The value of 2 ((π)/(8)) ((2π)/(8)) ((3π)/(8)) ((5π)/(8)) ((6π)/(8)) ((7π)/(8)) is :

Correct answer: B. (1)/(8) ⇒ 2 ((π)/(8)) ((2π)/(8)) ((3π)/(8)) ((5π)/(8)) ((6π)/(8)) ((7π)/(8)) ⇒ 2 ((π)/(8)) ((2π)/(8)) ((3π)/(8)) (π - (3π)/(8)) (π - (2π)/(8)) (π - (π)/(8)) ⇒ 2 ^ 2 (π)/(8) ^ 2 ((2π)/(8)) ^ 2 ((3π)/(8)) ⇒ 2 ^ 2 (π)/(8) ^ 2 ((2π)/(8)) ^ 2 ((3π)/(8)) ⇒ 2 ^ 2 (π)/(8) ^ 2 ((π)/(4)) ^ 2 ((3π)/(8)) ⇒ ^ 2 (π)/(8) ^ 2 ((3π)/(8)) ⇒ ^ 2 (π)/(8) ^ 2 ((π)/(2) - (π)/(8)) ⇒ ^ 2 (π)/(8) ^ 2 (π)/(8) ⇒ (1)/(4) (2sin(π)/(8)cos(π)/(8)) ^ 2 (1)/(4) ^ 2 (π)/(4) = (1)/(8)

JEE Main 202126 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 26 August
Shift: Evening Shift
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main 2021 · 26 August · Evening Shift · Mathematics · Trigonometric Ratios & Identities

The value of 2 ((π)/(8)) ((2π)/(8)) ((3π)/(8)) ((5π)/(8))...

The value of

2\sin \left(\dfrac{\pi}{8}\right)\sin \left(\dfrac{2\pi}{8}\right)\sin \left(\dfrac{3\pi}{8}\right)\sin \left(\dfrac{5\pi}{8}\right)\sin \left(\dfrac{6\pi}{8}\right)\sin \left(\dfrac{7\pi}{8}\right)

is :

Options

A $\dfrac{1}{4\sqrt{2}}$
B $\dfrac{1}{8}$
C $\dfrac{1}{8\sqrt{2}}$
D $\dfrac{1}{4}$

Correct answer

B. $\dfrac{1}{8}$

Step-by-step solution

$⇒ 2\sin \left(\dfrac{\pi}{8}\right)\sin \left(\dfrac{2\pi}{8}\right)\sin \left(\dfrac{3\pi}{8}\right)\sin \left(\dfrac{5\pi}{8}\right)\sin \left(\dfrac{6\pi}{8}\right)\sin \left(\dfrac{7\pi}{8}\right)$

$⇒ 2\sin \left(\dfrac{\pi}{8}\right)\sin \left(\dfrac{2\pi}{8}\right)\sin \left(\dfrac{3\pi}{8}\right)\sin \left(\pi - \dfrac{3\pi}{8}\right)\sin \left(\pi - \dfrac{2\pi}{8}\right)\sin \left(\pi - \dfrac{\pi}{8}\right)$

$⇒ 2\sin^{2}\dfrac{\pi}{8}\sin^{2}\left(\dfrac{2\pi}{8}\right)\sin^{2}\left(\dfrac{3\pi}{8}\right)$

$⇒ 2\sin^{2}\dfrac{\pi}{8}\sin^{2}\left(\dfrac{\pi}{4}\right)\sin^{2}\left(\dfrac{3\pi}{8}\right)$

$⇒ \sin^{2}\dfrac{\pi}{8}\sin^{2}\left(\dfrac{3\pi}{8}\right)$

$⇒ \sin^{2}\dfrac{\pi}{8}\sin^{2}\left(\dfrac{\pi}{2} - \dfrac{\pi}{8}\right)$

$⇒ \sin^{2}\dfrac{\pi}{8}\cos^{2}\dfrac{\pi}{8}$

$⇒ \dfrac{1}{4}{\left(2sin\dfrac{\pi}{8}cos\dfrac{\pi}{8}\right)}^{2}$

$\dfrac{1}{4}\sin^{2}\dfrac{\pi}{4} = \dfrac{1}{8}$

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