Jordan's Inequality

From ProofWiki
Jump to navigation Jump to search


$\dfrac 2 {\pi} x \le \sin x \le x$

for all $x$ in the interval $\left[{0 \,.\,.\, \dfrac {\pi} 2}\right]$


The right inequality is true by Sine Inequality.

The left inequality is true for $x = 0$ and $x = \dfrac {\pi} 2$, where we have equality.

Now consider $x \in \left({0 \,.\,.\, \dfrac {\pi} 2}\right)$.

From Shape of Sine Function, $\sin x$ is concave on the interval $\left[{0 \,.\,.\, \pi }\right]$.

Letting $x_1 = 0$, $x_2 = x$ and $x_3 = \dfrac {\pi} 2$ in the definition of Concave Real Function we obtain:

$\dfrac {\sin x} x \ge \dfrac {1 - \sin x} {\dfrac {\pi} 2 - x}$

Rearranging gives the desired inequality.


Source of Name

This entry was named for Marie Ennemond Camille Jordan.