# Simple formula for integer polynomial with $2\sin(2\pi/n)$ as a root?

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 $\zeta_n - \zeta_n^{-1} = 2i\sin(2\pi i/n)$. –  Arturo Magidin Apr 1 '11 at 15:35

A similar, but much simpler answer than Douglas's results if one uses the Chebyshev polynomials of the second kind

$$U_n(x)=\frac{\sin((n+1)\arccos(x))}{\sin(\arccos(x))}$$

in which case, the (monic) polynomial with integer coefficients you need is $U_{n-1}\left(\frac{x}{2}\right)$.

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If you have an integer polynomial $p(x)$ with $\cos(2\pi/n)$ as root, then you can construct an integer polynomial $q(y)$ with $\sin(2\pi/n)$ as root. Namely, you can write

$$p(\sqrt{1-y^2}) = q_1(y)\sqrt{1-y^2} + q_2(y)$$

and we can set $q(y) := q_2(y)^2 - q_1(y)^2(1-y^2)$. Of course, this polynomial is not the minimal polynomial of $\sin(2\pi/n)$ as it is invariant under $y \to -y$, but it's better than nothing.

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I hadn't thought of this. Very nice. –  admchrch Apr 2 '11 at 0:41

$\zeta_n=\cos (2\pi/n)+i\sin(2\pi/n)$ so $i\zeta_n=i\cos (2\pi/n)-\sin(2\pi/n)$ and adding conjugates of both gives $i(\zeta_n-\zeta_n^{-1})=2\sin(2\pi/n)$. you can probably work from there

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I'll work with $\sin \frac{2 \pi}n$ instead of $2 \sin \frac{2 \pi}n$.

Since $\sin x = \cos (\frac \pi 2 - x)$, $\sin \frac{2 \pi}n = \cos (\frac {\pi}2 - \frac {2\pi}n) = \cos \frac{n-4}{2n}\pi$.

The Chebyshev polynomials of the first kind $T_k(x)$ have the property that $T_k(\cos x) = \cos kx$. They have integer coefficients since they satisfy the recurrence $T_0(x)=1, T_1(x)=x, T_{k+1}(x) = 2xT_k(x)-T_{k-1}(x)$.

$T_{4n}(\sin \frac {2 \pi}n) = T_{4n}(\cos \frac{n-4}{2n}\pi) = \cos((n-4) \times 2\pi) = 1$.

Therefore, $\sin \frac{2\pi}n$ is a root of $T_{4n}(x)-1$.

$2 \sin \frac{2 \pi}n$ is a root of $T_{4n}(x/2)-1$, but you may have to clear denominators to get a polynomial over the integers. Of course, this is far from the minimal polynomial.

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