Question

Find the value of $\frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}$.

Answer 1

$\frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}$

$=\frac{({\sin }^{4}x-{\cos }^{4}x)({\sin }^{4}x+{\cos }^{4}x)}{1-2{\sin }^{2}x{\cos }^{2}x}$

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

$=\frac{(-\cos 2x)\left(1\right)\{1-2{\sin }^{2}x.{\cos }^{2}x\}}{1-2{\sin }^{2}x{\cos }^{2}x}$ [Since ${\sin }^{2}x+{\cos }^{2}x=1$ and $\cos 2x={\cos }^{2}x-{\sin }^{2}x$]

$=-\cos 2x$.