sin ^4 frac{pi}{8}+sin ^4 frac{2 pi}{8}+sin ^ | vedclass

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(C) Let $S = \sum_{k=1}^{7} \sin^4 \frac{k\pi}{8}$.Using the property $\sin(\pi - 	heta) = \sin 	heta$,we have $\sin \frac{7\pi}{8} = \sin \frac{\pi

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let $S = \sum_{k=1}^{7} \sin^4 \frac{k\pi}{8}$.Using the property $\sin(\pi - 	heta) = \sin 	heta$,we have $\sin \frac{7\pi}{8} = \sin \frac{\pi}{8}$,$\sin \frac{6\pi}{8} = \sin \frac{2\pi}{8}$,and $\sin \frac{5\pi}{8} = \sin \frac{3\pi}{8}$.Thus,$S = 2(\sin^4 \frac{\pi}{8} + \sin^4 \frac{2\pi}{8} + \sin^4 \frac{3\pi}{8}) + \sin^4 \frac{4\pi}{8}$.Since $\sin \frac{2\pi}{8} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{4\pi}{8} = \sin \frac{\pi}{2} = 1$,we have $\sin^4 \frac{2\pi}{8} = (\frac{1}{\sqrt{2}})^4 = \frac{1}{4}$ and $\sin^4 \frac{4\pi}{8} = 1^4 = 1$.Now,$\sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} = \sin^4 \frac{\pi}{8} + \cos^4 \frac{\pi}{8} = (\sin^2 \frac{\pi}{8} + \cos^2 \frac{\pi}{8})^2 - 2\sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8} = 1 - \frac{1}{2}\sin^2 \frac{\pi}{4} = 1 - \frac{1}{2}(\frac{1}{2}) = 1 - \frac{1}{4} = \frac{3}{4}$.Substituting these values back into the expression:$S = 2(\frac{3}{4} + \frac{1}{4}) + 1 = 2(1) + 1 = 3$.

$\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{2 \pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{4 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{6 \pi}{8}+\sin ^4 \frac{7 \pi}{8} = ?$

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