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

Answer: (A)

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