The expression 2cos frac{pi }{13} cos frac{9p | vedclass

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(B) Given expression: $2\cos \frac{\pi }{13} \cos \frac{9\pi }{13} + \cos \frac{3\pi }{13} + \cos \frac{5\pi }{13}$Using the sum-to-product formula $\

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(B) Given expression: $2\cos \frac{\pi }{13} \cos \frac{9\pi }{13} + \cos \frac{3\pi }{13} + \cos \frac{5\pi }{13}$Using the sum-to-product formula $\cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2}$ for the last two terms:$\cos \frac{3\pi }{13} + \cos \frac{5\pi }{13} = 2\cos \frac{8\pi / 13}{2} \cos \frac{-2\pi / 13}{2} = 2\cos \frac{4\pi }{13} \cos \frac{\pi }{13}$Substituting this back into the expression:$= 2\cos \frac{\pi }{13} \cos \frac{9\pi }{13} + 2\cos \frac{4\pi }{13} \cos \frac{\pi }{13}$$= 2\cos \frac{\pi }{13} \left( \cos \frac{9\pi }{13} + \cos \frac{4\pi }{13} \right)$Using $\cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2}$ again:$= 2\cos \frac{\pi }{13} \left( 2\cos \frac{13\pi / 13}{2} \cos \frac{5\pi / 13}{2} \right)$$= 2\cos \frac{\pi }{13} \left( 2\cos \frac{\pi }{2} \cos \frac{5\pi }{26} \right)$Since $\cos \frac{\pi }{2} = 0$,the entire expression equals $0$.

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

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