Question

Prove that ${\sin }^{8}A-{\cos }^{8}A=({\sin }^{2}A-{\cos }^{2}A)(1-2{\sin }^{2}A{\cos }^{2}A)$.

Answer 1

L.H.S

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

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

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

Hence, proved.