Does $\cos (\pi/5)$ belong to $\mathbb{Q} (\sin(\pi/5))$?

Question

I need to know if $$\cos(\pi/5) \in \mathbb{Q} (\sin(\pi/5))?$$

I can compute explicitly such $\cos$ and $\sin$, but I have some difficulties how to deduce from this an answer.

Answer 1

Hint : show that $c=\frac{3}{2}-2s^2$, where $c=\cos(\frac{\pi}{5})$ and $s=\sin(\frac{\pi}{5})$. You can check it using Euler's identities for example.

Answer 2

What you’re missing is the observation that $c=\cos(\pi/5)$ is a quadratic irrationality over $\Bbb Q$. Once you know that $c^2+Ac+B=0$ for well-chosen rational numbers $A$ and $B$ with $A$ nonzero, the equation $c=(B-c^2)/A$ becomes $c=(s^2+B-1)/A$, where $s=\sin(\pi/5)$, and you have it.

Answer 3

As it turns out ${\displaystyle \cos\bigg({\pi \over 5}\bigg) = {1 + \sqrt{5} \over 4}}$, which can be seen for example by using the equation $\cos(3x) = \cos(2x)$ for ${\displaystyle x = {\pi \over 5}}$, writing everything in terms of $\cos(x)$ using double and triple angle formulas, and then solving for $\cos(x)$. So we have $$\sin^2\bigg({\pi \over 5}\bigg) = 1 - \bigg({1 + \sqrt{5} \over 4}\bigg)^2$$ This is of the form $a + b \sqrt{5}$ for rationals $a$ and $b$ with $b \neq 0$. Thus ${\displaystyle \sqrt{5} \in Q\big(\sin{\pi \over 5}\big)}$, as is ${\displaystyle \cos({\pi \over 5})}$ since it is just ${\displaystyle {1 + \sqrt{5} \over 4}}$.