TITLE: Proof that a trigonometric function of a rational angle must be non-transcendental
QUESTION [55 upvotes]: Suppose I was working in radians and took
$$\sin \left(\frac{a\pi}{b}\right)$$
where both $a$ and $b$ are integers.
Is there a proof that the output of this function cannot be transcendental? Or, conversely, that 
$$\arcsin \left(t \right)$$
where $t$ is some transcendental number, e.g., $e$, cannot be a rational angle $\left(\frac{a\pi }{b}\right)$?
Sadly, I couldn't get very far with this problem with my maths teacher so any information on the topic would be very useful.

REPLY [10 votes]: For each positive integer $b,$ 
\begin{align}
\cos b\theta &= \frac{e^{ib\theta}+e^{-ib\theta}}{2}
\\&= \frac{(\cos\theta + i \sin \theta)^b + (\cos \theta - i \sin \theta)^b}{2}
\\&= \frac{\sum_{k=0}^b \binom{b}{k} (\cos^{b-k} \theta) (i^{k})(\sin^{k}\theta)+\sum_{k=0}^b\binom{b}{k} (\cos^{b-k} \theta) (-1)^{k}(i^{k})(\sin^{k}\theta)}{2}
\\&= \frac{\sum_{\substack{0\le k\le b
\\k \text{ even}}} 2 \binom{b}{k} (\cos^{b-k} \theta) ((-1)^{k/2})(\sin^{k}\theta)}{2}
\\&= \sum_{\substack{0\le k\le b
\\k \text{ even}}}(-1)^{k/2} \binom{b}{k} \cos^{b-k}\theta \sin^{k}\theta
\\&= \sum_{\substack{0\le k\le b
\\k \text{ even}}}(-1)^{k/2} \binom{b}{k} \cos^{b-k} \theta (1-\cos^{2}\theta)^{k/2}.
\end{align}
Write $P_b(x) = \sum_{\substack{0\le k\le b
\\k \text{ even}}}(-1)^{k/2} \binom{b}{k} x^{b-k} (1-x^2)^{k/2}.$
$P_b$ is a polynomial in one variable with integer coefficients, with the property that
$$\cos(b\theta) = P_b(\cos \theta)$$
for all $\theta.$
Now, if $\theta = a \pi/b$ where $a$ is an integer and $b$ is, without loss of generality, a positive integer, then
$$P_{b}(\cos \theta) = \cos (a \pi) = (-1)^a.$$
$P_{b}(x)-(-1)^a$ is not identically zero (since, if it were, the identity $P_{b}(\cos \theta)=\cos(b \theta)$ could not hold).
It follows that $\cos (\frac{a}{b}\pi) = \cos \theta$ is algebraic, since it's the root of a non-zero polynomial with integer coefficients.
Finally, $\sin (\frac{a}{b}\pi) = \cos (\frac{a}{b} \pi - \frac{\pi}{2}) = \cos(\frac{2a-b}{2b} \pi),$ which we already know is algebraic by the argument above, since $2a-b$ and $2b$ are integers.
If you want to exhibit a specific polynomial with integer coefficients that  $\sin (\frac{a}{b}\pi)$ is a root of, $P_{2b}(x)-(-1)^b$ will work.