Question

$\text{ If }x=\cos 2\theta -2{\cos }^{2}2\theta +3{\cos }^{3}2\theta -4{\cos }^{4}2\theta +\dots \dots \infty$
$\text{ and }y=\cos 2\theta +2{\cos }^{2}2\theta +3{\cos }^{3}2\theta +4{\cos }^{4}2\theta +\dots \dots \infty$
$\text{ where }\theta \in \left(0,\frac{\pi }{4}\right)\text{, then find least integral value of }\left(\frac{1}{x}+\frac{1}{y}\right)$ .

Answer 1

Answer: 5

$x=\cos 2\theta -2{\cos }^{2}2\theta +3{\cos }^{3}2\theta \dots \dots \infty$
$\frac{-x\cos 2\theta =-{\cos }^{2}2\theta +2{\cos }^{3}2\theta +\dots \dots \infty }{x(1+\cos 2\theta )=\cos 2\theta -{\cos }^{2}2\theta +{\cos }^{3}2\theta \dots \dots \infty }$
$=\frac{\cos 2\theta }{1+\cos 2\theta }$
$\therefore x=\frac{\cos 2\theta }{(1+\cos 2\theta {)}^2}$
$\text{ Similarly, }y=\frac{\cos 2\theta }{(1-\cos 2\theta {)}^2}$
$\therefore \frac{1}{x}+\frac{1}{y}=\frac{(1+\cos 2\theta {)}^2+(1-\cos 2\theta {)}^2}{\cos 2\theta }=\frac{2\left(1+{\cos }^{2}2\theta \right)}{\cos 2\theta }=2(\sec 2\theta +\cos 2\theta )$
$\frac{1}{x}+\frac{1}{y}>4$
$\therefore$ least integral value of $\frac{1}{x}+\frac{1}{y}$ is 5