Question

Prove that: ${\cos }^{3}x\sin 3x+{\sin }^{3}x\cos 3x=\frac{3}{4}\sin 4x$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}, \\ \cos 3x=4{\cos }^{3}x-3\cos x \\ \Rightarrow {\cos }^{3}x=\frac{\cos 3x+3\cos x}{4}...\left(\mathrm{i}\right) \\ \mathrm{Also}, \\ \sin 3x=3\sin x-4{\sin }^{3}x \\ \Rightarrow {\sin }^{3}x=\frac{3\sin x-\sin 3x}{4}...\left(\mathrm{ii}\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Now}, \\ \mathrm{LHS}={\cos }^{3}x\sin 3x+{\sin }^{3}x\cos 3x \\ =\left(\frac{\cos 3x+3\cos x}{4}\right)\sin 3x+\left(\frac{3\sin x-\sin 3x}{4}\right)\cos 3x \\ \left[\mathrm{Using}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\right] \\ =\frac{1}{4}\left[3\left(\sin 3x\cos x+\sin x\cos 3x\right)+\left(\cos 3x\sin 3x-\sin 3x\cos 3x\right)\right] \\ =\frac{1}{4}\left[3\sin \left(3x+x\right)+0\right] \\ =\frac{3}{4}\sin 4x \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$