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The value of \( \Large 2 cos\ x - \cos 3x - \cos 5x \) is equal to

A) \( \Large 16 \cos^{3} x \sin^{2} x \)
B) \( \Large 16 \sin^{3} x \cos^{2} x \)
C) \( \Large 4 \cos^{3} x \sin^{2} x \)
D) \( \Large 4 \sin^{3} x \cos^{2} x \)

Correct Answer:

A) \( \Large 16 \cos^{3} x \sin^{2} x \)

Description for Correct answer:

\( \Large 2 \cos x - \cos 3x - \cos 5x = 2 \cos x - 2 \cos x \cos 4x \)

= \( \Large 2 \cos x \left(1- \cos 4x\right) = 2 \cos x 2 \sin^{2} 2x \)

= \( \Large 4 \cos x \left(2 \sin x \cos\ x\right)^{2} = 16 \sin^{2}x \cos^{3}x \)

Part of solved Trigonometric ratio questions and answers : >> Elementary Mathematics >> Trigonometric ratio

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