Help understanding derivation of identity $\sum_{k=1}^{n}\cot^4\left({k\pi\over 2n+1}\right)=\frac{1}{45}n(2n-1)(4n^2+10n-9)$

Question

This question regards understanding some of the steps in the derivation of the identity for $\sum_{k=1}^{n}\cot^4\left({k\pi\over 2n+1}\right)$

It is shown 1 that using Vieta's formula that

$\sum_{k=1}^{n}\cot^4\left({k\pi\over 2n+1}\right)=\frac{1}{45}n(2n-1)(4n^2+10n-9)$

Since $\cot\left({k\pi\over 2n+1}\right)=\frac{i (\omega^k+1)}{\omega^k-1} .\tag{1}$ It follows that $\sum_{k=1}^{n}\cot^4\left({k\pi\over 2n+1}\right)=\sum_{k=1}^{n}(\frac{ \omega^k+1}{\omega^k-1})^4 .\tag{2}$

The source 1 goes on to say $P(x)=\frac{x^{2n+1}-1}{x-1} .\tag{3}$

will have roots $\omega,\omega^2..\omega^{2n}$ but I cannot understand why and how (4) is derived using $y=\frac{x+1}{x-1},x=\frac{y+1}{y-1}$?

$P(x)=\frac{(\frac{y+1}{y-1})^{2n+1}-1}{\frac{y+1}{y-1}-1} .\tag{4}$

1 brilliant

Answer 1

That is the polynomial transformation of the roots to evaluate the sum of the fourth powers of $\frac{w^k+1}{w^k-1}=\frac{x_k+1}{x_k-1}=y_k$, where $k=1,2,\ldots,n$. To apply Vieta's formulas for the sum, it is necessary to transform the polynomial and roots. If $x=w,\ w^2,\ w^3,\ldots$ are the roots of the polynomial $P(x)$, then $y=\frac{w+1}{w-1},\ \frac{w^2+1}{w^2-1},\ \frac{w^3+1}{w^3-1},\ldots$ will be the roots of the polynomial $\mathcal P(y)=P\left(\frac{y+1}{y-1}\right)$ as $\frac{y+1}{y-1}=x\implies y=\frac{x+1}{x-1}$. You can apply Vieta's formulas to the polynomial $\mathcal P(y)$ to evaluate the required sum. I hope it clarifies and helps you understand it.