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

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(B) Let the expression be $E = 2 \cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$.Using the product-to-sum form

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(B) Let the expression be $E = 2 \cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$.Using the product-to-sum formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have:$2 \cos \frac{\pi}{13} \cos \frac{9\pi}{13} = \cos(\frac{10\pi}{13}) + \cos(-\frac{8\pi}{13}) = \cos(\frac{10\pi}{13}) + \cos(\frac{8\pi}{13})$.So,$E = \cos \frac{10\pi}{13} + \cos \frac{8\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$.Using $\cos 	heta = \cos(\pi - 	heta)$,we note:$\cos \frac{10\pi}{13} = \cos(\pi - \frac{3\pi}{13}) = -\cos \frac{3\pi}{13}$.$\cos \frac{8\pi}{13} = \cos(\pi - \frac{5\pi}{13}) = -\cos \frac{5\pi}{13}$.Substituting these into the expression:$E = -\cos \frac{3\pi}{13} - \cos \frac{5\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0$.

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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