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I have come across the expression

$$ \sum_{k = 0}^{N - 1} \cos \left( \frac{2 \pi}{N}\left(k + \frac{1}{2}\right)\right) $$

and I happened to notice that it equals $0$ for all integer values of $N > 1$. This it very easy to prove for $N$ even, by noticing some symmetries in the angles that appear in the arguments of the cosines. This does not work for $N$ odd. When trying to prove it for $N$ odd, I split it as

$$ \sum_{k = 0}^{N - 1} \cos \left( \frac{2 \pi}{N}\left(k + \frac{1}{2}\right)\right) = \cos\left(\frac{\pi}{N} \right)\sum_{k = 0}^{N-1} \cos \left(\frac{2 \pi}{N}k\right) - \sin\left(\frac{\pi}{N} \right)\sum_{k = 0}^{N-1} \sin \left(\frac{2 \pi}{N}k\right)$$

and noticed (by checking it numerically) that the two terms are individually $0$ for all integer $N > 1$

$$\sum_{k = 0}^{N-1} \cos \left(\frac{2 \pi}{N}k\right) = 0 = \sum_{k = 0}^{N-1} \sin \left(\frac{2 \pi}{N}k\right).$$

From here I don't know to proceed. I have thought of applying some ideas from Fourier series, maybe thinking of the sums as the series of a function evaluated at $x = 1$, but I've got nothing concrete.

How to prove the statement for $N$ odd? Or even better, how to prove it for generic $N$?

TeneT
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    Are you open to proofs using complex numbers? If so, take the real part of the sum of the roots of $z^N+1$. – J.G. May 06 '22 at 09:29

1 Answers1

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Hint

You face sums of sines and cosines of angles in arithmetic progression $$\sum_{k = 0}^{N-p} \cos \left(\frac{2 \pi}{N}k\right) =\frac{1}{2} \left(1+\csc \left(\frac{\pi }{N}\right) \sin \left(\frac{\pi -2 \pi p}{N}\right)\right)$$ $$\sum_{k = 0}^{N-p} \sin \left(\frac{2 \pi}{N}k\right) =\frac{1}{2} \csc \left(\frac{\pi }{N}\right) \left(\cos \left(\frac{\pi (N+1-2 p)}{N}\right)+\cos \left(\frac{\pi }{N}\right)\right)$$

Claude Leibovici
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