Answer
See the explanation below.
Work Step by Step
$\sin 4t=4\sin t{{\cos }^{3}}t-4{{\sin }^{3}}t\cos t$
Recall the sum Trigonometric Identities,
$\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b$
Consider the left side of the given expression and apply the above formula,
$\begin{align}
& \sin 4t=\sin \left( 2t+2t \right) \\
& =\sin 2t\cos 2t+\cos 2t\sin 2t \\
& =\cos 2t\left( \sin 2t+\sin 2t \right) \\
& =\cos 2t\left( 2\sin 2t \right)
\end{align}$
Recall the double angle formula,
$\begin{align}
& \cos 2t={{\cos }^{2}}t-{{\sin }^{2}}t \\
& \sin 2t=2\sin t\cos t \\
\end{align}$
Apply the double angle formula
$\begin{align}
& \sin 4t=\left( {{\cos }^{2}}t-{{\sin }^{2}}t \right)\left( 2 \right)2\sin t\cos t \\
& =4\sin t{{\cos }^{3}}t-4{{\sin }^{3}}t\cos t
\end{align}$
Hence, it is proved that the given identity $\sin 4t=4\sin t{{\cos }^{3}}t-4{{\sin }^{3}}t\cos t$ holds true.
