The value of cos ((2π)/(7)) + cos ((4π)/(7)) + ((6π)/(7)) is equal to

Correct answer: B. - (1)/(2) Given, cos(2π)/(7) + cos(4π)/(7) + cos(6π)/(7) = (1)/(2 ((π)/(7)))\ 2 ((2π)/(7)) ((π)/(7)) + 2 ((4π)/(7)) ((π)/(7)) + 2 ((6π)/(7)) ((π)/(7))\ = (1)/(2sin(π)/(7))\ sin(3π)/(7) - sin(π)/(7) + sin(5π)/(7) - sin(3π)/(7) + sin(7π)/(7) - sin(5π)/(7)\ Since, 2cosAsinB = (A + B) - (A - B) = (1)/(2sin(π)/(7))\ - sin(π)/(7) + sin π\ = (- 1)/(2), (∵ sin π = 0)

JEE Main 202227 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 27 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main 2022 · 27 June · Morning Shift · Mathematics · Trigonometric Ratios & Identities

The value of cos ((2π)/(7)) + cos ((4π)/(7)) + ((6π)/(7)) is equal...

The value of

cos

\left(\dfrac{2\pi}{7}\right) + cos

\left(\dfrac{4\pi}{7}\right) + \cos \left(\dfrac{6\pi}{7}\right)

is equal to

Options

A $- 1$
B $- \dfrac{1}{2}$
C $- \dfrac{1}{3}$
D $- \dfrac{1}{4}$

Correct answer

B. $- \dfrac{1}{2}$

Step-by-step solution

Given, $cos\dfrac{2\pi}{7} + cos\dfrac{4\pi}{7} + cos\dfrac{6\pi}{7}$

$= \dfrac{1}{2\sin \left(\dfrac{\pi}{7}\right)}\left\{2\cos \left(\dfrac{2\pi}{7}\right)\sin \left(\dfrac{\pi}{7}\right) + 2\cos \left(\dfrac{4\pi}{7}\right)\sin \left(\dfrac{\pi}{7}\right) + 2\cos \left(\dfrac{6\pi}{7}\right)\sin \left(\dfrac{\pi}{7}\right)\right\}$

$= \dfrac{1}{2sin\dfrac{\pi}{7}}\left\{sin\dfrac{3\pi}{7} - sin\dfrac{\pi}{7} + sin\dfrac{5\pi}{7} - sin\dfrac{3\pi}{7} + sin\dfrac{7\pi}{7} - sin\dfrac{5\pi}{7}\right\}$

Since, $2cosAsinB = \sin \left(A + B\right) - \sin \left(A - B\right)$

$= \dfrac{1}{2sin\dfrac{\pi}{7}}\left\{- sin\dfrac{\pi}{7} + sin \pi\right\}$

$= \dfrac{- 1}{2}, \left(∵ sin \pi = 0\right)$

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