By considering the function $$f(x)=\sin x - x$$ for $0<x<\frac{\pi}{2}$, show that $$\frac{2}{\pi}x<\sin x<x$$ for $0<x<\frac{\pi}{2}$
My first step was to differentiate $f(x)$, upon which I get $\cos x - 1$. However, I do not know how to procede from there.
Let $f(x)=\frac{\sin x}{x}$. Note $$ f(x)'=\frac{x\cos x-\sin x}{x^2}=\frac{x-\tan x}{x^2\cos x}<0,x\in(0,\frac{\pi}{2}) $$ and hence $f(x)$ is strictly decreasing in $(0,\frac{\pi}{2})$. So for $0<x<\frac{\pi}{2}$, one has $$\frac{2}{\pi}=f(\frac{\pi}{2})< f(x)< \lim_{x\to0^+}f(x)=1 $$ or $$\frac{2}{\pi}x< \sin x< x. $$ Here $\tan x>x$ for $x\in(0,\frac{\pi}{2})$ is used.
$\sin$ is a concave function on $[0,\frac{\pi}{2}]$.
Thus, $\sin{x}\geq\frac{2}{\pi}x$ because graph of $\sin$ and the line $y=\frac{2}{\pi}x$
have two common points $(0,0)$ and $\left(\frac{\pi}{2},1\right)$.
Thus, for all $x\in\left(0,\frac{\pi}{2}\right)$ we obtain $\sin{x}>\frac{2}{\pi}x$ .
$\sin{x}<x$ for all $x\in\left(0,\frac{\pi}{2}\right)$ follows from the definition of $\sin$
Maybe I can help you. $$\frac{2}{\pi}x<\sin(x)<x, 0<x<\frac{\pi}{2}$$Let's use Cauchy's Mean Value Theorem: $$\frac{\sin\frac{\pi}{2}-\sin(x)}{1-\frac{2}{\pi}x}=\frac{-\cos(c)}{-\frac{2}{\pi}}<1$$ $$1-\sin(x)<1-\frac{2}{\pi}x$$ $$\sin(x)>\frac{2}{\pi}$$
