Prove that:

Question:

Show that: $3(\sin x-\cos x)^{4}+6(\sin x+\cos )^{2}+4\left(\sin ^{6} x+\cos ^{6} x\right)=13$

Solution:

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

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

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

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

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

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

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

$=3\left(1-2 \sin 2 x+\sin ^{2} 2 x\right)+6+6 \sin 2 x+4-20 \sin ^{2} x \cos ^{2} x$

$=-6 \sin 2 x+3 \sin ^{2} 2 x+13+6 \sin 2 x-5(2 \sin x \cos x)^{2}$

$=3 \sin ^{2} 2 x+13-5 \sin ^{2} 2 x$

$=13-2 \sin ^{2} 2 x=\mathrm{RHS}$

Hence proved.