Question

Prove that: $\sin 5x=5\sin x-20{\sin }^{3}x+16{\sin }^{5}x$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sin 5x \\ =\sin \left(3x+2x\right) \\ =\sin 3x\times \cos 2x+\cos 3x\times \sin 2x \\ =\left(3\sin x-4{\sin }^{3}x\right)\left(1-2{\sin }^{2}x\right)+\left(4{\cos }^{3}x-3\cos x\right)\times 2\sin x\cos x \\ =3\sin x-6{\sin }^{3}x-4{\sin }^{3}x+8{\sin }^{5}x+\left(8{\cos }^{4}x-6{\cos }^{2}x\right)\sin x \\ =3\sin x-10{\sin }^{3}x+8{\sin }^{5}x+\left\{8\sin x{\left(1-{\sin }^{2}x\right)}^2-6\sin x\left(1-{\sin }^{2}x\right)\right\} \\ =3\sin x-10{\sin }^{3}x+8{\sin }^{5}x+\left\{8\sin x\left(1-2{\sin }^{2}x+{\sin }^{4}x\right)-6\sin x+6{\sin }^{3}x\right\} \\ =3\sin x-10{\sin }^{3}x+8{\sin }^{5}x+8\sin x-16{\sin }^{3}x+8{\sin }^{5}x-6\sin x+6{\sin }^{3}x \\ =5\sin x-20{\sin }^{3}x+16{\sin }^{5}x \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$