I feel like such kind of things does not exist, but the only thing I could find is this question:
Is sin(x) necessarily irrational where x is rational?
According to this if $x$ is nonzero rational, then $\sin{x}$ will be irrational. But this does not help at all.
Any help?
Let's try Schanuel's conjecture, namely
If $n$ complex numbers $z_1,\dots,z_n$ are linearly independent over $\mathbb Q$, then the field $\mathbb Q(z_1,\dots,z_n,e^{z_1},\dots,e^{z_n})$ has transcendence degree at least $n$ over $\mathbb Q$.
Now suppose $p,q$ are nonzero rationals and $\sin q = \pi/p$. Now the two complex numbers $i q$ and $i\pi/p$ are linearly independent over $\mathbb Q$, since we know $\pi$ is irrational. So according to Schanuel, $$ \mathbb Q\left(iq,\frac{i\pi}{p},e^{iq}, e^{i\pi/p} \right) $$ has transcendence degree at least $2$. But \begin{align} iq\qquad &\text{is algebraic (its square is rational)} \\ e^{i\pi/p}\qquad &\text{is algebraic (some integer power of it is $1$)} \end{align} Next, $\sin q = \pi/p$ so $$ \frac{e^{iq}-e^{-iq}}{2i} = \frac{\pi}{p} $$ and thus $$ e^{iq}\text{ and }\frac{i\pi}{p}\qquad\text{are algebraically dependent}. $$ Therefore $$ \mathbb Q\left(iq,\frac{i\pi}{p},e^{iq}, e^{i\pi/p} \right) $$ has transdcendence degree at most $1$. This contradicts Schanuel's conjecture.
So this is not a proof for the OP, just showing that the "yes" answer follows from a well-known conjecture.
