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UPSC CDS 2 2020 Math Paper

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Question : 55 of 100

Marks: +1, -0

If

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

\sin^6 \theta + \cos^6 \theta + 6 \sin^2 \theta \cos^2 \theta

equal to?

$\frac{1}{4}$
$\frac{3}{4}$
1
$\frac{7}{4}$

Solution:

Concept:

Use algebraic identities and the given condition to simplify the trigonometric expression.

Explanation:

Given

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

.

Square both sides:

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

.

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

.

Since

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

, we get

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

.

\Rightarrow 2\sin\theta\cos\theta = 1 \Rightarrow \sin\theta\cos\theta = \frac{1}{2}

.

Hence

\sin^2\theta\cos^2\theta = \frac{1}{4}

.

Now use the identity

(x+y)^3 = x^3 + y^3 + 3xy(x+y)

with

x = \sin^2\theta

,

y = \cos^2\theta

.

(\sin^2\theta + \cos^2\theta)^3 = \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 have

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

.

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

.

The required expression is

\sin^6\theta + \cos^6\theta + 6\sin^2\theta\cos^2\theta

.

Substitute:

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

.

Plug

\sin^2\theta\cos^2\theta = \frac{1}{4}

:

1 + 3 \times \frac{1}{4} = 1 + \frac{3}{4} = \frac{7}{4}

.

Answer:

\frac{7}{4}

(Option D).

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