Proof that $\frac{2\theta}{\pi} < \sin \theta < \theta$ The following inequality hold: if $\theta $ is in radians and $ 0 < \theta  <\pi/2$, then
$\frac{2\theta}{\pi} < \sin \theta < \theta$
How can be proved this inequality?

Sometimes a picture is worth a hundred words.

$\sin\theta$ is the vertical line.  It is the shortest distance to the x-axis.
$\theta$ is the arclength, which is clearly bigger than $\sin\theta$
And the diagonal line $(0,1)$ to $(1,0)$ are points that are $\frac {2\theta}{\pi}$ above the x-axis for $0<\theta<\frac{\pi}2.$
