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This is completely for my own curiosity.

Does $y = \sin(n)$ have rational solutions for $n$, an integer number of radians.

I know that this is strange because usually integers are only used in degrees. I know that there will be convergent sub series of $n$'s to any rational number in $[-1,1]$. Would we ever get an algebraic integer? It would have something like $$ e^\left(n i\right)=e^\left(\frac{n}{\pi}\pi i\right) $$

I suspect that it will never have a rational solution but I don't know how to prove it.

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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.

For example:

$$\sin(\pi 0)=0$$

$$\sin(\pi 1)=0$$

$$\sin(\pi2)=0$$

And so on. But, through the polynomial method, we also know that if $n$ is rational, then the result is algebraic.

$$\sin(\pi \frac12)=1$$

$$\cos(\pi\frac74)=\cos^7(\pi\frac14)-7\cos^5(\pi\frac14)-16\cos^3(\pi\frac14)+40\cos(\pi\frac14)$$

$$=(\frac{\sqrt{2}}2)^7-7(\frac{\sqrt{2}}2)^5-16(\frac{\sqrt{2}}2)^3+40(\frac{\sqrt{2}}2)$$

And that is indeed algebraic. (It's just very tedious to calculate.)

EDIT

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 $n>4$.

EDIT

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.

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