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?

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: $m_{13}(x)=-1 + 6 x + 24 x^2 - 32 x^3 - 80 x^4 + 32 x^5 + 64 x^6$
