 Hi, I'm Zor. Welcome to InDesert Education. We continue solving problems which I qualify as non-standard problems, not related to just to check your theoretical knowledge, but just to force you to think about certain things. And the purpose is actually to give you some material to practice this analytical approach to different problems. Math presents a perfect tool to develop your analytical and creative abilities. And obviously you will be using these everywhere in all kinds of industries wherever you work. So mathematics is just a gym for your brain, so to speak. And these problems are the perfect tool to exercise your creative abilities. Okay, so today we will consider a couple of problems. Well, actually it's a one general problem with two different variations and three different ways to solve this problem. Okay, so it's trigonometry. And what is very important is to have certain theoretical knowledge before you approach these problems. So that was the purpose of the course Math 14s presented on the same site Unisor.com where this course is now. This course is Math Plus and Problems and the prerequisite is Math 14s. Over there I do have certain theoretical material about trigonometry and we will also use complex numbers here and others formula. So these are things which you have to basically know from your regular theoretical course of mathematics. And again if you don't anything about these, these topics go back to whatever the source of information including Math 14s course on Unisor.com. So this is for those people who understand perfectly well the trigonometry, the basic course of trigonometry and basic course of complex numbers whatever is supposed to be touched here. Okay, so trigonometry and here is two problems which are actually variations of the same problem. Consider you have a trigonometric series, sine of x plus sine of x plus y plus sine of x plus 2y plus etcetera plus sine of x plus ny. This is the sum which I would like to evaluate or present it in more compact form. It's sine of x plus k times y where k is from 0 to n. So I have to evaluate that for now. In the regular course of school mathematics you probably learned how to summarize the arithmetic series when there is a constant difference between consecutive members or geometric series when there is a constant factor from one to the next one. This is neither of those. So what do we do? Now before you proceed I do suggest you to think about this, obviously. Now I will present two different approaches to this problem. The second problem actually is the sum of cosines which will be analogous. But in any case I will present two different approaches, one trigonometry related and another related to complex numbers and others formula. So right now I'm going to present the first solution related to pure trigonometry and it's kind of a trick. So whenever you need some kind of a trick to solve the problem question is how you come up with this trick. Well maybe some smart guy did but and since then it's a known trick so to speak. But if you don't know this you have to really come up yourself with some very artificial kind of a trick. I use the word trick to basically simplify this problem to solve it. Now obviously if this is the first problem of this type that's very difficult. I mean you might but anyway again I can actually suggest you a hint. You have to multiply this by something which will drastically reduce the whole thing and basically all members will will cancel each other except two of them which are the very first one and the very last one. Okay that's a hint so to speak right. So with this hint again you can stop the video and try to come up with this particular approach. Again if you don't I will continue talking about this but my problem is that the more problems of this type you will solve the greater will be your repertoire of tricks so to speak. So the next one would be maybe easier or faster for you to come up with certain artificial steps which you have to which you have to make to solve this particular problem. So in this particular case an artificial kind of a step which you have to make artificial in terms of it's not really obvious it's not like according to the theory you have to do this there is no theory but there is this particular trick. So what I'm going to do is I'm going to multiply my sum by sine of y over half. Why? Well again as I said this is a trick which helps and you will see that this helps. How did I come up with this? I didn't quite frankly. Long long time ago when I was probably in high school I read about this or something like this and since that it just came into my repertoire of tricks as I said. So now I basically know that it's supposed to be something like this. I remember that I have to multiply by something from my school years. Now when I was preparing this lecture I was just thinking about what should I really do. I was thinking maybe I should multiply by sine or by cosine by y or by y over half etc. So I just figured around and found that this particular thing does work. So what happens if you will multiply it by this? Well in the short description that would be sum of k from 0 to n, sum of x plus k y times sine of y over half. Okay fine. Now what is this? Okay now remember cosine of alpha plus beta is equal to cosine alpha times cosine beta minus sine alpha times sine beta. If I will put minus here I will put plus here, right? How can I get sine times sine from this? So I would like to cosine