Let P = \(cos\frac{\pi}{5}cos\frac{2\pi}{5}cos\frac{4\pi}{5}cos\frac{8\pi}{5}\)

⇒ P \(sin\frac{\pi}{5}\) = \(sin\frac{\pi}{5}\) \(cos\frac{\pi}{5}\) \(cos\frac{2\pi}{5}\) \(cos\frac{4\pi}{5}\) \(cos\frac{8\pi}{5}\) 

⇒ P \(sin\frac{\pi}{5}\) = \(\frac{1}{2}\)\(sin\frac{2\pi}{5}\) \(cos\frac{2\pi}{5}\) \(cos\frac{4\pi}{5}\) \(cos\frac{8\pi}{5}\) 

[∵ sin 2A = 2 sin A cos A] 

⇒ P \(sin\frac{\pi}{5}\) = \(\frac{1}{4}[2sin(\frac{2\pi}{5})cos(\frac{2\pi}{5})]cos(\frac{4\pi}{5})cos(\frac{8\pi}{5})\)

⇒  \(sin\frac{\pi}{5}\) = \(\frac{1}{4}[sin(\frac{4\pi}{5})cos(\frac{4\pi}{5})]cos(\frac{8\pi}{5})\)

⇒ P \(sin\frac{\pi}{5}\) = \(\frac{1}{8}[2sin(\frac{4\pi}{5})cos(\frac{4\pi}{5})]cos(\frac{8\pi}{5})\)

⇒ P \(sin\frac{\pi}{5}\) =\(\frac{1}{8}sin(\frac{8\pi}{5})cos(\frac{8\pi}{5})\)

⇒ P \(sin\frac{\pi}{5}\) =\(\frac{1}{16}sin(\frac{16\pi}{5})\)

⇒ P \(sin\frac{\pi}{5}\) = \(\frac{1}{16}sin(3\pi+\frac{\pi}{5})\)

⇒ P \(sin\frac{\pi}{5}\) = \(-\frac{1}{16}sin(\frac{\pi}{5})\)

⇒ P =\(-\frac{1}{16}\)

∴ \(cos\frac{\pi}{5}cos\frac{2\pi}{5}cos\frac{4\pi}{5}cos\frac{8\pi}{5}\) = \(-\frac{1}{16}\) Hence Proved