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?



Minimal polynomial of cos(π/n)
[The minimal polynomial of $\cos(2\pi/n)$](https://www.jstor.org/stable/2324301?seq=1#page_scan_tab_contents) (by William Watkins and Joel Zeitlin, The American Mathematical Monthly
Vol. 100, No. 5 (May, 1993), pp. 471-474) has full clarity on this matter (just take their result for even $n$ to resolve your case). 


