sine inequality

First, note $\sin{x}=\frac{2}{\pi}\cdot x$ for $x=0$ . Since $\frac{d}{dx}\sin{x}\Big|_{x=0}=\cos{0}=1 > \frac{d}{dx}\frac{2}{\pi}x = \frac{2}{\pi}$ , and the first intersection of the derivatives is at $x=\cos^{-1}{\frac{2}{\pi}}$ , we have $\frac{d}{dx}\sin(x)\Big|_{x=a} =\cos{a} > \frac{d}{dx}\frac{2}{\pi}x\Big|_{x=a}=\frac{2}{\pi},$ for any $a\in [0,\cos^{-1}{\frac{2}{\pi}})$ .

This implies $\sin{x}>\frac{2}{\pi}x$ , for any $x\in (0,\cos^{-1}{\frac{2}{\pi}}]$ .

Further, since the following two facts hold:

$\frac{d}{dx}\sin{x}\Big|_{x=\cos^{-1}{\frac{2}{\pi}}} = \frac{d}{dx}\frac{2}{\pi}x\Big|_{x=\cos^{-1}{\frac{2}{\pi}}}$ ,

and $\frac{d^2}{dx^2}\sin{x}\Big|_{x=b} = -\sin{b} < \frac{d^2}{dx^2}\frac{2}{\pi}x\Big|_{x=b} = 0$ , for any $b\in (\cos^{-1}{\frac{2}{\pi}},\frac{\pi}{2}]$ ,

we must have:

$\frac{d}{dx}\sin(x)\Big|_{x=b} < \frac{d}{dx}\frac{2}{\pi}x,$ for any $b\in (\cos^{-1}{\frac{2}{\pi}},\frac{\pi}{2}]$ .

(Also note that we have already shown $\sin{x}>\frac{2}{\pi}x$ for $x = \cos^{-1}{\frac{2}{\pi}}$ .)

This means that $\sin{x}$ is increasing less rapidly than $\frac{2}{\pi}x$ on the entire interval $(\cos^{-1}{\frac{2}{\pi}},\frac{\pi}{2}]$ . So, if $y=\sin{x}$ intersects $y=\frac{2}{\pi}x$ for some $x=c\in (\cos^{-1}{\frac{2}{\pi}},\frac{\pi}{2})$ , then we'd have $\sin{x}\geq\frac{2}{\pi}x$ for all $x \in [0,c]$ and $\sin{x} < \frac{2}{\pi}x$ for all $x\in (c,\frac{\pi}{2}]$ . However, $\sin{x} = \frac{2}{\pi}x$ for $x=\frac{\pi}{2}$ . Thus, the final result, $\sin{x}\geq\frac{2}{\pi}x$ (for $x\in [0,\frac{\pi}{2}]$ ) is shown.

Note a complete solution

$\DeclareMathOperator{\sinc}{sinc}$

Assume that $\sinc$ is decreasing on $[0;π/2]$ where $\sinc$ if defined like that: $\sinc(x)=\frac{\sin(x)}{x}$ .

At $x=π/2,$ $\sinc(x)=2/π$

Therefore $\forall x \in [0,π/2], \ \frac{\sin(x)}{x}\geq 2/π$ which is the inequality we wanted !

in the region [o,pi/2], sin(theta) is an increasing function, as is 2 theta/pi. We note that when theta = pi/2(the upper end of the interval), we have equality since sin(pi/2) = 1 and 2 (pi/2)/pi = 1. All we need do, is choose a theta less than pi/2 and evaluate both expressions. For example let theta = 1.5. Then sin(1.5) = .997494987, while 2*theta/pi = 3/pi = .954929659. Ed Gray