Answer
$\sin (7t)-\sin (3t)=2\cos 5t\sin 2t.$
Work Step by Step
Consider the function,
\begin{align*}
\sin (7t)-\sin (3t).
\end{align*}
According to the formula for sum of two cosines, we have
\begin{align*}
\sin A- \sin B=2\cos \dfrac{A+B}{2}\sin \dfrac{A-B}{2}.
\end{align*}
Let $A=7t, B=3t$. Thus, we get
\begin{align*}
\sin (7t)-\sin (3t)&=2\cos \dfrac{7t+3t}{2}\sin \dfrac{7t-3t}{2}\\
&=2\cos \dfrac{10t}{2}\sin \dfrac{4t}{2}\\
&=2\cos 5t\sin 2t.
\end{align*}
Hence, $\sin (7t)-\sin (3t)=2\cos 5t\sin 2t.$
