Question

Let $3 \cos \theta+5 \cos ^{2} \theta+7 \cos ^{3} \theta+\ldots \infty=-\frac{1}{2},$ where $|\cos \theta|<1 .$ The value of $\left(1+2 \sin ^{2} \frac{\theta}{2}\right)^{2}$ is equal to___

Answer 1

$S=3 \cos \theta+5 \cos ^{2} \theta+7 \cos ^{3} \theta+\cdots \infty$
$\Rightarrow \cos \theta S=3 \cos ^{2} \theta+5 \cos ^{3} \theta+\cdots \infty$
So, $(1-\cos \theta) S=3 \cos \theta+2\left[\cos ^{2} \theta+\cos ^{3} \theta+\ldots \infty\right]$
$\Rightarrow (1-\cos \theta) S=3 \cos \theta+\frac{2 \cos ^{2} \theta}{1-\cos \theta}$
$\Rightarrow (1-\cos \theta)^{2}\left(\frac{-1}{2}\right)=3 \cos \theta(1-\cos \theta)+2 \cos ^{2} \theta$
$\Rightarrow -\cos ^{2} \theta+2 \cos \theta-1=6 \cos \theta-6 \cos ^{2} \theta+4 \cos ^{2} \theta$
$\Rightarrow \cos ^{2} \theta-4 \cos \theta-1=0$
$\Rightarrow \cos \theta=2 \pm \sqrt{5}$
$\Rightarrow \cos \theta=2-\sqrt{5}$
$\left(1+2 \sin ^{2} \frac{\theta}{2}\right)^{2}=(1+1-\cos \theta)^{2}=(2-(2-\sqrt{5}))^{2}=5$