IIT-JEE 2003Question 1
Using the relation $2(1-\cos x)<x^2$, $x\neq0$ or otherwise, prove that $\sin(\tan x)\ge x$, $\forall x\in\left[0,\dfrac{\pi}{4}\right]$.
Differentiation and its Application
Answerproved
Solution
Concept. Show $h(x)=\sin(\tan x)-x$ is increasing on $\left[0,\frac\pi4\right]$; since $h(0)=0$, $h\ge0$. Formulas used. (1) $2(1-\cos t)<t^2$, i.e. $\cos t>1-\dfrac{t^2}{2}$ for $t\ne0$(2) $\sec^2x=1+\tan^2x$Step 1 - Derivative. $h'(x)=\cos(\tan x)\sec^2x-1$. Step 2 - Bound. Let $u=\tan^2x\in(0,1]$ for $x\in\left(0,\frac\pi4\right]$. By (1) with $t=\tan x$ and (2): $$h'(x)>\left(1-\frac u2\right)(1+u)-1=\frac{u}{2}-\frac{u^2}{2}=\frac{u(1-u)}{2}\ge0.$$ Step 3 - Conclude. $h$ is increasing on $\left[0,\frac\pi4\right]$ and $h(0)=0$, so $h(x)\ge0$, i.e. $\sin(\tan x)\ge x$. Hence proved.