Jordan's Inequality

Theorem

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

for all $x$ in the interval $\closedint 0 {\dfrac \pi 2}$

Proof

The right hand side inequality is true by Sine Inequality.

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

Now consider $x \in \openint 0 {\dfrac \pi 2}$.

From Shape of Sine Function, $\sin x$ is concave on the interval $\closedint 0 \pi$.

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.

$\blacksquare$

Source of Name

This entry was named for Marie Ennemond Camille Jordan.

Sources

• Weisstein, Eric W. "Jordan's Inequality." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/JordansInequality.html