Sum of a cosine series I have a short question:
It turns out that the following holds:
$\sum_{i=1}^{k-1} \cos(\frac{2\pi i}{k}) = - 1$. 
Why is that? 
Thank you!
 A: From here, observe that the expansion of $\cos rx$ does not involve $\cos^{r-1}x$
So, using Vieta's formulas, the sum of the $r$ roots of $\displaystyle\cos^r x+\cdots-\cos rx=0$ is $0$
Now let $\displaystyle\cos rx=1\implies \cos rx=\cos2k\pi$ where $k$ is any integer
$\displaystyle\implies rx=2k\pi\implies x=\frac{2k\pi}r$ where $k=0,1,2,\cdots,r-1$
$\displaystyle\implies \sum_{0\le k\le r-1}\cos\frac{2k\pi}r=0$ 
Now, for $\displaystyle k=0,\cos\frac{2k\pi}r=?$
A: Clearly, the Series vanishes if $k=1$
If $k>1, \sin\frac{\pi}k\ne0$
As $2\cos A\sin B=\sin(A+B)-\sin(A-B),$ using this,  $$2\cos\frac{2i\pi}k\sin\frac{\pi}k=\sin\frac{(2i+1)\pi}k-\sin\frac{(2i-1)\pi}k $$
Putting $i=1,2,\cdots,k-1$ and summing we get  $$2\sin\frac{\pi}k\sum_{1\le i\le k-1}\cos\frac{2i\pi}k=\sin\frac{(2k-1)\pi}k-\sin\frac{\pi}k=2\sin\frac{(k-1)\pi}k\cos\frac{k\pi}k$$
Using $\displaystyle\sin C-\sin D=2\sin\frac{C-d}2\cos\frac{C+D}2,$
$\displaystyle\sin\frac{(2k-1)\pi}k-\sin\frac{\pi}k=2\sin\frac{(k-1)\pi}k\cos\frac{k\pi}k$
Now, $\displaystyle \cos\frac{k\pi}k=\cos\pi=-1$ and $\displaystyle \sin\frac{(k-1)\pi}k=\sin\left(\pi-\frac\pi k\right)=\sin\frac\pi k\ne0$
