Question

Prove the following identity:
${\sin }^{8}x-{\cos }^{8}x=({\sin }^{2}x-{\cos }^{2}x)(1-2{\sin }^{2}x{\cos }^{2}x)$

Answer 1

${\sin }^{8}x-{\cos }^{8}x=({\sin }^{4}x-{\cos }^{4}x)({\sin }^{4}x+{\cos }^{4}x)$

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

$=({\sin }^{2}x-{\cos }^{2}x)(1-2{\sin }^{2}x{\cos }^{2}x)$ (since, ${\sin }^{2}x+{\cos }^{2}x=1$)

Therefore, ${\sin }^{8}x-{\cos }^{8}x=({\sin }^{2}x-{\cos }^{2}x)(1-2{\sin }^{2}x{\cos }^{2}x)$