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UPSC NDA I 2024 Mathematics Paper

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Question : 37 of 120

Marks: +1, -0

What is (sin 9° - cos 9°) equal to ?

$-\frac{\sqrt{5-\sqrt{5}}}{2}$
$-\frac{\sqrt{5-\sqrt{3}}}{2}$
$\frac{\sqrt{5-\sqrt{5}}}{2}$
$\frac{\sqrt{5-\sqrt{5}}}{4}$
None of these

Solution:

Concept:

Use trigonometric identities:

\sin A \cos B - \cos A \sin B = \sin(A - B)

and

\cos 36^\circ = \frac{\sqrt{5}+1}{4}

.

Explanation:

Start with

\sin 9^\circ - \cos 9^\circ

.

Factor out

\sqrt{2}

:

\sin 9^\circ - \cos 9^\circ = \sqrt{2}\left(\frac{1}{\sqrt{2}}\sin 9^\circ - \frac{1}{\sqrt{2}}\cos 9^\circ\right)

.

Note

\frac{1}{\sqrt{2}} = \sin 45^\circ = \cos 45^\circ

.

Rewrite:

\sqrt{2}(\sin 9^\circ\cos 45^\circ - \cos 9^\circ\sin 45^\circ)

.

Apply identity:

\sin A\cos B - \cos A\sin B = \sin(A - B)

.

So it equals

\sqrt{2}\sin(9^\circ - 45^\circ) = \sqrt{2}\sin(-36^\circ)

.

Since

\sin(-\theta) = -\sin\theta

, we get

-\sqrt{2}\sin 36^\circ

.

Express

\sin 36^\circ

using

\cos 36^\circ

:

\sin 36^\circ = \sqrt{1 - \cos^2 36^\circ}

.

Given

\cos 36^\circ = \frac{\sqrt{5}+1}{4}

, compute:

\cos^2 36^\circ = \frac{5 + 2\sqrt{5} + 1}{16} = \frac{6 + 2\sqrt{5}}{16}

.

Then

1 - \cos^2 36^\circ = 1 - \frac{6 + 2\sqrt{5}}{16} = \frac{16 - 6 - 2\sqrt{5}}{16} = \frac{10 - 2\sqrt{5}}{16}

.

So

\sin 36^\circ = \sqrt{\frac{10 - 2\sqrt{5}}{16}} = \frac{\sqrt{10 - 2\sqrt{5}}}{4}

.

Therefore

-\sqrt{2}\sin 36^\circ = -\sqrt{2} \cdot \frac{\sqrt{10 - 2\sqrt{5}}}{4} = -\frac{\sqrt{2(10 - 2\sqrt{5})}}{4}

.

Simplify inside:

2(10 - 2\sqrt{5}) = 20 - 4\sqrt{5}

. Factor:

4(5 - \sqrt{5})

.

Thus

-\frac{\sqrt{4(5 - \sqrt{5})}}{4} = -\frac{2\sqrt{5 - \sqrt{5}}}{4} = -\frac{\sqrt{5 - \sqrt{5}}}{2}

.

Answer:

-\frac{\sqrt{5-\sqrt{5}}}{2}

, which corresponds to option A.

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