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Homeproof of Jordan's Inequality

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# 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 unit circle (circle with radius = 1 unit). 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

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

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## Recent Activity

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

## Corrections

Some nitpicking by mathwizard ✓

Small things by igor ✓

broken by yark ✓

capitalization of title by Mathprof ✓

## Comments

## the problem with geometric proofs

Nice, but how do you know the length of arc

PAQ is less than the length of arc PBQ?

Note: length of arc PAQ > PMQ

by http://planetmath.org/encyclopedia/StraightLineIsShortestCurveBetweenTwo...

but there is some work involved in that result:

(either you need to know the sup definition of arc length (= integral definition)

or you compare the integrals directly)