Question

Prove that

${\sin }^{8}x-{\cos }^{8}x=\left({\sin }^{2}x-{\cos }^{2}x\right)\left(1-2{\sin }^{2}x{\cos }^{2}x\right)$.

Answer 1

L.H.S

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

$\Rightarrow ({\sin }^{4}x{)}^2-({\cos }^{4}x{)}^2$

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

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

$\Rightarrow \left(\right({\sin }^{2}x-{\cos }^{2}x\left)\right({\sin }^{2}x+{\cos }^{2}x\left)\right)\left(\right({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x)$

$\Rightarrow \left(\right({\sin }^{2}x-{\cos }^{2}x\left)\right(1\left)\right)\left(\right(1{)}^2-2{\sin }^{2}x{\cos }^{2}x)$ $\because {\sin }^{2}x+{\cos }^{2}x=1$

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

R.H.S

Hence, proved.