• Aug 27th 2008, 08:34 PM
kevek
topic is 3<π<2√3
Hint: apply mean value theorem on function Sin on the interval [0, π/6]
• Aug 28th 2008, 04:57 AM
kalagota
Quote:

Originally Posted by kevek
topic is 3<π<2√3
Hint: apply mean value theorem on function Sin on the interval [0, π/6]

MVT: If a function f is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$

so, $\sin$ is continuous on $[0,\pi/6]$ and differentiable on $(0,\pi/6)$. thus, there exist a $c$ in $(0,\pi/6)$ such that $\cos c = \frac{\sin \pi/6 - \sin 0}{\pi/6}$

$\frac{3}{\pi} = \cos c$

since $c$ in $(0,\pi/6)$, $\frac{\sqrt{3}}{2} < \cos c < 1$.. thus, $1 < \frac{1}{\cos c} < \frac{2}{\sqrt{3}}$

therefore, we get $1 < \frac{\pi}{3} < \frac{2}{\sqrt{3}}$ which gives you the desired result..
• Aug 28th 2008, 05:03 AM
kevek
Quote:

Originally Posted by kalagota
MVT: If a function f is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$

so, $\sin$ is continuous on $[0,\pi/6]$ and differentiable on $(0,\pi/6)$. thus, there exist a $c$ in $(0,\pi/6)$ such that $\cos c = \frac{\sin \pi/6 - \sin 0}{\pi/6}$

$\frac{3}{\pi} = \cos c$

since $c$ in $(0,\pi/6)$, $\frac{\sqrt{3}}{2} < \cos c < 1$.. thus, $1 < \frac{1}{\cos c} < \frac{2}{\sqrt{3}}$

therefore, we get $1 < \frac{\pi}{3} < \frac{2}{\sqrt{3}}$ which gives you the desired result..

you are brilliant Thank you very much !!!