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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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    This can be reduced to a geometric series just noting that $\cos nx =\frac{e^{inx}+e^{-inx}}{2}$.2012-07-27
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    @Jon Thank you very much!2012-07-27
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    @Amanda You [should not change](http://math.stackexchange.com/review/suggested-edits/168405) other people's questions in the way which changes the meaning. If you want to ask for solution of some question using some specific method, you can ask a new question. (However, in this case there already is an answer using complex numbers.) This has been also discussed on meta: [Why was this edit approved?](http://meta.math.stackexchange.com/questions/12808/why-was-this-edit-approved).2014-02-21
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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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    Thank you Norbert, and $\sum_{n=0}^N $ should be changed to $ \sum_{n=1}^N$ above.2012-07-27
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    Interesting how you managed to write an answer, even if there was no question... :D2012-07-27
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    @J.M. I read the thought2012-07-27