Jack d'Aurizio's exercise on Chebyshev polynomials

Question

I am working through Jack D'Aurizio's “Superior Mathematics from an Elementary point of view”, and I found (Lemma 61) the following lemma: $\sum_{k=1}^{n-1}\frac{1}{\sin^2(\pi k/n)}=(n^2-1)/3$. He does not provide a proof, but says that it follows by considering the roots of the Chebyshev polynomial $U_n$ or $T_n$. I know that this is about relating the roots to the coefficients, but what what bothers me is the that the sin terms (closely related to the roots) appears in the denumerator - I can solve the other problems in Lemma 61. Could someone point out (as a hint) how to reformulate the sum in a more manageable way? Thanks!

Answer 1

Hint

Recall that if $p(x)=x^n+p_1 x^{n-1} + \cdots + x p_{n-1} + p_n$ and $(r_i)_{i=0}^{n-1}$ are the roots of $p$, then $$\sum_{i=0}^{n-1} \frac{1}{r_i}= \frac{p_{n-1}}{p_n}$$

Of course provided that none of the roots is $0$, that is $p_n \neq 0$.