When is $\sin(x)$ rational? Obviously, there are some points (e.g. $\pi$, $30^\circ$) but I am unsure if there are more. How can it be proved that there are no more points, or, if there are, what those points will be?
EDIT: I largely meant to ask what is supposed in the first comment. Put another way, are there numbers such that $\sin x$ and $\arcsin x$ are both rational?
 A: I assume you mean to ask: when $x$ is in whole degrees, when is $\sin(x)$ rational?
If $x$ is in whole degrees, then $x^\circ=\pi x/180\text{ radians}=\pi p/q\text{ radians}$, so we wish to find all rational multiples of $\pi$ so that $\sin(\pi p/q)$ is rational.
If $p/q\in\mathbb{Q}$, then $e^{\pm i\pi p/q}$ is an algbraic integer since $\left(e^{\pm i\pi p/q}\right)^q-(-1)^p=0$. Thus, $2\sin(\pi p/q)=-i\left(e^{i\pi p/q}-e^{-i\pi p/q}\right)$ is the difference and product of algebraic integers, and therefore an algebraic integer. However, the only rational algebraic integers are normal integers.  Thus, the only values of $\sin(\pi p/q)$ which could be rational, are those for which $2\sin(\pi p/q)$ is an integer, that is $\sin(\pi p/q)\in\{-1,-\frac{1}{2},0,\frac{1}{2},1\}$.
A: You might be looking for Niven's theorem:
If $\sin(r \pi) = q$ where $r,q$ are rationals, then $q$ is $0$, $\pm 1/2$, $\pm 1$.
A: $\sin(x)$ is locally a bijection (with the exception of maxima/minima), so it's rational whenever evaluated on $\arcsin(r)$ for rational $r$ unless this is maximum/minimum.
A: There are infinitely many points where both the sine and the cosine are rational, namely
$$
\left( \frac{n^2-m^2}{n^2+m^2}, \frac{2mn}{n^2+m^2} \right).
$$
You can see that if $(x,y)=\text{that pair}$ then $x^2+y^2=1$, so it's on the unit circle.
Maybe a more interesting question is when do you have an angle that is a rational number of degrees (or, equivalently, a rational multiple of $\pi$ radians) for which the sine is rational.  I think in that case you get only the obvious ones: $\sin 0^\circ=0$, $\sin 30^\circ=1/2$, $\sin 90^\circ=1$, and the counterparts in the other quadrants.  I'm not sure right now how to prove that.
