Question

Prove the following identity : 

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

Answer 1

LHS = ${\displaystyle {\sin }^8\theta -{\cos }^8\theta }$

 = ${\displaystyle {\left({\sin }^4\theta \right)}^2-{\left({\cos }^4\theta \right)}^2}$

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

= ${\displaystyle ({\sin }^2\theta -{\cos }^2\theta )({\sin }^2\theta +{\cos }^2\theta )({\sin }^4\theta +{\cos }^4\theta )}$

= ${\displaystyle ({\sin }^2\theta -{\cos }^2\theta )({\sin }^4\theta +{\cos }^4\theta )}$

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

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

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