Inequality involving theta and sin(theta)

[Sistine]

Homework Statement

I'm trying to prove the following inequality for 0\leq \theta\leq\frac{\pi}{2}

\frac{2}{\pi} \theta \leq\sin\theta

Homework Equations

0 \leq \theta\leq\frac{\pi}{2}

0\leq \frac{2}{\pi}\theta\leq 1

The Attempt at a Solution

I've looked at the limit

\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1

I've also looked at other inequalities involving \sin\theta

\sin\theta<\theta

I've also tried to take derivatives to see how fast \sin\theta grows in comparison to \theta, but I have not managed to prove the inequality.

 

[Mark44]

That limit won't do you much good, since it is for x near zero.

The graphs of y = 2x/\pi and y = sin(x), you'll see that the sine curve is above the line, and that the two intersect at the origin and at (\pi, 1), and at no other points on the interval you're interested in. This isn't a proof, though, but you should be able to convey what the graph shows through calculus.

You should be able to establish your inequality by showing that the graph of y = sin(x) is concave down on the interval (0, \pi), meaning that the graph of y = sin(x) will be above the graph of y = 2x/\pi except at the endpoints of your interval.