Irrational $x$ such that $\sin(\pi x)$ is algebraic

Question

It's well-known that $\sin(\pi \frac pq)$ is always algebraic. In particular, as I understand, it can always be expressed in terms of radicals, because it can be connected to the abelian group of $e^{i\pi \frac pq}$. Because these can always be expressed in terms of radicals -- this means there must be some other algebraic numbers (for instance, the roots of irreducible quintics) whose inverse sines are not a rational multiple of pi. So what form do these take? Is $\sin(\sqrt 2 \pi)$ known to be transcendental, for instance? That would be the "first place to start looking", to me, but I have little basis for that.

Answer 1

$\sin \pi \alpha$, for $\alpha$ irrational but algebraic, is transcendental by the Gelfond-Schneider theorem.

To start, write

$$\sin \pi \alpha = \frac{e^{i \pi \alpha} - e^{-i \pi \alpha}}{2i}.$$

We will actually show that $e^{i \pi \alpha}$ is transcendental. This follows from writing it as $(e^{i \pi})^{\alpha} = (-1)^{\alpha}$. (This notation may look funny; for complex numbers, $a^b$ is multivalued, and Gelfond-Schneider applies to all of the possible values.)

So I don't know what there is to say beyond "take the inverse sine of some algebraic number, then divide by $\pi$."