The value of sin^{2}left(frac{pi}{8}right) = | vedclass

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(D) We use the identity $\sin^{2}(	heta) = \frac{1 - \cos(2	heta)}{2}$.Substituting $	heta = \frac{\pi}{8}$:$\sin^{2}\left(\frac{\pi}{8}ight) = \

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$\frac{\sqrt{2}-1}{2\sqrt{2}}$

Vedclass · Answer

(D) We use the identity $\sin^{2}(	heta) = \frac{1 - \cos(2	heta)}{2}$.Substituting $	heta = \frac{\pi}{8}$:$\sin^{2}\left(\frac{\pi}{8}ight) = \frac{1 - \cos\left(2 	imes \frac{\pi}{8}ight)}{2}$$= \frac{1 - \cos\left(\frac{\pi}{4}ight)}{2}$Since $\cos\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$,we have:$= \frac{1 - \frac{1}{\sqrt{2}}}{2} = \frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{2} = \frac{\sqrt{2}-1}{2\sqrt{2}}$.

The value of $\sin^{2}\left(\frac{\pi}{8}\right) = $

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