If cos(a) is algebraic over QQ, so is cos(a/3)

Suppose \cos(\alpha) is algebraic over \mathbb{Q}. Prove that \cos(\alpha/3) is also algebraic over \mathbb{Q}.

Recall from the triple angle formula for cosine that \cos(\alpha) = 4\cos(\alpha/3)^3 - 3\cos(\alpha/3). (To see this, expand e^{i(3\theta)} = (e^{i\theta})^3 in two ways using Euler’s formula and compare real and imaginary parts.) Letting \theta = \alpha/3, we see that 0 = 4\cos(\alpha/3)^3 - 3\cos(\alpha/3) - \cos(\alpha). In particular, \cos(\alpha/3) is a root of p(x) = 4x^3 - 3x - \cos(\alpha). Since \cos(\alpha) is algebraic over \mathbb{Q}, and \cos(\alpha/3) is algebraic over \mathbb{Q}(\cos(\alpha), \cos(\alpha/3) is algebraic over \mathbb{Q}.

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