Question

$\sin 5x=5{\cos }^{4}x\sin x-10{\cos }^{2}x{\sin }^{3}x+{\sin }^{5}x$

Answer 1

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

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