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I found out that this equality holds by accident,$$\sum_{k=1}^{\frac{n-1}{2}}\cos\left(\frac{2\pi k}{n}\right)=-\frac{1}{2}$$ if $n=1\mod 2$. However, I am not able to prove this directly with rules of trigonometry etc. Does anyone know that this indeed holds always and does anyone know a proof? Thank you very much.

Martin Sleziak
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Badshah
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  • You can use $\cos \varphi = \operatorname{Re} e^{i\varphi}$ and the formula for geometric sums. Or $\cos\varphi = \frac{1}{2} \left(e^{i\varphi} + e^{-i\varphi}\right)$. – Daniel Fischer Aug 18 '14 at 12:22
  • http://math.stackexchange.com/questions/117114/sum-cos-when-angles-are-in-arithmetic-progression – lab bhattacharjee Aug 18 '14 at 12:30

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Let $\zeta = \exp(2\pi \textrm{i}/n)$ be a primitive root of unity. Then your sum equals the real part of $$s = \sum_{k=1}^{\frac{n-1}{2}} \zeta^k.$$ Now $$s + \overline{s} + 1 = \sum_{k=0}^{n-1} \zeta^k = 0$$ and $\textrm{Re}(s) = \textrm{Re}(\overline{s})$ so $2 \textrm{Re}(s) + 1 = 0$.

WimC
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