Proof that $\frac{2\theta}{\pi} < \sin \theta < \theta$ The following inequality hold: if $\theta $ is in radians and $ 0 < \theta  <\pi/2$, then
$\frac{2\theta}{\pi} < \sin \theta < \theta$
How can be proved this inequality?
 A: Sometimes a picture is worth a hundred words.

$\sin\theta$ is the vertical line.  It is the shortest distance to the x-axis.
$\theta$ is the arclength, which is clearly bigger than $\sin\theta$
And the diagonal line $(0,1)$ to $(1,0)$ are points that are $\frac {2\theta}{\pi}$ above the x-axis for $0<\theta<\frac{\pi}2.$
A: There is a formal way to show it.
The fact we will use is 

if $f(x_0) = g(x_0)$ and $f'(x) \ge g'(x)$ when $x \in [x_0, x_1]$ then $f(x) \ge g(x)$ when $x\in [x_0, x_1]$.

First let's prove that $\sin\vartheta \leq \vartheta$ when $\vartheta \in \left[0,\frac{\pi}{2}\right]$. Indeed, $\sin 0 = 0$ and $\sin'\vartheta = \cos\vartheta \leq 1 = \vartheta '$.
Then let's prove that $\sin\vartheta \geq \frac{2\vartheta}{\pi}$. Let $f(\vartheta) = \sin\vartheta $ and $g(\vartheta) = \frac{2\vartheta}{\pi}$. Note that $f(0) = g(0)$ and $f\left(\frac{\pi}{2}\right) = g\left(\frac{\pi}{2}\right)$. Suppose that there is one more point $\varphi \in \left[0,\frac{\pi}{2}\right]$ such that $f(\varphi) = g(\varphi)$. Then we get using Rolle's theorem that function $f(\vartheta) - g(\vartheta)$ has at least two stationary points on $\left[0,\frac{\pi}{2}\right]$. But equation
$$
f'(\vartheta) - g'(\vartheta) = \cos\vartheta - \frac{2}{\pi} = 0
$$
leads to $\vartheta = \pm\arccos\left(\frac{2}{\pi}\right) + 2\pi n$, $n \in \mathbb{Z}$ and it has exactly one solution in the interval $\left[0,\frac{\pi}{2}\right]$. Thus there is no solutions of equation $\frac{2\vartheta}{\pi} = \sin \vartheta$ on the interval $\left[0,\frac{\pi}{2}\right]$ (except $0$ and $\frac{\pi}{2}$) and as, for example, 
$$
\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} > \frac{1}{2} = \frac{2}{\pi}\cdot\frac{\pi}{4}
$$
this inequality holds on the whole interval. QED.
A: Consider an unit circle centered at the origin $O$. Draw an arc $AB$ of angle $\theta$ beggining at $(1,0)$ and going counter clockwise. Now draw a paralell to $X$ axis through $B$, that intersects the $Y$ axis at $C$.
Now draw a $90$ degree arc with radius $OC$.
Compare the longitudes of the arc $AB$, trhe segment $OC$, and the lastly drawn arc.
A: Assuming you meant $0 < \theta <\frac{\pi}{2}$, try 
$$
1) f_1 = \theta - \sin \theta \\
2) f_2 = \pi \sin \theta - 2 \theta
$$
The second equation is a bit more challenging, but if you take the second derivative, you notice that $f'_2$ is monotone decreasing, so by intermediate value theorem it has only one root, hence $f_2$ first increases and then decreases, and in $\frac{\pi}{2}$ is equal to $0$, so by Rolle's theorem it has roots only at $0$ and $\frac{\pi}{2}$. 
