If 47^{} + 47^{} = k, then what is the value of ² 47^{} - ² 47^{}?

Correct Answer is : −k2−k2-k{2 - k²−k2−k2

Correct Answer is : −k2−k2-k{2 - k²−k2−k2

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UPSC CDS 2 2020 Math Paper

Question : 57 of 100

Marks: +1, -0

\cos 47^{\circ} + \sin 47^{\circ} = k

\cos^{2} 47^{\circ} - \sin^{2} 47^{\circ}

Solution:

Concept:

Use the identity

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

to relate the required expression to

\cos 94^{\circ}

, then find

\cos 94^{\circ}

from the given

k

using

\sin^2 + \cos^2 = 1

Explanation:

Given:

\cos 47^{\circ} + \sin 47^{\circ} = k

Square both sides:

(\cos 47^{\circ} + \sin 47^{\circ})^2 = k^2

Expand:

\cos^2 47^{\circ} + \sin^2 47^{\circ} + 2\cos 47^{\circ}\sin 47^{\circ} = k^2

Since

\cos^2 47^{\circ} + \sin^2 47^{\circ} = 1

and

2\sin\theta\cos\theta = \sin 2\theta

, we get

1 + \sin 94^{\circ} = k^2

Thus

\sin 94^{\circ} = k^2 - 1

Use

\sin^2 94^{\circ} + \cos^2 94^{\circ} = 1

(k^2 - 1)^2 + \cos^2 94^{\circ} = 1

Simplify:

\cos^2 94^{\circ} = 1 - (k^4 - 2k^2 + 1) = 2k^2 - k^4 = k^2(2 - k^2)

Take square root:

\cos 94^{\circ} = \pm k\sqrt{2 - k^2}

94^{\circ}

lies in the second quadrant where cosine is negative, so

\cos 94^{\circ} = -k\sqrt{2 - k^2}

Now

\cos^2 47^{\circ} - \sin^2 47^{\circ} = \cos(2 \times 47^{\circ}) = \cos 94^{\circ} = -k\sqrt{2 - k^2}

Answer:

Option B:

-k\sqrt{2 - k^2}

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