I have recently found this excercise and was not able to solve it so far. Show that
$\sum_{i=1}^N\sin^2\frac{m\pi i}{N+1}=\frac{N+1}{2}\;,$ where $m \in \lbrace1,2,...,N\rbrace$.
This was one of my attempts:
$\sum_{i=1}^N\sin^2\frac{m\pi i}{N+1}=\sum_{i=1}^N\frac{1}{2}\left[\cos(0)-\cos\frac{2m\pi i}{N+1}\right]=\frac{1}{2}-\frac{1}{2}\sum_{i=1}^N\cos\frac{2m\pi i}{N+1}$
It seems the last sum should be equal to $-N$. Can anyone give me a clue?