I wasn't sure how to start this but I thought it had something to do with a Taylor series.Let $\displaystyle f:[0,1] \rightarrow \mathbb{R}$ be a continuous function.

If also f is differentiable on (0,1) prove that $\displaystyle \exists \ \theta \in (0,1)$ such that $\displaystyle \frac{f'(\theta)}{\cos \left( \frac{\pi}{2} \theta \right)}=\frac{2}{\pi}$.

So I did this:

$\displaystyle \frac{f'(\theta)}{\cos \left( \frac{\pi}{2} \theta \right)}=\frac{2}{\pi} \Rightarrow f'(\theta)=\frac{2}{\pi} \cos \frac{\pi}{2} \theta$

$\displaystyle f'(\theta)=\frac{2}{\pi} \left( 1- \frac{ \left( \frac{\pi}{2} \theta \right)^2}{2!} + \frac{\left( \frac{\pi}{2} \theta \right)^4}{4!}-....\right)$

and I know that this is only valid for $\displaystyle \theta \in (0,1)$.

However, this isn't a proof!

Can someone help me finish this?