When is cos(2*pi/n) rational?

June 25, 2010

When is cos(2*pi/n) rational?

Filed under: Miscellaneous — Wilson Ong @ 5:05 pm

For which positive integers $n$ is $\cos\frac{2\pi}{n}$ rational? We present a simple solution to this problem using Galois Theory. We claim $\cos\frac{2\pi}{n}$ is rational if and only if $n=1,2,3,4,6$ . Indeed, first note that for any automorphism $\sigma\in Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ (where $\zeta_n=e^\frac{2\pi i}{n}$ ), we have $\sigma(\cos\frac{2\pi}{n})=\sigma(\frac{\zeta_n+\zeta_n^{-1}}{2})=\frac{\zeta_n^j+\zeta_n^{-j}}{2}=\cos\frac{2\pi j}{n}$ for some $0\leq j\leq n-1$ . But it is geometrically clear (from the unit-circle definition of cosine) that $\cos\frac{2\pi j}{n}=\cos\frac{2\pi}{n}$ if and only if $j=1,n-1$ . Thus $\cos\frac{2\pi}{n}$ is fixed by at most two automorphisms. Now if $\cos\frac{2\pi}{n}$ is rational then it is fixed by every automorphism, so we must have $|Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})|=\phi(n)\leq 2$ . This is only satisfied for $n=1,2,3,4,6$ , and it is clear $\cos\frac{2\pi}{n}$ is rational for these $n$ .

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Comments (2)

2 Comments »

pretty straightforward…

Comment by Grant — July 13, 2010 @ 10:07 am

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I feel that there should be a slightly more elementary way to show this, although it should be technically the same.
If we let
j_{n} = cos(2\pi/n)
for each natural number k, we have a “k-angle” formula
cos(k x) = P_k( cos(x) )
where P_k is some polynomial, i.e. P_2(x) = 2x^2-1.
This means that
j_{n} = P_k( j_{kn} )
For all naturals k,n. mmmm. maybe you can use this system of polynomial relations and some results about rational solutions of integral polynomial eqns to show the results (which is essentially galois theory!). not sure

Comment by Ryan Mickler — January 18, 2011 @ 3:29 am

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June 25, 2010