Question

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

• A. $2$
• B. $0$
• C. $1$
• D. $3$

Answer 1

The correct option is B $0$

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

using,

$\cos (A+B)+\cos (A-B)=2\cos A\cos B$

Here, $A=\frac{\pi }{13}$ $B=\frac{9\pi }{13}$

So,

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

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

$\cos (-A)=\cos A$

So,

$=\cos \left(\frac{10\pi }{13}\right)+\cos \left(\frac{8\pi }{13}\right)+\cos \left(\frac{3\pi }{13}\right)+\cos \left(\frac{5\pi }{13}\right)$

using, $\cos A=-\cos (\pi -A)$

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

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

$=0$