Simplifying a summation involving "cos".

Question

$\sum_{r=1}^{n-1}\cos ^{2}\left ( \frac{r\pi }{n} \right )$

How can I simplify this summation when I do not know whether "n" is odd or even?

Answer 1

If $\displaystyle S=\sum_{r=1}^{n-1}\cos ^2\left ( \frac{r\pi }n \right )$

$\displaystyle 2S=\sum_{r=1}^{n-1}[1+\cos \frac{2r\pi }n ]=n-1+\sum_{r=1}^{n-1}\cos \frac{2r\pi }n=n-1-1+\sum_{r=0}^{n-1}\cos \frac{2r\pi }n$

Now if $\displaystyle\cos(nx)=1,nx=2m\pi$ where $m$ is any integer

$\displaystyle x=\frac{2m\pi}n$ where $0\le m\le n-1$

Now using the pattern here, $\displaystyle\cos(nx)=2^{n-1}\cos^nx+0\cdot\cos^{n-1}x\cdots=0$

If we set $\cos(nx)=1,$ the roots of $\displaystyle2^{n-1}\cos^nx+0\cdot\cos^{n-1}x\cdots-1=0$ are $\displaystyle\cos\frac{2m\pi}n$ where $0\le m\le n-1$

So, using Vieta's formula, $\displaystyle\sum_{m=0}^{n-1}\cos\frac{2m\pi}n=\frac0{2^{n-1}}$