Proving $\sum_{k=0}^{n-1} \text{cos}\left(\frac{2\pi k}{n} \right) = \sum_{k=0}^{n-1} \text{sin}\left(\frac{2\pi k}{n} \right) = 0$
Proving $\sum_{x=0}^{n-1} \cos\left(k +x{2\pi\over n}\right) =\sum_{x=0}^{n-1} \sin\left(k +x{2\pi\over n}\right) =0. $