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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.

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    Doesn't a similar argument hold, using the fact that $2\sin(\pi/n) = iz^{-1}-iz$, with $z$ satisfying a cyclotomic polynomial? Use the fact that $\zeta_n=e^{2\pi i/n} = \cos(2\pi/n) + i\sin(2\pi i/n)$, where $\zeta_n$ is a primitive $n$-th root of unity, so $\zeta_n - \zeta_n^{-1} = 2i\sin(2\pi i/n)$.2011-04-01

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