Question

If $sin \theta = cos^{2} ⁡ \theta $ , then the value of $cos^{12} \theta + 3 cos^{10} ⁡ \theta + 3 cos^{8} ⁡ \theta + cos^{6} ⁡ \theta - 1$ is equal to

Answer 1

$\sin \theta=\cos ^{2} \theta \Rightarrow \sin \theta+\sin ^{2} \theta=1$
Now, $\cos ^{6} \theta\left(\cos ^{6} \theta+3 \cos ^{4} \theta+3 \cos ^{2} \theta+1\right)-1$
$=\cos ^{6} \theta\left(\cos ^{2} \theta+1\right)^{3}-1$
$=\left(\cos ^{2} \theta\right)^{3}(\sin \theta+1)^{3}-1$
$=\sin ^{3} \theta(\sin \theta+1)^{3}-1$
$=\left(\sin ^{2} \theta+\sin \theta\right)^{3}-1=1-1=0$