Question

Given $\sin \theta +{\sin }^2\theta +{\sin }^3\theta =1$, $\text{Find}{\cos }^6\theta -4{\cos }^4\theta +8{\cos }^2\theta$

Answer 1

Given,

$\sin \theta +{\sin }^2\theta +{\sin }^3\theta =1$

$\Rightarrow \sin \theta (1+{\sin }^2\theta )=1-{\sin }^2\theta$

$\Rightarrow \sin \theta (2-{\cos }^2\theta )={\cos }^2\theta$

$\Rightarrow {\sin }^2\theta (2-{\cos }^2\theta {)}^2={\cos }^4\theta$

$\Rightarrow (1-{\cos }^2\theta )(4-4{\cos }^2\theta +{\cos }^4\theta )={\cos }^4\theta$

$\Rightarrow 4-8{\cos }^2\theta +5{\cos }^4\theta -\cos 6\theta ={\cos }^4\theta$

$\therefore {\cos }^6\theta -4{\cos }^4\theta +8{\cos }^2\theta =4$