TITLE: Proving that ${n+3\choose 3} =\frac{n+2}{2}\sum_{k=1}^{n+1}\csc^2\frac{k\pi}{n+2}$
QUESTION [8 upvotes]: Fancy physics predicts the equality
$${n+3\choose 3} =\frac{n+2}{2}\;\sum_{k=1}^{n+1}\csc^2\frac{k\pi}{n+2}$$
which I can check (numerically and symbolically) for small $n$, but cannot prove for every $n$.
Does someone see an elegant way of doing this? Expressing $\csc$ in terms of exponentials allows one to see this as a sum involving roots of unity, but  I don't see how to proceed.

REPLY [5 votes]: Using parts of this question and answers
$$\sum_{k=1}^{n+1}\frac{1}{\sin^2\left(\frac{k\pi}{n+2}\right)}=
\sum_{k=1}^{n+1}\left(1+\cot^2\left(\frac{k\pi}{n+2}\right)\right)=\\
n+1+\sum_{k=1}^{n+1}\left(\cot^2\left(\frac{k\pi}{n+2}\right)\right)=n+1+\frac{(n+1)n}{3}=\frac{(n+3)(n+1)}{3}$$
Note: Jack's answer, with a link to wikipedia, touches this question/answers as well.