# proof of Jordan’s Inequality

To prove that

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

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 $x=\angle POM.$

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

 $\sin(x)\leq x$

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

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

Thus we have

 $\frac{2}{\pi}x\leq\sin(x)\leq x,\forall\;x\in[0,\frac{\pi}{2}]$
Title proof of Jordan’s Inequality ProofOfJordansInequality 2013-03-22 13:08:48 2013-03-22 13:08:48 mathcam (2727) mathcam (2727) 17 mathcam (2727) Proof msc 26D05