Question

Let \[ \sum_{k=1}^{15} \sin (t_k) = 0 \quad {and} \quad \sum_{k=1}^{15} \sin (3t_k) = \frac{-24}{5}, \] where \( t_1, t_2, t_3, \dots \) are real numbers. Then the value of the sum \[ \sum_{k=1}^{15} \sin^3 (t_k) \] is equal to

• A. \( \frac{4}{5} \)
• B. \( \frac{6}{5} \)
• C. \( \frac{3}{10} \)
• D. \( \frac{24}{5} \)
• E. \( \frac{96}{5} \)

Hint:

Use trigonometric sum identities to simplify summations involving cube terms.

Answer 1

The Correct Option is B

We use the identity: \[ \sum \sin^3 x = \frac{3}{4} \sum \sin x - \frac{1}{4} \sum \sin(3x) \] Since \( \sum_{k=1}^{15} \sin (t_k) = 0 \), we get: \[ \sum_{k=1}^{15} \sin^3 (t_k) = \frac{-1}{4} \sum_{k=1}^{15} \sin (3t_k) \] Substituting \( \sum_{k=1}^{15} \sin (3t_k) = \frac{-24}{5} \): \[ \sum_{k=1}^{15} \sin^3 (t_k) = \frac{-1}{4} \times \left( \frac{-24}{5} \right) = \frac{6}{5} \] Final Answer: \[ \boxed{\frac{6}{5}} \]