$1)$ For $ 0\le \theta \le\frac{\pi}{2}$, show that $\sin \theta \ge \frac{2}{\pi} \theta$ Not sure how to go about this:
$1)$ For $ 0\le \theta \le\frac{\pi}{2}$, show that $\sin \theta  \ge \frac{2}{\pi} \theta$
$2)$ By using Part 1, or by any other method, show that if $\lambda \le 1,$ then 
$$\lim_{R \to \infty} R^{\lambda} \int_{0}^{\frac{\pi}{2}}e^{-R\sin\theta}d\theta=0$$
The other method part threw me off a bit. 
EDIT:
After working on it I have Two questions:
1) If I were to use the integral inequality such that:
$$ J=\int_{0}^\frac{\pi}{2}e^{-Rsin\theta}Rd\theta \le \int_{0}^\frac{\pi}{2}e^{-2Rsin\theta}Rd\theta =-\pi e^{-2Rsin\theta}|^\frac{\pi}{2}_{0} \le \pi$$ 
Is that correct?
2)How would I go about finishing this second part using Jordan's Lemma :
$$ R^{\lambda} \int_{0}^{\frac{\pi}{2}} e^{-Rsin\theta}d\theta= R^{\lambda} \int_{0}^{\frac{\pi}{3}} e^{-Rsin\theta}d\theta+R^{\lambda} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} e^{-Rsin\theta}d\theta$$

For (1), just draw the graph of $y=\sin x$ and $y=\frac {2}{\pi}x$. Since the graph of $y=\sin x$ is concave when $0<x< \pi/2$, it must be drawn above the line $y=\frac 2{\pi}x$ when $0<x<\frac {\pi}2$.
For (2), just use the result of (1).
