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.
 A: 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
$\displaystyle \sum_0^{n - 1} \cos \left ( \dfrac{2 \pi k}{n} \right ) = \sum_0^{n - 1} \sin \left ( \dfrac{2 \pi k}{n} \right ) = 0. \tag 8$
A: (Too long for a comment.)  The algebraic proof using complex numbers is the most direct, yet...

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 each other out to $\textbf{0}$.

...this is too good a geometric insight to let go to waste. The following sketches how a proof could be built around this intuition (though it doesn't formalize it in full detail).
Let the respective vectors be $\,\mathbf{u_1}+\mathbf{u_2}+\ldots+\mathbf{u_n} = \mathbf{u}\,$, and assuming $\,n \gt 1\,$.
If $\,n\,$ is even then the regular $n$-gon has central symmetry i.e. $\,\mathbf{u_{j+n/2}}=-\mathbf{u_j}\,$ and therefore the $\,n\,$ vectors cancel out pairwise, so $\mathbf{u}=\mathbf{0}$.
Otherwise if $\,n\,$ is odd, let $\,T\,$ be the rotation of $\,\mathbb{R}^2\,$ by $\,\pi / n\,$. It is geometrically obvious that the endpoints of $\,\mathbf{u_1}, T(\mathbf{u_1}), \mathbf{u_2}, T(\mathbf{u_2}), \ldots \mathbf{u_n}, T(\mathbf{u_n})\,$ are the vertices of a regular $2n$-gon, thus by the previous observation $\,\mathbf{u_1} + T(\mathbf{u_1}) + \mathbf{u_2} + T(\mathbf{u_2}) + \ldots + \mathbf{u_n} + T(\mathbf{u_n})=\mathbf{0}\,$. On the other hand, $\,T\,$ is a linear transformation, so the latter implies $\,\mathbf{u} + T(\mathbf{u})=\mathbf{0}\,$, which in turn implies $\,\mathbf{u}=\mathbf{0}\,$.
