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UPSC CDS 1 2022 Math Paper

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Question : 48 of 100

Marks: +1, -0

What is the minimum value of

\sin^{4} \theta + \cos^{4} \theta - 2\sin^{2} \theta \cos^{2}\theta

?

0
1
2
Minimum value does not exist

Solution:

Concept:

Rewrite the trigonometric expression using algebraic identity

(a-b)^2 = a^2 + b^2 - 2ab

and the double-angle formula

\cos^2\theta - \sin^2\theta = \cos 2\theta

.

Explanation:

We have

\sin^4 \theta + \cos^4 \theta - 2\sin^2 \theta \cos^2 \theta

.

Write it as

(\cos^2 \theta)^2 + (\sin^2 \theta)^2 - 2(\sin^2 \theta)(\cos^2 \theta)

.

This is of the form

a^2 + b^2 - 2ab

, with

a = \cos^2 \theta

,

b = \sin^2 \theta

.

Thus, it equals

(\cos^2 \theta - \sin^2 \theta)^2

.

Using the identity

\cos^2 \theta - \sin^2 \theta = \cos 2\theta

, we get

(\cos 2\theta)^2 = \cos^2 2\theta

.

The minimum value of

\cos^2 2\theta

is

0

because

\cos^2

of any angle lies between

0

and

1

.

Answer:

0

(Option A)

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