Question

Prove that $y=\frac{4\sin \theta }{(2+\cos \theta )}-\theta$ is an increasing function of $\theta$ in $[0,\frac{\pi }{2}]$

Answer 1

Given:

$y=\frac{4\sin \theta }{(2+\cos \theta )}-\theta$

Differentiating w.r.t. $\theta$,
we get

$\frac{dy}{d\theta }=\frac{4\cos \theta (2+\cos \theta )-4\sin \theta (-\sin \theta )}{(2+\cos \theta {)}^2}-1$

$\Rightarrow \frac{dy}{d\theta }=\frac{8\cos \theta +4{\cos }^2\theta +4{\sin }^2\theta }{(2+\cos \theta {)}^2}-1$

$\Rightarrow \frac{dy}{d\theta }=\frac{8\cos \theta +4}{(2+\cos \theta {)}^2}-1$

$\Rightarrow \frac{dy}{d\theta }=\frac{8\cos \theta +4-(2+\cos \theta {)}^2}{(2+\cos \theta {)}^2}$

$\Rightarrow \frac{dy}{d\theta }=\frac{8\cos \theta +4-(4+{\cos }^2\theta +4\cos \theta )}{(2+\cos \theta {)}^2}$

$\Rightarrow \frac{dy}{d\theta }=\frac{4\cos \theta -{\cos }^2\theta }{(2+\cos \theta {)}^2}$

$\Rightarrow \frac{dy}{d\theta }=\frac{\cos \theta (4-\cos \theta )}{(2+\cos \theta {)}^2}$

For $\theta ϵ[0,\frac{\pi }{2}]$

$\frac{\cos \theta (4-\cos \theta )}{(2+\cos \theta {)}^2}>0$

$\Rightarrow \frac{dy}{d\theta }>0$

so, given function is increasing in $[0,\frac{\pi }{2}]$

Hence proved.