Question

Find the value of 3 $(sinX-cosX{)}^4+6(sinX+cos{)}^2+4(si{n}^{6}X+co{s}^{6}X).$

Answer 1

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

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

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

(since, $a^3+b^3=(a+b)(a^2-ab+b^2))$

$=3[1+4{\sin }^{2}x{\cos }^{2}x-4\sin x\cos x]+6+12\sin x\cos x+4\left(1\right)[{\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x]$

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

$=9+12{\sin }^{2}x{\cos }^{2}x+4\left[\right({\sin }^{2}x+{\cos }^{2}x{)}^2-3{\sin }^{2}x{\cos }^{2}x]$

$=9+12{\sin }^{2}x{\cos }^{2}x+4-12{\sin }^{2}x{\cos }^{2}x$

$=9+4=13$.