Find the regions where sinx < x
Maybe this will help. $\displaystyle sinx\leq{x}\Rightarrow{sinx}-x\leq0$
I took this image from http://webgraphing.com/
sinx is always between -1 and 1. It follows then, that when |x|>1, the graph will either become more and more negative, or more and more positive. Know that, all you must do is analyze the behavior of the graph for -1<x<1 and determine the behavior of the function for those values.
Let $\displaystyle f(x)=\sin x-x$
$\displaystyle f'(x)=\cos x-1=-2\sin^2\frac{x}{2}\leq 0$
Thus $\displaystyle f(x)$ is a decreasing funtion(i.e. if $\displaystyle x_{1}<x_{2}$ ,then $\displaystyle f(x_{1})>f(x_{2}))$
Thus for $\displaystyle x\geq 0$
$\displaystyle f(x)\leq f(0)$
$\displaystyle f(x)\leq 0$
$\displaystyle \sin x-x\leq 0$
$\displaystyle \sin x\leq x$ for $\displaystyle x\in [0,\infty)$
O.K. then.
Now consider acircle of radius 1 unit.Let a chord AB of this circle subtend an angle x at the centre O of the circle.
Clearly,Ar($\displaystyle \triangle AOB$)<Area(\sector AOB).
Ar($\displaystyle \triangle AOB$)=$\displaystyle \frac{1}{2}\sin x$
Area of sector AOB=$\displaystyle \frac{1}{2}.1^2.x=\frac{1}{2}.x$
$\displaystyle \frac{1}{2}\sin x\leq \frac{1}{2}x$
$\displaystyle
\sin x\leq x
$
which is true for all non-negative $\displaystyle x$.If $\displaystyle x<0$,then $\displaystyle \sin x>x$