Degree of field extension

Question

If $p$ is odd prime and $c=\cos(\frac{2\pi}{p})$, $s=\sin(\frac{2\pi}{p})$ then for which values of $p$ does $\mathbb{Q}(s,c)=\mathbb{Q}(c)$?

Answer 1

If $p \geq 3$ is an integer then the degree of $K = \mathbb{Q}(\cos(2\pi/p))$ over $\mathbb{Q}$ is $\phi(p)/2$. Now suppose that $p$ is odd let $L = K(\sin(2 \pi/p))$. By induction it follows from the sum formulas for $\sin$ and $\cos$ that $\sin(2 \pi k /p) \in L$ and $\cos(2 \pi k /p) \in L$ for all integral $k$.

In particular if $p \equiv 1 \pmod 4$ then

$$ L \ni \sin(\frac{2(p-1)\pi}{4 p}) = \sin(\frac{\pi}{2} - \frac{\pi}{2 p}) = \cos(\frac{\pi}{2 p}) $$

and if $p \equiv 3 \pmod 4$ then

$$ L \ni \sin(\frac{2(p+1)\pi}{4 p}) = \sin(\frac{\pi}{2} + \frac{\pi}{2 p}) = \cos(\frac{\pi}{2 p}). $$

So in all cases $\cos(\pi/{2p}) \in L$. This implies that $[L:\mathbb{Q}] \geq \phi(4p)/2 = \phi(p)$ and therefore $L \neq K$.