Sum of a cosine series

Question

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!

Answer 1

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 $\displaystyle k=0,\cos\frac{2k\pi}r=?$

Answer 2

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 $\displaystyle \sin\frac{(k-1)\pi}k=\sin\left(\pi-\frac\pi k\right)=\sin\frac\pi k\ne0$