The value of left(1+cos frac{pi}{8}right)left | vedclass

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Question

(A) We have the expression: $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac

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$\frac{1}{8}$

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(A) We have the expression: $\left(1+\cos \frac{\pi}{8}ight)\left(1+\cos \frac{3 \pi}{8}ight)\left(1+\cos \frac{5 \pi}{8}ight)\left(1+\cos \frac{7 \pi}{8}ight)$ Since $\cos(\pi - 	heta) = -\cos 	heta$,we have $\cos \frac{7\pi}{8} = -\cos \frac{\pi}{8}$ and $\cos \frac{5\pi}{8} = -\cos \frac{3\pi}{8}$. Substituting these,the expression becomes: $\left(1+\cos \frac{\pi}{8}ight)\left(1-\cos \frac{\pi}{8}ight)\left(1+\cos \frac{3 \pi}{8}ight)\left(1-\cos \frac{3 \pi}{8}ight)$ $= \left(1-\cos^2 \frac{\pi}{8}ight)\left(1-\cos^2 \frac{3 \pi}{8}ight)$ $= \sin^2 \frac{\pi}{8} \sin^2 \frac{3 \pi}{8}$ $= \frac{1}{4} \left(2 \sin \frac{\pi}{8} \sin \frac{3 \pi}{8}ight)^2$ Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$: $= \frac{1}{4} \left(\cos \frac{2\pi}{8} - \cos \frac{4\pi}{8}ight)^2$ $= \frac{1}{4} \left(\cos \frac{\pi}{4} - \cos \frac{\pi}{2}ight)^2$ $= \frac{1}{4} \left(\frac{1}{\sqrt{2}} - 0ight)^2 = \frac{1}{4} 	imes \frac{1}{2} = \frac{1}{8}$

The value of $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$ is

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