{sin ^2}frac{pi }{8} + {sin ^2}frac{{3pi }}{8 | vedclass

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Question

$\sin \frac{7 \pi}{8}=\sin \left(\pi-\frac{\pi}{8}\right)=\sin \frac{\pi}{8}$

$\sin \frac{5 \pi}{8}=\sin \left(\pi-\frac{3 \pi}{8}\right)=\sin \fra

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\sin \frac{7 \pi}{8}=\sin \left(\pi-\frac{\pi}{8}ight)=\sin \frac{\pi}{8}$

$\sin \frac{5 \pi}{8}=\sin \left(\pi-\frac{3 \pi}{8}ight)=\sin \frac{3 \pi}{8}$

$	herefore $ $	ext { The given value } =2\left[\sin ^{2} \frac{\pi}{8}+\sin ^{2} \frac{3 \pi}{8}ight] $

$=2\left[\sin ^{2} \frac{\pi}{8}+\cos ^{2} \frac{\pi}{8}ight] $

$\left[\because \sin \frac{3 \pi}{8}ight.\left.=\sin \left(\frac{\pi}{2}-\frac{\pi}{8}ight)=\cos \frac{\pi}{8}ight] $

$=2(1)=2$

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

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