मान ज्ञात कीजिए: sin ^4 frac{pi}{8} + sin ^4 | vedclass

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

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$\frac{3}{2}$

Vedclass · Answer

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

मान ज्ञात कीजिए: $\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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