Question

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

Answer 1

The given function in $\theta$ is

$f\left(\theta \right)=\frac{4\sin \theta }{2+\cos \theta }-\theta$
Now differentiating the function w.r.t. $\theta$,

$f^{\prime }\left(\theta \right)=\frac{(2+\cos \theta )4\cos \theta -4\sin \theta (-\sin \theta )}{(2+\cos \theta {)}^2}-1$

$\Rightarrow f^{\prime }\left(\theta \right)=\frac{8\cos \theta +4{\cos }^2\theta +4{\sin }^2\theta }{(2+\cos \theta {)}^2}-1$

$\Rightarrow f^{\prime }\left(\theta \right)=\frac{8\cos \theta +4({\sin }^2\theta +{\cos }^2\theta )-(2+\cos \theta {)}^2}{(2+\cos \theta {)}^2}$

$\Rightarrow f^{\prime }\left(\theta \right)=\frac{8\cos \theta +4-4-{\cos }^2\theta -4\cos \theta }{(2+\cos \theta {)}^2}$

$\Rightarrow f^{\prime }\left(\theta \right)=\frac{4\cos \theta -{\cos }^2\theta }{(2+\cos \theta {)}^2}$

$\Rightarrow f^{\prime }\left(\theta \right)=\frac{\cos \theta (4-\cos \theta )}{(2+\cos \theta {)}^2}$

$\Rightarrow$ Here, $f\left(\theta \right)$ is increasing when $f^{\prime }\left(\theta \right)>0$

i.e., $\frac{\cos \theta (4-\cos \theta )}{(2+\cos \theta {)}^2}>0$
$\Rightarrow \cos \theta >0[\because \frac{4-\cos \theta }{(2+\cos \theta {)}^2}>0\forall \theta \in R]$
$\Rightarrow \theta \in [0,\frac{\pi }{2}]$