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UPSC NDA I 2019 Mathematics Solved Paper

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Question : 38 of 120

Marks: +1, -0

If

\sin \alpha + \cos \alpha = p

\cos^{2}(2\alpha)

equal to?

$p^{2}$
$p^{2} - 1$
$p^{2}(2- p^{2})$
$p^{2} + 1$

Solution:

Concept:

Use the identity

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

and

\sin 2x = 2 \sin x \cos x

.

Explanation:

Square both sides of

\sin \alpha + \cos \alpha = p

.

This gives

\sin^2 \alpha + 2 \sin \alpha \cos \alpha + \cos^2 \alpha = p^2

.

Replace

\sin^2 \alpha + \cos^2 \alpha

with

1

and

2 \sin \alpha \cos \alpha

with

\sin 2\alpha

.

So

1 + \sin 2\alpha = p^2

, hence

\sin 2\alpha = p^2 - 1

.

Now

\cos^2 2\alpha = 1 - \sin^2 2\alpha = 1 - (p^2 - 1)^2

.

Simplify:

1 - (p^4 - 2p^2 + 1) = 2p^2 - p^4 = p^2 (2 - p^2)

.

Answer:

\cos^2 2\alpha = p^2 (2 - p^2)

(Option C).

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