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

Question

How would you prove $$\sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}\right)=0=\sum_{k=0}^{n-1}\sin\left(\frac{2\pi k}{n}\right)$$

We tried induction first, but it doesnt seem to work. Any help would be appreciated!

The centroid of a regular polyhydron is at its circumcenter. — Batominovski, Dec 02 '17 at 09:46

score 2 · Answer 1 · answered Dec 02 '17 at 08:05

2

Guide:

Try to use geometric sum to prove that $$\sum_{k=0}^{n-1} \exp\left(\frac{2\pi k i}{n}\right)=0$$

answered Dec 02 '17 at 08:05

Siong Thye Goh

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score 1 · Answer 2 · answered Dec 02 '17 at 08:14

A simple geometric interpretration comes if we divide by $n$ so we're sayng averages, not sums, are $0$. The cosines and sines are respectively the $x$ and $y$ coordinates of a regular $n$-gon of circumradius 1, centred on the origin with a vertex at $(1,\,0)$. By symmetry the vertices' coordinates average to the origin.

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

2 Answers2