Question

Prove that y=4sinθ/(2+cosθ)-θ is an increasing function of θ in [0,π/2]

Answer 1

given, $\bf{y=\frac{4sin\theta}{2+cos\theta}-\theta}$
differentiate with respect to $\theta$,
$\bf{\frac{dy}{d\theta}=\frac{(2+cos\theta)\frac{d}{d\theta}(4sin\theta)-4sin\theta\frac{d}{d\theta}(2+cos\theta)}{(2+cos\theta)^2}-\frac{d\theta}{d\theta}}$

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

$=\bf{\frac{8cos\theta+4cos^2\theta+4sin^2\theta-(2+cos\theta)^2}{(2+cos\theta)^2}}$

$=\bf{\frac{(8cos\theta+4(sin^2\theta+cos^2\theta)-4-4cos\theta-cos^2\theta}{(2+cos\theta)^2}}$

$=\bf{\frac{(cos\theta)(4- cos\theta)}{(2+cos\theta)^2}}$

we know one thing , cosx < 1
then, (4 - cos$\theta$) > 0 for all real value of $\theta$ .
so, dy/d$\theta$ depends on cos$\theta$
in [ 0, π/] , cos$\theta$ > 0
therefore , f(x) is increasing in [0, π/2]
$\textbf{\underline{hence proved}}$