Question

If $\sin \,x+\sin ^{2} x=1$, then $\cos ^{6} x+\cos ^{12} x+3 \cos ^{10} x+3 \cos ^{8} x$ is equal to :

• A. 1
• B. $\cos ^{3} \,x \sin ^{3} \,x$
• C. 0
• D. $\infty$

Answer 1

Answer: (A)

Given $\sin \,x+\sin ^{2} x=1$
$\Rightarrow \, \sin x=1-\sin ^{2} x=\cos ^{2} x$
$\Rightarrow \, \sin x=\cos ^{2} x$
$\therefore \cos ^{6} x+\cos ^{12} x+3 \cos ^{10} x+3 \cos ^{8} x$
$=\sin ^{3} x+\sin ^{6} x+3 \sin ^{5} x+3 \sin ^{4} x$
$=\left(\sin x+\sin ^{2} x\right)^{3}=1^{3}=1$