Prove that: \[ \cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4}\sin 4x \]
Solution
Start with the left hand side:
\[
\cos^3 x \sin 3x + \sin^3 x \cos 3x
\]
Using the identities
\[
\sin 3x = 3\sin x – 4\sin^3 x
\]
\[
\cos 3x = 4\cos^3 x – 3\cos x
\]
Substitute these values:
\[
= \cos^3 x (3\sin x – 4\sin^3 x)
+ \sin^3 x (4\cos^3 x – 3\cos x)
\]
Expand the terms:
\[
= 3\sin x \cos^3 x
-4\sin^3 x \cos^3 x
+4\sin^3 x \cos^3 x
-3\sin^3 x \cos x
\]
Middle terms cancel out:
\[
= 3\sin x \cos^3 x
-3\sin^3 x \cos x
\]
\[
= 3\sin x \cos x (\cos^2 x – \sin^2 x)
\]
Using
\[
2\sin x \cos x = \sin 2x
\]
\[
\cos^2 x – \sin^2 x = \cos 2x
\]
we get
\[
= 3\left(\frac{\sin 2x}{2}\right)\cos 2x
\]
\[
= \frac{3}{2}\sin 2x \cos 2x
\]
Now using
\[
2\sin A \cos A = \sin 2A
\]
\[
2\sin 2x \cos 2x = \sin 4x
\]
Therefore,
\[
\frac{3}{2}\sin 2x \cos 2x
=
\frac{3}{4}\sin 4x
\]
Hence proved.