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.
Best Answer
The easiest proof is via Liouville's criterion.
Lemma. Suppose $\alpha$ is an irrational algebraic number. There exist integers $C,n$ such that for all integers $p/q$, $$ \left| \alpha - \frac{p}{q} \right| \geq \frac{C}{q^n}. $$
You can find a proof of the lemma on Wikipedia, and in many other sources. Now consider the number $$ \alpha = \sum_{m=0}^\infty \frac{1}{10^{m!}}, $$ sometimes known as Liouville's number. Since the decimal expansion of $\alpha$ is aperiodic, it is irrational. On the other hand, for all $r$ we have $$ \left| \alpha - \frac{\sum_{m=0}^r 10^{r!-m!}}{10^{r!}} \right| \leq \sum_{m=n+1}^\infty \frac{1}{10^{m!}} \leq \sum_{m=r+1}^\infty \frac{1}{10^{(r+1)!+m-(r+1)}} \leq \frac{2}{10^{(r+1)!}}. $$ Since $2/10^{(r+1)!}$ is so much smaller than $10^{r!}$ for large $r$, it is not hard to check that $\alpha$ doesn't satisfy the conclusion of the lemma. Indeed, suppose that it did, for some $C,n$. Then for all $r$ we must have $$ \frac{2}{10^{(r+1)!}} \geq \frac{C}{10^{r!n}} \Longleftrightarrow 10^{r!n} \geq (C/2) 10^{(r+1)!} \Longleftrightarrow 1 \geq (C/2) 10^{r! (r+1-n)}, $$ which fails for large enough $r$.
Since the conclusion of the lemma doesn't hold, one of its premises must be false. Since $\alpha$ is definitely irrational, it must be not algebraic. In other words, $\alpha$ is transcendental.