How to prove $\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$ for any n>1? I can show for any given value of n that the equation
$$\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$$
is true and I can see that geometrically it is true. However, I can not seem to prove it out analytically. I have spent most of my time trying induction and converting the cosine to a sum of complex exponential functions 
$$\frac{1}{2}\sum_{k=1}^n [\exp(\frac{i 2 \pi k}{n})+\exp(\frac{-i 2 \pi k}{n})] = 0$$
and using the conversion for finite geometric sequences
$$S_n = \sum_{k=1}^n r^k = \frac{r(1-r^n)}{(1-r)}$$
I have even tried this this suggestion I have seen on the net by pulling out a factor of $\exp(i \pi k)$ but I have still not gotten zero.
Please assist.
 A: We have $$\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{n}\right)=\textrm{Re}\left(\sum_{k=1}^{n}e^{2\pi ik/n}\right)
 $$ and so $$\sum_{k=1}^{n}e^{2\pi ik/n}=\frac{e^{2\pi i/n}\left(1-e^{2\pi i}\right)}{1-e^{2\pi i/n}}
 $$ and notice that $$e^{2\pi i}=\cos\left(2\pi\right)+i\sin\left(2\pi\right)=1
 $$ so the claim follows.
A: One approach is to write
$$\cos\left(\frac{2\pi k}{n}\right)=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(k+1)}{n}\right)-\sin\left(\frac{2\pi(k-1)}{n}\right)\right)$$
Now, we have converted the sum into a telescoping sum, which we can evaluate directly as 
$$\begin{align}
\sum_{k=1}^n\cos\left(\frac{2\pi k}{n}\right)&=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(n+1)}{n}\right)+\sin\left(\frac{2\pi(n)}{n}\right)\right)\\\\
&-\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(2-1)}{n}\right)+\sin\left(\frac{2\pi(1-1)}{n}\right)\right)\\\\
&=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(n+1)}{n}\right)-\sin\left(\frac{2\pi(2-1)}{n}\right)\right)\\\\
&=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi}{n}\right)-\sin\left(\frac{2\pi}{n}\right)\right)\\\\
&=0
\end{align}$$
A: You are considering the real part of the sum of the roots of $x^n-1$. By Viète's theorem the sum of the roots of $x^n-1$ is the coefficient of $x^{n-1}$, i.e. zero.
