Possible Duplicate:
Prove an inequality with a $\sin$ function
prove that $$\sin\theta\geq \frac{2}{\pi}\theta$$ for $0 \leq \theta \leq \dfrac{\pi}{2}$ My idea was to divide by $\theta$ take a $\lim$ when theta goes to $0$. But it only works for small $\theta$, not for the entire interval...
Consider $$f(x)=\sin x-\frac{2}{\pi}x$$
Take it's derivative and find regions of increase/decrease in the region to find its minimum value and use this to prove your result.
It's true that $$\frac{\pi}{2}\frac{\sin\theta}{\theta}\geq 1$$ since $\frac{\sin\theta}{\theta}$ is decreasing on $\theta \in (0,\pi/2]$.
