cos^4 frac{pi}{8} + cos^4 frac{3pi}{8} + cos | vedclass

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(C) Given expression: $ S = \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \cos^4 \frac{7\pi}{8} $Since $ \cos(\pi - 	heta) =

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$ \frac{3}{2} $

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(C) Given expression: $ S = \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \cos^4 \frac{7\pi}{8} $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} $.Thus,$ \cos^4 \frac{7\pi}{8} = \cos^4 \frac{\pi}{8} $ and $ \cos^4 \frac{5\pi}{8} = \cos^4 \frac{3\pi}{8} $.Substituting these into the expression:$ S = 2 \left( \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} ight) $Using $ a^2 + b^2 = (a+b)^2 - 2ab $:$ S = 2 \left[ \left( \cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} ight)^2 - 2 \cos^2 \frac{\pi}{8} \cos^2 \frac{3\pi}{8} ight] $Since $ \frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8} $,$ \cos \frac{3\pi}{8} = \sin \frac{\pi}{8} $. Thus $ \cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} = \cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8} = 1 $.$ S = 2 \left[ 1^2 - 2 \cos^2 \frac{\pi}{8} \sin^2 \frac{\pi}{8} ight] = 2 \left[ 1 - \frac{1}{2} (2 \sin \frac{\pi}{8} \cos \frac{\pi}{8})^2 ight] $Using $ 2 \sin 	heta \cos 	heta = \sin 2	heta $:$ S = 2 \left[ 1 - \frac{1}{2} \sin^2 \frac{\pi}{4} ight] = 2 \left[ 1 - \frac{1}{2} \left( \frac{1}{\sqrt{2}} ight)^2 ight] = 2 \left[ 1 - \frac{1}{4} ight] = 2 \left( \frac{3}{4} ight) = \frac{3}{2} $.

$ \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \cos^4 \frac{7\pi}{8} = $

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