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.
 A: 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 $\cos \frac{2\pi}{13}$.
A: Let $\zeta = \exp(2i\pi/13)$.
It is well known that $\Bbb Q(\zeta)$ is a Galois extension of $\Bbb Q$ whose Galois group is isomorphic to $(\Bbb Z/13\Bbb Z)^*$ via the maps induced by $a \mapsto (\sigma_a : \zeta \mapsto \zeta^a)$
By the fundamental theorem of Galois theory, its subgroups correspond to subfields of $\Bbb Q(\zeta)$. 
For starters, $x = 2\cos(2\pi/13) = \zeta+\zeta^{12}$ is invariant by $\{\sigma_1,\sigma_{12}\}$ so the Galois group of $L = \Bbb Q(x)$ is the quotient $G = (\Bbb Z/13\Bbb Z)^*/\{\pm1\}$, which is a cyclic group of order $6$. And so $x$ is a root of a degree $6$ polynomial over $\Bbb Q$.
In order to "decompose" the computation we can find an intermediate subfield of $L$ / subgroup of $G$ and split it in two steps.
For example, let $H = \{\pm 1, \pm 5\}$.
The conjugates of $x$ over $L^H$ are $x$ and $\sigma_5(x) = \zeta^5 + \zeta^8$.
So $x$ is a root of a polynomial of degree $2$ over $L^H$, which is $(X-x)(X-\sigma_5(x)) = X^2 - (\zeta+\zeta^5+\zeta^8+\zeta^{12})X + (\zeta^4+\zeta^6+\zeta^7+\zeta^9)$
Let $y_1 = \zeta+\zeta^5+\zeta^8+\zeta^{12} \in L^H$. Since $G/H$ has order $3$, it is a root of a polynomial of degree $3$ over $\Bbb Q$.
Its two conjugates are $y_2 = \sigma_2(y_1) = \zeta^2 + \zeta^3+\zeta^{10}+\zeta^{11}$ and $y_3 = \sigma_4(y_1) = \zeta^4+\zeta^6+\zeta^7+\zeta^9$
Now, they are a root of $(Y-y_1)(Y-y_2)(Y-y_3) = Y^3 - (\sum \zeta^k)Y^2 + 4(\sum \zeta^k)Y-(4+5\sum \zeta^k) = Y^3 + Y^2 - 4Y + 1$.
Furthermore, $y_3$ is in $\Bbb Q(y_1)$ so it has to be a polynomial expression in $y_1$. 
We compute $y_1^2 = 4 + y_2 + 2y_3$ and $y_2+y_3 = -1-y_1$,
which gives $y_3 = y_1^2+y_1-3$.
Putting everything together, this gives a way to compute $x$ by solving a degree $3$ and a degree $2$ equation :


*

*find a root $y_1$ of $Y^3+Y^2-4Y+1$.

*compute $y_3 = y_1^2+y_1-3$

*find a root $x$ of $X^2-y_1X+y_3$



Note that we could have chosen another subgroup $H' = \{\pm 1, \pm 3, \pm 4\}$ and it would give another method where first we solve a degree $2$ equation, and then a degree $3$ one. 
A third method (the best ?) is to use both of them at once :
We already know $y_1 = \sum_{a \in H} \sigma_a(\zeta)$ has minimal polynomial $Y^3+Y^2-4Y+1$.
Let $z_1 = \sum_{a \in H'} \sigma_a(\zeta)$. Its conjugate over $\Bbb Q$ is $z_2 = \sigma_2(z_1)$,
and $(Z-z_1)(Z-z_2) = Z^2-(\sum \zeta^k)Z+(3\sum \zeta^k) = Z^2+Z-3$
Since $\Bbb Q(y_1,z_1)$ is a subfield of $L$ and is not fixed by any element of $G$, it has to be $G$.
By computing $(1,y_1,y_1^2,z_1,y_1z_1,y_1^2z_1)$ in the basis $(\sigma_a(x))_{x\in G}$ of $L$
and then solving a linear system in $6$ dimensions, you get the expression of $x$ as a polynomial in $y_1,z_1$.
