Question

For any angle $\theta$ , prove that $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

Answer 1

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

$=\cos (2\theta +\theta )$

$=\cos 2\theta \cos \theta -\sin 2\theta \sin \theta$

$=({\cos }^2\theta -{\sin }^2\theta )\cos \theta -2{\sin }^2\theta \cos \theta$

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

$=\cos \theta ({\cos }^2\theta -{\sin }^2\theta -2{\sin }^2\theta )$

$=\cos \theta ({\cos }^2\theta -3{\sin }^2\theta )$

$={\cos }^3\theta -3{\sin }^2\theta \cos \theta ={\cos }^3\theta -3(1-{\cos }^2\theta )\cos \theta )$

$={\cos }^3\theta -3\cos \theta +3{\cos }^3\theta$

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