Question

Prove that ${\sin }^6\theta +{\cos }^6\theta =1-3{\sin }^2\theta .{\cos }^2\theta$

Answer 1

L.H.S. $={\sin }^6\theta +{\cos }^6\theta$
$=({\sin }^2\theta {)}^3+({\cos }^2\theta {)}^3$

Put ${\sin }^2\theta =a$ and ${\cos }^2\theta =b$

$\therefore$ L.H.S. $=a^3+b^3$
$=(a+b{)}^3-2ab(a+b)$
$=({\sin }^2\theta -{\cos }^2\theta {)}^3-3{\sin }^2\theta {\cos }^2\theta ({\sin }^2\theta +{\cos }^2\theta )$
$=(1{)}^3-3{\sin }^2\theta {\cos }^2\theta [\because {\sin }^2\theta +{\cos }^2\theta =1]$
$=1-3{\sin }^2\theta {\cos }^2\theta$
$=$ R.H.S.