Proof that a certain subset of the reals is not a ring

Let $A = \{x \sin x : x \in \mathbb{Z}\} \subset \mathbb{R}$. Is $A$ a ring under the usual addition and multiplication operations of $\mathbb{R}$? It looks like it's not, but I can't find something concrete to justify this.

Suppose $A$ were a ring; then in particular, we have $2\sin 1=n\sin n$ for some $n$. Now, use the duplication formulas for $\sin$ to write $\sin n$ as $(\sin 1)\cdot P_n(\cos 1)$, where $P_n(x)$ is a polynomial of degree $n$ (this can be shown straightforwardly via induction; these polynomials are known as the Chebyshev polynomials of the second kind). This yields an identity of the form $nP_n(\cos 1)-2=0$, contradicting the fact that $\cos 1$ is transcendental. In fact, this shows that no non-trivial linear relation can hold among any finite number of members of $A$.

There is no 1 (multiplicative identity) so it is not a ring. x sin(x) is either 0 (when x=0) or is a transcendental number due to Lindemann–Weierstrass theorem when x <> 0