Question

If $\sin \theta +{\sin }^2\theta +{\sin }^3\theta =1$ then prove that ${\cos }^6\theta -4{\cos }^4\theta +8{\cos }^2\theta =4$.

Answer 1

Given : $sin\theta +si{n}^2\theta +si{n}^3\theta =1$

$sin\theta (1+si{n}^2\theta )=1-si{n}^2\theta =co{s}^2\theta$

$\Rightarrow \sqrt{1-co{s}^2\theta }(2-co{s}^2\theta )=co{s}^2\theta$

Squaring both sides, we have

$(1-co{s}^2\theta )(4+co{s}^4\theta -4co{s}^2\theta )=co{s}^4\theta$

$\Rightarrow co{s}^6\theta -4co{s}^4\theta +8co{s}^2\theta =4$

Hence proved