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}.

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