When is sin(r \pi) expressible in radicals for r rational?

As $\cos x=\pm\sqrt{1-\sin^2 x}$ and $e^{ix}=\cos x +i\sin x$, and $\sin x=(e^{ix}-1/e^{ix})/2i$ then $\sin x$ is in a radical extension of $\mathbb{Q}$ iff $e^{ix}$ is. For rational $r$ with denominator $d$, $e^{2\pi i r}$ is a primitive $d$-th roots of unity. The extension of $\mathbb{Q}$ generated by a root of unity is a cyclotomic field. Every cyclotomic field is an abelian extension of $\mathbb{Q}$. (By the Kronecker-Weber theorem any abelian extension is contained in a cyclotomic field.) The Galois group of the $n$-th cyclotomic field is isomorphic to the multplicative group $(\mathbb{Z}/n\mathbb{Z})^*$

So we can obtain any $\sin r\pi$ for $r$ rational in terms of radicals, both in a trivial way if you allow an $n$-th root of unity as $1^{1/n}$, and also in a stricter sense, if you insist that you ascend through a chain of fields by adjoining at each stage a root of $x^m-a$ where this polynomial is irreducible over the previous field.

However if you insist on this more exacting definition, you will need radicals of non-real numbers unless $n$ is a product of distinct Fermat primes. All this is well-known.

Your question has been thoroughly answered, so I don't really have much to add, except that there is a paper where you can see examples of these ideas in action with a very elementary presentation (i.e. suitable for undergraduates who have a basic knowledge of Galois theory).

Skip Garibaldi
Somewhat more than governors need to know about trigonometry
Mathematics Magazine 81 (2008) #3, 191-200.