The sine inequality $\frac2\pi x \le \sin x \le x$ for $0<x<\frac\pi2$ [duplicate]
Solution 1:
For $x \in \left[0, \frac{\pi}{2}\right]$, we have $\sin''(x) = -\sin(x) \le 0$. So the sine function is concave on $\left[0, \frac{\pi}{2}\right]$. So the inequality follows from the principle (I suggest drawing the graph to see it clearly) :
$$\textrm{secant} \le \textrm{function} \le \textrm{tangent}$$