Question

Prove that ${\sin }^8\theta -{\cos }^8\theta =({\sin }^2\theta -{\cos }^2\theta )(1-2{\sin }^2\theta {\cos }^2\theta )$

Answer 1

L.H.S$={\left({\sin }^4\theta \right)}^2-{\left({\cos }^4\theta \right)}^2$

$=({\sin }^4\theta +{\cos }^4\theta )({\sin }^4\theta -{\cos }^4\theta )$

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

$=[{\left(1\right)}^2-2{\sin }^2\theta {\cos }^2\theta ]({\sin }^2\theta -{\cos }^2\theta )\left(1\right)$

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

$=$R.H.S

Hence proved.