# .

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

$$\frac{2}{\pi}x\leq \sin{x} \leq x\text{ for }x \in \left[0,\frac{\pi}{2}\right]$$.

It can be proven through the geometry of circles (see drawing).[1]

unit circle with angle x and a second circle with radius |$$EG|=\sin(x)$$ around E. \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}

Notes

http://planetmath.org/encyclopedia/ProofOfJordansInequality.html