Can we show a sum of symmetrical cosine values is zero by using roots of unity? Can we show that 
$$\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{5\pi}{7}+\cos\frac{6\pi}{7}=0$$
by considering the seventh roots of unity? If so how could we do it?
Also I have observed that
$$\cos\frac{\pi}{5}+\cos\frac{2\pi}{5}+\cos\frac{3\pi}{5}+\cos\frac{4\pi}{5}=0$$
as well, so just out of curiosity, is it true that $$\sum_{k=1}^{n-1} \cos\frac{k\pi}{n} = 0$$
for all $n$ odd?
 A: I think you can use Euler's Formula. 
The Nth roots of unity = $e^{2\pi k i/N}$ for values of k between $0$ and $N-1$ inclusive. 
There sum from k to $N-1$ is a geometric series. 
$S= \sum_{k=0}^{N-1} \ e^{2\pi i k/N}=\frac{1\cdot e^{(2\pi i /N)N}-1}{e^{2\pi i /N}-1}$
The numerator is zero for any N. 
But the real part of $S$ is the real part of the individual terms of the sum, i.e. the cosines. The real part of the sum is zero so the sum of the real parts of the roots of unity is 0. 
A: Pointing at the link I left in the comments 

$$1 + \cos\theta + \cos2\theta +... + \cos n\theta = \frac{1}{2} + \frac{\sin[(n+\frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})}$$

Then for $\forall n\in\mathbb{N}, n>0$
$$\cos\frac{\pi}{n+1}+ \cos\frac{2\pi}{n+1} +... + \cos \frac{n\pi}{n+1} =  \frac{\sin\left[(n+\frac{1}{2})\frac{\pi}{n+1}\right]}{2\sin\left(\frac{\pi}{2(n+1)}\right)}-\frac{1}{2}=\\
\frac{\sin\left[\frac{2n+1}{2(n+1)}\pi\right]}{2\sin\left(\frac{\pi}{2(n+1)}\right)}-\frac{1}{2}=
\frac{\sin\left[\pi-\frac{\pi}{2(n+1)}\right]}{2\sin\left(\frac{\pi}{2(n+1)}\right)}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0$$
A: Note that  $$\cos(\pi - \alpha)= - \cos(\alpha)$$ Therefore $$\cos(\frac{\pi}{7})+\cos(\frac{2\pi}{7})+\cos(\frac{3\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{5\pi}{7})+\cos(\frac{6\pi}{7})=$$
$$\cos(\frac{\pi}{7})+\cos(\frac{2\pi}{7})+\cos(\frac{3\pi}{7})-\cos(\frac{3\pi}{7})-\cos(\frac{2\pi}{7})-\cos(\frac{\pi}{7})=0$$
The same goes for other natural numbers $n$ instead of $7$.
