Perhaps this question will not be considered appropriate for MO - so be it. But hear me out before you dismiss it as completely elementary.

As the question suggests, I would like to know when $\sin(p\pi/q)$ can be expressed in radicals (in the way that $\sin(\pi/4) = \sqrt{2}/2$ and $\sin(\pi/3) = \sqrt{3}/2$ can). Let $\alpha = \sin(x)$, and consider the field extension $\mathbb{Q}[\alpha]$. Using $(\cos(x) + i\sin(x))^k = \cos(kx) + i \sin(kx)$ together with the binomial formula and the Pythagorean identity relating sine and cosine, we can see that $\sin(kx)$ lies in a solvable extension of $\mathbb{Q}[\alpha]$. Thus $\sin(p\pi/q )$ is expressible in radicals if $\sin(\pi/q)$ is.

To handle $\sin(\pi/q)$, we start by using the same trick (which most people also learn in an elementary trig class). Write $-1 = (\cos(\pi/q) + i\sin(\pi/q))^q$, use the binomial theorem to expand, compare imaginary parts, and express the right-hand-side in terms of sine using the Pythagorean identity. This gives an explicit equation for any $q$ one of whose solutions is $\sin(\pi/q)$. This equation is not a polynomial in $\sin(\pi/q)$ since it involves terms of the form $\sqrt{1 - \sin^2(\pi/q)}$, but it is enough to prove that $\sin(\pi/q)$ is algebraic.

So I am curious about the number theoretic properties of this equation. What can be said about the Galois group of its "splitting field" over $\mathbb{Q}$? Can we at least determine when it is solvable? Note that if the prime factors of $q$ are $p_1, \ldots p_k$ and we can express each $\sin(\pi/p_j)$ in radicals, then the same is true for $\sin(\pi/q)$. So it suffices to consider the case where $q$ is prime. That's about all the progress I have made.