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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)} $

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  • $\begingroup$ 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. $\endgroup$ Aug 26, 2015 at 7:20
  • $\begingroup$ In other words, you should show your own thoughts. $\endgroup$ Aug 26, 2015 at 7:22
  • $\begingroup$ I'm quite sure that this is a duplicate but I didn't find the link yet. $\endgroup$ Aug 26, 2015 at 7:25

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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.

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  • $\begingroup$ Did you search for duplicates first? $\endgroup$ Aug 26, 2015 at 7:23

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