Question

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

Answer 1

To prove :

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

LHS ${\sin }^6\theta +{\cos }^6\theta$

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

$=({\sin }^2\theta +{\cos }^2\theta )({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta )[a^3+b^3=(a+b\left)\right(a^2+b^2-ab\left)\right]$

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

$=({\sin }^2\theta +{\cos }^2\theta {)}^2-2{\sin }^2\theta {\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta )\left(\right(a+b{)}^2-2ab=a^2+b^2)$

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

= R.H.S