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Proof

$1+2\cos(\theta)+2\cos(2\theta)+2\cos(3\theta)+...+2\cos((n-1)\theta)=\frac{2\sin((n-\frac{1}{2})\theta)}{2\sin(\frac{1}{2}\theta)} $

Omar Barrera Valentin
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  • Why don't you multiply the sum of cosines on the left by $\sin(\theta/2)$ and use the formula $$2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta).$$ First do this for small values of $n=1,2,\ldots$. Then for full proof either induct on $n$ or see if there's a trick. – Jyrki Lahtonen Aug 26 '15 at 07:20
  • In other words, you should show your own thoughts. – Jyrki Lahtonen Aug 26 '15 at 07:22
  • I'm quite sure that this is a duplicate but I didn't find the link yet. – Marco Cantarini Aug 26 '15 at 07:25

1 Answers1

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One of such proofs is: $1+e^{i\theta} + e^{2i\theta}+\cdots + e^{(n-1)i\theta} = \dfrac{e^{ni\theta}-1}{e^{i\theta}-1}$. Then take the real part of both and equate them.

DeepSea
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