Question

If $y={\sin }^{4}x+{\cos }^{4}x$ then $y_4\left(0\right)$ is equal to

Answer 1

$y={\sin }^{4}x+{\cos }^{4}x=({\sin }^{2}x{)}^2+({\cos }^{2}x{)}^2=({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x\cdot {\cos }^{2}x$
$y=1-\frac{1}{2}{\sin }^{2}2x$
$y_1=-2\sin 2x\cdot \cos 2x=-\sin 4x$
$y_2=-4\cos 4x$
$y_3=16\sin 4x$
$y_4=64\cos 4x$
$y_4\left(0\right)=64$