Question

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

Answer 1

$L.H.S.=2\cos \frac{\pi }{13}\cos \frac{9\pi }{13}+\cos \frac{3\pi }{13}+\cos \frac{5\pi }{13}=\cos (\frac{9\pi }{13}+\frac{\pi }{13})+\cos (\frac{9\pi }{13}-\frac{\pi }{13})+\cos \frac{3\pi }{13}+\cos \frac{5\pi }{13}=\cos \left(\frac{10\pi }{13}\right)+\cos \left(\frac{8\pi }{13}\right)+\cos \frac{3\pi }{13}+\cos \frac{5\pi }{13}=\cos \left(\frac{10\pi }{13}\right)+\cos \left(\frac{8\pi }{13}\right)+\cos (\pi -\frac{10\pi }{13})+\cos (\pi -\frac{8\pi }{13})$
$=\cos \left(\frac{10\pi }{13}\right)+\cos \left(\frac{8\pi }{13}\right)-\cos \left(\frac{10\pi }{13}\right)-\cos \left(\frac{8\pi }{13}\right)\{\because \cos (\pi -x)=-\cos x\}$
$=0=R.H.S.$

Hence proved.