Question

If $\sin \theta +{\sin }^2\theta +{\sin }^3\theta =1$ then the value of ${\cos }^6\theta -4{\cos }^4\theta +8{\cos }^2\theta$ is equal to ___

Answer 1

Answer: 4

$\text{ Given: }\sin \theta +{\sin }^2\theta +{\sin }^3\theta =1$
$\Rightarrow \sin \theta +{\sin }^3\theta =1-{\sin }^2\theta$
$\Rightarrow \sin \theta \left(1+{\sin }^2\theta \right)={\cos }^2\theta$
$\Rightarrow \sin \theta \left(1+1-{\cos }^2\theta \right)={\cos }^2\theta$
$\left\{\because \sin {\theta }^2=1-{\cos }^2\theta \right\}$
$\sin \theta \left(2-{\cos }^2\theta \right)={\cos }^2\theta$
Squaring both sides:
${\sin }^2\theta \left(2-{\cos }^2\theta \right)2={\cos }^4\theta$
$\left(1-{\cos }^2\theta \right)\left(4+{\cos }^4\theta -4{\cos }^2\theta \right)={\cos }^4\theta$
$4-4{\cos }^2\theta +{\cos }^4\theta -{\cos }^6\theta -4{\cos }^2\theta +4{\cos }^4\theta ={\cos }^4\theta$
Cancel ${\cos }^4\theta$ on both sides
$4=8{\cos }^2\theta +{\cos }^6\theta -4{\cos }^4\theta$