Nathen Peterson

2022-03-31

${\mathrm{cot}}^{2}\frac{\pi }{2m+1}+{\mathrm{cot}}^{2}\frac{2\pi }{2m+1}+\dots +{\mathrm{cot}}^{2}\frac{m\pi }{2m+1}=\frac{m\left(2m-1\right)}{3}$
m is a positive integer.

### Answer & Explanation

Mason Knight

Divide this equality by ${\mathrm{sin}}^{2m+1}\theta$ to get
$\frac{\mathrm{sin}\left(2m+1\right)\theta }{{\mathrm{sin}}^{2m+1}\theta }=\sum _{k=0}^{m}{\left(-1\right)}^{k}\left(\begin{array}{c}2m+1\\ 2k+1\end{array}\right){\mathrm{cot}}^{2k}\theta$ (1)
Denote
${P}_{m}\left(x\right)=\sum _{k=0}^{m}{\left(-1\right)}^{k}\left(\begin{array}{c}2m+1\\ 2k+1\end{array}\right){x}^{m}$
This is polynomial of degree m. From (1) we see that ${P}_{m}$ have m roots ${x}_{k}={\mathrm{cot}}^{2}\frac{\pi k}{2m+1}$. By Vietas formula their sum equals to
$-\frac{\left(\begin{array}{c}2m+1\\ 3\end{array}\right)}{\left(\begin{array}{c}2m+1\\ 1\end{array}\right)}=\frac{m\left(2m-1\right)}{3}$
So we get
$\sum _{k=1}^{m}{\mathrm{cot}}^{2}\frac{\pi k}{2m+1}=\frac{\left(2m-1\right)m}{3}$

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