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why does multiplying by $2sin(b/2)$ cancel out most of the terms,what is the intuition behind it?how does multiplying by it creates a chain of terms that get cancelled?

edit:1 I have attached an image about what I tried to simplify using $sin(d)$enter image description here

Dheeraj Gujrathi
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    Makes more sense if you think at it as $;\text{Im}\left(e^{ia}\left(1+e^{ib}+\dots+e^{i(n-1)b}\right)\right),$. – dxiv Aug 04 '21 at 17:55
  • @dxiv,sorry but I am still a student in learning phase,so I have not studied complex numbers yet – Dheeraj Gujrathi Aug 04 '21 at 18:54
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    The trick is similar in a way to multiplying $,1+x+x^2+\dots+x^{n-1},$ by $,1-x,$ in order to get a telescoping sum which reduces to $,1-x^n,$ in the end. However, that similarity is not obvious without understanding the complex representation. – dxiv Aug 04 '21 at 19:35
  • @dxiv,yes,I will be learning complex numbers very soon,"1+x" expression helped me to get a logic,though i have used another method to find the sum,is it correct? – Dheeraj Gujrathi Aug 05 '21 at 09:01
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    That looks correct, and can be further simplified to the other form using the sum-product identities. – dxiv Aug 05 '21 at 22:04

1 Answers1

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  1. In the given example, it is observed that the original series can be converted into a telescoping series, which is a series whose general term is expressible as the difference of two consecutive terms of some sequence.

    This is achieved by multiplying҂ it by $$2\sin\frac b2,$$ then applying the product-to-sum identity $$2\sin\theta\sin\phi=\cos(\theta-\phi)-\cos(\theta+\phi).$$ The resulting telescoping series is $$\sum_{r=1}^n\left[\cos\left(a+\frac{(2r-3)b}2\right)-\cos\left(a+\frac{(2r-1)b}2\right)\right].$$

  2. Next, the method of differences is used to cancel out all the terms, except for the first and last ones, of this telescoping series.

  3. Finally, the sum-to-product identity $$\cos\theta-\cos\phi=-2\sin\frac{\theta+\phi}2\sin\frac{\theta-\phi}2$$ is applied to express the result in terms of $\sin$ again.

҂Answer to the OP's further question:

Multiplying the series by $(2\sin b)$ instead of $(2\sin\frac b2)$ will result in the series $$\sum_{r=1}^n\left[\cos\left(a+(r-2)b\right)-\cos\left(a+rb\right)\right],$$ which isn't telescoping (refer the italicised text above); using the method of differences on this series will leave four instead of two terms.

ryang
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