Question

Prove that:

Answer 1

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