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I was struggling with finding a proof to prove: $$ \sum_{0\leq n \leq \alpha -1} \cos{\frac{2n\pi}{\alpha}} = -1 $$

I know it's true but I cannot figure out how to prove it,

Sorry to ask such a simple question,

And many thanks to whoever can help.

Adrian Salt
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    First notice that $1=\cos 2\alpha\pi/\alpha$ so you can move the $-1$ to the other side and make the sum run to $\alpha.$ Second, half of your angles are in the 2nd and 3rd quadrants, so their cosines are negatives of those in the 1st and 4th quadrants. – B. Goddard Nov 14 '22 at 15:27
  • Presumably $\alpha$ is an integer greater than 1. One way would be to use the complex number definition of $\cos$. Another would be to pair up terms as per @Goddard. – Blitzer Nov 14 '22 at 17:03
  • Actually - it's not true for $\alpha=2$ either! – Blitzer Nov 14 '22 at 17:04
  • Maybe double check the question. Should the saum start at n=1? – Blitzer Nov 14 '22 at 17:05

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