Question

Prove that ${\cos }^{2}2\theta -{\sin }^2\theta =\cos \theta .\cos 3\theta .$

Answer 1

Now

${\cos }^{2}2\theta -{\sin }^2\theta$

$=(2{\cos }^2\theta -1{)}^2-1+{\cos }^2\theta$ [Since $\cos 2\theta =2{\cos }^2\theta -1$]

$=4{\cos }^4\theta -3{\cos }^2\theta$

$=\cos \theta (4{\cos }^3\theta -3\cos \theta )$

$=\cos \theta .\cos 3\theta$ [Since $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$]