sin^4 frac{pi}{8} + sin^4 frac{3pi}{8} + sin^ | 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) = \s

Vedclass · Accepted Answer

$\frac{3}{2}$

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 $2 \sin^2 	heta = 1 - \cos 2	heta$ का उपयोग करने पर,$\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$.इसलिए,$\sin^4 	heta = \left( \frac{1 - \cos 2	heta}{2} ight)^2 = \frac{1}{4} (1 - 2 \cos 2	heta + \cos^2 2	heta)$.$S = 2 \left[ \frac{1}{4} (1 - 2 \cos \frac{\pi}{4} + \cos^2 \frac{\pi}{4}) + \frac{1}{4} (1 - 2 \cos \frac{3\pi}{4} + \cos^2 \frac{3\pi}{4}) ight]$.$S = \frac{1}{2} \left[ (1 - 2(\frac{1}{\sqrt{2}}) + \frac{1}{2}) + (1 - 2(-\frac{1}{\sqrt{2}}) + \frac{1}{2}) ight]$.$S = \frac{1}{2} \left[ (1 - \sqrt{2} + 0.5) + (1 + \sqrt{2} + 0.5) ight] = \frac{1}{2} [1.5 + 1.5] = \frac{1}{2} (3) = \frac{3}{2}$.

$\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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