[gon]

$\sin (\pi/14) = \sin[(7\pi - 6\pi)/14] = \cos (6\pi/14) = -\cos(4\pi/7)$

$\sin (3\pi/14) = \cos(2\pi/7)$

$\sin(5\pi/14) = \cos (\pi/7) = -\cos(8\pi/7)$

ดังนั้น โจทย์ = $\cos A \cos 2A \cos 4A $ เมื่อ $A = 2\pi/7$

$\cos A \cos 2A \cos 4A = \frac{2\sin A}{2\sin A}(\cos A \cos 2A \cos 4A) = \frac{\sin 2A \cos 2A \cos 4A}{2\sin A} = ... = \frac{\sin 8A}{8\sin A} = \frac{\sin (16\pi/7)}{8\sin(\pi/7)} = \frac{1}{8}$