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

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Question

(A) Given $L = \sin^{2}\left(\frac{\pi}{16}ight) - \sin^{2}\left(\frac{\pi}{8}ight)$.Using the identity $\sin^{2} 	heta = \frac{1 - \cos 2	heta}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $L = \sin^{2}\left(\frac{\pi}{16}ight) - \sin^{2}\left(\frac{\pi}{8}ight)$.Using the identity $\sin^{2} 	heta = \frac{1 - \cos 2	heta}{2}$,we get:$L = \left(\frac{1 - \cos(\pi/8)}{2}ight) - \left(\frac{1 - \cos(\pi/4)}{2}ight)$$L = \frac{1}{2} \left[ \cos(\pi/4) - \cos(\pi/8) ight] = \frac{1}{2\sqrt{2}} - \frac{1}{2} \cos\left(\frac{\pi}{8}ight)$.Given $M = \cos^{2}\left(\frac{\pi}{16}ight) - \sin^{2}\left(\frac{\pi}{8}ight)$.Using the identity $\cos^{2} 	heta = \frac{1 + \cos 2	heta}{2}$ and $\sin^{2} 	heta = \frac{1 - \cos 2	heta}{2}$,we get:$M = \left(\frac{1 + \cos(\pi/8)}{2}ight) - \left(\frac{1 - \cos(\pi/4)}{2}ight)$$M = \frac{1}{2} \cos\left(\frac{\pi}{8}ight) + \frac{1}{2} \cos(\pi/4) = \frac{1}{2} \cos\left(\frac{\pi}{8}ight) + \frac{1}{2\sqrt{2}}$.Thus,option $A$ is correct.

If $L = \sin^{2}\left(\frac{\pi}{16}\right) - \sin^{2}\left(\frac{\pi}{8}\right)$ and $M = \cos^{2}\left(\frac{\pi}{16}\right) - \sin^{2}\left(\frac{\pi}{8}\right)$,then which of the following is correct?

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