Question

Find $m$ : ${\sin }^8\theta -{\cos }^8\theta =({\sin }^m\theta -{\cos }^m\theta )(1-2{\sin }^m\theta {\cos }^m\theta )$

Answer 1

LHS

${\sin }^8\theta -{\cos }^8\theta$

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

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

$\{\because a^2+b^2=(a+b)(a-b)\}$

$\Rightarrow ({\sin }^4\theta +{\cos }^4\theta )[{\left({\sin }^2\theta \right)}^2-{\left({\cos }^2\theta \right)}^2]$

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

$\{\because a^2+b^2={(a+b)}^2-2ab{\sin }^{2}x+{\cos }^{2}x=1\}$

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

$\Rightarrow [1-2{\sin }^2\theta {\cos }^2\theta ][{\sin }^2\theta -{\cos }^2\theta ]$

on comparing with RHS

we get $m=2$