Test Index

UPSC NDA I 2015 Mathematics Solved Paper

© examsnet.com

Question : 22 of 120

Marks: +1, -0

What is

\left[\frac{\sin\frac{\pi}{6}+i\left(1-\cos\frac{\pi}{6}\right)}{\sin\frac{\pi}{6}-i\left(1-\cos\frac{\pi}{6}\right)}\right]^3

i=\sqrt{-1}

$1$
$-1$
$i$
$-i$

Solution:

Simplifying the above complex number and by using Euler’s formula we get

\left[\frac{\sin 30^{\circ}+i(1-\cos 30^{\circ})}{\sin 30^{\circ}-i(1-\cos 30^{\circ})}\right]^3

=\left[\frac{\sin 30^{\circ}+2i\sin^{2}15^{\circ}}{\sin 30^{\circ}-2i\sin^{2}15^{\circ}}\right]^3

=\left[\frac{2\sin15^{\circ}\cos15^{\circ}+2i\sin^{2}15^{\circ}}{2\sin15^{\circ}\cos15^{\circ}-2i\sin^{2}15^{\circ}}\right]^3

=\left[\frac{2\sin15^{\circ}(\cos15^{\circ}+i\sin15^{\circ})}{2\sin15^{\circ}(\cos15^{\circ}-i\sin15^{\circ})}\right]^3

=\left[\frac{\cos15^{\circ}+i\sin15^{\circ}}{\cos15^{\circ}-i\sin15^{\circ}}\right]^3

=\left(\frac{e^{i\pi/12}}{e^{i(-\pi)/12}}\right)^3

=\left(e^{i\pi/6}\right)^3

=e^{i\frac{3\pi}{6}}=e^{i\pi/2}

=i

© examsnet.com

Go to Question:

Prev Question

Next Question