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Let $m\geq 2$ be an integer, then there is the well known formula $$\sum\limits_{j=1}^{m-1}\frac{1}{\sin^2(\frac{j\pi}{m})}=\frac{m^2-1}{3},$$

I'm interested in similar equations for the following expression

$$\sum\limits_{j=1}^{m-1}\frac{1}{\sin^{2p}(\frac{j\pi}{m})}=(...),$$

for any $p\in\mathbb{N}$. Does someone know whether there are well known formulas like for $p=1$?

Best wishes

Braten
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    Each side is of the form $$\sum_1^{\infty}{c_r\over r^2}$$ for some integer coefficient $c_r$ which counts the (net) number of times the term $1/r^2$ appears on that side. So maybe you can show that $c_r$ is the same on the left side as on the right. – Gerry Myerson Jun 19 '15 at 11:00
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    Taking $\frac{d^2}{dx^2}\log$ of the Weierstrass sine product gives $\sum_{k \in \mathbb{z}} (mk-j)^{-2}=\pi^2 m^{-2} \csc^2(\pi j/m)$ so it suffices to prove $$\sum_{j=1}^{m-1} \frac{1}{\sin^2(\frac{\pi j}{m})} = \frac{m^2-1}{3}$$ – Noam Shalev - nospoon Jun 19 '15 at 11:02
  • @Gerry Myerson: I had this idea, but I could not prove it by this idea. But I will try it again, maybe I had a mistake in my calculations. – Braten Jun 19 '15 at 11:15
  • @nospoon: actually the equality which I asked for is used in my book to prove the equality you wrote down. By the way, I'm interested in equations of the type $\sum\limits_{j=1}^{m-1}\frac{1}{sin^(2p)(\frac{j\pi}{m})}=(...)$ for any $p\in\mathbb{N}$ are there formulas known for that? – Braten Jun 19 '15 at 11:19
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    A related question http://math.stackexchange.com/questions/544228/is-sum-k-1m-1-frac1-sin2-frack-pim-fracm2-13-true-for-m/544488#544488 – user91500 Jun 19 '15 at 11:50

1 Answers1

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Your sums may be seen (with a minor fix) as $$ \sum_{\zeta\in Z} \zeta^{-2p} $$ where $Z$ is the set given by the roots of a Chebyshev polynomial of the second kind. By Newton's identities, in order to compute our power sums it is enough to compute the right number of coefficients of a Chebyshev polynomial of the second kind. In order to do that, we just need to exploit the binomial theorem, giving: $$ U_n(x)=\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n-k}{k}(-1)^k (2x)^{n-2k}.$$

Jack D'Aurizio
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