I know that $\cos(\pi/n)$ is a root of the Chebyshev polynomial $(T_n + 1)$, in fact it is the largest root of that polynomial, but often that polynomial factors. For example, if $n = 2 k$ then $\cos(\pi/n)$ is the largest root of $T_k$, which is a polynomial of lower degree, and if $n = 3$ then $\cos(\pi/n)$ is a root of $2 x - 1$, again lower degree than $T_3 + 1$.

How can I compute, for a given $n$, a polynomial of minimal degree that $\cos(\pi/n)$ is a root of?

I know that $\cos(\pi/n)$ is a root of the Chebyshev polynomial $(T_n + 1)$, in fact it is the largest root of that polynomial, but often that polynomial factors. For example, if $n = 2 k$ then $\cos(\pi/n)$ is the largest root of $T_k$, which is a polynomial of lower degree, and if $n = 3$ then $\cos(\pi/n)$ is a root of $2 x - 1$, again lower degree than $T_3 + 1$.

How can I compute, for a given $n$, a polynomial in $\mathbb{Q}[x]$ of minimal degree that $\cos(\pi/n)$ is a root of?