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

Nathen Peterson

Nathen Peterson

Answered question


m is a positive integer.

Answer & Explanation

Mason Knight

Mason Knight

Beginner2022-04-01Added 11 answers

Divide this equality by sin2m+1θ to get
sin(2m+1)θsin2m+1θ=k=0m(1)k(2m+12k+1)cot2kθ (1)
This is polynomial of degree m. From (1) we see that Pm have m roots xk=cot2πk2m+1. By Vietas formula their sum equals to
So we get

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