I want to find radical representation of $\cos\dfrac {2\pi}{11}$.
My attempt
Consider the 11th root of unity:
$$ \begin{aligned} ω&= e^{2πi/11}\\ ω^n&=ω^{n\bmod 11} \end{aligned} $$
From Euler's formula:
$$ \begin{aligned} \cos\frac{2\pi}{11}&=\frac{1}{2}\left(ω^1+ω^{10}\right)\\ \cos\frac{4\pi}{11}&=\frac{1}{2}\left(ω^2+ω^9\right)\\ \cos\frac{6\pi}{11}&=\frac{1}{2}\left(ω^3+ω^8\right)\\ \cos\frac{8\pi}{11}&=\frac{1}{2}\left(ω^4+ω^7\right)\\ \cos\frac{10\pi}{11}&=\frac{1}{2}\left(ω^5+ω^6\right)\\ \end{aligned} $$
Define
$$ω^{a,b,c, \cdots} = ω^a + ω^b + ω^c + \cdots$$
Let
$$ \begin{aligned} σ_{0} &=ω^{\{1,2,3,4,5,6,7,8,9,10\}}=-1 \end{aligned} $$
and
$$ \begin{aligned} σ_{1}&=ω^{\{1,10\}}\\ σ_{2}&=ω^{\{2,9\}}\\ σ_{3}&=ω^{\{3,8\}}\\ σ_{4}&=ω^{\{4,7\}}\\ σ_{5}&=ω^{\{5,6\}}\\ \end{aligned} $$
With
$$ \begin{aligned} σ_1+σ_2+σ_3+σ_4+σ_5&=σ_0 \\ \sum_{\rm{cycle}}{σ_i σ_j}&=4σ_0 \\ \sum_{\rm{cycle}}{σ_i σ_j σ_k}&=10+7σ_0 \\ \sum_{\rm{cycle}}{σ_i σ_j σ_k σ_m}&=10+7σ_0 \\ σ_1 σ_2 σ_3 σ_4 σ_5&=2+3σ_0 \\ \end{aligned} $$
According to Vieta's formulas, $(σ_{1},σ_{2},σ_{3},σ_{4},σ_{5})$ is the root of $x^5+x^4-4 x^3-3 x^2+3 x+1$.
This is a solvable quintic equation, but I don't know how to solve the radical.
Consider
$$ \sum_{k=1}^5\left(10\cos\frac{2 k π i}{11}+1\right)=0 $$
Equivalent to solving the equation:
$$\frac{x^5}{11}=10x^3+5x^2-210x-89=0$$
Its 5 roots $x_i$ can be obtained by combining the 4 roots $ζ_i$ of the resolvent.
$$ x^4+89⋅11x^3+351⋅11^3x^2+89⋅11^6⋅x+11^{10}=0 $$
Let
$$ \begin{aligned} ω&= e^{2πi/5}\\ ω^n&=ω^{n\bmod 5} \end{aligned} $$
The roots can be obtained by:
$$ \begin{bmatrix} ζ_1\\ζ_2\\ζ_3\\ζ_4\\ \end{bmatrix}= 11\begin{bmatrix} 1 & ω & ω^4 & ω^2 \\ 1 & ω^2 & ω^3 & ω^4 \\ 1 & ω^3 & ω^2 & ω \\ 1 & ω^4 & ω & ω^3 \\ \end{bmatrix}⋅ \begin{bmatrix} -6\\35\\10\\20\\ \end{bmatrix} $$
So the solution of the original quintic equation are:
$$ \begin{bmatrix} x_1\\x_2\\x_3\\x_4\\x_5\\ \end{bmatrix}= \begin{bmatrix} ω^4 & ω & ω^4 & ω \\ ω & 1 & 1 & ω^4 \\ 1 & ω^3 & ω^2 & 1 \\ ω^3 & ω^4 & ω & ω^2 \\ ω^2 & ω^2 & ω^3 & ω^3 \\ \end{bmatrix}⋅ \begin{bmatrix} \sqrt[5]{ζ_1} \\ \sqrt[5]{ζ_2} \\ \sqrt[5]{ζ_3} \\ \sqrt[5]{ζ_4} \\ \end{bmatrix} $$
So
$$ \begin{aligned} \cos\left(\frac{2kπ}{11}\right) = \frac{1}{10}\left(x_k-1\right) \end{aligned} $$
where
$$ ω^k = \cos\left(\frac{2kπ}{5}\right) +\sqrt{\cos ^2\left(\frac{2kπ}{5}\right)-1} $$
$$ \begin{aligned} ω^0 &= 1 \\ ω^1 &=+\frac{1}{4} \left(\sqrt{5}-1+i\sqrt{10+2\sqrt{5}}\right) \\ ω^2 &=-\frac{1}{4} \left(\sqrt{5}+1-i\sqrt{10-2\sqrt{5}}\right) \\ ω^3 &=-\frac{1}{4} \left(\sqrt{5}+1+i\sqrt{10-2\sqrt{5}}\right) \\ ω^4 &=+\frac{1}{4} \left(\sqrt{5}-1-i\sqrt{10+2\sqrt{5}}\right) \\ \end{aligned} $$
$$ \begin{aligned} ζ_1&=-\frac{11}{4} \left(89-25 \sqrt{5}-5 i \sqrt{410+178 \sqrt{5}}\right) \\ ζ_2&=-\frac{11}{4} \left(89+25 \sqrt{5}+5 i \sqrt{410-178 \sqrt{5}}\right) \\ ζ_3&=-\frac{11}{4} \left(89+25 \sqrt{5}-5 i \sqrt{410-178 \sqrt{5}}\right) \\ ζ_4&=-\frac{11}{4} \left(89-25 \sqrt{5}+5 i \sqrt{410+178 \sqrt{5}}\right) \\ \end{aligned} $$