Question

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

Answer 1

Answer: 5

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