Calculate the degree of the extension $[\mathbb{Q}(\cos(\frac{2\pi}{p})):\mathbb{Q}]$

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.