Question

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

Answer 1

Hint:in the given question we have to find that the given function is increasing in the given interval. So, in order to do so we have to find the first derivative of the function. As the given function $y=\dfrac{4\sin \theta }{2+\cos \theta }$ is a function of $\theta $, so we have to find $\dfrac{dy}{d\theta }$. In order to find the value of $\dfrac{dy}{d\theta }$ we have to use the formula $\dfrac{d\left( \dfrac{u}{v} \right)}{d\theta }=\dfrac{v\dfrac{du}{d\theta }-u\dfrac{dv}{d\theta }}{{{v}^{2}}}$. Here we take the value of $u=4\sin \theta $ and the value $v=2+\cos \theta $. Once we find the first derivative of $y$, that is ${f}'(\theta )$, we have to look the value of ${f}'(\theta )$in the given interval. If ${f}'(\theta )>0$in the given interval then the function is increasing in the given interval.

Complete step by step answer:
It is given from question
$y=\dfrac{4\sin \theta }{2+\cos \theta }$
Now we have to differentiate the given function with respect to $\theta $
\[\dfrac{dy}{d\theta }=\dfrac{d}{d\theta }\left( \dfrac{4\sin \theta }{2+\cos \theta } \right)\]
As we know that
$\dfrac{d\left( \dfrac{u}{v} \right)}{d\theta }=\dfrac{v\dfrac{du}{d\theta }-u\dfrac{dv}{d\theta }}{{{v}^{2}}}$,
Let us assume that $u=4\sin \theta \text{ and }v=(2+\cos \theta )$, so, using the above property we can write further
\[\dfrac{dy}{d\theta }=\dfrac{\left( 2+\cos \theta \right)\dfrac{d}{d\theta }\left( 4\sin \theta \right)-4\sin \theta \dfrac{d}{d\theta }\left( 2+\cos \theta \right)}{{{\left( 2+\cos \theta \right)}^{2}}}\]
Now as we know that $\dfrac{d}{d\theta }\left( \sin \theta \right)=\cos \theta \text{ and }\dfrac{d}{d\theta }\left( \cos \theta \right)=-\sin \theta ,\text{ }\dfrac{d}{d\theta }\left( c \right)=0$, where $c$is a constant. Using the above formula, we can write
\[\dfrac{dy}{d\theta }=\dfrac{\left( 2+\cos \theta \right)(4\cos \theta )-4\sin \theta (-\sin \theta )}{{{\left( 2+\cos \theta \right)}^{2}}}\]
On further simplification we have
\[\dfrac{dy}{d\theta }=\dfrac{8\cos \theta +4\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)}{{{\left( 2+\cos \theta \right)}^{2}}}\]
As we know that
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, so, we have
\[\dfrac{dy}{d\theta }=\dfrac{8\cos \theta +4}{{{\left( 2+\cos \theta \right)}^{2}}}\]
Now as we know that square of any real number is positive so in the above the denominator is positive. Also, in the interval $\left[ 0,\dfrac{\pi }{2} \right]$, $\cos \theta $is non negative, so $8\cos \theta +4$is also positive.
Hence, we conclude that
\[\dfrac{dy}{d\theta }>0\]
Hence the function $y=\dfrac{4\sin \theta }{2+\cos \theta }$ is an increasing function of $\theta $ on $\left[ 0,\dfrac{\pi }{2} \right]$.

Note:
It should be important to note that if we plot the graph for an increasing function and if we draw a tangent at any point, the angle made by tangent with x-axis in counter clock wise direction, the angle so formed is always positive. In other words, the slope at any point of an increasing function is non negative.