When does a rational portion of a circle have a rational cosine?

If *r* is a rational number, cos(2π*r*) rational if and only if the denominator of *r* is 1, 2, 3, 4, or 6.

This means that the special values of cosine you learn in a trig class, with a simple argument and simple value, are the only ones possible. (Here simple argument means an angle with an integer number of degrees and simple value means a rational number.) And if you see a result such as cos(π/7) = 837/929, you know it can’t be exactly correct, though in this case it’s very close.

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Proof: de Moivre’s formula and rational root test, I presume?

When does an algebraic portion of a circle have an algebraic cosine?

f(r) = cos(2 pi r)

1/31 is rational. The values of r that make f(r) closest to 1/31 are 2/6 & 3/6, with f(2/6) = 0.5 and f(3/6) = 0. No rational value of r with denominator in {1,2,3,4,6} can make f(r) = 1/31, therefore 1/31 is not in the range of cos and cos has a discontinuity at this point.

Somehow this doesn’t seem like a good conclusion.

Joe: You’re restricting your inputs to discrete possibilities by restricting the denominators. That’s what’s creating the discontinuity. The cosine function is continuous. There is an infinite sequence of r’s, even rational r’s, that cause f(r) to converge to 1/31.

Sorry, I had misunderstood your original question ‘When is both r and cos(2πr) rational?’ to mean something like ‘When is cos rational?’, and my comment was meant as a counter-proof. So nevermind. :)

I’d be interested in the cases where both Θ and cos(Θ) are rational, or if that’s impossible.

(Sorry for the late response.)

Could the proof of Niven’s theorem also be a proof of Fermat’s Last Theorem with powers greater than 3?