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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NDA Math Complex Numbers

Question : 52 of 66

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\left(\frac{\pi}{2} - x\right) + i\sin\left(\frac{\pi}{2} - x\right)

= e^{i\left(\frac{\pi}{2} - x\right)}

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

z^3

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

= \left(e^{i\left(\frac{\pi}{2} - x\right)}\right)^3

= \left(e^{i\frac{\pi}{2}} e^{-ix}\right)^3

= \left(i e^{-ix}\right)^3

= -i e^{-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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