Question

If $sin\theta +si{n}^2\theta +si{n}^3\theta =1$,then find the value of $co{s}^6\theta -4co{s}^4\theta +8co{s}^2\theta$.

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Answer 1

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

$\Rightarrow sin\theta +si{n}^3\theta =1-si{n}^2\theta$

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

$\Rightarrow si{n}^2\theta (1+si{n}^2\theta {)}^2=co{s}^4\theta$

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

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

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

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

$\Rightarrow -co{s}^6\theta +4co{s}^4\theta -8co{s}^2\theta +4=0$

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