Given that $x$ is a rational number, is $\sin(x\pi)$ always expressible through radicals? This is a theory I just thought of and I am wondering if there is truth to it. 
Here is the logic that I am working upon:
Using Euler's formula, you can deduce that
$$ (-1)^x = i\sin(x\pi)+\cos(x\pi)$$
and thus,
$$\sin(x\pi) = Im((-1)^x)$$
$$\cos(x\pi) = Re((-1)^x)$$
Now, let us let x be a rational number expressible as $\frac pq$ where $p$ and $q$ are integers
Using this, we can calculate $(-1)^x$ by doing
$$ a+bi=(-1)^{\frac pq} $$
$$(a+bi)^q=(-1)^p$$
Now, $(a+bi)^q$ can be expanded into a polynomial and grouped into real parts, and since $p$ is an integer, $(-1)^p$ will be either $1$ or $-1$.
We now group it into 2 polynomial parts, the real and imaginary, and set the imaginary part equal to $0$ and the real part equal to $1$ if $p$ is even and $-1$ if $p$ is odd.
Here is my example:
$$x = 2/3$$
$$(a+bi)^3=(-1)^2=1$$
$$a^3-3ab^2+i(3a^2b-b^3)=1$$
which gives us the two equations:
$$a^3-3ab^2=1$$ $$3a^2b-b^3=0$$
Now we have a system two equations with two variables and so they can just be solved using algebra, and since $Im((-1)^x)=Im(a+bi)=b$ and $\sin(x\pi) = Im((-1)^x)=b$, then can't we just solve for b with the system of equations and then that is our answer for $\sin(\pi x)$
Since it is a polynomial then shouldn't be be expressible through radicals and thus, shouldn't all $\sin(x\pi)$ be expressible through radicals?
 A: Yes, the numbers $\sin(\pi x)$ (for rational $x$) are always expressible by radicals, but in a somewhat trivial sense.
Namely, if we write $x = m/n$, and let $\zeta_{2n} = \cos \pi x + i \sin \pi x,$
then $$\sin (\pi x) = \frac{1}{2 i} (\zeta_{2n} - \zeta_{2n}^{-1}),$$
and $\zeta_{2n}$ is a radical, since $\zeta_{2n}^{2n} = 1$.
You can see here for
a more detailed discussion.
A: Although such real numbers $\sin(\pi x)$ are algebraic, they are not necessarily expressible by radicals on that account.  In other words, being a root of a polynomial does not imply that root is expressible by radicals, as Galois and Abel showed us.
To conclude that they are expressible by radicals amounts to showing that the Galois group of the field extension $\mathbb{Q}(2 \cos \frac{\pi}{n} )$ over $\mathbb{Q}$ is solvable, for each integer $n \ge 2$.  An argument that this Galois group is actually Abelian (hence solvable) is found in Sec. 4 of Wolfdieter Lang's recent paper The field $\mathbb{Q}(2 cos \frac{\pi}{n} )$, its Galois group, and length ratios in the regular n-gon, as summarized on page 22 there.
