# Prove that:

Question:

Prove that: $\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}=\frac{-1}{16}$

Solution:

$\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}$

$=\frac{1}{2 \sin \frac{\pi}{5}}\left(2 \sin \frac{\pi}{5} \cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}\right) \quad$ (Multiplying and dividing by $\frac{1}{2 \sin \frac{\pi}{5}}$ )

$=\frac{1}{2 \sin \frac{\pi}{2}}\left(\sin \frac{2 \pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}\right) \quad(\sin 2 A=2 \sin A \cos A)$

$=\frac{1}{4 \sin \frac{\pi}{5}}\left(2 \sin \frac{2 \pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}\right) \quad$ (Multiplying and dividing by 2 )

$=\frac{1}{4 \sin \frac{\pi}{5}}\left(\sin \frac{4 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}\right)$

$=\frac{1}{8 \sin \frac{\pi}{5}}\left(2 \sin \frac{4 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}\right) \quad$ (Multiplying and dividing by 2 )

$=\frac{1}{8 \sin \frac{\pi}{5}}\left(\sin \frac{8 \pi}{5} \cos \frac{8 \pi}{5}\right)$

$=\frac{1}{16 \sin \frac{\pi}{5}}\left(2 \sin \frac{8 \pi}{5} \cos \frac{8 \pi}{5}\right) \quad$ (Multiplying and dividing by 2)

$=\frac{\sin \frac{16 \pi}{5}}{16 \sin \frac{\pi}{5}}$

$=\frac{\sin \left(3 \pi+\frac{\pi}{5}\right)}{16 \sin \frac{\pi}{5}}$

$=\frac{-\sin \frac{\pi}{5}}{16 \sin \frac{\pi}{5}}$ $[\sin (3 \pi+\theta)=-\sin \theta]$

$=\frac{-1}{16}$