Prove that:

Solution:

$\mathrm{LHS}=\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{7 \pi}{15}$

$=\frac{2 \sin \frac{\pi}{15} \cos \frac{\pi}{15}}{2 \sin \frac{\pi}{15}} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{7 \pi}{15}$

$\left[\right.$ On dividing and multiplying by $\left.2 \sin \frac{\pi}{15}\right]$

$=\frac{2 \sin \frac{2 \pi}{15} \times \cos \frac{2 \pi}{15}}{2 \times 2 \sin \frac{\pi}{15}} \cos \frac{4 \pi}{15} \cos \frac{7 \pi}{15}$

$=\frac{2 \sin \frac{4 \pi}{15} \times \cos \frac{4 \pi}{15}}{2 \times 2 \times 2 \sin \frac{\pi}{15}} \cos \frac{7 \pi}{15}$

$=\frac{\sin \frac{8 \pi}{15}}{2 \times 2 \times 2 \sin \frac{\pi}{15}} \cos \frac{7 \pi}{15}$

$=\frac{2 \sin \frac{8 \pi}{15} \cos \frac{7 \pi}{15}}{2 \times 2 \times 2 \times 2 \sin \frac{\pi}{15}}$

$=\frac{2 \sin \frac{8 \pi}{15} \cos \frac{7 \pi}{15}}{16 \sin \frac{\pi}{15}}$

$=\frac{\sin \left(\frac{8 \pi}{15}+\frac{7 \pi}{15}\right)+\sin \left(\frac{8 \pi}{15}-\frac{7 \pi}{15}\right)}{16 \sin \frac{\pi}{15}} \quad[\because 2 \sin A \cos B=\sin (A+B)+\sin (A-B)]$

$=\frac{\sin \pi+\sin \frac{\pi}{15}}{16 \sin \frac{\pi}{15}}$

$=\frac{0+\sin \frac{\pi}{15}}{16 \sin \frac{\pi}{15}}$

$=\frac{\sin \frac{\pi}{15}}{16 \sin \frac{\pi}{15}}$

$=\frac{1}{16}$

= RHS

Hence proved.