If cosA = 3/4, then what is the value of ≤ft( {A/2} ) ≤ft( {{{3A}}{2}} )?

Correct Answer is : 5/165/165/16

Correct Answer is : 5/165/165/16

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UPSC NDA I 2019 Mathematics Solved Paper

Question : 42 of 120

Marks: +1, -0

cosA = \frac{3}{4},

\sin \left( {\frac{A}{2}} \right)\sin \left( {\frac{{3A}}{2}} \right)?

Solution:

Concept:

Use trigonometric identities:

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

, the product-to-sum formula

2 \sin P \sin Q = \cos(P - Q) - \cos(P + Q)

, and the double-angle formula

\cos 2A = \cos^2 A - \sin^2 A

Explanation:

Given

\cos A = \frac{3}{4}

First, find

\sin A

using

\sin^2 A = 1 - \cos^2 A = 1 - \frac{9}{16} = \frac{7}{16}

Thus

\sin A = \frac{\sqrt{7}}{4}

(positive since

A

is acute in standard contexts).

We need

\sin\left(\frac{A}{2}\right) \sin\left(\frac{3A}{2}\right)

Use the identity:

2 \sin P \sin Q = \cos(P - Q) - \cos(P + Q)

Let

P = \frac{A}{2}

Q = \frac{3A}{2}

. Then

P - Q = -A

\cos(-A) = \cos A

, and

P + Q = 2A

2 \sin\left(\frac{A}{2}\right) \sin\left(\frac{3A}{2}\right) = \cos A - \cos 2A

Therefore

\sin\left(\frac{A}{2}\right) \sin\left(\frac{3A}{2}\right) = \frac{1}{2}(\cos A - \cos 2A)

Now find

\cos 2A

using

\cos 2A = \cos^2 A - \sin^2 A = \frac{9}{16} - \frac{7}{16} = \frac{2}{16} = \frac{1}{8}

Substitute values:

\frac{1}{2}\left(\frac{3}{4} - \frac{1}{8}\right) = \frac{1}{2}\left(\frac{6}{8} - \frac{1}{8}\right) = \frac{1}{2} \times \frac{5}{8} = \frac{5}{16}

Answer:

\frac{5}{16}

(Option B).

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