proof of Jordan’s Inequality

To prove that

$$\frac{2}{\pi }x\le \sin (x)\le x,\forall x\in [0,\frac{\pi }{2}]$$

consider a circle (circle with radius = 1 ). Take any point $P$ on the circumference of the circle.

Drop the perpendicular from $P$ to the horizontal line, $M$ being the foot of the perpendicular and $Q$ the reflection of $P$ at $M$.(refer to figure)

Let $x=\mathrm{\angle }POM.$

For $x$ to be in $[0,\frac{\pi }{2}]$, the point $P$ lies in the first quadrant, as shown.

The length of line segment $PM$ is $\sin (x)$.Construct a circle of radius $MP$, with $M$ as the center.

Length of line segment $PQ$ is $2\sin (x)$.

Length of arc $PAQ$ is $2x$.

Length of arc $PBQ$ is $\pi \sin (x)$.

Since $PQ\le$ length of arc $PAQ$ (equality holds when $x=0$)we have $2\sin (x)\le 2x$.This implies

$$\sin (x)\le x$$

Since length of arc $PAQ$ is $\le$ length of arc $PBQ$ (equality holds true when $x=0$ or $x=\frac{\pi }{2}$),we have $2x\le \pi \sin (x)$.This implies

$$\frac{2}{\pi }x\le \sin (x)$$

Thus we have

$$\frac{2}{\pi }x\le \sin (x)\le x,\forall x\in [0,\frac{\pi }{2}]$$