Question

Find the value of $3(\sin x-\cos x{)}^4+6(\sin x+\cos x{)}^2+4({\sin }^{6}x+{\cos }^{6}x)$.

Answer 1

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

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

${\sin }^{6}x+{\cos }^{6}x=\frac{1}{8}[(1-\cos 2x{)}^3+(1+\cos 2x{)}^3]=\frac{1}{4}[1+3{\cos }^{2}2x]$

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

$=10+3=13$