# If L = sin squer ( π/16 ) - sin squer ( π/8) and

Question:

If $L=\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$ and

$\mathrm{M}=\cos ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$, then :

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

2. $L=\frac{1}{4 \sqrt{2}}-\frac{1}{4} \cos \frac{\pi}{8}$

3. $M=\frac{1}{4 \sqrt{2}}+\frac{1}{4} \cos \frac{\pi}{8}$

4. $L=-\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}$

Correct Option: 1

Solution:

$L=\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$

$\left(\because \sin ^{2} \theta=\frac{1-\cos 2 \theta}{2}\right)$

$\Rightarrow \mathrm{L}=\left(\frac{1-\cos (\pi / 8)}{2}\right)-\left(\frac{1-\cos (\pi / 4)}{2}\right)$

$\mathrm{L}=\frac{1}{2}\left[\cos \left(\frac{\pi}{4}\right)-\cos \left(\frac{\pi}{8}\right)\right]$

$\mathrm{L}=\frac{1}{2 \sqrt{2}}-\frac{1}{2} \cos \left(\frac{\pi}{8}\right)$

$\mathrm{M}=\cos ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$

$M=\frac{1+\cos (\pi / 8)}{2}-\frac{1-\cos (\pi / 4)}{2}$

$\mathrm{M}=\frac{1}{2} \cos \left(\frac{\pi}{8}\right)+\frac{1}{2 \sqrt{2}}$