Is $\mathbb{Q}(\cos\frac{2\pi}{13})/\mathbb{Q}$ a Galois extension? I've been solving problems from my Galois Theory course, and I got stuck in this problem:

Calculate the number of subfields of $\mathbb Q(\cos\frac{2\pi}{13})$

What I considered doing is finding the Galois group of $\mathbb Q(\cos\frac{2\pi}{13})/\mathbb Q$, to the find its subgroups (and that number equals the number of subfields of $\mathbb Q(\cos\frac{2\pi}{13})$). First I proved that
$$\left[\mathbb Q(\cos\frac{2\pi}{13}):\mathbb{Q}\right]=6,$$
it was easy using that $\mathbb{Q}(e^{\frac{2\pi i}{13}})/\mathbb Q$ has degree $12$. Now, what I wanted to check is if the extension $\mathbb Q(\cos\frac{2\pi}{13})/\mathbb Q$ is normal, to then conclude it's a Galois extension (and that would let me with just the work of proving if it's isomorphic to $\mathbb Z_6$ or to $S_3$). My problem is that I don't know how can I prove this extension is normal (if it is, I'm not really sure). Maybe finding the minimal polynomial of $\cos\frac{2\pi}{13}$ would make it easy, but I don't know how to find this polynomial.
How can I prove if this extension is normal or not? How can I find the minimal polynomial of $\cos\frac{2\pi}{13}$ over $\mathbb Q$ to find it's roots? Any help will be appreciated, thanks in advance.

Let $\zeta:=e^{\frac{2\pi i}{13}}$ and consider the Galois extension
$\mathbb Q(\zeta)/\mathbb Q$. It has cyclic Galois group of order 12:
$Gal(\mathbb Q(\zeta)/\mathbb Q)= \mathbb Z/ 12 \mathbb Z$ (https://en.wikipedia.org/wiki/Cyclotomic_field).
Since $\mathbb Z/ 12 \mathbb Z$ is abelian, its subgroups are all normal, whence
by the fundamental theorem of Galois theory every subextension $\mathbb Q(\zeta)/K/\mathbb Q$ is normal. In particular,
$\mathbb Q(\cos\frac{2\pi i}{13})/\mathbb Q$ is Galois with
Galois-group cyclic of order 6, which has 4 subgroups (2 proper).
