<s>[INST] Explain how to compute $\cos(2\pi/13)$ by solving quadratic and cubic equations only I know that we can express $2\pi/13$ as a root of unity on the unit circle taking $z^{13} = 1$ and $z=\cos(2\pi/13)+i\sin(2\pi/13)$ and that we should be able to find a polynomial for this over the rational numbers, and because its not constructible it will consist of irreducible polynomials of degrees 2 and 3. But I'm not sure of any method to go about doing this.
 [/INST] You can use the triple angle and the quadruple angle formula for cosine.
\begin{align}
&\cos x = \cos \frac{2\pi}{13}\\
&\cos 4x = 8 \cos^4 x - 8 \cos^2 x + 1\\
&\cos 3x = 4 \cos^3 x - 3 \cos x\\
&\cos 9x = 4 (4 \cos^3 x - 3 \cos x)^3 - 3 (4 \cos^3 x - 3 \cos x)\\
&\cos (4x+9x) = \cos 4x \cos 9x - \sin 4x \sin 9x \\
&= \cos 4x \cos 9x - \sqrt{1-\cos^2 4x} \sqrt{1-\cos^2 9x} = 1\\
&(\cos 4x \cos 9x -1)^2= (1-\cos^2 4x)(1-\cos^2 9x)\\
&\cos^2 4x + \cos^2 9x - 2\cos 4x \cos 9x = (\cos 4x -\cos 9x)^2=0\\
&\cos 4x = \cos 9x\
\end{align}
It was only after I reached the above line that I realized
$$4\times \frac{2\pi}{13}+9\times \frac{2\pi}{13}=2\pi$$
So we could have obtained $\cos 4x = \cos (2\pi-4x)= \cos 9x$ directly.
Anyway, let's keep going to find the equation
\begin{align}
&\cos x \mapsto y\\
&8y^4-8y^2+1=4(4y^3-3y)^3-3(4y^3-3y)\\
&(y-1)(4y^2+2y-1)(64y^6+32y^5-80y^4-32y^3+24y^2+6y-1)=0
\end{align}
$y-1=0$ is for the trivial solution of $x=0$.
 And $4y^2+2y-1$ must be coming from one of those steps of squaring and tripling polynomials to get $\cos 3x$ and $\cos 4x$.
So the final equation you're looking for is
$$64y^6+32y^5-80y^4-32y^3+24y^2+6y-1=0$$
that has 6 real roots and one of them is the $\cos \frac{2\pi}{13}$.
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