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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1$\begingroup$ This can be reduced to a geometric series just noting that $\cos nx =\frac{e^{inx}+e^{-inx}}{2}$. $\endgroup$– JonCommented Jul 27, 2012 at 7:57
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$\begingroup$ @Jon Thank you very much! $\endgroup$– BamilyCommented Jul 27, 2012 at 8:01
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1$\begingroup$ @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?. $\endgroup$– Martin SleziakCommented Feb 21, 2014 at 7:40
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$\begingroup$ If you have a look at math.stackexchange.com/questions/225941/… and the question linked there, you can find several similar question. (Maybe some of these questions should be closed as duplicates?) $\endgroup$– Martin SleziakCommented Feb 21, 2014 at 8:09
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1 Answer
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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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$\begingroup$ Thank you Norbert, and $\sum_{n=0}^N $ should be changed to $ \sum_{n=1}^N$ above. $\endgroup$– BamilyCommented Jul 27, 2012 at 9:25
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$\begingroup$ Interesting how you managed to write an answer, even if there was no question... :D $\endgroup$ Commented Jul 27, 2012 at 15:16
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