How prove that $\frac{1}{\sin^2\frac{\pi}{2n}}+\frac{1}{\sin^2\frac{2\pi}{2n}}+\cdots+\frac{1}{\sin^2\frac{(n-1)\pi}{2n}} =\frac{2}{3}(n-1)(n+1)$

Question

How prove that sum $$\frac{1}{\sin^2\frac{\pi}{2n}}+\frac{1}{\sin^2\frac{2\pi}{2n}}+\cdots+\frac{1}{\sin^2\frac{(n-1)\pi}{2n}} =\frac{2}{3}(n-1)(n+1)$$

This will be helpful: http://math.stackexchange.com/questions/544228/is-sum-k-1m-1-frac1-sin2-frack-pim-fracm2-13-true-for-m?rq=1 — Jean-Claude Arbaut, Apr 25 '15 at 10:48

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

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To develop further on Jean-Claude Arbaut's comment, you'll find here the proof that: $$\sum_{k=1}^{n-1}{\frac{1}{\sin^2\left(\frac{k\pi}{n}\right)}}=\dfrac{(n-1)(n+1)}{3}$$ All you have to do now, is show that: $$\sum_{k=1}^{n-1}{\frac{1}{\sin^2\left(\frac{k\pi}{\color{red}{2n}}\right)}}=\color{red}{2}\sum_{k=1}^{n-1}{\frac{1}{\sin^2\left(\frac{k\pi}{\color{red}{n}}\right)}}$$

edited Apr 13 '17 at 12:21

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answered Apr 25 '15 at 11:11

Demosthene

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How prove that $\frac{1}{\sin^2\frac{\pi}{2n}}+\frac{1}{\sin^2\frac{2\pi}{2n}}+\cdots+\frac{1}{\sin^2\frac{(n-1)\pi}{2n}} =\frac{2}{3}(n-1)(n+1)$

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