Prove that:

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

We have:

$\left|\cos x \cos \left(60^{\circ}-x\right) \cos \left(60^{\circ}+x\right)\right|$

$=\left|\cos x\left(\cos ^{2} 60^{\circ}-\sin ^{2} x\right)\right| \quad\left[\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~B}=\cos (\mathrm{A}-\mathrm{B}) \cos (\mathrm{A}+\mathrm{B})\right]$

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

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

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

$=\left|\frac{1}{4} \cos x\left\{-3+4 \cos ^{2} x\right\}\right|$

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

$=\left|\frac{1}{4} \cos 3 x\right| \quad\left[\because \cos 3 x=4 \cos ^{3} x-3 \cos x\right]$

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

$\therefore\left|\cos x \cos \left(60^{\circ}-x\right) \cos \left(60^{\circ}+x\right)\right| \leq \frac{1}{4}$