to cancel out. I will do the following. Sine alpha times sine beta is equal to, okay I have to take difference that would be plus here and I will subtract sum. It will be minus but with a minus it will be plus and cosine will cancel each other and that would be double so I have to put one half. So this is the formula which is derivation of a simple formula which you have to know from trigonometry cosine of sum or difference of two angles. So I will use this formula like this. So sine times sine this is alpha this is beta so I have to equate it to one half well actually yeah sum k from 0 to n this is one half and here I will put cosine of the difference so it's a cosine of x plus ky minus one half of y minus cosine of x plus ky plus y is it better than it was before? Yes and here is y. This is my sine this is s sine times sine of half of y, right? That's what it is. Now why is it better? Well think about it this way. Let me just get rid of this too. I'll put it here, okay? So let me just consider this sum. The first member is with k is equal to 0 would be cosine of x minus y over half minus if k is equal to 0 cosine of x plus y half. Next with k is equal to 1 what will be x plus y minus half of y, right? So it's cosine of x plus y over half minus if k is equal to 1 would be 3 seconds so cosine of x plus 3 y over 2 plus next one k is equal to 2. 2 minus half is 3 seconds okay so it will be cosine of x plus 3 y over 2 minus 2 and half so it's 5 half. I think by now you realize what happens minus plus minus plus minus plus etc. So what will be couple of last members? Couple of last members will be when k is equal to minus 1 it will be cosine of x plus n minus 1 minus 1 so it will be n minus 3 halves of y minus cosine of x plus n minus 1 plus half that will be n minus half n minus 1 half of y plus and the last one when the k is equal to n I will have cosine x plus n minus 1 half y minus cosine of x plus n plus 1 half of y so that happens this would be the previous one and what remains is this one and this one minus cosine of x plus n plus 1 half and this is equal to this from which our sum of sines is equal to we have to have this one but I can put it 2n plus 1 over 2 y doesn't really matter divided by 2 sine of y over half okay so this is a compact expression of this trigonometric series I call it trigonometric series because it's sines and sines and sines okay so a trick multiplied by this particular factor again you cannot just easily come up nobody if nobody tells you about this you have to come up with this yourself and it's difficult if it's the first time of this particular problem actually there are many problems of this kind when you have to do this kind of a trick to come up with a very easy solution well again the more tricks you learned you read about or you heard about from something the easier will be for you to come up with something new because in in practical life you will always face certain problems which nobody taught you how to solve which means you have to solve them yourself and come up with certain creative solution this is a creative solution so try to do as many of these as possible and mathematics has millions of these particular kind of problems and the more you do the better you're prepared for real life now the second problem which is very similar to this one is with the cosines so I will just use the same approach just slightly different formulas will be so these will be all cosines cosine cosine cosine and this will be also cosine cosine so what do we do to calculate this particular thing well artificial approach is exactly the same you multiply by sine of y over half but now we have slightly different formulas now the first time it was sine by sine which we converted into a difference so every member was after multiplication sine of some sum times sine of this and we converted into difference now we have cosine and sine so we have to you know develop slightly different approach we have to convert cosine times sine into a difference of something okay so here is basically the formula so you have sine of alpha plus beta is equal to sine of alpha times cosine beta plus cosine alpha the sine of beta so this is cosine and sine so we have to extract it somehow from from this well if I will put minus here it will be minus here right so if I will subtract from sine of alpha plus beta if I will subtract sine of alpha minus beta this will cancel each other and this will be plus minus minus the way how I want it actually so I need to subtract it will be two of those so I have to have half of this and that would be cosine alpha times sine beta exactly what we need so this would be so I multiply it by s cos times sine of y over two would be equal to sigma k from zero to n cosine x plus k times y times sine of y over two so this would be alpha this would be beta so it's equal to sigma k is from zero to n cosine times sine cosine times sine so it's a sine okay it's one half sine of their sum x plus k plus one half y minus sine of their difference x plus k minus one half y and now what do we have same thing all the members except except two will cancel each other okay with k is okay this is equal to s cosine times sine of y half all right so this is again I will get rid of this too I'll put it here all right the first member would be with k is equal to zero it will be sine of x plus one half of y minus sine of x minus y over two plus okay with k is equal to one I will have one and a half three quarter