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

Nathen Peterson

Nathen Peterson

Answered question

2022-03-31

cot2π2m+1+cot22π2m+1++cot2mπ2m+1=m(2m1)3
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)
Denote
Pm(x)=k=0m(1)k(2m+12k+1)xm
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
(2m+13)(2m+11)=m(2m1)3
So we get
k=1mcot2πk2m+1=(2m1)m3

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