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

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(C) We know that $\cos(\pi - 	heta) = -\cos 	heta$. Therefore,$\cos \frac{{7\pi }}{8} = \cos(\pi - \frac{\pi }{8}) = -\cos \frac{\pi }{8}$ and $\cos

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$\frac{1}{8}$

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(C) We know that $\cos(\pi - 	heta) = -\cos 	heta$. Therefore,$\cos \frac{{7\pi }}{8} = \cos(\pi - \frac{\pi }{8}) = -\cos \frac{\pi }{8}$ and $\cos \frac{{5\pi }}{8} = \cos(\pi - \frac{{3\pi }}{8}) = -\cos \frac{{3\pi }}{8}$. Substituting these into the expression: $E = \left( {1 + \cos \frac{\pi }{8}} ight)\left( {1 + \cos \frac{{3\pi }}{8}} ight)\left( {1 - \cos \frac{{3\pi }}{8}} ight)\left( {1 - \cos \frac{\pi }{8}} ight)$ Grouping the terms: $E = \left( {1 + \cos \frac{\pi }{8}} ight)\left( {1 - \cos \frac{\pi }{8}} ight)\left( {1 + \cos \frac{{3\pi }}{8}} ight)\left( {1 - \cos \frac{{3\pi }}{8}} ight)$ Using the identity $(a+b)(a-b) = a^2 - b^2$: $E = \left( {1 - \cos^2 \frac{\pi }{8}} ight)\left( {1 - \cos^2 \frac{{3\pi }}{8}} ight)$ $E = \sin^2 \frac{\pi }{8} \sin^2 \frac{{3\pi }}{8}$ Using the identity $2\sin A \sin B = \cos(A-B) - \cos(A+B)$: $E = \left( \sin \frac{\pi }{8} \sin \frac{{3\pi }}{8} ight)^2 = \left( \frac{1}{2} \left( \cos \frac{{2\pi }}{8} - \cos \frac{{4\pi }}{8} ight) ight)^2$ $E = \left( \frac{1}{2} \left( \cos \frac{\pi }{4} - \cos \frac{\pi }{2} ight) ight)^2$ Since $\cos \frac{\pi }{4} = \frac{1}{\sqrt{2}}$ and $\cos \frac{\pi }{2} = 0$: $E = \left( \frac{1}{2} \left( \frac{1}{\sqrt{2}} - 0 ight) ight)^2 = \left( \frac{1}{2\sqrt{2}} ight)^2 = \frac{1}{8}$.

$\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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