## 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 $\mathbb{Q}$.