# Given $\alpha$, can we always find $\beta$ such that both $\sin(\alpha+\beta)$ and $\sin(\alpha-\beta)$ are rational?

Given $\alpha$, can we always find $\beta$ such that both $\sin(\alpha+\beta)$ and $\sin(\alpha-\beta)$ are rational?

• The α and β are suppposed to be real numbers? Rationals? – mvggz Oct 16 '14 at 8:59

## 1 Answer

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.