Using the Dirichlet approximation theorem I managed to show that $\sin(\mathbb{N})$ is dense in the unit circle. Then, using the symmetry of the circle we may deduce that:
$$ \begin{equation} \forall k \in \mathbb{Z}, \sin(\mathbb{N}+k) \text{ is dense in $[-1,1]$}\end{equation} \tag{1}$$
$$ \begin{equation} \forall k \in \mathbb{Z}, \sin(-\mathbb{N}+k) \text{ is dense in $[-1,1]$}\end{equation} \tag{2}$$
I also managed to deduce that:
$$ \begin{equation} \forall n,m \in \mathbb{N}, m*\sin^n(-\mathbb{N}+m) \cap [-1,1] \text{ is dense in $[-1,1]$}\end{equation} \tag{3}$$
However, I had trouble finding elementary arguments to show that:
$$ \begin{equation} \forall k \in \mathbb{Z}, \sin(k\mathbb{N}) \text{ is dense in $[-1,1]$}\end{equation} \tag{*}$$
I managed to show that it must be true using the Weyl equidistribution theorem where $(n\theta)_{n=1}^{\infty}$ is equidistributed mod 1, by choosing $\theta=\frac{1}{2\pi}$. But, I wonder whether there's a simpler method that escaped me.
Basically, my question is whether I could have deduced this result with little more than Dirichlet's approximation theorem.
Instead of Weyl Equidsitribution theorem you can use Kronecker's Approximation theorem (source: Kroneckers Approximation Theorem) which can be simply proved by pigeonhole principle.