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.
To start with, let's notice that we might just as well prove this fact for the cosines of rational multiples of $\pi$ as the sines. We just need to use the identity $\sin x=\cos\left(\frac{\pi}{2} - x\right)$, which gives us: $$\sin \frac{a\pi}{b}=\cos\left(\frac{\pi}{2} - \frac{a\pi}{b}\right)=\cos\left(\frac{(b-2a)\pi}{2b}\right)$$ As this is the cosine of a rational multiple of $\pi$, we get the result for sines more or less for free from the result for cosines.
Once this preliminary is out of the way, the key thing to notice is that $\cos(nx)$ can always be written as a polynomial function of $\cos x$. For example: \begin{align} \cos 2x &= 2 \cos^2 x - 1 \\ \cos 3x &= 4 \cos^3 x - 3 \cos x\\ \cos 4x &= 8 \cos^4 x - 8 \cos^2 x + 1 \\ \cos 5x &= 16 \cos^5 x - 20 \cos^3 x + 5 \cos x \end{align} and so on. (Formulas taken from here.)
Why does this matter? It means that we can find an explicit polynomial which has $\cos \frac{a\pi}{b}$ as a root. To see an example of this, let's look at $\alpha = \cos \frac{2\pi}{5}$. Using the last formula above, we know that $$ \cos (2\pi)=\cos\left(5 \cdot \frac{2\pi}{5}\right)=16 \cos^5 \frac{2\pi}{5}- 20 \cos^3 \frac{2\pi}{5} + 5 \cos \frac{2\pi}{5}=16\alpha^5-20\alpha^3+5\alpha $$ Since $\cos(2\pi)=1$, we know that $\alpha$ is a solution to the equation $16x^5-20x^3+5x=1$, or, to put it another way, a root of the polynomial $16x^5-20x^3+5x-1$. So, by definition, $\alpha$ is algebraic (i.e., not transcendental).
You can use multiple-angle formulas to do this in general. If you want to prove that $\beta=\cos\frac{a\pi}{b}$ is algebraic, you just need to use the formula for $\cos bx$. This will give you a polynomial expression for $\cos(a\pi)$ in terms of $\beta$; since we know that $\cos(a\pi)=\pm 1$, that's enough to prove that $\beta$ is algebraic.
So we just need to prove this key fact, which can be done by induction. For the base case, notice that $\cos x$ and $\cos 2x=2\cos^2 x -1$ are both polynomial functions of $\cos x$. For the inductive step, we'll suppose that $\cos nx$ and $\cos (n-1)x$ are polynomials, and use the identity given here for $\cos a + \cos b$, with $a=(n+1)x$, $b=(n-1)x$:
$$\cos(n+1)x + \cos(n-1)x=2\cos nx \cos x$$
This can be rearranged into the formula $\cos(n+1)x = 2 \cos x \cos nx - \cos(n-1)x$, which means that if $\cos nx$ and $\cos(n-1)x$ are polynomials in $\cos x$, then so is $\cos(n+1)x$. So the proof by induction is finished.
If you're interested in knowing more details about this kind of polynomial formula for $\cos nx$ in terms of $\cos x$, a useful keyword to search on would be "Chebyshev polynomials."
As a final note, we could try to do this proof directly for $\sin \frac{a\pi}{b}$, as there are multiple-angle formulas for $\sin$ which are similar to the ones for $\cos$. But it'd be a little bit trickier, because those formulas often involve both $\sin$ and $\cos$. For example, $\sin 2x=2\cos x \sin x$, and getting rid of the $\cos$ would involve taking a square root and doing a bunch of annoying case analysis involving signs.
Write:
$$z=e^{a\pi i/b}=\cos(a\pi/b)+i\cdot\sin(a\pi/b)$$
Then,
$$z^{2b}=e^{2a\pi i}=(e^{2\pi i})^{a}=1^a=1$$
$z$ is an algebraic number, as a root of the equation $x^{2b}-1=0$. Thus its real and imaginary parts ($\cos(a\pi/b)$ and $\sin(a\pi/b)$ respectively) are both algebraic.
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.
