Prove that $ \sum\limits_{n=1}^N \cos(2\pi n/N)= 0 $?

Question

Is there a way to algebraically prove that $ \sum\limits_{n=1}^N \cos(2 \pi n/N) = 0 $ for any $N > 0$? (And if so, how?)

Answer 1

Hint :

$$\sum_{n=1}^N\cos\left(\frac{2\pi n}{N}\right)=\Re\left(\sum_{n=1}^N\exp\left(\frac{2i\pi n}{N}\right)\right)$$

and also :

$$\forall\theta\in\mathbb{R},\,\forall n\in\mathbb{N},\,\exp(i n\theta)=\left[\exp(i\theta)\right]^n$$

Answer 2

Another approach: note that

$$\sum_{n=1}^{2N}\cos(2\pi n/2N)=\sum_{n=1}^{N}\cos(2\pi n/2N)+\sum_{n=1}^{N}\cos(2\pi (n+N)/2N)$$

Since $\cos(x+\pi)=-\cos(x)$, we have $\sum_{n=1}^{N}\cos(2\pi (n+N)/2N)=-\sum_{n=1}^{N}\cos(2\pi n/2N)$ and done. The odd case is similiar; use also $\cos(\pi)=0$.