# $1)$ For $0\le \theta \le\frac{\pi}{2}$, show that $\sin \theta \ge \frac{2}{\pi} \theta$

$1)$ 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 (2), I know the answer: when $\lambda=1$, the limit is 1, when $\lambda <1$, the limit is 0. see link – 王李远 May 24 '18 at 13:44

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 $0<x<\frac {\pi}2$.