Answer
$$\frac{-3}{4}$$
Work Step by Step
\begin{align*}
\lim _{\theta \rightarrow 0} \frac{ \sin (-3\theta)}{\sin (4\theta)}&=\lim _{\theta \rightarrow 0} \frac{ \sin (-3\theta)}{\sin 4\theta}\\
&=\lim _{\theta \rightarrow 0} \frac{-3}{4}\frac{ \sin (-3\theta)}{-3\theta}\frac{ 4\theta}{\sin 4\theta}\\
&= \frac{-3}{4}\lim _{-3\theta \rightarrow 0}\frac{ \sin (-3\theta)}{-3\theta}\lim _{4\theta \rightarrow 0}\frac{ 4\theta}{\sin 4\theta}\\
&= \frac{-3}{4}.
\end{align*}
Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1. $
