Question

If \( 0 < x < \pi \), then \( \frac{\sin 8x + 7\sin 6x + 18\sin 4x + 12\sin 2x}{\sin 7x + 6\sin 5x + 12\sin 3x} = \)

• A. \( 2\sin x \)
• B. \( \sin x \)
• C. \( \sin 2x \)
• D. \( 2\cos x \)
• E. \( \cos x \)

Hint:

When coefficients like 1, 7, 18, 12 appear, look for Pascal-like patterns or grouping strategies that allow you to factor out a common trigonometric term like $2\cos x$.

Answer 1

The Correct Option is D

Concept: The problem involves simplifying a trigonometric fraction by grouping terms and using the sum-to-product identities: \[ \sin C + \sin D = 2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right) \] The coefficients in the numerator and denominator suggest we should break apart the terms to form pairs that can be simplified.

Step 1: Rewrite the numerator to group terms.
The numerator is \( \sin 8x + 7\sin 6x + 18\sin 4x + 12\sin 2x \). We can split the coefficients to match the structure of the denominator: \[ (\sin 8x + \sin 6x) + 6(\sin 6x + \sin 4x) + 12(\sin 4x + \sin 2x) \]

Step 2: Apply the sum-to-product formula to each pair.
Using \( \sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2}) \):
• \( \sin 8x + \sin 6x = 2\sin 7x \cos x \)
• \( 6(\sin 6x + \sin 4x) = 6(2\sin 5x \cos x) = 12\sin 5x \cos x \)
• \( 12(\sin 4x + \sin 2x) = 12(2\sin 3x \cos x) = 24\sin 3x \cos x \)

Step 3: Factorize and simplify the fraction.
Substitute these back into the numerator: \[ \text{Numerator} = 2\cos x (\sin 7x + 6\sin 5x + 12\sin 3x) \] The expression is: \[ \frac{2\cos x (\sin 7x + 6\sin 5x + 12\sin 3x)}{\sin 7x + 6\sin 5x + 12\sin 3x} \] Canceling the common term in the numerator and denominator, we get: \[ 2\cos x \]