![2
p
x
<=
sin
[?]
(
x
)
<=
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for
x
[?]
[
0
,
p
2
]
{\displaystyle {\frac {2}{\pi }}x\leq \sin(x)\leq x{\text{ for }}x\in \left[0,{\frac {\pi }{2}}\right]} Jordan inequality.svg](http://upload.wikimedia.org/wikipedia/commons/thumb/7/7d/Jordan_inequality.svg/330px-Jordan_inequality.svg.png)
![unit circle with angle x and a second circle with radius
|
E
G
|
=
sin
[?]
(
x
)
{\displaystyle |EG|=\sin(x)}
around E.
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D
E
|
<=
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D
C
^
|
<=
|
D
G
^
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=
sin
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(
x
)
<=
x
<=
p
2
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(
x
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=
2
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(
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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}}} Jordans inequality.svg](http://upload.wikimedia.org/wikipedia/commons/thumb/a/a7/Jordans_inequality.svg/330px-Jordans_inequality.svg.png)
In mathematics, Jordan's inequality, named after Camille Jordan, states that [1]
Contents
It can be proven through the geometry of circles (see drawing). [2]
In mathematics, Jordan's inequality, named after Camille Jordan, states that [1]
It can be proven through the geometry of circles (see drawing). [2]