Question

Prove $\cos 3x=4\cos 3x-3\cos x$

Answer 1

Proof.

Consider,

$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}=\cos 3x \\ =\cos (2x+x)\left[\because \cos (A+B)=\cos A\cos B-\sin A\sin B\right] \\ =\cos 2x\cos x-\sin 2x\sin x\left[\because \cos 2x=2{\cos }^{2}x-1\mathrm{and}\mathrm{Sin}2x=2\sin x\cos x\right] \\ =(2{\cos }^{2}x-1)\cos x-2\cos x\sin x\sin x \\ =2{\cos }^{3}x-\cos x-2{\sin }^{2}x\cos x \\ =-\cos x+2{\cos }^{3}x-2{\sin }^{2}x\cos x \\ =-\cos x+2{\cos }^{3}x-2{\sin }^{2}x\cos x\left[\because {\sin }^{2}x=1-{\cos }^{2}x\right] \\ =-\cos x+2{\cos }^{3}x-2(1-{\cos }^{2}x)\cos x \\ =-\cos x+2{\cos }^{3}x-2\cos x+2{\cos }^{3}x \\ =4{\cos }^{3}x-3\cos x \\ =\mathrm{R}.\mathrm{H}.\mathrm{S} \end{gathered}$$

Hence, proved.