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I am trying to prove that
$\underset{N\to \mathrm{\infty }}{lim}\frac{1}{2N}\sum _{k=1}^{N}\left[\mathrm{cos}\left(\left(x-kL\right)\cdot q\right)+\mathrm{cos}\left(\left(x+kL\right)\cdot q\right)\right]=0$
for every $x,L\in {\mathbb{R}}^{+}$ with $x\le L$ and for every $q\ne \frac{2n\pi }{L}$, $n\in \mathbb{Z}$.
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America Barrera
$\mathrm{cos}\left(\alpha ±\beta \right)=\mathrm{cos}\alpha \mathrm{cos}\beta \mp \mathrm{sin}\alpha \mathrm{sin}\beta .$
Plugging this into your expression yields
$\mathrm{cos}\left(xq\right)\frac{1}{N}\sum _{k=1}^{N}\mathrm{cos}\left(kLq\right).$
Also,
$\mathrm{sin}\left(\alpha ±\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\beta ±\mathrm{sin}\beta \mathrm{cos}\alpha .$
Therefore,
$2\mathrm{sin}\left(\frac{\alpha }{2}\right)\mathrm{cos}\left(k\alpha \right)=\mathrm{sin}\left(\left(k+\frac{1}{2}\right)\alpha \right)-\mathrm{sin}\left(\left(k-\frac{1}{2}\right)\alpha \right).$
This allows for a telescopic sum argument.
$\frac{1}{N}\sum _{k=1}^{N}\mathrm{cos}\left(kLq\right)=\frac{\mathrm{sin}\left(\left(N+\frac{1}{2}\right)Lq\right)-\mathrm{sin}\left(\frac{1}{2}Lq\right)}{2N\mathrm{sin}\left(\frac{Lq}{2}\right)}\underset{N\to \mathrm{\infty }}{\to }0$