Q (1)  Prove that $\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+\cos\frac{3\pi }{13}+\cos\frac{5\pi }{13}=0$

We know that

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

we use this in our problem

$\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{(\frac{3\pi }{13}+\frac{5\pi}{13})}{2}\cos\frac{(\frac{3\pi}{13}-\frac{5\pi }{13})}{2}$

$\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{4\pi }{13}\cos\frac{-\pi}{13}$                             (   we know that          cos(-x) = cos x )

$\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{4\pi }{13}\cos\frac{\pi}{13}$
$\small 2\cos\frac{\pi }{13}(\cos\frac{9\pi }{13}+\cos\frac{4\pi }{13})$
again use the above identity

$\small 2\cos\frac{\pi }{13}(2\cos(\frac{\frac{9\pi }{13}+\frac{4\pi }{13}}{2})\cos(\frac{\frac{9\pi }{13}-\frac{4\pi }{13}}{2})$
$\small 2\cos\frac{\pi }{13}2\cos\frac{\pi }{2}\cos\frac{5\pi }{26}$
we know that

$\small \cos\frac{\pi }{2}$  = 0
So,
$\small 2\cos\frac{\pi }{13}2\cos\frac{\pi }{2}\cos\frac{5\pi }{26}$   = 0  = R.H.S.

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