As @lulu states in the comments of the question, the Lindermann-Weierstrass Theorem is a theorem that goes along these lines:
If I have $\sin(n)$ where $n$ is algebraic e.g. it is a solution to a polynomial with rational coefficients, then $\sin(n)$ is transcendental.
And, furthermore, transcendental numbers are not rational.
However, I can tell you that $\sin(\pi n)$ is algebraic, but usually not rational.
We know this because $\sin(\pi n)$ can be rewritten as a polynomial $P_n(\sin(\pi))$ where the degree of this polynomial is $n$ and the coefficients are all whole numbers.
And so on. But, through the polynomial method, we also know that if $n$ is rational, then the result is algebraic.
And that is indeed algebraic. (It's just very tedious to calculate.)
My mistake, we can only get algebraic solutions to $\sin(\pi n)$ if $n=\frac pq$ where $q=2^a3^b$ and $a,b$ are whole numbers.
This is a direct result of the inability to find the inverse of a polynomial of degree $n>4$.
I messed up with the last example of $\cos(\pi\frac72)$, it is rather tedious and prone to mistakes. But you can apply trig identities ($\cos(\pi\frac62+\pi\frac12)$) seven times.