# proof of Jordan’s Inequality

To prove that

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

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 $\mathrm{sin}(x)$.
Construct a circle of radius $MP$, with $M$ as the center.

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

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

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

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

$$\mathrm{sin}(x)\le x$$ |

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

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

Thus we have

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

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 |