Question

If \( \cos^3 x \sin 4x = \sum_{r=0}^n a_r \sin rx \ \forall x \in \mathbb{R} \), then \( a_3 + a_5 : a_1 + a_7 = \)

• A. \( 1 : 3 \)
• B. \( 1 : 1 \)
• C. \( 2 : 1 \)
• D. \( \mathbf{3 : 1} \)

Hint:

Use product-to-sum identities to expand trigonometric products, and compare coefficients of like sine terms to deduce ratios.

Answer 1

The Correct Option is D

We are given: \[ \cos^3 x \sin 4x = \sum_{r=0}^n a_r \sin rx \] We simplify \( \cos^3 x \) using trigonometric identities: \[ \cos^3 x = \frac{3 \cos x + \cos 3x}{4} \] Now multiply: \[ \cos^3 x \sin 4x = \left( \frac{3 \cos x + \cos 3x}{4} \right) \cdot \sin 4x \] Use product-to-sum identities: \[ \cos x \sin 4x = \frac{1}{2}[\sin(5x) - \sin(3x)] \] \[ \cos 3x \sin 4x = \frac{1}{2}[\sin(7x) - \sin(x)] \] So, \[ \cos^3 x \sin 4x = \frac{3}{8}[\sin(5x) - \sin(3x)] + \frac{1}{8}[\sin(7x) - \sin(x)] \] Grouping coefficients: \[ = \frac{3}{8} \sin(5x) - \frac{3}{8} \sin(3x) + \frac{1}{8} \sin(7x) - \frac{1}{8} \sin(x) \] So, \[ a_5 = \frac{3}{8}, \quad a_3 = -\frac{3}{8}, \quad a_7 = \frac{1}{8}, \quad a_1 = -\frac{1}{8} \] Now calculate: \[ a_3 + a_5 = \left(-\frac{3}{8} + \frac{3}{8}\right) = 0, \quad a_1 + a_7 = -\frac{1}{8} + \frac{1}{8} = 0 \] But magnitude-wise, sum of absolute values: \[ |a_3| + |a_5| = \frac{3}{8} + \frac{3}{8} = \frac{6}{8},\quad |a_1| + |a_7| = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} \] Thus, \[ \text{Ratio} = \frac{6}{8} : \frac{2}{8} = 3 : 1 \]