Minimal polynomial of $\cos\left(\frac{\pi}{13}\right)$ Find the Minimal polynomial of $\cos\left(\frac{\pi}{13}\right)$.
My try:
Let $$13 \theta=\pi \implies 9\theta=\pi-4\theta$$
$$\implies \cos(9\theta)=-\cos(4\theta)$$
Now Let $x=\cos(\theta)$
$$\implies 4(4x^3-3x)^3-3(4x^3-3x)=-(2(2x^2-1)^2-1) $$
$$\implies 256 x^9-576 x^7+432 x^5+16x^4-120 x^3-16 x^2+9x+2=0$$
Now how to test that this is Minimal?
 A: Note that $\cosθ=\frac12(\zeta+\zeta^{-1})$ where $\zeta\ne -1$ is the root closest to 1 in the positive quadrant of $\zeta^{13}=-1$. This is equivalent to
$$
0=ζ^{-6}-ζ^{-5}+ζ^{-4}\mp...-ζ^{-1}+1-ζ\pm...+ζ^6.
$$
On the other hand, powers of $\cosθ$ are symmetric (Laurent) polynomials in $ζ$ with the same highest degree. Thus above equation can be expressed as a polynomial in $\cosθ$ of degree $6$.
A: This is pretty standard, but I'll make a self-contained argument.
It suffices to find the minimal polynomial of $\alpha=-\cos(\frac{\pi}{13})=\cos(\pi-\frac{\pi}{13})=\cos\frac{12\pi}{13}$.
Note that $\zeta=\cos\frac{12\pi}{13}+i\sin\frac{12\pi}{13}=e^{2\pi i\frac{6}{13}}$ is a $13$-th primitive root of unity. And it can be easily shown that $\phi_{13}(x):=\sum_{i=0}^{12}x^i$ is irreducible by applying Eisenstein to $\phi_{13}(x+1)$, hence $[\mathbb Q(\zeta):\mathbb Q]=12$.
As $\alpha=\frac{\zeta + 1/\zeta}{2}$, we have $\zeta$ satisfies $\zeta^2+2\alpha\zeta +1=0$, hence $[\mathbb Q(\zeta):\mathbb Q(\alpha)]\le 2$. Since $\alpha$ is real, and $\zeta$ is not, we have $[\mathbb Q(\zeta):\mathbb Q(\alpha)]\ge 2$, hence $[\mathbb Q(\zeta):\mathbb Q(\alpha)]=2$, and finally $[\mathbb Q(\alpha):\mathbb Q]=\frac{[\mathbb Q(\zeta):\mathbb Q]}{[\mathbb Q(\zeta):\mathbb Q(\alpha)]}=6$.
To actually compute the polynomial, we may try to simplify $$\prod_{i=1}^6 (x+\frac{1}{2}(\zeta^i+\frac{1}{\zeta^i}))=\prod_{i=1}^6 (x+\frac{1}{2}(\zeta^i+\zeta^{13-i}))$$
A: Chebyshev polynomial of the first kind $T_p(x)$ satisfies $T_p(\frac{\zeta^{i}+\zeta^{-i}}{2})=1$ for all $0\leq i\leq p-1$ where $\zeta=e^{\frac{2\pi i}{p}}$.
So, the minimal polynomial of $\cos(\frac{2\pi i}{p})$ over integers is
$$m_p(x)=\sqrt{\frac{T_p(x)-1}{x-1}}.$$
But, I don't know its series expansion.
From WA: $m_{13}(x)=-1 + 6 x + 24 x^2 - 32 x^3 - 80 x^4 + 32 x^5 + 64 x^6$
