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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

Answers (1)

best_answer

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.

Posted by

Gautam harsolia

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