TITLE: Given $\alpha$, can we always find $\beta$ such that both $\sin(\alpha+\beta)$ and $\sin(\alpha-\beta)$ are rational?
QUESTION [5 upvotes]: Given $\alpha$, can we always find $\beta$ such that both $\sin(\alpha+\beta)$ and $\sin(\alpha-\beta)$ are rational?

REPLY [7 votes]: Suppose $\sin a$ is transcendental. Then from the addition and subtraction formulas for sine, the two values $\sin(a+b),\ \sin(a-b)$ are both rational iff each of $r=\sin a \cos b$ and $s=\cos a \sin b$ are rational. Supposing $r$ is rational, from the first of these  $\cos b=r/\sin a.$ Then 
$$s=\cos a \sin b = (\pm \sqrt{1-\sin^2 a})(\pm \sqrt{1-r^2/\sin^2 a}).\tag{2}$$
If this were rational, squaring it would lead to a polynomial (with rational coefficients) having $\sin a$ as a zero, against the choice of $\sin a$ as transcendental. 
Added: If we denote $\sin a$ by $x,$ then squaring $(2)$ leads to the quartic equation $$x^4 +(s^2-r^2-1)x^2+r^2=0.$$
This is quadratic in $x^2$ with rational coefficients, so that instead of the rather strong assumption that $\sin a$ is transcendental, we can get the contradiction by just assuming it is not constructible.