Question

If ${\cos }^{6}A+{\sin }^{6}A=1-k{\sin }^{2}2A$, then the value of $4k$ is

Answer 1

${\cos }^{6}A+{\sin }^{6}A=({\cos }^{2}A{)}^3+({\sin }^{2}A{)}^3=({\cos }^{2}A+{\sin }^{2}A)({\sin }^{4}A+{\cos }^{4}A-{\sin }^{2}A{\cos }^{2}A)=1\cdot \left[\right({\sin }^{2}A+{\cos }^{2}A{)}^2-3{\sin }^{2}A{\cos }^{2}A]=1^2-3\times \frac{4{\sin }^{2}A{\cos }^{2}A}{4}=1^2-\frac{3}{4}(2\sin A\cos A{)}^2=1-\frac{3}{4}{\sin }^{2}2A$
So,
$k=\frac{3}{4}$
$\therefore 4k=3$