The equation sin⁻¹x = 2 sin⁻¹a has a solution for:
The equation sin⁻¹x = 2 sin⁻¹a has a solution for:
- A. |a| ≤ 1/√2
- B. |a| ≥ 1/√2
- C. all real a
- D. |a| ≤ 1/2
Answer: A) |a| ≤ 1/√2
Explanation: sin⁻¹x = 2 sin⁻¹a → x = sin(2 sin⁻¹a) = 2a√(1−a²). Also sin⁻¹x ∈ [−π/2, π/2] → 2 sin⁻¹a ∈ [−π/2, π/2] → sin⁻¹a ∈ [−π/4, π/4] → |a| ≤ sin π/4 = 1/√2.
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