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For $x \neq 0$, $ 1 + 2 \sum_{n=1}^N \cos n x = \frac{ \sin (N + 1/2) x }{\sin \frac{x}{2}} $

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    If you have a look at http://math.stackexchange.com/questions/225941/proving-sum-limits-k-0n-coskx-frac12-frac-sin-frac2n12x and the question linked there, you can find several similar question. (Maybe some of these questions should be closed as duplicates?)2014-02-21

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Here is a well known trigonometric trick $ 1+2\sum\limits_{n=1}^N\cos (nx)= 1+\frac{1}{\sin(x/2)}\sum\limits_{n=1}^N 2\cos (nx)\sin (x/2)=\\ 1+\frac{1}{\sin (x/2)}\sum\limits_{n=1}^N(\sin (nx+x/2)-\sin (nx-x/2))=\\ 1+\frac{1}{\sin (x/2)}(\sin (Nx+x/2)-\sin (x/2))=\\ 1+\frac{\sin (Nx+x/2)}{\sin (x/2)}-1= \frac{\sin (N+1/2)x}{\sin (x/2)} $ And this is a complex analysis approach $ 1+2\sum\limits_{n=1}^N\cos(nx)= e^{i0x}+\sum\limits_{n=1}^N(e^{inx}+e^{-inx})= $ $ \sum\limits_{n=-N}^N e^{inx}= \frac{e^{-iNx}(e^{i(2N+1)x}-1)}{e^{ix}-1}= \frac{e^{i(N+1)x}-e^{-iNx}}{e^{ix}-1}= $ $ \frac{e^{i(N+1/2)x}-e^{-i(N+1/2)x}}{e^{ix/2}-e^{-ix/2}}= \frac{2i\sin(N+1/2)x}{2i\sin(x/2)}= \frac{\sin(N+1/2)x}{\sin(x/2)} $

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    @J.M. I read the thought2012-07-27