Prove that (cos^{-1}(1/3) )/π is irrational.

**Solution**

Let x = cos^{-1}(1/3). If x = m/n π for some integers m, n, then cos nx = cos mπ = ±1. But we show that cos nx cannot be ±1. It follows that x/π must be irrational as required.

As usual, we have cos nx = nC0 c^{n} - nC2 c^{n-2}s^{2} + nC4 c^{n-4}s^{4} - ... , where c = cos x, s = sin x. We may put s^{2} = 1 - c^{2} to get cos nx = a polynomial of degree n in c with integer coefficients. The coefficient of c^{n} = nC0 + nC2 + nC4 + ... = 2^{n-1}. But c = 1/3, so cos nx = 2^{n-1}/3^{n} + k/3^{n-1} = (2^{n-1} + 3k)/3^{n} for some integer k. This must be in its lowest terms since 2^{n-1} is not divisible by 3. In particular, it cannot be ±1.

[A variant on this is to consider cos(2^{n}x). By a simple induction using cos 2y = 2 cos^{2}y - 1, we show that cos(2^{n}x) = a_{n}/b_{n}, where a_{n} is not a multiple of 3 and b_{n}is 3 to the power of 2^{n}. It follows that as n runs through the natural numbers, all the values cos(2^{n}x) are distinct. But if x/π was rational, there would only be finitely many distinct values.]

© John Scholes

jscholes@kalva.demon.co.uk

18 Aug 2001