Question

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

Answer 1

We know that

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

Cubing both sides

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

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

${\sin }^6\theta +{\cos }^6\theta +3{\sin }^2\theta {\cos }^2\theta \left(1\right)=1$

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

Hence proved.