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UPSC NDA II 2024 Mathematics Paper

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Question : 29 of 120

Marks: +1, -0

What is sin12° sin48° equal to?

$\frac{\sqrt5-1}{4}$
$\frac{\sqrt5+1}{4}$
$\frac{\sqrt5-1}{8}$
$\frac{\sqrt5+1}{8}$

Solution:

Concept:

Use the product-to-sum formula for sines to convert the product into a sum.

Explanation:

The formula is:

\sin A \sin B = \frac{1}{2}[\cos(A-B)-\cos(A+B)]

.

Here

A = 12^\circ

,

B = 48^\circ

.

Then

A-B = -36^\circ

and

\cos(-36^\circ) = \cos 36^\circ

.

Also

A+B = 60^\circ

and

\cos 60^\circ = \frac{1}{2}

.

Thus

\sin 12^\circ \sin 48^\circ = \frac{1}{2}[\cos 36^\circ - \frac{1}{2}]

.

Known value:

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

.

Substitute:

\frac{1}{2}\left[\frac{\sqrt{5}+1}{4} - \frac{1}{2}\right] = \frac{1}{2}\left[\frac{\sqrt{5}+1-2}{4}\right] = \frac{1}{2}\left[\frac{\sqrt{5}-1}{4}\right] = \frac{\sqrt{5}-1}{8}

.

Answer:

Option C:

\frac{\sqrt{5}-1}{8}

is correct.

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