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I found exercise in "Introduction to algebra" Part I (A.I. Kostrikin)

Check expression $\sum_{k=1}^n\cot^2\frac{k\pi}{2n+1}=\frac{n(2n-1)}{3}$ for $n=1,2,3,4,5$.

For $n=1,2$ it is simple.

$\cot\frac{\pi}{3}=\frac{1}{\sqrt{3}}$, so $\left(\frac{1}{\sqrt{3}}\right)^2=\frac{1}{3}=\frac{1(2-1)}{3}$.

$\cot\frac{\pi}{5}=\sqrt{1+\frac{2}{\sqrt{5}}}$ and $\cot\frac{2\pi}{5}=\sqrt{1-\frac{2}{\sqrt{5}}}$, so $\left(\sqrt{1+\frac{2}{\sqrt{5}}}\right)^2+\left(\sqrt{1-\frac{2}{\sqrt{5}}}\right)^2=2=\frac{2\cdot3}{3}$.

But for bigger values i have a lot of problems. This is begin of book and I feel that should be simple way to resolve that. I found proof for one case here (Proof for $n=3$), but this is much more complicated than cases $n=1,2$. Is there any basic way to resolve that exercise? Without Chebyshev polynomials or Teoplitz matrixes?

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TheKwiatek666
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1 Answers1

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See this proof of the Basel problem; it proves your expression for ALL natural $n$, as well as the same expression with $\csc$ instead of $\cot$. I imagine if you used $(\cos x)^n$ in the denominator instead of $(\sin x)^n$ in the proof, you could also determine the value of the expression for $\tan$ and $\sec$ as well.

Pauly B
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