Prove that:

$\frac{\pi}{3}=60^{\circ}$

We have:

$|\sin x \sin (60-x) \sin (60+x)|$

$=\left|\sin x\left(\sin ^{2} 60-\sin ^{2} x\right)\right|$

$\left[\because \sin (A+B) \sin (A-B)=\sin ^{2} A-\sin ^{2} B\right]$

$=\left|\sin x\left(\frac{3}{4}-\sin ^{2} x\right)\right|$

$=\left|\frac{1}{4} \sin x\left(3-4 \sin ^{2} x\right)\right|$

$=\left|\frac{1}{4} 3 \sin x-4 \sin ^{3} x\right|$

$=\frac{1}{4}|\sin 3 \mathrm{x}| \quad\left(\because 3 \sin x-4 \sin ^{3} x=\sin 3 x\right)$

$\leq \frac{1}{4} \quad(\because|\sin x| \leq 1$ for all $x)$

$\therefore|\sin x \sin (60-x) \sin (60+x)| \leq \frac{1}{4}$

Hence proved.