Prove that:

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

Vedclass pdf generator app on play store
Vedclass iOS app on app store

$L.H.S.$ $=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$

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

$\left[\cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)\right]$

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

$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \frac{\pi}{13}$

$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \frac{\pi}{13}$

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

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

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

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

$=0=R . H . S.$

Similar Questions

The equation ${(a + b)^2} = 4ab\,{\sin ^2}\theta $ is possible only when

Prove that $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$

If ${\rm{cosec }}A + \cot A = \frac{{11}}{2},$ then $\tan A = $

$tan\,\, 20^o + tan\,\, 40^o + \sqrt 3\,\,  tan\,\, 20^o tan\,\, 40^o$ is equal to

Find the value of $\sin \frac{31 \pi}{3}$.