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  4. If sin θ+ sin 2 θ+ sin 3 θ=1 then the value of cos 6 θ-4 cos 4 θ+8 cos 2 θ is equal to

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Q. 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 ___

Trigonometry

Solution:

$\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 $
$ \quad\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
$4=8 \cos ^2 \theta+\cos ^6 \theta-4 \cos ^4 \theta$