Jordan's inequality

From HandWiki
[math]\displaystyle{ \frac{2}{\pi}x\leq \sin(x) \leq x\text{ for }x \in \left[0,\frac{\pi}{2}\right] }[/math]
unit circle with angle x and a second circle with radius [math]\displaystyle{ |EG|=\sin(x) }[/math] around E. [math]\displaystyle{ \begin{align}&|DE|\leq|\widehat{DC}|\leq|\widehat{DG}|\\ \Leftrightarrow &\sin(x) \leq x \leq\tfrac{\pi}{2}\sin(x)\\ \Rightarrow &\tfrac{2}{\pi}x \leq \sin(x)\leq x \end{align} }[/math]

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

[math]\displaystyle{ \frac{2}{\pi}x\leq \sin(x) \leq x\text{ for }x \in \left[0,\frac{\pi}{2}\right]. }[/math]

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

Notes

  1. Weisstein, Eric W.. "Jordan's inequality". http://mathworld.wolfram.com/JordansInequality.html. 
  2. Feng Yuefeng, Proof without words: Jordan`s inequality, Mathematics Magazine, volume 69, no. 2, 1996, p. 126

Further reading

External links