If L = ^ 2 ((π)/(16)) - ^ 2 ((π)/(8)) and M = ^ 2 ((π)/(16)) - ^ 2 ((π)/(8))

Correct answer: D. M = 1 2 2 + (1)/(2)cos(π)/(8) L = ((π)/(16) + (π)/(8)) ((π)/(16) - (π)/(8)) = sin(3π)/(16) · (- (π)/(16)) = (1)/(2)( ((3π)/(16) + (π)/(16)) - ((3π)/(16) - (π)/(16))) = (1)/(2)( 1 2 - cos(π)/(8)) M = ((π)/(16) + (π)/(8)) ((π)/(16) - (π)/(8)) = cos(3π)/(16) · (- (π)/(16)) = (1)/(2)( ((3π)/(16) + (π)/(16)) + ((3π)/(16) - (π)/(16))) = (1)/(2)( 1 2 + cos(π)/(8))

JEE Main 20205 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 5 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main 2020 · 5 September · Evening Shift · Mathematics · Trigonometric Ratios & Identities

If L = ^ 2 ((π)/(16)) - ^ 2 ((π)/(8)) and M = ^ 2 ((π)/(16)) - ^ 2...

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

Options

A $L = - \dfrac{1}{2\sqrt{2}} + \dfrac{1}{2}cos\dfrac{\pi}{8}$
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 $M = \dfrac{1}{2\sqrt{2}} + \dfrac{1}{2}cos\dfrac{\pi}{8}$

Correct answer

D. $M = \dfrac{1}{2\sqrt{2}} + \dfrac{1}{2}cos\dfrac{\pi}{8}$

Step-by-step solution

$L = \sin \left(\dfrac{\pi}{16} + \dfrac{\pi}{8}\right)\sin \left(\dfrac{\pi}{16} - \dfrac{\pi}{8}\right)$
$= sin\dfrac{3\pi}{16} \cdot \sin \left(- \dfrac{\pi}{16}\right)$

$= \dfrac{1}{2}\left(\cos \left(\dfrac{3\pi}{16} + \dfrac{\pi}{16}\right) - \cos \left(\dfrac{3\pi}{16} - \dfrac{\pi}{16}\right)\right) = \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}} - cos\dfrac{\pi}{8}\right)$
$M = \cos \left(\dfrac{\pi}{16} + \dfrac{\pi}{8}\right)\cos \left(\dfrac{\pi}{16} - \dfrac{\pi}{8}\right)$

$= cos\dfrac{3\pi}{16} \cdot \cos \left(- \dfrac{\pi}{16}\right)$
$= \dfrac{1}{2}\left(\cos \left(\dfrac{3\pi}{16} + \dfrac{\pi}{16}\right) + \cos \left(\dfrac{3\pi}{16} - \dfrac{\pi}{16}\right)\right) = \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}} + cos\dfrac{\pi}{8}\right)$

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