Prove that:

$\mathrm{LHS}=\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}$

On dividing and multiplying by $2 \sin \frac{2 \pi}{15}$, we get

$=\frac{1}{2 \sin \frac{2 \pi}{15}} \times\left(2 \sin \frac{2 \pi}{15} \times \cos \frac{2 \pi}{15}\right) \times \cos \frac{4 \pi}{15} \times \cos \frac{8 \pi}{15} \times \cos \frac{16 \pi}{15}$

$=\frac{1}{2 \times 2 \sin \frac{2 \pi}{15}} \times\left(2 \sin \frac{4 \pi}{15} \times \cos \frac{4 \pi}{15}\right) \times \cos \frac{8 \pi}{15} \times \cos \frac{16 \pi}{15}$

$=\frac{1}{2 \times 4 \sin \frac{2 \pi}{15}}\left(2 \sin \frac{8 \pi}{15} \times \cos \frac{8 \pi}{15}\right) \times \cos \frac{16 \pi}{15}$

$=\frac{1}{2 \times 8 \sin \frac{2 \pi}{15}}\left(2 \sin \frac{16 \pi}{15} \times \cos \frac{16 \pi}{15}\right)$

$=\frac{1}{16 \sin \frac{2 \pi}{15}}\left(\sin \frac{32 \pi}{15}\right)$

$=-\frac{1}{16 \sin \frac{2 \pi}{15}}\left(\sin 2 \pi-\frac{32 \pi}{15}\right) \quad[\because \sin (2 \pi-\theta)=-\sin \theta]$

$=-\frac{1}{16 \sin \frac{2 \pi}{15}} \sin \left(-\frac{2 \pi}{15}\right)$

$=\frac{1}{16}=\mathrm{RHS}$

Hence proved.