Are all values of $\sin(x)$ algebraic. Can we prove that for all $x$ in $(0,2\pi)$ $\sin(x)$ is an algebraic number?
I have seen people express various values of $\sin(x)$ like $\sin(3)$ and $\sin(30)$ using radicals so I suspect that all values of $\sin(x)$ must be algebraic. Is that correct? Can we prove it?

Let $\frac{m}{n}\in\mathbb{Q}$, then $\sin\left(\frac{m}{n}\pi\right)=\sin\left(\frac{m}{n}180^\circ\right)$ is always algebraic: put
$$
\alpha=e^{\frac{i\pi}{n}}=\cos\frac{\pi}{n}+i\sin\frac{\pi}{n}.
$$
Then $\alpha^n+1=0$, i.e. $\alpha$ is an algebraic number (it is a root of the polynomial $X^n+1$) and hence both
$$
\cos\frac{m\pi}{n}=\frac{\alpha^m+\alpha^{-m}}{2}
\qquad\text{and}\qquad
\sin\frac{m\pi}{n}=\frac{\alpha^m-\alpha^{-m}}{2i},
$$
are algebraic numbers. This shows that for a countable number of transcendental values of $x$, the value of  $\sin(x)$ is an algebraic number. Conversely, it can be shown that if $x$ is an algebraic non-zero number, $\sin(x)$ is transcendental! This follows from the famous Lindemann-Weierstrass Theorem.
