Question

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 :

• A. $M =\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}$
• B. $L=\frac{1}{4 \sqrt{2}}-\frac{1}{4} \cos \frac{\pi}{8}$
• C. $M =\frac{1}{4 \sqrt{2}}+\frac{1}{4} \cos \frac{\pi}{8}$
• D. $L=-\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}$

Answer 1

The Correct Option is A

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