Question:

$\frac{\sin 7 x + 6 \sin 5 x + 17 \sin 3 x + 12 \sin x}{\sin 6 x + 5 \sin 4 x + 12 \sin 2 x}$ equals:

Updated On: Jun 23, 2024

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The Correct Option is D

Explanation:
Given,

\frac{\sin 7 x + 6 \sin 5 x + 17 \sin 3 x + 12 \sin x}{\sin 6 x + 5 \sin 4 x + 12 \sin 2 x}

= \frac{\sin 7 x + \sin 5 x + 5 \sin 5 x + 5 \sin 3 x + 12 \sin 3 x + 12 \sin x}{\sin 6 x + 5 \sin 4 x + 12 \sin 2 x}

By using,

\sin x + \sin y = 2 \sin (\frac{x + y}{2}) \cos (\frac{x - y}{2})

Therefore,

= \frac{(2 \sin (\frac{7 x + 5 x}{2}) \cos (\frac{7 x - 5 x}{2}) + 5 \times (2 \sin (\frac{5 x + 3 x}{2}) \cos (\frac{5 x - 3 x}{2})) + 12 \times (2 \sin (\frac{3 x + x}{2}) \cos (\frac{3 x - x}{2}))}{(\sin 6 x + 5 \sin 4 x + 12 \sin 2 x)}

= \frac{2 \sin 6 x \cos x + 10 \sin 4 x \cos x + 24 \sin 2 x \cos x}{(\sin 6 x + 5 \sin 4 x + 12 \sin 2 x)}

= \frac{2 \cos x (\sin 6 x + 5 \sin 4 x + 12 \sin 2 x)}{(\sin 6 x + 5 \sin 4 x + 12 \sin 2 x)}

= 2 \cos x

Hence, the correct option is (D).

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