Question

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

Answer 1

L.H.S.

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

Using that,

$2\mathrm{cosAcosB}=cos(A+B)+\cos (A-B)$

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

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

$=\cos \frac{10\pi }{13}+\cos (-\frac{8\pi }{13})+\cos \frac{3\pi }{13}+\cos \frac{5\pi }{13}\therefore \cos (-\theta )=\cos \theta$

$=\cos \frac{10\pi }{13}+\cos \frac{8\pi }{13}+\cos \frac{3\pi }{13}+\cos \frac{5\pi }{13}$

$=\cos (\pi -\frac{3\pi }{13})+\cos (\pi -\frac{5\pi }{13})+\cos \frac{3\pi }{13}+\cos \frac{5\pi }{13}$

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

$=0$

R.H.S

Hence, proved.