Originally Posted by
kalagota MVT: If a function f is continuous on $\displaystyle [a,b]$ and differentiable on $\displaystyle (a,b)$, then there exists $\displaystyle c \in (a,b)$ such that $\displaystyle f'(c) = \frac{f(b)-f(a)}{b-a}$
so, $\displaystyle \sin$ is continuous on $\displaystyle [0,\pi/6]$ and differentiable on $\displaystyle (0,\pi/6)$. thus, there exist a $\displaystyle c$ in $\displaystyle (0,\pi/6)$ such that $\displaystyle \cos c = \frac{\sin \pi/6 - \sin 0}{\pi/6}$
$\displaystyle \frac{3}{\pi} = \cos c$
since $\displaystyle c$ in $\displaystyle (0,\pi/6)$, $\displaystyle \frac{\sqrt{3}}{2} < \cos c < 1$.. thus, $\displaystyle 1 < \frac{1}{\cos c} < \frac{2}{\sqrt{3}}$
therefore, we get $\displaystyle 1 < \frac{\pi}{3} < \frac{2}{\sqrt{3}}$ which gives you the desired result..