Question

$\mathrm{Prove}\mathrm{that}4\cos x\cos \left(\frac{\mathrm{\pi }}{3}+x\right)\cos \left(\frac{\mathrm{\pi }}{3}-x\right)=\cos 3x.$

Answer 1

$$\begin{gathered} \mathrm{LHS}=4\cos x\cos \left(\frac{\mathrm{\pi }}{3}+x\right)\cos \left(\frac{\mathrm{\pi }}{3}-x\right) \\ =2\cos x\left[2\cos \left(\frac{\mathrm{\pi }}{3}+x\right)\cos \left(\frac{\mathrm{\pi }}{3}-x\right)\right] \\ =2\cos x\left[\cos \left(\frac{\mathrm{\pi }}{3}+x+\frac{\mathrm{\pi }}{3}-x\right)+\cos \left(\frac{\mathrm{\pi }}{3}+x-\frac{\mathrm{\pi }}{3}+2x\right)\right]\left[\because 2\cos A\cos B=\cos (A+B)+\cos (A-B)\right] \\ =2\cos x\left[\cos \frac{2\mathrm{\pi }}{3}+\cos 2x\right] \\ =2\cos x\left[-\frac{1}{2}+\cos 2x\right] \\ =-\cos x+2\cos x\cos 2x \\ =-\cos x+\cos \left(x+2x\right)+\cos \left(x-2x\right) \\ =-\cos x+\cos 3x+\cos \left(-x\right) \\ =-\cos x+\cos 3x+\cos x \\ =\cos 3x \\ \mathrm{RHS}=\cos 3x \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$