Radical representation of $\cos\frac {2\pi}{11}$

Question

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:

$$ \begin{aligned} \cos\frac{2\pi}{11}&=\frac{1}{2}\left(ω^1+ω^{10}\right)\\ \cos\frac{4\pi}{11}&=\frac{1}{2}\left(ω^2+ω^9\right)\\ \cos\frac{6\pi}{11}&=\frac{1}{2}\left(ω^3+ω^8\right)\\ \cos\frac{8\pi}{11}&=\frac{1}{2}\left(ω^4+ω^7\right)\\ \cos\frac{10\pi}{11}&=\frac{1}{2}\left(ω^5+ω^6\right)\\ \end{aligned} $$

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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.

Answer 1

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:

$$ \begin{bmatrix} x_1\\x_2\\x_3\\x_4\\x_5\\ \end{bmatrix}= \begin{bmatrix} ω^4 & ω & ω^4 & ω \\ ω & 1 & 1 & ω^4 \\ 1 & ω^3 & ω^2 & 1 \\ ω^3 & ω^4 & ω & ω^2 \\ ω^2 & ω^2 & ω^3 & ω^3 \\ \end{bmatrix}⋅ \begin{bmatrix} \sqrt[5]{ζ_1} \\ \sqrt[5]{ζ_2} \\ \sqrt[5]{ζ_3} \\ \sqrt[5]{ζ_4} \\ \end{bmatrix} $$

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So

$$ \begin{aligned} \cos\left(\frac{2kπ}{11}\right) = \frac{1}{10}\left(x_k-1\right) \end{aligned} $$

where

$$ ω^k = \cos\left(\frac{2kπ}{5}\right) +\sqrt{\cos ^2\left(\frac{2kπ}{5}\right)-1} $$

$$ \begin{aligned} ω^0 &= 1 \\ ω^1 &=+\frac{1}{4} \left(\sqrt{5}-1+i\sqrt{10+2\sqrt{5}}\right) \\ ω^2 &=-\frac{1}{4} \left(\sqrt{5}+1-i\sqrt{10-2\sqrt{5}}\right) \\ ω^3 &=-\frac{1}{4} \left(\sqrt{5}+1+i\sqrt{10-2\sqrt{5}}\right) \\ ω^4 &=+\frac{1}{4} \left(\sqrt{5}-1-i\sqrt{10+2\sqrt{5}}\right) \\ \end{aligned} $$

$$ \begin{aligned} ζ_1&=-\frac{11}{4} \left(89-25 \sqrt{5}-5 i \sqrt{410+178 \sqrt{5}}\right) \\ ζ_2&=-\frac{11}{4} \left(89+25 \sqrt{5}+5 i \sqrt{410-178 \sqrt{5}}\right) \\ ζ_3&=-\frac{11}{4} \left(89+25 \sqrt{5}-5 i \sqrt{410-178 \sqrt{5}}\right) \\ ζ_4&=-\frac{11}{4} \left(89-25 \sqrt{5}+5 i \sqrt{410+178 \sqrt{5}}\right) \\ \end{aligned} $$

Answer 2

It is true that, since that quintic has a cyclic Galois group, $\cos \frac{2 \pi}{11}$ can be expressed in terms of radicals. However, the expression is quite complicated. Wolfram's has a function called $\texttt{ToRadicals[]}$ that does the job. In this case, $\cos\frac{2\pi}{11} = A/B$, where

