Jordan's inequality

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

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

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

unit circle with angle x and a second circle with radius ${\displaystyle |EG|=\sin(x)}$ around E. ${\displaystyle {\begin{aligned}&|DE|\leq |{\widehat {DC}}|\leq |{\widehat {DG}}|\\\Leftrightarrow &\sin(x)\leq x\leq {\tfrac {\pi }{2}}\sin(x)\\\Rightarrow &{\tfrac {2}{\pi }}x\leq \sin(x)\leq x\end{aligned}}}$

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