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

Martin Sleziak
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Bamily
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    This can be reduced to a geometric series just noting that $\cos nx =\frac{e^{inx}+e^{-inx}}{2}$. – Jon Jul 27 '12 at 07:57
  • @Jon Thank you very much! – Bamily Jul 27 '12 at 08:01
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    @Amanda You should not change 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?. – Martin Sleziak Feb 21 '14 at 07:40
  • 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?) – Martin Sleziak Feb 21 '14 at 08:09

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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 $$ 1+2\sum\limits_{n=1}^N\cos(nx)= e^{i0x}+\sum\limits_{n=1}^N(e^{inx}+e^{-inx})= $$ $$ \sum\limits_{n=-N}^N e^{inx}= \frac{e^{-iNx}(e^{i(2N+1)x}-1)}{e^{ix}-1}= \frac{e^{i(N+1)x}-e^{-iNx}}{e^{ix}-1}= $$ $$ \frac{e^{i(N+1/2)x}-e^{-i(N+1/2)x}}{e^{ix/2}-e^{-ix/2}}= \frac{2i\sin(N+1/2)x}{2i\sin(x/2)}= \frac{\sin(N+1/2)x}{\sin(x/2)} $$

Norbert
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