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