## If sin(a) is algebraic over QQ, then cos(a) and sin(a/2) are algebraic over QQ

Suppose $\sin(\alpha)$ is algebraic over $\mathbb{Q}$. Prove that $\cos(\alpha)$ and $\sin(\alpha/2)$ are also algebraic over $\mathbb{Q}$. Show that $\sin(\pi/12)$ is algebraic over $\mathbb{Q}$.

Recall from this previous exercise that if $\alpha$ is algebraic, then so are $\alpha^2$ and $\sqrt{\alpha}$.

Suppose $\sin(\alpha)$ is algebraic over $\mathbb{Q}$. Recall from the Pythagorean identity that $\sin(\alpha)^2 + \cos(\alpha)^2 = 1$; in particular, $\cos(\alpha)^2 = 1 - \sin(\alpha)^2$. Since $\sin(\alpha)$ is algebraic, so is $\sin(\alpha)^2$, and so is $1 - \sin(\alpha)^2$. So $\cos(\alpha)^2$ is algebraic, and thus $\cos(\alpha)$ is as well.

Similarly, from the half-angle identity we have $\sin(\alpha/2) = \sqrt{\frac{1}{2}(1 - \cos(\alpha)}$, so that $\sin(\alpha/2)$ is algebraic over $\mathbb{Q}$.

Recall that $\sin(\pi/6) = 1/2$, so that $\sin(\pi/6)$ is algebraic. Thus $\sin(\pi/12)$ is algebraic over $\mathbb{Q}$.