# Minimal polynomial of a trigonometric number

I am trying to calculate the minimal polynomials of $h_{1}=-\cos(\pi/n)-\sqrt{\cos(2\pi/n)}$ and $h_{2}=-\cos(\pi/n)+\sqrt{\cos(2\pi/n)}$ when $n$ is odd. I think (and numerical calculations suggest that) these two have the same minimal polynomial.

I tried the calculation $(x-h_{1})(x-h_{2})=x^{2}+2x\cos(\pi/n)+\sin^{2}(\pi/n)$ which upon simplification is just $$x^{2}+\left(\zeta_{n}+\zeta_{n}^{-1}\right)x-\frac{1}{4}\left(\zeta_{n}^{2}-2+\zeta_{n}^{-2}\right)$$ where $\zeta_{n}=e^{\pi i/n}.$

So the minimal polynomial $p_{n}(x)$ will have the above quadratic as a factor in $C[x]$. How do we get these minimal polynomials?

Calculations in mathematica suggest that the minimal polynomials are non-trivial factors of polynomials generated by the rational function $$J(x,t):=\frac{1+(4x-1)t+(4x-1)t^{2}+t^{3}}{1-4(1+x^{2})t+6(1+4x^{2})t^{2}-4(1+x^{2})t^{3}+t^{4}}$$

Notice that $-\cos(\pi/n)$ is a zero of $T_{2n}(x)-1$, where $T_{2n}$ is the $2n$-th Chebyshev polynomial of the first kind.
Similarly, $\sqrt{\cos(2\pi/n)}$ is a zero of $T_n(x^2)-1$.
Then the sum $-\cos(\pi/n)+\sqrt{\cos(2\pi/n)}$ is a zero of the resultant $$\mathrm{Res}_x(T_{2n}(x)-1,T_n((y-x)^2)-1).$$ Then it remains to factor the resultant and find the factor that has zero $-\cos(\pi/n)+\sqrt{\cos(2\pi/n)}$.
For example, for $n=7$, we'll get the minimal polynomial $$64y^6 + 64y^5 - 16y^4 + 48y^3 - 56y^2 + 56y + 7.$$
In the same vein as Max Alekseyev's answer: if $q = \exp(i \pi/n)$ and $z = -\cos(\pi/n) \pm \sqrt{\cos(2\pi/n)}$, we have ${q}^{4}-4\,{q}^{3}z-4\,{q}^{2}{z}^{2}-2\,{q}^{2}-4\,qz+1=0$. $q$ is a root of the cyclotomic polynomial $C_{2n}$, so $z$ is a root of the resultant of ${Q}^{4}-4\,{Q}^{3}Z-4\,{Q}^{2}{Z}^{2}-2\,{Q}^{2}-4\,QZ+1$ and $C_{2n}(Q)$ with respect to $Q$. For at least the first $200$ integers $n \ge 2$ (and I suspect for all of them), this resultant is of the form $p(Z)^2$ where $p$ is irreducible over the rationals (and thus $p(Z)$ is the minimal polynomial of $z$).