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UPSC NDA II 2015 Mathematics Paper

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Question : 56 of 120

Marks: +1, -0

On simplifying

\frac{{{{\sin }^3}{\rm{A}} + \sin 3{\rm{\;A}}}}{{\sin {\rm{A}}}} + \frac{{{{\cos }^3}{\rm{A}} - \cos 3{\rm{\;A}}}}{{\cos {\rm{A}}}}

we get

sin 3 A
cos 3 A
sin A + cos A
3

Solution:

Concept:

Use triple angle formulas:

\sin 3A = 3\sin A - 4\sin^3 A

and

\cos 3A = 4\cos^3 A - 3\cos A

.

Explanation:

Substitute the formulas into the given expression.

First term:

\frac{\sin^3 A + \sin 3A}{\sin A} = \frac{\sin^3 A + (3\sin A - 4\sin^3 A)}{\sin A}

.

Simplify numerator:

3\sin A - 3\sin^3 A

.

Divide by

\sin A

:

3 - 3\sin^2 A

.

Second term:

\frac{\cos^3 A - \cos 3A}{\cos A} = \frac{\cos^3 A - (4\cos^3 A - 3\cos A)}{\cos A}

.

Simplify numerator:

3\cos A - 3\cos^3 A

.

Divide by

\cos A

:

3 - 3\cos^2 A

.

Add both terms:

(3 - 3\sin^2 A) + (3 - 3\cos^2 A) = 6 - 3(\sin^2 A + \cos^2 A)

.

Since

\sin^2 A + \cos^2 A = 1

, we get

6 - 3(1) = 3

.

Answer:

3

(Option D).

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