How to prove that $\mathbb{Q}(2\cos40, 2\cos80, 2 \cos160)=\mathbb{Q}(\cos(40))$

Question

How to prove that $\mathbb{Q}(2\cos40, 2\cos80, 2 \cos160)=\mathbb{Q}(\cos(40))$ (Sorry for using degrees instead of radians) I got that $\mathbb{Q}\cos(40) \leq \mathbb{Q}(2\cos40, 2\cos80, 2 \cos160)$, and that $2\cos40 \in \mathbb{Q}(\cos(40))$. But how can I show that the other 2 are in $\mathbb{Q}(\cos(40))$?

Why is it true that $\mathbb{Q}(\sin(40)) \leq \mathbb{Q}(\cos(40))$? Or $\mathbb{Q}(\sin(40)) \leq \mathbb{Q}(i\sin(40))$

Can you help me solve these? Im not sure how I can finish from here. Thank you for your help!

Answer 1

Hint:

For each $n$, $\cos n\theta$ is a polynomial (of degree $n$) in $\cos\theta$ (these are the Chebyshev polynomials).