Show that $\sin(x)$ is a subset of $\sin^2(x)$ when they are Rational Numbers

Question

Consider the sets

$$ \begin{split} A &= \left\{x\in \mathbb R\mid\sin(x)\in \mathbb Q\right\} \\ B &= \left\{x\in\mathbb R\mid\sin^2(x)\in \mathbb Q\right\} \end{split} $$

Why is $A\subseteq B$?

The main confusion for me is that when $\sin(x)$ is a rational number it can be negative, but $\sin^2(x)$ only has positive value right?

Answer 1

Ok, so $A$ is the set of all real numbers with rational sine. Let $x \in A$, then $\sin x \in \mathbb{Q}$. Then, $\exists a,b \in \mathbb{Z}$ with $b\ne 0$ such that $\sin x = \frac{a}{b}$.

In that case, $$ \sin^2 x = \left( \sin x \right)^2 = \left(\frac{a}{b} \right)^2 = \frac{a^2}{b^2} \in \mathbb{Q}, $$ since $a^2, b^2 \in \mathbb{Z}$ and $b^2 \ne 0$ (can you prove it?)

Thus, $x \in B$ as well...

Answer 2

If $x\in A$ then $\sin(x)=a/b$ for some $a,b\in\mathbb{N}$. Thus $ \sin^2(x)=a^2/b^2 $ And since both $a^2,b^2\in\mathbb{N}$ we conclude that $\sin^2(x)\in\mathbb{Q}$ and so $x\in B$.