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Please I have tried to show that

1) $\sum_{j=1}^{N-1}\cos^2(\frac{2\pi}{N}\cdot j\cdot\Delta c)=N-1 $ for $\Delta c=\frac{N}{2} $

and that

2)$\sum_{j=1}^{N-1}\cos^2(\frac{2\pi}{N}\cdot j\cdot\Delta c)=\frac{N-2}{2} $ for $\Delta c\neq\frac{N}{2} $

I tried the first one this way, substituting $\Delta c=\frac{N}{2}$ results in

$\sum_{j=1}^{N-1}\cos^2(\pi\cdot j)=\sum_{j=1}^{N-1}\frac{1+\cos(\pi\cdot j)}{2}$ since $\cos^2\alpha=\frac{1+\cos(2\alpha)}{2}$, but I don't know how to continue.

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For the first part, $ \cos(\pi * j)=(-1)^j$.

Hence $ \cos^2(\pi * j)=1$, and the sum follows.

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  • $\begingroup$ Thank you so much. Please, what about the second one. Any clue? $\endgroup$ – Anthony Uwaechia Jul 5 '17 at 13:42

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