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UPSC NDA II 2025 Mathematics Paper

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Question : 48 of 120

Marks: +1, -0

If

x+\frac{1}{x}=2cos\theta

x^{3}+\frac{1}{x^{3}}

equal to?

$cos^{3}\theta$
$cos3\theta$
$2cos3\theta$
$3cos3\theta$

Solution:

Concept:

Use the identity

(a+b)^3 = a^3 + b^3 + 3ab(a+b)

and the triple-angle formula

\cos3\theta = 4\cos^3\theta - 3\cos\theta

.

Explanation:

Start with

x + \frac{1}{x} = 2\cos\theta

.

Cube both sides:

\left(x+\frac{1}{x}\right)^3 = (2\cos\theta)^3 = 8\cos^3\theta

.

Expand the left side:

x^3 + \frac{1}{x^3} + 3\left(x+\frac{1}{x}\right) = 8\cos^3\theta

.

Substitute

x+\frac{1}{x} = 2\cos\theta

:

x^3 + \frac{1}{x^3} + 3(2\cos\theta) = 8\cos^3\theta

.

Simplify:

x^3 + \frac{1}{x^3} + 6\cos\theta = 8\cos^3\theta

.

Isolate

x^3+\frac{1}{x^3}

:

x^3+\frac{1}{x^3} = 8\cos^3\theta - 6\cos\theta

.

Factor

2\cos\theta

:

x^3+\frac{1}{x^3} = 2\cos\theta (4\cos^2\theta - 3)

.

Recall the identity

4\cos^2\theta - 3 = \cos3\theta

? Actually,

\cos3\theta = 4\cos^3\theta - 3\cos\theta

, so

4\cos^2\theta - 3

is not the same. Wait, careful:

2\cos\theta(4\cos^2\theta - 3) = 2(4\cos^3\theta - 3\cos\theta) = 2\cos3\theta

. Yes.

Thus

x^3+\frac{1}{x^3} = 2\cos3\theta

.

Answer:

Option C:

2\cos3\theta

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