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UPSC NDA I 2015 Mathematics Paper

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Question : 27 of 120

Marks: +1, -0

What is the real part of

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

–cos 3x
–sin 3x
sin 3x
cos 3x

Solution:

Concept:

Use Euler's formula:

e^{ix} = \cos x + i \sin x

and

e^{-ix} = \cos x - i \sin x

.

Factor out

i

to convert the expression into a known form.

Explanation:

Start with

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

.

Take

i

common:

\left(i\left(\frac{\sin x}{i} + \cos x\right)\right)^3 = i^3 \left(\frac{\sin x}{i} + \cos x\right)^3

.

Note

i^3 = -i

and

\frac{1}{i} = -i

, so

\frac{\sin x}{i} = -i \sin x

.

Thus expression becomes

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

.

Using Euler:

\cos x - i \sin x = e^{-ix}

.

So

(\cos x - i \sin x)^3 = e^{-i3x}

.

Then

-i \cdot e^{-i3x} = -i (\cos 3x - i \sin 3x)

.

Expand:

-i\cos 3x + i^2 \sin 3x = -i\cos 3x - \sin 3x

.

This is

-\sin 3x - i\cos 3x

.

The real part is

-\sin 3x

.

Answer:

B.

-\sin 3x

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