Evaluate the expression: left(1+cos frac{pi}{ | vedclass

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(B) Let $P = \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

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$\frac{1}{16}$

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(B) Let $P = \left(1+\cos \frac{\pi}{8}ight)\left(1+\cos \frac{2 \pi}{8}ight)\left(1+\cos \frac{3 \pi}{8}ight)\left(1+\cos \frac{4 \pi}{8}ight)\left(1+\cos \frac{5 \pi}{8}ight)\left(1+\cos \frac{6 \pi}{8}ight)\left(1+\cos \frac{7 \pi}{8}ight)$.Since $\cos(\pi - 	heta) = -\cos 	heta$,we have $\cos \frac{7\pi}{8} = -\cos \frac{\pi}{8}$,$\cos \frac{6\pi}{8} = -\cos \frac{2\pi}{8}$,and $\cos \frac{5\pi}{8} = -\cos \frac{3\pi}{8}$.Also,$\cos \frac{4\pi}{8} = \cos \frac{\pi}{2} = 0$.Substituting these values:$P = \left(1+\cos \frac{\pi}{8}ight)\left(1+\cos \frac{\pi}{4}ight)\left(1+\cos \frac{3 \pi}{8}ight)(1+0)\left(1-\cos \frac{3 \pi}{8}ight)\left(1-\cos \frac{2 \pi}{8}ight)\left(1-\cos \frac{\pi}{8}ight)$.Grouping terms:$P = \left(1-\cos^2 \frac{\pi}{8}ight)\left(1-\cos^2 \frac{3 \pi}{8}ight)\left(1+\cos \frac{\pi}{4}ight)\left(1-\cos \frac{\pi}{4}ight)$.$P = \sin^2 \frac{\pi}{8} \cdot \sin^2 \frac{3 \pi}{8} \cdot \left(1-\cos^2 \frac{\pi}{4}ight)$.Using $\sin^2 	heta = \frac{1-\cos 2	heta}{2}$:$P = \left(\frac{1-\cos \frac{\pi}{4}}{2}ight) \left(\frac{1-\cos \frac{3\pi}{4}}{2}ight) \cdot \left(1-\frac{1}{2}ight)$.$P = \left(\frac{1-\frac{1}{\sqrt{2}}}{2}ight) \left(\frac{1+\frac{1}{\sqrt{2}}}{2}ight) \cdot \frac{1}{2} = \frac{1-\frac{1}{2}}{4} \cdot \frac{1}{2} = \frac{1}{8} \cdot \frac{1}{2} = \frac{1}{16}$.

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

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