Prove that \[ \sin4x=4\sin x\cos^3x-4\cos x\sin^3x \]
Proof:
Using the double angle identity:
\[
\sin4x=2\sin2x\cos2x
\]
Also,
\[
\sin2x=2\sin x\cos x
\]
and
\[
\cos2x=\cos^2x-\sin^2x
\]
Substituting these values:
\[
\sin4x
=
2(2\sin x\cos x)(\cos^2x-\sin^2x)
\]
\[
=
4\sin x\cos x(\cos^2x-\sin^2x)
\]
Multiplying:
\[
=
4\sin x\cos^3x
–
4\cos x\sin^3x
\]
Hence proved,
\[
\boxed{
\sin4x=4\sin x\cos^3x-4\cos x\sin^3x
}
\]