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UPSC CDS I 2019 Math Paper

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Question : 59 of 100

Marks: +1, -0

What is the value of

\sin^{6}\theta + \cos^{6}\theta + 3\sin^{2}\theta\cos^{2}\theta - 1

?

0
1
2
4

Solution:

Concept:

Use the identity

(A+B)^3 = A^3 + B^3 + 3AB(A+B)

and

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

.

Explanation:

Let

A = \sin^2\theta

and

B = \cos^2\theta

.

Then

(A+B)^3 = (\sin^2\theta + \cos^2\theta)^3 = 1^3 = 1

.

Expand:

A^3 + B^3 + 3AB(A+B) = \sin^6\theta + \cos^6\theta + 3\sin^2\theta\cos^2\theta(\sin^2\theta + \cos^2\theta)

.

Since

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

, we get

\sin^6\theta + \cos^6\theta + 3\sin^2\theta\cos^2\theta = 1

.

The given expression is

(\sin^6\theta + \cos^6\theta + 3\sin^2\theta\cos^2\theta) - 1 = 1 - 1 = 0

.

Answer:

0

(Option A).

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