Evaluate the product: left(1+cos frac{pi}{8}r | vedclass

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Question

(A) Let the given expression be $P = \prod_{k=1}^{7} \left(1+\cos \frac{k\pi}{8}ight)$.Using the identity $1+\cos 	heta = 2\cos^2 \frac{	heta}{2}$

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$\frac{1}{16}$

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(A) Let the given expression be $P = \prod_{k=1}^{7} \left(1+\cos \frac{k\pi}{8}ight)$.Using the identity $1+\cos 	heta = 2\cos^2 \frac{	heta}{2}$,we have $1+\cos \frac{k\pi}{8} = 2\cos^2 \frac{k\pi}{16}$.Alternatively,note that $\cos(\pi - 	heta) = -\cos 	heta$,so $1+\cos(\pi - 	heta) = 1-\cos 	heta$.Pairing terms: $\left(1+\cos \frac{\pi}{8}ight)\left(1+\cos \frac{7\pi}{8}ight) = (1+\cos \frac{\pi}{8})(1-\cos \frac{\pi}{8}) = 1-\cos^2 \frac{\pi}{8} = \sin^2 \frac{\pi}{8}$.Similarly,$\left(1+\cos \frac{2\pi}{8}ight)\left(1+\cos \frac{6\pi}{8}ight) = (1+\cos \frac{\pi}{4})(1-\cos \frac{\pi}{4}) = 1-\cos^2 \frac{\pi}{4} = \sin^2 \frac{\pi}{4} = \frac{1}{2}$.And $\left(1+\cos \frac{3\pi}{8}ight)\left(1+\cos \frac{5\pi}{8}ight) = (1+\cos \frac{3\pi}{8})(1-\cos \frac{3\pi}{8}) = 1-\cos^2 \frac{3\pi}{8} = \sin^2 \frac{3\pi}{8}$.The middle term is $\left(1+\cos \frac{4\pi}{8}ight) = 1+\cos \frac{\pi}{2} = 1+0 = 1$.Thus,$P = \sin^2 \frac{\pi}{8} \cdot \sin^2 \frac{3\pi}{8} \cdot \frac{1}{2} \cdot 1$.Since $\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}$,$P = \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8} \cdot \frac{1}{2} = \frac{1}{4} (2 \sin \frac{\pi}{8} \cos \frac{\pi}{8})^2 \cdot \frac{1}{2} = \frac{1}{4} (\sin \frac{\pi}{4})^2 \cdot \frac{1}{2} = \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{16}$.

Evaluate the product: $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{4 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{6 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$

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