Using de-Moivres formula it is easy to show that $\sin(\frac{\pi}{180})$ and $\cos(\frac{\pi}{180})$ are algebraic numbers.
But how can I get the degree of their minimal-polynomials (It is $48$, which is $\phi(180)$, where $\phi$ denotes the totient-function) and how can I prove that both numbers have the same minimal polynomial ?
How can I prove that $\tan(\frac{\pi}{180})$ has a symmetric minimal polynomial with degree $24$ ?
I think, the problem is related to roots of unity.
The number $\zeta= e^{2\pi i/n}$ is a primitive $n$th root of unity. Its degree is given by the Euler's phi function $\phi(n)$.
Now $2\cos(2\pi/n)=\zeta+\bar \zeta$, being real is in the fixed field of complex conjugation automorphism: its degree is half that of $\zeta$. Same way for $\sin(2\pi/n)$.
Now $\tan(\pi/180)=\tan (2\pi/360)$. It is the ratio of two numbers of degree $\frac12\phi(360)=48$. It can be expressed $(\zeta-\bar\zeta)/(\zeta+\bar\zeta)$. Using this expression counting distinct conjugates we can prove its degree is 24.