Is there a simple formula an integer polynomial that $2\sin(2\pi/n)$ satisfies?
For $2\cos(2\pi/n)$ the answer is relatively nice. For any given $n$, we have $2\cos(2\pi/n)= z + z^{-1}$ where $z = e^{2 \pi i/n}$ satisfies a cyclotomic polynomial of degree $\varphi(n) = 2k$, $ 0 = a_{2k}z^{2k} + a_{2k-1}z^{2k-1} + \ldots + a_{1}z + a_0, $ where $a_{i} = a_{2k-i}$. Dividing by $\zeta^k$ gives $ 0 = a_{2k}z^k + \ldots + a_k + a_0z^{-k} $ Using the symmetry of the coefficients lets us write this as $ 0 = a_0(z^k+z^{-k}) + \ldots + a_k. $ Then $ (z+z^{-1})^2 = z^2 + z^{-2} + \binom{2}{1} $ $ (z+z^{-1})^3 = z^3 + z^{-3} + \binom{3}{1}(z+z^{-1}) $ $ (z+z^{-1})^4 = z^4 + z^{-4} + \binom{4}{1}(z^2+z^{-2}) + \binom{4}{2} $ and so on, and in the end we get something fairly nice.
What happens with $2\sin(2\pi/n)$?
EDIT: I am aware that $2\sin(2\pi/n) = -i(z - z^{-1})$.
EDIT: Arturo's comment made me realize that dividing by $(-z)^k$ or $(iz)^k$ may be the way to go.