three seconds x plus three seconds of y minus sine of x plus one minus one half so it's this y we have minus sine of x is equal to two two and one half so it's x plus five y over half minus sine of x plus one and a half so what happens here it's slightly different but but anyway it's still all out this goes with this this goes with this this goes with this and the one which is remaining is this one with k is equal to n so the whole thing is equal to sine of x plus n plus one half y minus the very first one sine of x minus y over so this particular sum is equal to this divided by two sine of y half and that's the answer now I have suggested two different ways I said I would suggest two different ways to solve this problem so this first way is kind of a tricky you have to somehow come up with this thought to basically do whatever I just said to multiply by sine of y half here is another approach which arguably is more elegant and gives you again relatively fast the result but it's completely different okay now we go to complex numbers and Euler's formula which can which which connects trigonometric functions and exponential function with complex argument i is obviously imaginary one so called square root of minus one if you wish minus one so so imaginary number complex numbers again this is a theory which you have to know including this thing on the course mass routines I do spend some time basically introducing this formula in justification for this formula and Euler is by the way a very famous mathematician Swiss but he spent a lot of time in Russia it's one of the basically founders of Russian mathematics okay so the Euler's formula what does it give you well here it is let's consider let's consider e to the power i x plus k y and have sigma of this where k from 0 to n now according to this that's basically sigma of um cosine x plus k y plus i sine x plus k y from 0 to n now this is the real part this is imaginary part real part comes with real part with all different case and imaginary so I can actually write it as i as sigma of x x plus k y plus i sigma sine of x plus k y from 0 to n from 0 to n so i basically goes outside of sigma and I have basically this is the sum which I was considering the second time and this is the sum which I consider the first time now being as it may I will use this expression instead of this expression because this expression is equal to if I will consider just this one well it's sigma a to the power i x i x times e to the power k y right right because whenever you are multiplying with the same base your exponents are supposed to be i k y exponents are supposed to be added together now e to the power of i x is not dependent on k so I can put it here e to the power i x sigma a to the power i k y k from 0 to n now instead of putting this I will put it this way to the power of k because as you know whenever you are when you are raising something to the power the exponents are multiplying so basically it's the same thing and what is this this is a geometric series with a factor e to the power of i y i i i y and the factor the first member and the factor are e to the power i y so you can basically use the formula for geometric sequence to get the result of this so what you will have as a result you will have e to the power i x times this would be e to the power i n plus 1 minus 1 divided by e to the power of i y y minus 1 so I just used the formula for geometric series it's the last member next after last member minus 1 divided by factor minus 1 in this particular case so this is the answer now it's not looking actually like we had before right why because it's a complex number it should really separate the imaginary part from the real part the real part will be answer to this and the imaginary part will be the answer to this one now how to do this this is unfortunately a little bit more tricky process not tricky no it's more calculation involved which I definitely don't want to do it right now I did it in the notes for this lecture every lecture in on unisor.com has notes so in the notes I really converted into real and imaginary part and I will have and I had exactly the same answers as before here I'm not going to do it I suggest you to do it but the most important part is probably to get rid of the i in the denominator and for this I just multiply it by by e to the power of minus i y minus 1 e to the power minus i y minus 1 both numerator and denominator and if you will multiply this by this you will have only real only real number primarily 2 minus 2 cosine of y that will be in the denominator and the numerator you will just have to multiply and convert it into cosine and sine using the Euler's formula and it will eventually give you the answer so the more elegant this just because it's very fast gives you both sums it's more calculation incentive because it just at the very end when you already have a formula in a complex representation you have to go back to real and imaginary part separately and that will take some calculations again all calculations I suggest you to try to do yourself and again check it against my notes which which which gives you the same answer as I had before so that's that's it for today I suggest you to read the notes they are very detailed and I might actually make a mistake but probably I didn't but anyway the notes are correct so every lecture has both video and the notes on myunicero.com website that's it for today thank you very much and good luck