Let be an odd prime and let be the -th Chebyshev polynomial of the first kind. By Lemma 1 on page 5 of this paper, is irreducible over . Now
so the minimal polynomial of over is . (Note that we are slightly abusing terminology here. By definition, the minimal polynomial is a monic polynomial. But let us drop precision for the sake of brevity.)
By definition (and induction, if you may), . Hence we deduce the following:
Proposition 1. is an algebraic number of degree .
Since consists of only even powers of , can be expressed as a polynomial in . Therefore:
Proposition 2. is an algebraic number of degree .
A nice corollary of this is Niven’s theorem:
Corollary (Niven’s theorem). If is a rational multiple of such that is rational, then .
Proof. Note that if is rational, then so is for any positive integer . So, (reduction 1) without loss of generality, for some odd prime and positive integer . (The case is dealt with trivially.)
Further, multiplying by the inverse of , (reduction 2) one may assume without any loss of generality.
Now being rational is the same as being algebraic of degree . Hence, by the above, we must have and the conclusion follows immediately.
Exercise. Find all rational multiples of such that is a quadratic irrational.