Prove that:
Question:

Prove that:

$\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{3 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{5 \pi}{15} \cos \frac{6 \pi}{15} \cos \frac{7 \pi}{15}=\frac{1}{128}$

Solution:

$\mathrm{LHS}=\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{3 \pi}{15} \cos \frac{5 \pi}{15} \cos \frac{6 \pi}{15} \cos \frac{7 \pi}{15}$

$=\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15}\left(\cos \frac{3 \pi}{15} \cos \frac{6 \pi}{15}\right) \times\left(-\cos \frac{8 \pi}{15}\right)$

$=-\frac{1}{2}\left[\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15}\right] \times \frac{1}{2} \times\left(\cos \frac{3 \pi}{15} \cos \frac{6 \pi}{15}\right)$

$=-\frac{1}{2} \times \frac{2^{3}}{2^{4} \sin \frac{\pi}{15}}\left[2 \sin \frac{\pi}{15} \cos \frac{\pi}{15} \quad \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \quad \cos \frac{8 \pi}{15}\right] \times \frac{2}{2^{2} \times \sin \frac{3 \pi}{15}}\left(2 \sin \frac{3 \pi}{15} \cos \frac{3 \pi}{15} \cos \frac{6 \pi}{15}\right)$

$=-\frac{2^{3}}{132 \sin \frac{\pi}{3}}\left[\sin \frac{2 \pi}{15} \quad \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \quad \cos \frac{8 \pi}{15}\right] \times \frac{2}{4 \sin \frac{3 \pi}{2}}\left(\sin \frac{6 \pi}{15} \cos \frac{6 \pi}{15}\right)$

$=-\frac{2^{2}}{32 \sin \frac{\pi}{15}}\left[\begin{array}{llll}2 \sin \frac{2 \pi}{15} & \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} & \cos \frac{8 \pi}{15}\end{array}\right] \times \frac{1}{4 \sin \frac{3 \pi}{15}}\left(2 \sin \frac{6 \pi}{15} \cos \frac{6 \pi}{15}\right)$

$=-\frac{2}{32 \sin \frac{\pi}{15}}\left[\sin \frac{8 \pi}{15} \cos \frac{8 \pi}{15}\right] \times \frac{\sin \frac{12 \pi}{15}}{4 \sin \frac{3 \pi}{15}}$

$=-\frac{1}{32 \sin \frac{\pi}{15}}\left[\sin \frac{16 \pi}{15}\right] \times \frac{\sin \frac{12 \pi}{15}}{4 \sin \frac{3 \pi}{15}}$

$=-\frac{\sin \left(\pi+\frac{\pi}{15}\right)}{128 \sin \frac{\pi}{15}} \times \frac{\sin \left(\pi-\frac{3 \pi}{15}\right)}{\sin \frac{3 \pi}{15}}$

$=-\frac{-\sin \frac{\pi}{15}}{128 \sin \frac{\pi}{15}} \times \frac{\sin \frac{3 \pi}{15}}{\sin \frac{3 \pi}{15}}$

$=\frac{1}{128}$

= RHS

Hence proved.