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 :
Correct Option: , 4
Solution:
$\mathrm{L}+\mathrm{M}=1-2 \sin ^{2} \frac{\pi}{8}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$ ...(i)
and $\mathrm{L}-\mathrm{M}=-\cos \frac{\pi}{8}$ ...(ii)
From equation (i) and (ii),
$\mathrm{L}=\frac{1}{2}\left(\frac{1}{\sqrt{2}}-\cos \frac{\pi}{8}\right)=\frac{1}{2 \sqrt{2}}-\frac{1}{2} \cos \frac{\pi}{8}$ and
$M=\frac{1}{2}\left(\frac{1}{\sqrt{2}}+\cos \frac{\pi}{8}\right)=\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}$