Can both $x$ and $\sin(x)$ be rational at the same time?
Except, of course, trivial $x=0$ case ($\sin0=0$); $x$ is measured in radians.
The question turned out to be more complicated than it seemed to me at the first sight.
All I came up with, that posed question is equivalent to the question of chord and corresponding arc being together of rational length.
On
The Lindemann–Weierstrass theorem shows that if $\alpha$ is a non-zero algebraic number, then $e^{\alpha}$ is transcendental.
If $\alpha=ix$ when $x$ is rational, then $\alpha$ is algebraic, so $e^{ix}=\cos x + i\sin x$ is transcendental.
If $z$ is transcendental, then so is $w=z-\frac{1}{z}$. Otherwise then $z^2-w z -1 =0$ and thus $z$ is algebraic.
But if $z=e^{ix}$, with $x$ rational, then, $z$ is trascendental and hence so is $z-\frac{1}{z} = 2i\sin x$. So if $2i\sin x$ is transcendental, then so is $\sin x$.
If $\sin x$ is rational then $$ e^{ix} = \pm\sqrt{1-\sin^2 x}+i\sin x $$ is an algebraic number over $\mathbb{Q}$ of degree at most $4$. However, if $x\in\mathbb{Q}^+$ then $e^{ix}$ is a trascendental number, since $e^{i}$ is a trascendental number.