Proving $\sum_{k=0}^{n-1} \text{cos}\left(\frac{2\pi k}{n} \right) = \sum_{k=0}^{n-1} \text{sin}\left(\frac{2\pi k}{n} \right) = 0$ I would like to prove the following, $$\sum_{k=0}^{n-1} \text{cos}\left(\frac{2\pi k}{n} \right) = \sum_{k=0}^{n-1} \text{sin}\left(\frac{2\pi k}{n} \right) = 0$$ This is equivalent to showing that if we have a regular n-gon, all of the vectors in $\mathbb{R}^2$ pointing to every vertex cancel eachother out to $\textbf{0}$. It seems obvious enough and i've tried a number of examples for small $n$.  I am not sure how to start a proof though.

I assume $n \ge 2$.
Set
$\omega = \exp \left ( \dfrac{2\pi i}{n} \right ); \tag 1$
then
$\omega^n = \left ( \exp \left ( \dfrac{2\pi i}{n} \right ) \right )^n =  \exp \left ( \dfrac{2 \pi i n}{n} \right ) =  \exp ( 2 \pi i ) = 1; \tag 2$
we have
$( \omega - 1) \displaystyle \sum_0^{n - 1} \omega^k = \omega^n - 1 = 0; \tag 3$ 
since
$\omega \ne 1, \tag 4$
(3) yields
$ \displaystyle \sum_0^{n - 1} \omega^k = 0; \tag 5$ 
from (1)
$\omega^k =  \exp \left ( \dfrac{2\pi i k}{n} \right ) = \cos \left ( \dfrac{2 \pi k}{n} \right ) + i \sin \left ( \dfrac{2 \pi k}{n} \right ) ; \tag 6$
substituting (6) into (5) and separating out the real and imaginary parts gives us
$\displaystyle \sum_0^{n - 1} \cos \left ( \dfrac{2 \pi k}{n} \right ) + i  \sum_0^{n - 1} \sin \left ( \dfrac{2 \pi k}{n} \right ) = 0, \tag 7$
from which we immediately have
$\displaystyle \sum_0^{n - 1} \cos \left ( \dfrac{2 \pi k}{n} \right ) = \sum_0^{n - 1} \sin \left ( \dfrac{2 \pi k}{n} \right ) = 0. \tag 8$