\begin{align*} A &= 116\ 2^{4/5} 11^{2/5}-4\ 2^{4/5} \sqrt{5} 11^{2/5}-40 i 2^{3/10} 11^{2/5} \sqrt{5\left(5+\sqrt{5}\right)}\\ &-2\ 2^{3/5} 11^{4/5} \left(-178-50 \sqrt{5}-65 i \sqrt{2\left(5+\sqrt{5}\right)}+25 i \sqrt{10\left(5+\sqrt{5}\right)}\right)^{2/5}\\ &+2\ 2^{3/5}\sqrt{5} 11^{4/5} \left(-178-50 \sqrt{5}-65 i \sqrt{2 \left(5+\sqrt{5}\right)}+25 i\sqrt{10\left(5+\sqrt{5}\right)}\right)^{2/5}\\ &-8 \left(-178-50 \sqrt{5}-65 i \sqrt{2\left(5+\sqrt{5}\right)}+25 i \sqrt{10 \left(5+\sqrt{5}\right)}\right)^{3/5}\\ &-2^{2/5}\sqrt[5]{11} \left(-178-50 \sqrt{5}-65 i \sqrt{2\left(5+\sqrt{5}\right)}+25 i \sqrt{10\left(5+\sqrt{5}\right)}\right)^{4/5}\\ &+2^{2/5} \sqrt{5} \sqrt[5]{11} \left(-178-50\sqrt{5}-65 i \sqrt{2 \left(5+\sqrt{5}\right)}+25 i \sqrt{10\left(5+\sqrt{5}\right)}\right)^{4/5}\\ &-36\ 11^{3/5} \sqrt[5]{2 i \left(178 i+50 i \sqrt{5}-65 \sqrt{2 \left(5+\sqrt{5}\right)}+25 \sqrt{10 \left(5+\sqrt{5}\right)}\right)}\\ &+4 \sqrt{5} 11^{3/5} \sqrt[5]{2 i \left(178 i+50 i \sqrt{5}-65 \sqrt{2 \left(5+\sqrt{5}\right)}+25 \sqrt{10 \left(5+\sqrt{5}\right)}\right)}\\ &+i \sqrt[10]{2} \sqrt[5]{11} \sqrt{5+\sqrt{5}}\Bigg( \\ &16 \sqrt[5]{22}-12\ 2^{3/5} 11^{2/5} \sqrt[5]{-178-50 \sqrt{5}-65 i \sqrt{2 \left(5+\sqrt{5}\right)}+25 i \sqrt{10 \left(5+\sqrt{5}\right)}}\\ &-4\ 11^{3/5} \left(-178-50 \sqrt{5}-65 i \sqrt{2 \left(5+\sqrt{5}\right)}+25 i \sqrt{10 \left(5+\sqrt{5}\right)}\right)^{2/5}\\ &+\left(2 \left(-178-50 \sqrt{5}-65 i \sqrt{2 \left(5+\sqrt{5}\right)}+25 i \sqrt{10 \left(5+\sqrt{5}\right)}\right)\right)^{4/5}\\ &\Bigg)\\ \end{align*}

and

$B = 80 \left(-178-50 \sqrt{5}-65 i \sqrt{2 \left(5+\sqrt{5}\right)}+25 i \sqrt{10 \left(5+\sqrt{5}\right)}\right)^{3/5}.$

Answer 3

The strange thing I've observed about some cyclic $5$-degs and $7$-degs (I'm not yet certain about the extent of this phenomenon) is that their resolvents can be factored. To factor a small composite integer is trivial, but to factor one quartic root into two quartic roots (or one sextic root into three sextic roots) is not easy.

For example, given the OP's

$$Z_1=-\frac{11}{4} \left(89-25 \sqrt{5}-5 \sqrt{-(410+178 \sqrt{5})}\right)$$

then we have the factorization,

$$-Z_1 = \frac{ab}{11}$$

$$a=\frac{11}{4}\left(-1+5\sqrt{5}+\sqrt{-10(5+\sqrt{5})}\right)$$

$$b=\frac{11}{4}\left(-31+5\sqrt{5}-\sqrt{-10(85+31\sqrt{5})}\right)$$

and we can use this factorization to solve, in an alternative manner, the minimal polynomial of $x=2\cos(2\pi/11)$ namely,

$$x^5 +x^4 −4x^3 −3x^2 +3x+1 = 0$$

Let $\beta_k = \big({-Z_1}\big)^{1/5}\, \zeta_5^k,\;$ for $k = 0,1,2,3,4.$ Then the five roots are,

$$x_k = -\frac{1}{5}\left(\frac{1}{\beta_k^{-1}}+\frac{1}{\beta_k^0}+\frac{11}{\beta_k^1}+\frac{a}{\beta_k^2}+\frac{b}{\beta_k^3} \right)$$

with fifth root of unity $\zeta_5 = e^{2\pi\,i/5}.$

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P.S. Thus, we only need one root of the resolvent quartic, and one fifth root extraction instead of the four in the usual method,

$$x = \frac15\left(A+Z_1^{1/5}+Z_2^{1/5}+Z_3^{1/5}+Z_4^{1/5}\right)$$

which can lead to an ambiguous real value when all the $Z_i$ are complex. For a similar solution for the case $p=31$ of $2\cos(2\pi/p)$, see this post.