Question

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

Answer 1

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