Prove the following

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 :

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

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

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

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


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

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

 

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