polynomial of minimal degree $p(x)$ over $\mathbb{Q}$ such that $\beta=\cos(20)$ and $p(\beta)=0$.

Question

how does one find the polynomial of minimal degree $p(x)$ over $\mathbb{Q}$ such that $\beta=\cos(20)$ (in degrees, not in rad) and $p(\beta)=0$.

I know that the answer is $p(x)=8x^3-6x-1$, (irreducible in $\mathbb{Q}$) but I wonder how one would get this result.

Answer 1

hint: use the following trig identity $$ \cos 3\theta = 4\cos^3 \theta - 3 \cos \theta $$ if you set $\theta=\frac{\pi}{9}$ (20 degrees) then $\cos 3\theta = \frac12$