Question

Prove that: $(3+\cos 4\theta )\cos 2\theta =4({\cos }^8\theta -{\sin }^8\theta )$

Answer 1

$R.H.S.=({\cos }^2\theta +{\sin }^2\theta )({\cos }^2\theta -{\sin }^2\theta )({\cos }^4\theta +{\sin }^4\theta )$

$R.H.S.=4\times 1({\cos }^2\theta -{\sin }^2\theta )({\cos }^4\theta +{\sin }^4\theta )$

$=4\cos 2\theta (1-2{\sin }^2\theta {\cos }^2\theta )$

$=4\cos 2\theta [1-\frac{1}{2}{\sin }^{2}2\theta ]$

$=4\cos 2\theta [1-\frac{1}{2}\frac{(1-\cos 4\theta )}{2}]$

$=\cos 2\theta [3+\cos 4\theta ]$