Degree of extension $\mathbb{Q}(\cos(2\pi/p),\sin(2\pi/p))/\mathbb{Q}(\cos(2\pi/p))$. Let $p$ be an odd prime number. I want to compute the degree
$$
[\mathbb{Q}(\cos(2\pi/p),\sin(2\pi/p)):\mathbb{Q}(\cos(2\pi/p))].
$$
I already showed that
$$
[\mathbb{Q}(\cos(2\pi/p)):\mathbb{Q}] = \frac{p-1}{2},
$$
so it will be helpful to know
$$
[\mathbb{Q}(\cos(2\pi/p),\sin(2\pi/p)):\mathbb{Q}].
$$
We know that the polynomial
$$
X^2+\cos^2(2\pi/p)-1 \in \mathbb{Q}(\cos(2\pi/p))[X]
$$
vanish on $\sin(2\pi/p)$, but I don't how to show that this polynomial is irreducible over $\mathbb{Q}(\cos(2\pi/p))$. I believe that 
$$
[\mathbb{Q}(\cos(2\pi/p),\sin(2\pi/p)):\mathbb{Q}(\cos(2\pi/p))]=2,
$$
but I cannot see why it is possible or not to write
$$
\sin(2\pi/p) = \sum_{k=0}^{n} a_k\cos^k(2\pi/p)
$$
for some $a_0,a_1,\dots,a_n\in\mathbb{Q}$. 
Any help will be appreciated!
 A: With $\zeta_p$ the p-th primitive root of unity we have that $[\mathbb{Q}(\zeta_p,i):\mathbb{Q}]$ is equal to:
$$[\mathbb{Q}(\zeta_p,i):\mathbb{Q}(\cos2\pi/p,\sin2\pi/p)][\mathbb{Q}(\cos2\pi/p,\sin2\pi/p):\mathbb{Q}(\cos2\pi/p)][\mathbb{Q}(\cos2\pi/p):\mathbb{Q}]$$
Now, since $i$ cannot be in $\mathbb{Q}(\zeta_p)$ (why?), we have that  $[\mathbb{Q}(\zeta_p,i):\mathbb{Q}(\zeta_p)]=2$ because $i$ is the root of the an irreducible quadratic. Since $[\mathbb{Q}(\zeta_p):\mathbb{Q}]=p-1$, we see that the entire expression above is equal to $2(p-1)$. Note that $[\mathbb{Q}(\zeta_p,i):\mathbb{Q}(\cos2\pi/p,\sin2\pi/p)]$ is degree $2$ because $\mathbb{Q}(\zeta_p,i)=\mathbb{Q}(\cos2\pi/p,\sin2\pi/p,i)$ (why?) and $\mathbb{Q}(\cos2\pi/p,\sin2\pi/p)$ is real. 
Putting this all together, and by your previous computation we have that $2(p-1)=2[\mathbb{Q}(\cos2\pi/p,\sin2\pi/p):\mathbb{Q}(\cos2\pi/p)]\frac{p-1}{2}$ from which the result follows.
A: Let $\zeta = \exp(\frac{2\pi i}{4p})$. Then $\zeta$ is a primitive $(4p)$-th root of unity, and since $2p-1$ is coprime to $4p$, it follows that there is a Galois automorphism (let us call it $\sigma$) of ${\mathbb Q}(\zeta)$ such that $\sigma(\zeta)=\zeta^{2p-1}$.
From Euler's formula we deduce
$$
\cos\bigg(\frac{2\pi}{p}\bigg)=\frac{\zeta^4-\zeta^{2p-4}}{2}, \
\sin\bigg(\frac{2\pi}{p}\bigg)=\frac{\zeta^{p-4}-\zeta^{p+4}}{2} \tag{1}
$$
Also, straightforward computations show that
$$
\sigma(\zeta^4)=-\zeta^{2p-4},\sigma(\zeta^{2p-4})=-\zeta^{4},
\sigma(\zeta^{p-4})=\zeta^{p+4},\sigma(\zeta^{p+4})=\zeta^{p-4}, \tag{2}
$$
Combining (1) with (2), we see that $\sigma$ fixes $\cos\big(\frac{2\pi}{p}\big)$ and sends $\sin\big(\frac{2\pi}{p}\big)$ to its opposite. It follows that $\sin\big(\frac{2\pi}{p}\big)\not\in{\mathbb Q}\big(\cos\big(\frac{2\pi}{p}\big)\big)$.
