My teacher told us that :
$$\sum_{k=0}^{n-1} \cos\frac{2k\pi}{n}=0$$
Without giving any proof
If someone could tell me how to approach this that would be great !
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5Use the fact that $\cos y$ is the real part of $e^{iy}$ and us ethe formula for a geometric sum. – Kavi Rama Murthy Jan 24 '21 at 12:13
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1The average for the vertices of a regular $n$-gon is the center. In this case, take vertices $$\left(\cos\frac{2k\pi}{n},\sin\frac{2k\pi}{n}\right)$$ for $k=0,1,\dots,n-1$. This is the same as Kavi's comment, but appropriate for a student not familiar with complex numbers. – GEdgar Jan 24 '21 at 12:18