Question

Show that $\sin \theta <\theta <\tan \theta$ for $\theta ϵ(0,\pi /2)$

Answer 1

Consider the circle with unit radius shown in Fig.
From the figure, for any position of C on the circle
Area of $△OBC$< Area of sector OBC< Area of $△OBD$

using the corresponding formulae

$\frac{1}{2}\left(CE\right)\left(OB\right)<\frac{1}{2}\theta <\frac{1}{2}\left(OB\right)\left(BD\right)$
$\Rightarrow$ $\frac{1}{2}\left(\sin \theta \right)\left(1\right)<\frac{1}{2}\theta <\frac{1}{2}\left(1\right)\left(\tan \theta \right)$
$\Rightarrow$ $\sin \theta <\theta <\tan \theta$ for $\theta ϵ(0,\pi /2)$