Prove the sum equation

Question

Prove the following relation \begin{equation} 2\sum_{k=1}^{N}\cos^2\big(\frac{k\pi}{N+1}\big) = N-1. \end{equation} where $N$ is even. Any ideas?

Answer 1

We have

$$ S=\frac{1}{2}\sum_{k=1}^N\left(e^{i\frac{k\pi}{N+1}}+e^{-i\frac{k\pi}{N+1}}\right)^2=\frac{1}{2}\sum_{k=1}^N\left(e^{2i\frac{k\pi}{N+1}}+e^{-2i\frac{k\pi}{N+1}}+2\right)$$

Can you evaluate this using geometric series?

Answer 2

$$\cos^2 x=\frac{1+\cos 2x}{2}\Rightarrow2\sum_{k=1}^N\cos^2\frac{k\pi}{N+1} =N+\sum_{k=1}^N\cos\frac{2k\pi}{N+1}=N-1,$$

the last equality following from: $$\sum_{k=0}^N e^{\frac{2\pi k}{N+1}}=0\Rightarrow \sum_{k=0}^N\cos{\frac{2\pi k}{N+1}}=0\Rightarrow \sum_{k=1}^N\cos{\frac{2\pi k}{N+1}}=-1.$$

The first expression is zero as it is sum of all roots of the equation $z^{N+1}=1$.

Answer 3

The answer is yes. Use power reduce on $2\cos^2 (\pi k / (N+1)) = 1 + \cos (2 \pi k / (N+1))$. Then, it is straightforward to see $$\sum_{k=1}^N \cos ( 2 \pi k /(N+1)) = -1$$ Whence $$2 \sum_{k=1}^N \cos^2 (\pi k /(N+1)) = -1 + \sum_{k=1}^N 1 = N-1$$