Calculate the degree of the extension $[\mathbb{Q}(\cos(\frac{2\pi}{p})):\mathbb{Q}]$ 
Calculate the degree of the extension $[\mathbb{Q}(\cos(\frac{2\pi}{p})):\mathbb{Q}]$ where $p$ is a prime number.

My thoughts are:


*

*I am lost

*My intuition says it has to be $ \frac{p-1}{2}$ and that the base vectors are probably $\{\cos(\frac{2n\pi}{p}) \mid 0 \le n \le p-1 \}$, notice  $\cos(\frac{2n\pi}{p})=\cos(\frac{2(p-n)\pi}{p})$.


I don't know how to prove/calculate it.
 A: If we accept as given the fact that
$[\Bbb Q(\zeta): \Bbb Q] = p - 1, \tag{1}$
where $\zeta = \cos (2\pi / p) + i\sin (2\pi/p)$, we may proceed as follows:
since, as the comments point out,
$\cos \dfrac{2\pi}{p} = \dfrac{\zeta + \zeta^{-1}}{2}, \tag{2}$
we infer that $\cos (2\pi / p) \in \Bbb Q(\zeta)$; thus $Q(\cos (2\pi / p))$ is a real subfield of $\Bbb Q(\zeta)$; we have the tower of fields
$\Bbb Q \subset \Bbb Q(\cos \dfrac{2\pi}{p}) \subset \Bbb Q(\zeta); \tag{3}$
from (3) we conclude that
$[\Bbb Q(\zeta): \Bbb Q(\cos\dfrac{2\pi}{p})][\Bbb Q(\cos \dfrac {2\pi}{p}):\Bbb Q] = [\Bbb Q(\zeta): \Bbb Q] = p - 1; \tag{4}$
next, observe:
$\zeta = \cos \dfrac{2\pi}{p}  + i \sin \dfrac{2\pi}{p}. \tag{5}$
whence
$(\zeta - \cos \dfrac{2\pi}{p})^2 = -\sin^2 \dfrac{2\pi}{p} = \cos^2 \dfrac{2\pi}{p} - 1, \tag{6}$
or
$\zeta^2 - 2\cos \dfrac{2\pi}{p} \zeta + 1 = 0; \tag{7}$
$\zeta$ thus satisfies the quadratic polynomial
$q_\zeta(x) = x^2 -  2\cos \dfrac{2\pi}{p} x + 1 \in \Bbb Q(\cos \dfrac{2\pi}{p})[x]; \tag{8}$
the zeroes of $q_\zeta(x)$ are in fact $\zeta$ and $\bar \zeta$, as may easily seen by taking the complex conjugate of (7); neither lies in $\Bbb Q (\cos (2\pi/p))$ since it is a real subfield of $\Bbb Q(\zeta)$; thus $q_\zeta(x)$ is irreducible over $\Bbb Q(\cos (2\pi /p))$  and hence
$[\Bbb Q(\zeta): \Bbb Q(\cos \dfrac{2\pi}{p})] = 2; \tag{9}$
using this fact in (4) yields
$[\Bbb Q(\cos \dfrac{2\pi}{p}): \Bbb Q] = \dfrac{p - 1}{2}, \tag{10}$
the answer sought.
