Prove that: $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + \cos \frac{3 \pi}{13} + \cos \frac{5 \pi}{13} = 0$

$L.H.S. = 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + \cos \frac{3 \pi}{13} + \cos \frac{5 \pi}{13}$
Using the formula $\cos x + \cos y = 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$ for the last two terms:
$= 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + 2 \cos \left(\frac{\frac{3 \pi}{13} + \frac{5 \pi}{13}}{2}\right) \cos \left(\frac{\frac{3 \pi}{13} - \frac{5 \pi}{13}}{2}\right)$
$= 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + 2 \cos \frac{4 \pi}{13} \cos \left(\frac{-\pi}{13}\right)$
Since $\cos(-\theta) = \cos \theta$:
$= 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + 2 \cos \frac{4 \pi}{13} \cos \frac{\pi}{13}$
$= 2 \cos \frac{\pi}{13} \left[ \cos \frac{9 \pi}{13} + \cos \frac{4 \pi}{13} \right]$
Using the formula $\cos x + \cos y = 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$ again:
$= 2 \cos \frac{\pi}{13} \left[ 2 \cos \left(\frac{\frac{9 \pi}{13} + \frac{4 \pi}{13}}{2}\right) \cos \left(\frac{\frac{9 \pi}{13} - \frac{4 \pi}{13}}{2}\right) \right]$
$= 2 \cos \frac{\pi}{13} \left[ 2 \cos \frac{\pi}{2} \cos \frac{5 \pi}{26} \right]$
Since $\cos \frac{\pi}{2} = 0$:
$= 2 \cos \frac{\pi}{13} \times 2 \times 0 \times \cos \frac{5 \pi}{26} = 0 = R.H.S.$