Minimal polynomial of cos(π/n)

Question

I know that $\cos(\pi/n)$ is a root of the Chebyshev polynomial $(T_n + 1)$, in fact it is the largest root of that polynomial, but often that polynomial factors. For example, if $n = 2 k$ then $\cos(\pi/n)$ is the largest root of $T_k$, which is a polynomial of lower degree, and if $n = 3$ then $\cos(\pi/n)$ is a root of $2 x - 1$, again lower degree than $T_3 + 1$.

How can I compute, for a given $n$, a polynomial in $\mathbb{Q}[x]$ of minimal degree that $\cos(\pi/n)$ is a root of?

Answer 1

The minimal polynomial of $\cos(2\pi/n)$ (by William Watkins and Joel Zeitlin, The American Mathematical Monthly Vol. 100, No. 5 (May, 1993), pp. 471-474) has full clarity on this matter (just take their result for even $n$ to resolve your case).

Answer 2

I guess you mean a polynomial $p(x)$ with rational coefficients. Then, once $\cos {\pi/n}$ is a root of $p(x)$, $\deg p=d$, $e^{i\pi/n}$ is a root of a polynomial $t^dp((t+1/t)/2)$. But $e^{i\pi/n}$ is a root of a cyclotomic polynomial $g(t)=\Phi_{2n}(t)$, which is irreducible, thus $\Phi_{2n}(t)$ should divide $t^dp((t+1/t)/2)$, that is, for any $k\in \{0,1,\dots,2n-1\}$ coprime to $2n$ the number $\cos \pi k/n$ is also a root of $p(x)$. For $k$ and $2n-k$ we get the same value of a cosine, so we get $\varphi(2n)/2$ different roots. Actually the polynomial with all these $\varphi(2n)/2$ roots has rational coefficients. To see this observe that $\Phi_{2n}(t)=t^{\varphi(2n)/2}H(t+1/t)$ for some polynomial $H$, which of course has rational (even integer) coefficients, and this $H$ has roots $2\cos \pi k/n$ for $k$ coprime to $2n$.

Answer 3

There is a Wikipedia entry dedicated to this, which contains an alternative method to compute the minimal polynomial of $2\cos(\pi/n)$, which is essentially the same as for $\cos(π/n)$. In fact, denoting the minimal polynomial of $2\cos(π/n)$ by $\Psi_n(x)$, we have by the quoted AMM article in Vladimir Dotsenko's answer for odd $n=2k+1$ $$\prod_{d \mid n}\Psi_d(x) = 2\Big(T_{k + 1}(\frac x2) - T_k(\frac x2)\Big)$$ and for even $n=2k$ $$\prod_{d \mid n}\Psi_d(x) = 2\Big(T_{k + 1}(\frac x2) - T_{k-1}(\frac x2)\Big).$$ But there is in fact no need to calculate the Chebyshev polynomials. Defining for odd $n=2k+1$ $$\chi_{n}(x):= \sum_{k=0}^m (-1)^{\lfloor k/2\rfloor}\binom {m-\lfloor (k+1)/2\rfloor}{\lfloor k/2\rfloor} x^{m-k},$$ we have directly

$$\prod _{d\mid n}\Psi_{d}( x)=(x-2)\chi_{n}(x),$$ noting that $\Psi_{1}( x)=x-2$.

For even $n=2k$, we just need to sum up two of those polynomials: $$\prod _{d\mid n}\Psi_{d}( x)=(x-2)\Big(\chi_{n+1}(x)+\chi_{n-1}(x)\Big).$$

Answer 4

Maple 2017.3 helps you. For example,

convert(cos((1/7)*Pi), RootOf);
RootOf(8*_Z^3-4*_Z^2-4*_Z+1, .9009688679)
convert(cos((1/27)*Pi), RootOf)
RootOf(512*_Z^9-1152*_Z^7+864*_Z^5-240*_Z^3+18*_Z-1, .9932383577)

See convert and RootOf for info.

Addition. Also

convert(sin(7*Pi*(1/22)), RootOf);
RootOf(32*_Z^5+16*_Z^4-32*_Z^3-12*_Z^2+6*_Z+1, .8412535328)
convert(sin(7*Pi*(1/22))^2, RootOf);
-(1/4)*(RootOf(_Z^20-_Z^18+_Z^16-_Z^14+_Z^12-_Z^10+_Z^8-_Z^6+_Z^4-_Z^2+1, index = 1)^7+
RootOf(_Z^20-_Z^18+_Z^16-_Z^14+_Z^12-_Z^10+_Z^8-_Z^6+_Z^4-_Z^2+1, index = 1)^15)^2

Answer 5

In PARI/ GP :

r2n = Mod(x, polcyclo(2*n))

minpoly((r2n + 1/r2n) / 2)