Question

If $0<\theta <\frac{\pi }{2}$ then prove that $\frac{\sin \theta }{\theta }$ is decreasing.

Answer 1

Given,

$f\left(\theta \right)=\frac{\sin \theta }{\theta }$.

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

For $0<\theta <\frac{\pi }{2}$ we have $\tan \theta >\theta \Rightarrow \sin \theta >\theta \cos \theta$ or $\theta .\cos \theta -\sin \theta <0$.

Using this from (1) we get,

$f^{\prime }\left(\theta \right)<0$ for $0<\theta <\frac{\pi }{2}$.

This gives $f\left(\theta \right)$is decreasing in the interval $0<\theta <\frac{\pi }{2}$.