Question

If $L = {\sin^2}\left( {\dfrac{\pi }{{16}}} \right) - {\sin^2}\left( {\dfrac{\pi }{8}} \right)$ and $M = {\cos^2}\left( {\dfrac{\pi }{{16}}} \right) - {\sin^2}\left( {\dfrac{\pi }{8}} \right)$ , then:
A. $M = \dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{2}\cos \dfrac{\pi }{8}$
B. $L = \dfrac{1}{{4\sqrt 2 }} - \dfrac{1}{4}\cos \dfrac{\pi }{8}$
C. $M = \dfrac{1}{{4\sqrt 2 }} + \dfrac{1}{4}\cos \dfrac{\pi }{8}$
D. $L = - \dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{2}\cos \dfrac{\pi }{8}$

Answer 1

Hint: Use Trigonometric identity to solve $L$ and $M$ until you get any angle whose values are available and solve till last. Match the required value with all the options that are correct.

Formula Used:
Trigonometric formula –
${\sin^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$
${\cos^2}\theta = \dfrac{{1 + \cos 2\theta }}{2}$

Complete step by step solution:
Given that,
$L = {\sin^2}\left( {\dfrac{\pi }{{16}}} \right) - {\sin^2}\left( {\dfrac{\pi }{8}} \right)$
$L = \left( {\dfrac{{1 - \cos \left( {\dfrac{\pi }{8}} \right)}}{2}} \right) - \left( {\dfrac{{1 - \cos \left( {\dfrac{\pi }{4}} \right)}}{2}} \right)$
$L = \dfrac{1}{2}\left[ {\cos \left( {\dfrac{\pi }{4}} \right) - \cos \left( {\dfrac{\pi }{8}} \right)} \right]$
$L = \dfrac{1}{2}\left[ {\dfrac{1}{{\sqrt 2 }} - \cos \left( {\dfrac{\pi }{8}} \right)} \right]$
$L = \dfrac{1}{{2\sqrt 2 }} - \dfrac{1}{2}\cos \left( {\dfrac{\pi }{8}} \right)$
And $M = {\cos^2}\left( {\dfrac{\pi }{{16}}} \right) - {\sin^2}\left( {\dfrac{\pi }{8}} \right)$
$M = \left( {\dfrac{{1 + \cos \left( {\dfrac{\pi }{8}} \right)}}{2}} \right) - \left( {\dfrac{{1 - \cos \left( {\dfrac{\pi }{4}} \right)}}{2}} \right)$
$M = \dfrac{1}{2}\left[ {\cos \left( {\dfrac{\pi }{8}} \right) + \cos \left( {\dfrac{\pi }{4}} \right)} \right]$
$M = \dfrac{1}{2}\left[ {\cos \left( {\dfrac{\pi }{8}} \right) + \left( {\dfrac{1}{{\sqrt 2 }}} \right)} \right]$
$M = \dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{2}\cos \left( {\dfrac{\pi }{8}} \right)$

Option ‘A’ is correct

Note: The key concept involved in solving this problem is the good knowledge of Trigonometry formula, ratio, and identities. Students must remember that while applying any trigonometric formula do focus on its angle also.