For all real $x $, $-1 \le \sin x \le 1$. By the extreme value theorem, a continuous function achieves a maximum and a minimum value over a closed interval, so there exists $c_1,c_2$ such that for all $x \in \mathbb{R} $, $f(\sin c_1) \le f(\sin x) \le f(\sin c_2)$.
Now you can use the Squeeze Theorem.