Using the relation $2(1-\cos x)<x^2,x\neq 0$ or otherwise, prove that $$ \sin (\tan x)\geq x\;\forall x\in \left[0,\frac{\pi}{4}\right] $$
My Attempt:
Let $f(x) = \sin (\tan x)-x$. Then $f'(x) = \cos (\tan x)\cdot \sec^2 x-1$. Now using the given statement that $2(1-\cos x)<x^2$ for $x\neq0$, set $x=\tan x$ to get
$$ \begin{align} 2(1-\cos (\tan x))&<\tan^2 x\\ \left[1-\cos (\tan x)\right]&<\frac{\tan^2 x}{2}\\ \cos (\tan x)&>\left[\frac{2-\tan^2 x}{2}\right]\\ \cos (\tan x)\cdot \sec^2 x-1&>\left[\frac{2-\tan^2 x}{2}\right]\cdot \sec^2 x-1\\ f'(x)&>\left[\frac{2-\tan^2 x}{2}\right]\cdot \sec^2 x-1\\ f'(x)&>\tan^2 x-\frac{\tan^2 x}{2}\cdot \sec^2 x\\ f'(x) \color{red}{=} \frac{\tan^2 x}{2}\cdot \left[2-\sec^2 x\right]&>0\;\; \forall x\in \left(0,\frac{\pi}{4}\right)\\ f'(x)&>0\;\; \forall x\in \left(0,\frac{\pi}{4}\right) \end{align} $$
Thus $f(x)$ is a strictly increasing function for $x\in \left(0,\frac{\pi}{4}\right)$. Furthermore,
$$ \begin{align} f(x)&>f(0)\\ \sin (\tan x)-x&\geq 0\\ \sin (\tan x)-x&>0 \;\forall x\in \left(0,\frac{\pi}{4}\right) \end{align} $$
Now $f(0)=0$ and $\displaystyle f\left(\frac{\pi}{4}\right) = \sin (1)-\frac{\pi}{4}>0$. Therefore,
$$ f(x)\geq 0\;\forall x \in \left[0,\frac{\pi}{4}\right] $$
Can we solve this using any other method?
