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UPSC NDA I and II 2020 Math Solved Paper

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Question : 10 of 120

Marks: +1, -0

What is

\sin 3x + \cos 3x + 4 \sin^{3} x - 3 \sin x + 3 \cos x - 4 \cos^{3} x

equal to?

0
1
2 sin 2x
4 cos 4x

Solution:

Concept:

Use the trigonometric identities for

\sin 3x

and

\cos 3x

to simplify the expression.

Explanation:

Start with the given expression:

\sin 3x + \cos 3x + 4 \sin^{3} x - 3 \sin x + 3 \cos x - 4 \cos^{3} x

.

Recall the identities:

\sin 3x = 3 \sin x - 4 \sin^{3} x

and

\cos 3x = 4 \cos^{3} x - 3 \cos x

.

Substitute these into the expression:

(3 \sin x - 4 \sin^{3} x) + (4 \cos^{3} x - 3 \cos x) + 4 \sin^{3} x - 3 \sin x + 3 \cos x - 4 \cos^{3} x

.

Now group like terms:

\sin x

terms:

3 \sin x - 3 \sin x = 0

.

\cos x

terms:

-3 \cos x + 3 \cos x = 0

.

\sin^{3} x

terms:

-4 \sin^{3} x + 4 \sin^{3} x = 0

.

\cos^{3} x

terms:

4 \cos^{3} x - 4 \cos^{3} x = 0

.

All terms cancel, leaving

0

.

Answer:

0

(Option A).

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