Question

The value of the expression
$3(\sin x-\cos x{)}^4+4({\sin }^{6}x+{\cos }^{6}x)+6(\sin x+\cos x{)}^2$ is

Answer 1

$3(\sin x-\cos x{)}^4+4({\sin }^{6}x+{\cos }^{6}x)+6(\sin x+\cos x{)}^2$

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

${\sin }^{6}x+{\cos }^{6}x=({\sin }^{2}x{)}^3+({\cos }^{2}x{)}^3=\left(1\right)({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x)=({\sin }^{2}x+{\cos }^{2}x{)}^2-3{\sin }^2{\cos }^{2}x=1-3{\sin }^{2}x{\cos }^{2}x$

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

So, $3(\sin x-\cos x{)}^4+4({\sin }^{6}x+{\cos }^{6}x)+6(\sin x+\cos x{)}^2=13$