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UPSC CDS I 2025 Mathematics Paper

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Question : 31 of 100

Marks: +1, -0

If

\sin \alpha + \cos \alpha = \sqrt{2}

0 < \alpha < \frac{\pi}{2}

\sin^3 \alpha - \cos^3 \alpha

equal to?

1
$\frac{1}{2}$
$\frac{1}{4}$
0

Solution:

Concept:

Use the identity

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

and find

\sin\alpha - \cos\alpha

from the given sum.

Explanation:

Given

\sin\alpha + \cos\alpha = \sqrt{2}

. Square both sides:

(\sin\alpha+\cos\alpha)^2 = 2

.

This gives

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

. Since

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

, we get

1 + 2\sin\alpha\cos\alpha = 2

.

Thus

2\sin\alpha\cos\alpha = 1

, so

\sin\alpha\cos\alpha = \frac{1}{2}

.

Now

(\sin\alpha-\cos\alpha)^2 = \sin^2\alpha+\cos^2\alpha - 2\sin\alpha\cos\alpha = 1 - 2\left(\frac{1}{2}\right) = 0

.

Hence

\sin\alpha - \cos\alpha = 0

(for

0 < \alpha < \frac{\pi}{2}

, the difference is zero).

Using the identity:

\sin^3\alpha - \cos^3\alpha = (\sin\alpha-\cos\alpha)(\sin^2\alpha + \sin\alpha\cos\alpha + \cos^2\alpha)

.

Substitute the values:

0 \times \left(1 + \frac{1}{2}\right) = 0

.

Answer:

The value is 0. Hence the correct option is D.

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