Question

Prove that ${\sin }^{4}A+{\cos }^{4}A=1-2{\sin }^{2}A\cdot {\cos }^{2}A$

Answer 1

${\sin }^{4}A+{\cos }^{4}A$

We have ${(a^2+b^2)}^2=a^4+b^4+2a^2{b}^2$

$\Rightarrow a^4+b^4={(a^2+b^2)}^2-2a^2{b}^2$

Take $a=\sin A$ and $b=\cos A$ we get

${\sin }^{4}A+{\cos }^{4}A={({\sin }^{2}A+{\cos }^{2}A)}^2-2{\sin }^{2}A{\cos }^{2}A$

$=1-2{\sin }^{2}A{\cos }^{2}A$ proved.