The value of 1 + sin 2 π 9 + i cos 2 π 9 1 + sin 2 π 9 − i cos 2 π 9 3 is

1 + sin 2 π 9 + i cos 2 π 9 1 + sin 2 π 9 − i cos 2 π 9 3 = 1 + cos 5 π 18 + i sin 5 π 18 1 + cos 5 π 18 - i sin 5 π 18 3 = 2 cos 2 5 π 36 + i 2 sin 5 π 36 · cos 5 π 36 2 cos 2 5 π 36 - i 2 sin 5 π 36 · cos 5 π 36 3 = cos 5 π 36 + i sin 5 π 36 cos 5 π 36 - i sin 5 π 36 3 = cos 5 π 36 + i sin 5 π 36 6 = cos 6 × 5 π 36 + i sin 6 × 5 π 36 = cos 5 π 6 + i sin 5 π 6 = - 3 2 + i 1 2

Home
PYQ PracticeFree
JEE Rank Predictor
Answer Key

Mathematics - Complex Number Question with Solution | TestHub

MathematicsComplex NumberDe-Moivres theoremHard2 minPYQ_2020

MathematicsHardsingle choice

The value of ${(\frac{1 + \sin \frac{2 π}{9} + i \cos \frac{2 π}{9}}{1 + \sin \frac{2 π}{9} - i \cos \frac{2 π}{9}})}^{3}$ is

Options:

Answer:

Solution:

${(\frac{1 + \sin \frac{2 π}{9} + i \cos \frac{2 π}{9}}{1 + \sin \frac{2 π}{9} - i \cos \frac{2 π}{9}})}^{3}$

$= {(\frac{1 + \cos \frac{5 π}{18} + i \sin \frac{5 π}{18}}{1 + \cos \frac{5 π}{18} - i \sin \frac{5 π}{18}})}^{3} = {(\frac{2 \cos^{2} \frac{5 π}{36} + i 2 \sin \frac{5 π}{36} \cdot \cos \frac{5 π}{36}}{2 \cos^{2} \frac{5 π}{36} - i 2 \sin \frac{5 π}{36} \cdot \cos \frac{5 π}{36}})}^{3}$

$= {(\frac{\cos \frac{5 π}{36} + i \sin \frac{5 π}{36}}{\cos \frac{5 π}{36} - i \sin \frac{5 π}{36}})}^{3} = {(\cos \frac{5 π}{36} + i \sin \frac{5 π}{36})}^{6}$

$= \cos (6 \times \frac{5 π}{36}) + i \sin (6 \times \frac{5 π}{36}) = \cos \frac{5 π}{6} + i \sin \frac{5 π}{6}$

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

Stream:JEESubject:MathematicsTopic:Complex NumberSubtopic:De-Moivres theorem

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...