Algebraic numbers on both sides of trigonometric functions

In this answer I learned about Niven's theorem. As I understand it, it says

$$\left\{t\in\pi\mathbb Q\mid 0\le t\le\frac\pi2\wedge\sin(t)\in\mathbb Q\right\} =\left\{0,\frac\pi6,\frac\pi2\right\}$$

In words and degrees: the angles between $0°$ and $90°$ inclusively for which a rational number of degrees results in a rational value of the sine function are exactly $0°,30°,90°$.

Now I wonder, is there anything like this for algebraic numbers? Sure, a rational angle will always lead to an algebraic sine. But what about irrational algebraic angles? Is anything known about

$$\left\{t\in\pi\left(\mathbb A\setminus\mathbb Q\right)\mid 0\le t\le\frac\pi2\wedge\sin(t)\in\mathbb A\right\}$$

In particular, are there any practical approaches to judge whether a given algebraic value can be written as the sine of an algebraic angle? In the past I've looked at high precision numeric computations and tried to reconstruct algebraic numbers from these, but so far I failed in each such case. I wouldn't be surprised if irrational algebraic angles leading to algebraic sines would be exceptionally rare or outright impossible.

By the Lindemann-Weierstrass theorem, if $\alpha\in\mathbb{A}\setminus\mathbb{Q}$ then $e^{i\alpha}$ is trascendental, so $\sin(\alpha)=\frac{e^{i\alpha}-e^{-i\alpha}}{2i}$ is trascendental as well. If $\alpha\in\pi\left(\mathbb{A}\setminus\mathbb{Q}\right)$ then $e^{i\alpha}$ is trascendental due to the Gelfond-Schneider theorem.
• Nice, thanks! Applying the Gelfond-Schneider theorem I found it useful to write $e^{i\alpha}=\left(e^{i\pi}\right)^\beta=(-1)^\beta$ for $\alpha=\pi\beta$ and $\beta\in(\mathbb A\setminus\mathbb Q)$. I'm not perfectly sure how $e^{i\alpha}$ being transcendental implies $e^{i\alpha}-e^{-i\alpha}$ being transcendental, but this appears plausible enough that I take your word for it. – MvG Dec 17 '17 at 10:40
• Ah, I see an argument for that propagation of the transcendental property now. If $y=\sin\alpha$ were algebraic, then $\sqrt{1-y^2}+iy=e^{i\alpha}$ would be algebraic as well, since this is only algebraic operations applied to it. – MvG Dec 17 '17 at 12:15