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?


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).

  • $\begingroup$ It's really surprising this was published in 1993. If you had asked me before, I would have wagered it should have been done by 1950. Do you know if there is any substantial difficulty involved, or is it just that no one bothered before? $\endgroup$
    – tomasz
    Nov 28 '17 at 10:54
  • 3
    $\begingroup$ @tomasz: A slightly less explicit recipe (more or less the same as the one Fedor Petrov gives in another answer) was given already by D.H.Lehmer (A Note on Trigonometric Algebraic Numbers, The American Mathematical Monthly Vol. 40, No. 3 (Mar., 1933), pp. 165-166, jstor.org/stable/2301023) in 1933, 60 years before Watkins and Zeitlin. So in a way it has been known for a very long time, but Watkins and Zeitlin found a nice formula relating it to Chebyshev's polynomials. $\endgroup$ Nov 28 '17 at 13:27
  • $\begingroup$ I see. That seems more reasonable, thanks for clearing that up. $\endgroup$
    – tomasz
    Nov 28 '17 at 18:19

I guess you mean a polynomial $p(x)$ with rational coefficients. Then, once $\cos {\pi/n}$ is a root of $p(x)$, $\deg p=d$, $e^{ip/n}$ 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 $2n$.


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

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