If $\sin x+\sin ^{2} x=1, x \in\left(0, \frac{\pi}{2} ight)$, then $\left(\cos ^{12} x+ an ^{12} x

Question

If $\sin x+\sin ^{2} x=1, x \in\left(0, \frac{\pi}{2}ight)$, then $\left(\cos ^{12} x+	an ^{12} xight) +3\left(\cos ^{10} x+	an ^{10} x+\cos ^{8} x+	an ^{8} xight)+\left(\cos ^{6} x+	an ^{6} xight)$ is equal to :\

Professor PC Thomas Classes · Answer

(a): We have, $\sin x+\sin ^{2} x=1$\ $\Rightarrow \quad \sin x=1-\sin ^{2} x \Rightarrow \sin x=\cos ^{2} x$\ $\Rightarrow 	an x=\cos x$\ Now, $\left(\cos ^{12} x+	an ^{12} xight)+3\left(\cos ^{10} x+	an ^{10} xight. \left.+\cos ^{8} x+	an ^{8} xight)+\left(\cos ^{6} x+	an ^{6} xight)$\ $=\left(\cos ^{12} x+\cos ^{12} xight)+3\left(\cos ^{10} x+\cos ^{10} xight. \left.+\cos ^{8} x+\cos ^{8} xight)+\left(\cos ^{6} x+\cos ^{6} xight)$\ $=2 \cos ^{12} x+6 \cos ^{10} x+6 \cos ^{8} x+2 \cos ^{6} x$\ $=2 \sin ^{6} x+6 \sin ^{5} x+6 \sin ^{4} x+2 \sin ^{3} x \left[\because \sin x=\cos ^{2} xight]$\ $=2 \sin ^{3} x\left[\sin ^{3} x+3 \sin ^{2} x+3 \sin x+1ight]$\ $=2 \sin ^{3} x\left[(1+\sin x)^{3}ight]$\ $=2\left[\left(\sin x+\sin ^{2} xight)^{3}ight]=2\left\{\because \sin x+\sin ^{2} x=1ight\}$\