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

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Question

(C) हमारे पास व्यंजक है: $S = \sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}$ चूंकि $\sin(\pi - 	heta)

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) हमारे पास व्यंजक है: $S = \sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}$ चूंकि $\sin(\pi - 	heta) = \sin 	heta$,इसलिए $\sin \frac{5 \pi}{8} = \sin \frac{3 \pi}{8}$ और $\sin \frac{7 \pi}{8} = \sin \frac{\pi}{8}$ है। अतः,$S = 2 \left[ \sin ^4 \frac{\pi}{8} + \sin ^4 \frac{3 \pi}{8} ight]$। $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$ का उपयोग करने पर,$\sin^4 	heta = \left( \frac{1 - \cos 2	heta}{2} ight)^2$ प्राप्त होता है। $S = 2 \left[ \left( \frac{1 - \cos(\pi/4)}{2} ight)^2 + \left( \frac{1 - \cos(3\pi/4)}{2} ight)^2 ight]$ $S = 2 \left[ \left( \frac{1 - 1/\sqrt{2}}{2} ight)^2 + \left( \frac{1 - (-1/\sqrt{2})}{2} ight)^2 ight]$ $S = \frac{2}{4} \left[ (1 - 1/\sqrt{2})^2 + (1 + 1/\sqrt{2})^2 ight] = \frac{1}{2} \left[ (1 + 1/2 - \sqrt{2}) + (1 + 1/2 + \sqrt{2}) ight]$ $S = \frac{1}{2} \left[ 3 ight] = \frac{3}{2}$.

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

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