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COMEDK 2025 Afternoon Solved Paper

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Question : 126 of 180

Marks: +1, -0

If

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

\left(32 \sin \frac{A}{2} \sin \frac{5 A}{2}\right)=

7
16
11
8

Solution:

Concept:

Use product-to-sum identity to simplify the product of sines.

Explanation:

Apply

\sin\alpha \sin\beta = \frac{\cos(\alpha-\beta) - \cos(\alpha+\beta)}{2}

with

\alpha = \frac{A}{2}

and

\beta = \frac{5A}{2}

.

\sin\frac{A}{2} \sin\frac{5A}{2} = \frac{\cos(2A) - \cos(3A)}{2}

.

Then

32\sin\frac{A}{2}\sin\frac{5A}{2} = 16(\cos2A - \cos3A)

.

Given

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

, compute:

\cos2A = 2\cos^2 A - 1 = 2\cdot\frac{9}{16} - 1 = \frac{1}{8}

.

\cos3A = 4\cos^3 A - 3\cos A = 4\cdot\frac{27}{64} - 3\cdot\frac{3}{4} = \frac{108}{64} - \frac{9}{4} = \frac{27}{16} - \frac{36}{16} = -\frac{9}{16}

.

Thus

\cos2A - \cos3A = \frac{1}{8} - \left(-\frac{9}{16}\right) = \frac{2}{16} + \frac{9}{16} = \frac{11}{16}

.

Multiply by 16:

16 \cdot \frac{11}{16} = 11

.

Answer:

11 (Option C).

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