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

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 $\cos nx =\frac{e^{inx}+e^{-inx}}{2}$. – Jon Jul 27 '12 at 7:57
@Jon Thank you very much! – Bamily Jul 27 '12 at 8:01

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 $$1+2\sum\limits_{n=1}^N\cos(nx)= e^{i0x}+\sum\limits_{n=1}^N(e^{inx}+e^{-inx})=$$ $$\sum\limits_{n=-N}^N e^{inx}= \frac{e^{-iNx}(e^{i(2N+1)x}-1)}{e^{ix}-1}= \frac{e^{i(N+1)x}-e^{-iNx}}{e^{ix}-1}=$$ $$\frac{e^{i(N+1/2)x}-e^{-i(N+1/2)x}}{e^{ix/2}-e^{-ix/2}}= \frac{2i\sin(N+1/2)x}{2i\sin(x/2)}= \frac{\sin(N+1/2)x}{\sin(x/2)}$$
 Thank you Norbert, and $\sum_{n=0}^N$ should be changed to $\sum_{n=1}^N$ above. – Bamily Jul 27 '12 at 9:25 Interesting how you managed to write an answer, even if there was no question... :D – Ｊ. M. Jul 27 '12 at 15:16 @J.M. I read the thought – Norbert Jul 27 '12 at 15:23