In mathematics, Jordan's inequality, named after Camille Jordan, states that[1] 2 π x ≤ sin ( x ) ≤ x for x ∈ [ 0 , π 2 ] . {\displaystyle {\frac {2}{\pi }}x\leq \sin(x)\leq x{\text{ for }}x\in \left[0,{\frac {\pi }{2}}\right].} 2 π x ≤ sin ( x ) ≤ x for x ∈ [ 0 , π 2 ] {\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 | E G | = sin ( x ) {\displaystyle |EG|=\sin(x)} around E. | D E | ≤ | D C ^ | ≤ | D G ^ | ⇔ sin ( x ) ≤ x ≤ π 2 sin ( x ) ⇒ 2 π x ≤ sin ( x ) ≤ x {\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] NotesLoading content...Further readingLoading content...External linksLoading content...Loading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.
![{\displaystyle {\frac {2}{\pi }}x\leq \sin(x)\leq x{\text{ for }}x\in \left[0,{\frac {\pi }{2}}\right].}](http://79.170.44.75/hostdoctordemo.co.uk/downloads/vpn/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9hNmNhNGZlOGJiZDNiZTQ5NTIzM2U3MzNhZGEzOTM2N2NjMzc4NDQw)
![{\displaystyle {\frac {2}{\pi }}x\leq \sin(x)\leq x{\text{ for }}x\in \left[0,{\frac {\pi }{2}}\right]}](http://79.170.44.75/hostdoctordemo.co.uk/downloads/vpn/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9hNGE2ODBlZDY1MjQzMGNjMWUyNGE4ZWVhMDI1OTQ3NGQzN2IyNTUx)
unit circle with angle x and a second circle with radius 
around E. 
