Question

Solve:
${\cos }^6\theta -{\sin }^6\theta =\frac{1}{4}\left({\cos }^{3}2\theta +3\cos 2\theta \right)$

Answer 1

$R.H.S.$

$\frac{1}{4}({\cos }^{3}2\theta +3\cos 2\theta )$

$=\frac{1}{4}\cos 2\theta ({\cos }^{2}2\theta +3)$

$=\frac{1}{4}\cos 2\theta (1-{\sin }^{2}2\theta +3)$

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

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

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

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

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

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

$=\cos 2\theta [{\sin }^4\theta +{\cos }^4\theta +{\sin }^2\theta {\cos }^2\theta ]$

$=({\cos }^2\theta -{\sin }^2\theta )[{\left({\sin }^2\theta \right)}^2+{\left({\cos }^2\theta \right)}^2+{\sin }^2\theta {\cos }^2\theta ]$

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

$={\cos }^6\theta -{\sin }^6\theta$

$L.H.S.$