proof of Jordan’s Inequality

To prove that


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

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

Let x=POM.

For x to be in [0,π2], the point P lies in the first quadrantMathworldPlanetmath, as shown.

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

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

Length of arc PAQ is 2x.

Length of arc PBQ is πsin(x).

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


Since length of arc PAQ is length of arc PBQ (equality holds true when x=0 or x=π2), we have 2xπsin(x). This implies


Thus we have

Title proof of Jordan’s Inequality
Canonical name ProofOfJordansInequality
Date of creation 2013-03-22 13:08:48
Last modified on 2013-03-22 13:08:48
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 17
Author mathcam (2727)
Entry type Proof
Classification msc 26D05