The value of ((1 + sin(2π)/(9) + icos(2π)/(9))/(1 + sin(2π)/(9) - icos(2π)/(9))) ^ 3 is

Correct answer: C. - (1)/(2) ( 3 - i ) ((1 + sin(2π)/(9) + icos(2π)/(9))/(1 + sin(2π)/(9) - icos(2π)/(9))) ^ 3 = ((1 + cos(5π)/(18) + isin(5π)/(18))/(1 + cos(5π)/(18) - isin(5π)/(18))) ^ 3 = ( 2 ^ 2 (5π)/(36) + i2sin(5π)/(36) · cos(5π)/(36) 2 ^ 2 (5π)/(36) - i2sin(5π)/(36) · cos(5π)/(36) ) ^ 3 = ((cos(5π)/(36) + isin(5π)/(36))/(cos(5π)/(36) - isin(5π)/(36))) ^ 3 = (cos(5π)/(36) + isin(5π)/(36)) ^ 6 = (6 × (5π)/(36)) + isin(6 × (5π)/(36)) = cos(5π)/(6) + isin(5π)/(6) = - 3 2 + i(1)/(2)

JEE Main 20202 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 2 September
Shift: Morning Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2020 · 2 September · Morning Shift · Mathematics · Complex Number

The value of ((1 + sin(2π)/(9) + icos(2π)/(9))/(1 + sin(2π)/(9) -...

The value of ${\left(\dfrac{1 + sin\dfrac{2\pi}{9} + icos\dfrac{2\pi}{9}}{1 + sin\dfrac{2\pi}{9} - icos\dfrac{2\pi}{9}}\right)}^{3} is

Options

A $\dfrac{1}{2} ( 1 - i\sqrt{3} )$
B $\dfrac{1}{2} ( \sqrt{3} - i )$
C $- \dfrac{1}{2} ( \sqrt{3} - i )$
D $- \dfrac{1}{2}\left(1 - i\sqrt{3}\right)$

Correct answer

C. $- \dfrac{1}{2} ( \sqrt{3} - i )$

Step-by-step solution

${\left(\dfrac{1 + sin\dfrac{2\pi}{9} + icos\dfrac{2\pi}{9}}{1 + sin\dfrac{2\pi}{9} - icos\dfrac{2\pi}{9}}\right)}^{3}$

$= {\left(\dfrac{1 + cos\dfrac{5\pi}{18} + isin\dfrac{5\pi}{18}}{1 + cos\dfrac{5\pi}{18} - isin\dfrac{5\pi}{18}}\right)}^{3} = {\left(\dfrac{2\cos^{2}\dfrac{5\pi}{36} + i2sin\dfrac{5\pi}{36} \cdot cos\dfrac{5\pi}{36}}{2\cos^{2}\dfrac{5\pi}{36} - i2sin\dfrac{5\pi}{36} \cdot cos\dfrac{5\pi}{36}}\right)}^{3}$

$= {\left(\dfrac{cos\dfrac{5\pi}{36} + isin\dfrac{5\pi}{36}}{cos\dfrac{5\pi}{36} - isin\dfrac{5\pi}{36}}\right)}^{3} = {\left(cos\dfrac{5\pi}{36} + isin\dfrac{5\pi}{36}\right)}^{6}$

$= \cos \left(6 \times \dfrac{5\pi}{36}\right) + isin\left(6 \times \dfrac{5\pi}{36}\right) = cos\dfrac{5\pi}{6} + isin\dfrac{5\pi}{6}$

$= - \dfrac{\sqrt{3}}{2} + i\dfrac{1}{2}$

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