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

solution:

L.H.S.

$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$

$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \left(\frac{\frac{3 \pi}{13}+\frac{5 \pi}{13}}{2}\right) \cos \left(\frac{\frac{3 \pi}{13}-\frac{5 \pi}{13}}{2}\right)\left[\cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)\right]$

$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \left(\frac{-\pi}{13}\right)$

$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \frac{\pi}{13}$ $\pi\left[\begin{array}{cc}9 \pi & 4 \pi\end{array}\right]$

$=2 \cos \frac{\pi}{13}\left[\cos \frac{9 \pi}{13}+\cos \frac{4 \pi}{13}\right]$

$=2 \cos \frac{\pi}{13}\left[2 \cos \left(\frac{\frac{9 \pi}{13}+\frac{4 \pi}{13}}{2}\right) \cos \left(\frac{\frac{9 \pi}{13}-\frac{4 \pi}{13}}{2}\right)\right]$

$=2 \cos \frac{\pi}{13}\left[2 \cos \frac{\pi}{2} \cos \frac{5 \pi}{26}\right]$

$=2 \cos \frac{\pi}{13} \times 2 \times 0 \times \cos \frac{5 \pi}{26}$

$=0=$ R.H.S