Jordan's inequality

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2πx≤sin⁡(x)≤x for x∈[0,π2]
unit circle with angle x and a second circle with radius |EG|=sin⁡(x) around E. |DE|≤|DC^|≤|DG^|⇔sin⁡(x)≤x≤π2sin⁡(x)⇒2πx≤sin⁡(x)≤x

In mathematics, Jordan's inequality, named after Camille Jordan, states that[1]

2πx≤sin⁡(x)≤x for x∈[0,π2].

It can be proven through the geometry of circles (see drawing).[2]

Notes

  1. ↑ Template:MathWorld
  2. ↑ Feng Yuefeng, Proof without words: Jordan`s inequality, Mathematics Magazine, volume 69, no. 2, 1996, p. 126

Further reading