minus 1. So that would be cosine of 5 plus. This is my general... If N is equal to... Lower case is equal to N minus 1, it's 2N minus 2... No, this one will remain. This one will be reduced with the previous one. So this one will remain. So plus 2N minus 2 plus 1, minus 1 over 2 delta. That's the one which will not be reduced. Right. So whenever I multiply this by this, I will have only two members, cosine of 5 plus 2N minus 1 over 2 delta with a minus sign. Yes, minus sign. Because the previous one with a plus sign will be reduced. And the one which in the beginning would be with a plus sign, cosine of 5 plus delta over 2. So that's the result of summation in a concise form. Because now you can say that since S times sine of delta over 2 is equal to this, then S is equal to cosine of 5 plus delta over 2 minus cosine of 5 plus 2N minus 1 over 2 delta over sine of delta over 2. So this is the final formula. And again, it can be converted into a product using the same basically philosophy as I was using before. Because the final formula should look like a product of two sons. Now the final formula as I wrote it here is sine of 5 plus N minus 1 delta over 2 times sine of N delta over 2 divided by sine of delta over 2. Now the question is, is the product of these signs exactly the same as this? Again, it's probably very easy to prove using exactly the same formulas as before. So 2 sine U times sine V is equal to what? How can I get sine by sine? For sine I have to have cosine of difference and minus cosine of sum. Now this is the cosine times cosine minus sine times sine, I mean plus sine by sine. And this is also cosine cosine but in this case minus. If all signs will be reduced, signs will remain, this minus and this minus will be plus sine. So that's the formula. So my question is, if U is equal to 5 plus N minus 1 delta over 2 and V is equal to N delta over 2. U minus V is equal to minus delta over 2, am I right? And U plus V is equal to 5 plus 2 N minus 1 over 2. Is that right? Here, I actually made mistake here. That should be minus. So cosine of this and cosine of this. Yes, I think I made a mistake when I was transferring the formulas from one to another. So basically that corresponds to the formula which I have derived in the previous lecture. And the purpose of this was, if you are experienced enough solving all these problems, different methodologies will probably come to your mind just by themselves. So the one methodology which I was presenting on the lecture, in the previous lecture, was about using the complex numbers and geometric signals. Now in this particular case, I actually came up with a different solution, multiplied by sine of delta over 2 and then transform product of sines or product of sine times cosine into a difference and all consecutive members will be reduced. Well, that's another approach. You just have to be equipped, so to speak, that you can approach it this way, you can approach it that way. There are only a finite number of approaches. So the more problems you solve, the richer your repertoire is basically. So I will probably present some more problems in the same kind of area. And most likely either one of these two approaches would work. The only thing is you have to recognize, you have to find out, okay, I really have to multiply by delta over 2, the sine of delta over 2. Why? Well, in this case, I can actually give you a hint why. You see the difference between all consecutive members is delta. So whenever you have delta over 2, that's half of the difference. Basically what you're doing is you're replacing two consecutive members with the middle one minus delta and middle one plus delta. Well, actually delta over 2 if this is delta. So you have the middle one, the delta is a difference, and you replace this one with middle one minus half delta and this one with middle one plus delta. And then whenever you're doing other calculations, they are reducing each other. Each consecutive member would be reducing. Right? Because this one is middle minus delta over 2. But in the previous step, you had middle plus. So that's why. All right. So again, this is all I wanted to say about this particular problem, these particular two problems. And again, I'll try maybe to put some more on the web. But don't forget that everything would be in the notes on Unisor.com for every corresponding lecture. That's it. Thank you very much and good luck.