Compute $\sum_{k=1}^n \frac{1}{\sin^2(\frac{k\pi}{2n+1})}$ in terms of $n$.
Here's my way of doing it: using the classic expansion of $\sin((2n+1)x)$ as a polynomial in $\sin(x)$ we have: $$\sin((2n+1)x) = \sum_{k=0}^n \binom{2n+1}{2k+1}(-1)^k\cos^{2(n-k)}(x)\sin^{2k+1}(x)$$
As a result, the $\sin(\frac{k\pi}{2n+1})$ with $k\in\{-n,\ldots,n\}$ are roots of the polynomial $$\displaystyle P(X) = \sum_{k=0}^n \binom{2n+1}{2k+1}(-1)^k(1-X^2)^{n-k}X^{2k+1}$$
Note that $\deg P=2n+1$ and we have found $2n+1$ distinct roots.
Consequently, the $\displaystyle \frac{1}{\sin(\frac{k\pi}{2n+1})}$ with $k\in\{-n,\ldots,n\}\setminus\{0\}$ are roots of $Q$, the reciprocal polynomial of $P$ (Note that $\deg Q = 2n $).
Since we're interested in the sum of squares of the roots of $Q$, it is enough to look for the coefficients of $X^{2n}$ and $X^{2n-2}$ in $Q$, which are the coefficients of $X$ and $X^3$ in $P$.
A tedious computation shows that these coefficients are respectively $2n+1$ and $-\binom{2n+1}{3}-n(2n+1)$.
Noting that the roots of $Q$ are symmetric and using the standard expansion $\displaystyle \left(\sum_i x_i\right)^2 = \sum_i x_i^2 + 2\sum_{i<j} x_i$ yields $$\sum_{k=1}^n \frac{1}{\sin^2(\frac{k\pi}{2n+1})} = \frac 23 n(n+1) $$
The path I followed looks like it can be shortened. Is there a quicker way to find a suitable polynomial that annihilates all the $\displaystyle \frac{1}{\sin^2(\frac{k\pi}{2n+1})}$ ?
Is there a shorter way altogether ?
