# Inequality involving theta and sin(theta)

1. Jun 3, 2009

### Sistine

1. The problem statement, all variables and given/known data
I'm trying to prove the following inequality for $$0\leq \theta\leq\frac{\pi}{2}$$

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

2. Relevant equations

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

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

3. 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.

2. Jun 3, 2009

### Staff: Mentor

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.