Question

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

Answer 1

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

$\Rightarrow L.H.S={\sin }^6\theta +{\cos }^6\theta$

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

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

$[\because {(a+b)}^2=a^3+b^3+3ab(a+b)]=1-3{\sin }^2\theta {\cos }^2\theta$

Hence, the answer is $L.H.S=R.H.S.$