Question

Prove: $({\sin }^6\theta +{\cos }^6\theta )=1-3{\sin }^2\theta {\cos }^2\theta$

Answer 1

We need to prove $({\sin }^6\theta +{\cos }^6\theta )=1-3{\sin }^2\theta .{\cos }^2\theta$

Taking L.H.S. $={\sin }^6\theta +{\cos }^6\theta$
$=\left({\sin }^2\theta \right)+({\cos }^2\theta {)}^3$
$=({\sin }^2\theta +{\cos }^2\theta )({\sin }^4\theta -{\sin }^2\theta {\cos }^2\theta +{\cos }^4\theta )$
$=1({\sin }^4\theta -{\sin }^2\theta {\cos }^2\theta +{\cos }^4\theta )$
$={\sin }^4\theta -{\sin }^2\theta {\cos }^2\theta +{\cos }^4\theta$
$=({\sin }^2\theta {)}^2+({\cos }^2\theta {)}^2-{\sin }^2\theta {\cos }^2\theta$
$[\therefore a^2+b^2=(a+b{)}^2-2ab]$
$=({\sin }^2\theta +{\cos }^2\theta {)}^2-2\sin \theta {\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta$
$=1-3{\sin }^2\theta {\cos }^2\theta$

$=$ R.H.S.