Evaluate: left(1+cos frac{pi}{8}right)left(1+ | vedclass

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(C) Let the expression be $E = \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 \fr

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$\frac{1}{8}$

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(C) Let the expression be $E = \left(1+\cos \frac{\pi}{8}ight)\left(1+\cos \frac{3 \pi}{8}ight)\left(1+\cos \frac{5 \pi}{8}ight)\left(1+\cos \frac{7 \pi}{8}ight)$.Using the identity $\cos(\pi - 	heta) = -\cos 	heta$,we have $\cos \frac{5\pi}{8} = -\cos \frac{3\pi}{8}$ and $\cos \frac{7\pi}{8} = -\cos \frac{\pi}{8}$.Substituting these values,we get:$E = \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 \frac{\pi}{8}ight)$$E = \left(1-\cos^2 \frac{\pi}{8}ight)\left(1-\cos^2 \frac{3 \pi}{8}ight)$$E = \sin^2 \frac{\pi}{8} \sin^2 \frac{3 \pi}{8}$Since $\frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$,we have $\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}$.$E = \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8} = \left(\sin \frac{\pi}{8} \cos \frac{\pi}{8}ight)^2$Using $2 \sin 	heta \cos 	heta = \sin 2	heta$,we get $\sin \frac{\pi}{8} \cos \frac{\pi}{8} = \frac{1}{2} \sin \frac{\pi}{4} = \frac{1}{2} 	imes \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.$E = \left(\frac{1}{2\sqrt{2}}ight)^2 = \frac{1}{8}$.

Evaluate: $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$

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