Question

$3{(sinx-cosx)}^4+6{(sinx+cosx)}^2+4({sin}^{6}x+{cos}^{6}x)$ equal to

• A. $11$
• B. $12$
• C. $13$
• D. $14$

Answer 1

The correct option is A $11$

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

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

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

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

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

$=1-3{\sin }^{2}x{\cos }^{2}x$

$3(\sin x-\cos x{)}^4+6(\sin x+\cos x{)}^2+4({\sin }^{6}x+{\cos }^{6}x)=3+12{\sin }^{2}x{\cos }^{2}x-12\sin x\cos x+6+12\sin x\cos x+4-12{\sin }^{2}x{\cos }^{2}x$

$=3+6+4=13$