Question

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

Answer 1

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

$=cos\frac{10\pi }{13}+cos\frac{8\pi }{13}+cos\frac{3\pi }{13}+cos\frac{5\pi }{13}$ $[\because 2\cos A\cos B=\cos (A+B)+\cos (A-B\left)\right]$

$=\cos (\pi -\frac{3\pi }{13})+\cos (\pi -\frac{5\pi }{13})+cos\frac{3\pi }{13}+cos\frac{5\pi }{13}$ $[\because \cos (\pi -\theta )=-\cos \theta ]$

$=-cos\frac{3\pi }{13}-cos\frac{5\pi }{13}+cos\frac{3\pi }{13}+cos\frac{5\pi }{13}=0=RHS$

Hence proved