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UPSC NDA I and II 2020 Math Solved Paper

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Question : 13 of 120

Marks: +1, -0

What is the value of cos 48° - cos 12° ?

$\frac{\sqrt{5}-1}{4}$
$\frac{1-\sqrt{5}}{4}$
$\frac{\sqrt{5}+1}{2}$
$\frac{1-\sqrt{5}}{8}$

Solution:

Concept:

Use the sum-to-product identity

\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)

and the known value of

\sin 18^\circ

.

Explanation:

Apply the identity with

A = 48^\circ

and

B = 12^\circ

.

\cos 48^\circ - \cos 12^\circ = -2 \sin\left(\frac{48+12}{2}\right) \sin\left(\frac{48-12}{2}\right)

.

This simplifies to

-2 \sin 30^\circ \sin 18^\circ = -2 \times \frac{1}{2} \times \sin 18^\circ = -\sin 18^\circ

.

Now find

\sin 18^\circ

. Let

\theta = 18^\circ

. Then

2\theta = 36^\circ

and

\sin 2\theta = \cos 54^\circ = \cos 3\theta

.

Write

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

and

\cos 3\theta = 4\cos^3\theta - 3\cos\theta

.

Equate:

2 \sin\theta \cos\theta = \cos\theta (4\cos^2\theta - 3)

. Divide by

\cos\theta

(nonzero):

2 \sin\theta = 4\cos^2\theta - 3

.

Replace

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

:

2 \sin\theta = 4(1 - \sin^2\theta) - 3

.

Simplify:

2 \sin\theta = 4 - 4\sin^2\theta - 3

→

2 \sin\theta = 1 - 4\sin^2\theta

.

Bring all terms:

4\sin^2\theta + 2\sin\theta - 1 = 0

.

Solve the quadratic in

\sin\theta

:

\sin\theta = \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}

.

Since

18^\circ

is in the first quadrant,

\sin 18^\circ

is positive, so

\sin 18^\circ = \frac{\sqrt{5} - 1}{4}

.

Thus

\cos 48^\circ - \cos 12^\circ = -\sin 18^\circ = -\frac{\sqrt{5} - 1}{4} = \frac{1 - \sqrt{5}}{4}

.

Answer:

\frac{1 - \sqrt{5}}{4}

(Option B).

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