यदि L =sin ^{2}left(frac{pi}{16}right)-sin ^{ | vedclass

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Question

$L =\sin ^{2}\left(\frac{\pi}{16}ight)-\sin ^{2}\left(\frac{\pi}{8}ight)$

$\left(\because \sin ^{2} 	heta=\frac{1-\cos 2 	heta}{2}ight)$

Vedclass · Accepted Answer

$M =\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}$

Vedclass · Answer

$L =\sin ^{2}\left(\frac{\pi}{16}ight)-\sin ^{2}\left(\frac{\pi}{8}ight)$

$\left(\because \sin ^{2} 	heta=\frac{1-\cos 2 	heta}{2}ight)$

$\Rightarrow L =\left(\frac{1-\cos (\pi / 8)}{2}ight)-\left(\frac{1-\cos (\pi / 4)}{2}ight)$

$L =\frac{1}{2}\left[\cos \left(\frac{\pi}{4}ight)-\cos \left(\frac{\pi}{8}ight)ight]$

$L =\frac{1}{2 \sqrt{2}}-\frac{1}{2} \cos \left(\frac{\pi}{8}ight)$

$M =\cos ^{2}\left(\frac{\pi}{16}ight)-\sin ^{2}\left(\frac{\pi}{8}ight)$

$M =\frac{1+\cos (\pi / 8)}{2}-\frac{1-\cos (\pi / 4)}{2}$

$M =\frac{1}{2} \cos \left(\frac{\pi}{8}ight)+\frac{1}{2 \sqrt{2}}$

यदि $L =\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$ तथा $M =\cos ^{2}$$\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$ है, तो

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