Question

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

Answer 1

$LHS=({\sin }^{4}x{)}^2-({\cos }^{4}x{)}^2\because a^2-b^2=(a+b)(a-b)$

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

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

$=({\sin }^{2}x-{\cos }^{2}x)({\sin }^{2}x+{\cos }^{2}x)\left[\right({\sin }^{2}x{)}^2+({\cos }^{2}x{)}^2+2{\cos }^{2}x.{\sin }^{2}x-2{\cos }^{2}x.{\sin }^{2}x]$

$=({\sin }^{2}x-{\cos }^{2}x)\left[\right({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\cos }^{2}x.{\sin }^{2}x]$

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

$=RHS$