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Prove that $$ \sum_{k=1}^m \cot^2 \frac{k\pi}{2m+1}=\frac{m(2m-1)}{3} $$

I have tried to use $$\sin\left((2m+1)x\right)= \left(\sin^{2m+1}x\right) \cdot \left(\sum_{j=0}^m (-1)^j \binom{2m+1}{2j+1}\left(\cot^2x\right)^{m-j}\right)$$ and induction without any success. Thanks for any help!

rtybase
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user64066
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    A few good hints here – rtybase Jan 07 '19 at 23:13
  • Another hint contained in this question (the polynomial development part, see the question, not the answer). – rtybase Jan 07 '19 at 23:32
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    I guess you can find a copy of Cauchy's Cours d'Analyse - this is the key ingredient of Cauchy's proof of $\zeta(2)=\frac{\pi^2}{6}$. It can be proved in many ways, for instance by applying Vieta's formulas to Chebyshev polynomials of the second kind. You may have a look at my notes too, in the section about the Basel problem. – Jack D'Aurizio Jan 07 '19 at 23:40

2 Answers2

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Hint: The $-i\cot\left(\frac{k\pi}{2m+1}\right)$, for $1 \leq |k| \leq m$, are the distinct roots of $(X+1)^{2m+1}-(X-1)^{2m+1}$.

Aphelli
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Further to my last comment and given you used $$\sin\left((2m+1)x\right)= \left(\sin^{2m+1}x\right) \left(\sum_{j=0}^m (-1)^j \binom{2m+1}{2j+1}(\cot^2x)^{m-j}\right)= \left(\sin^{2m+1}x\right) \cdot P\left(\cot^2{x}\right)\tag{1}$$ where $$P(x)=\sum_{j=0}^m (-1)^j \binom{2m+1}{2j+1}x^{m-j}=\binom{2m+1}{1}x^m-\binom{2m+1}{3}x^{m-1}+...$$ is a polynomial of degree $m$, with (easy to see from $(1)$ since $\sin\left((2m+1)\frac{k\pi}{2m+1}\right)=0$) $\cot^2{\frac{k\pi}{2m+1}}, k=1..m$ as roots. Using Vieta's formulas $$\sum_{k=1}^m \cot^2{\frac{k\pi}{2m+1}}=-\frac{-\binom{2m+1}{3}}{\binom{2m+1}{1}}=\frac{(2m+1)(2m)(2m-1)}{(2m+1)\cdot 2 \cdot 3}=\frac{m(2m-1)}{3}$$

rtybase
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