Is sine injective over rationals?

We will show, that the function $\sin \colon \mathbb{Q} \to \mathbb{R}$ is injective. Indeed, suppose that $\sin \alpha = \sin \beta$ for some $\alpha, \beta \in \mathbb{Q}$ . In other words: $\sin \alpha - \sin \beta = 0.$ Recall that $\sin x - \sin y = 2 \sin \frac{x-y}{2} \cos \frac{x+y}{2}$ , for all $x, y \in \mathbb{R}$ , therefore $\sin \frac{\alpha-\beta}{2} \cos \frac{\alpha+\beta}{2} = 0.$ Furthermore, for $x \in \mathbb{R}$ :

• $\sin x = 0$ if and only if $x = \mu \pi$ , for some $\mu \in \mathbb{Z}$ ,
• $\cos x = 0$ if and only if $x = \frac{\pi}{2}+\nu \pi$ , for some $\nu \in \mathbb{Z}$ ,

hence $\alpha-\beta = 2\mu\pi \text{ or } \alpha+\beta = \pi+2\nu\pi,$ for some $\mu, \nu \in \mathbb{Z}$ .

Because $\pi$ is irrational, we have that:

• for all $k \in \mathbb{Z}$ the number $k \pi \in \mathbb{Q}$ if and only if $k = 0$ ,
• for all $k \in \mathbb{Z}$ the number $\pi + 2k\pi$ is irrational.

Using above and the assumption that $\alpha, \beta \in \mathbb{Q}$ , we obtain that $\alpha = \beta$ , which proves the claim.