Question

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

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $0$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $0$

LHS :

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

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

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

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

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