Question

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

Answer 1

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

We have, $y=\frac{4sin\theta }{(2+cos\theta )}-\theta$
On differentiating w.r.t $\theta$, we get
$\frac{dy}{d\theta }=\frac{d}{d\theta }[\frac{4sin\theta }{2+cos\theta }-\theta ]=\frac{(2+cos\theta )\frac{d}{d\theta }\left(4sin\theta \right)-4sin\theta \frac{d}{d\theta }(2+cos\theta )}{(2+cos\theta {)}^2}-1=\frac{4cos\theta (2+coscos\theta )-4sin\theta (-sin\theta )}{(2+cos\theta {)}^2}-1=\frac{8cos\theta +4co{s}^2\theta +4si{n}^2\theta }{(2+cos\theta {)}^2}-1=\frac{8cos\theta +4(co{s}^2\theta +si{n}^2\theta )}{(2+cos\theta {)}^2}-1=\frac{8cos\theta +4}{(2+cos\theta {)}^2}-1=\frac{8cos\theta +4-(2+cos\theta {)}^2}{(2+cos\theta {)}^2}=\frac{8cos\theta +4-4-co{s}^2\theta -4cos\theta }{(2+cos\theta {)}^2}=\frac{4cos\theta -co{s}^2\theta }{(2+cos\theta {)}^2}=\frac{cos\theta (4-cos\theta )}{(2+cos\theta {)}^2}Ininterval[0,\frac{\pi }{2}],wehavecos\theta \ge 0.Also,4>cos\theta \Rightarrow (4-cos\theta )>0\therefore cos\theta (4-cos\theta )\ge 0andalso(2+cos\theta {)}^2>0\Rightarrow \frac{cos\theta (4-cos\theta )}{(2+cos\theta {)}^2}\ge 0\Rightarrow \frac{dy}{d\theta }\ge 0$
Hence, given function is increasing in interval $[0,\frac{\pi }{2}]$
Note since, it is a continuous function,
(i) for increasing function , $f^{\prime }\ge 0$, (ii) for strictly increasing function, $f^{\prime }>0$