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

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(C) Let the given expression be $E = \left( {1 + \cos \frac{\pi }{8}} \right)\left( {1 + \cos \frac{{3\pi }}{8}} \right)\left( {1 + \cos \frac{{5\pi }

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$\frac{1}{8}$

Vedclass · Answer

(C) Let the given expression be $E = \left( {1 + \cos \frac{\pi }{8}} ight)\left( {1 + \cos \frac{{3\pi }}{8}} ight)\left( {1 + \cos \frac{{5\pi }}{8}} ight)\left( {1 + \cos \frac{{7\pi }}{8}} ight)$.Since $\cos \frac{{7\pi }}{8} = \cos \left( \pi - \frac{\pi }{8} ight) = -\cos \frac{\pi }{8}$ and $\cos \frac{{5\pi }}{8} = \cos \left( \pi - \frac{{3\pi }}{8} ight) = -\cos \frac{{3\pi }}{8}$,we can rewrite the expression as:$E = \left( {1 + \cos \frac{\pi }{8}} ight)\left( {1 - \cos \frac{\pi }{8}} ight)\left( {1 + \cos \frac{{3\pi }}{8}} ight)\left( {1 - \cos \frac{{3\pi }}{8}} ight)$Using the identity $(1 + \cos 	heta)(1 - \cos 	heta) = 1 - \cos^2 	heta = \sin^2 	heta$,we get:$E = \left( {1 - \cos^2 \frac{\pi }{8}} ight)\left( {1 - \cos^2 \frac{{3\pi }}{8}} ight) = \sin^2 \frac{\pi }{8} \sin^2 \frac{{3\pi }}{8}$Using the identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$:$E = \left( \sin \frac{\pi }{8} \sin \frac{{3\pi }}{8} ight)^2 = \left( \frac{1}{2} \left( \cos \left( \frac{3\pi }{8} - \frac{\pi }{8} ight) - \cos \left( \frac{3\pi }{8} + \frac{\pi }{8} ight) ight) ight)^2$$E = \left( \frac{1}{2} \left( \cos \frac{\pi }{4} - \cos \frac{\pi }{2} ight) ight)^2 = \left( \frac{1}{2} \left( \frac{1}{\sqrt{2}} - 0 ight) ight)^2 = \left( \frac{1}{2\sqrt{2}} ight)^2 = \frac{1}{8}$.

$\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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