What is the real part of ( x + i x)³ where i={-1} ? [NDA I 2015]

Correct Answer is : −sin⁡3x- 3x−sin3x

Correct Answer is : −sin⁡3x- 3x−sin3x

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UPSC NDA I 2015 Mathematics Solved Paper

Question : 27 of 120

Marks: +1, -0

(\sin x + i\cos x)^3

i=\sqrt{-1}

Solution:

Applying Euler’s equation to the above question, we get

z = sin x + i cos x

=\cos(\pi/2 - x) + i\sin(\pi/2 - x)

=e^{i(\pi/2 - x)}

Now by substituting the above value in the given question, we get

z^3

=(\sin x + i\cos x)^3

=(e^{i(\pi/2 - x)})^3

=(e^{i\pi/2}e^{-ix})^3

=(ie^{-ix})^3

=-ie^{-i3x}

=-i(\cos 3x - i\sin 3x)

=i(-\cos 3x + \sin 3x)

=-\sin 3x - i\cos 3x

\operatorname{Re}(z)=-\sin 3x

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