Question

Find the principal value of \( \sin^{-1}\left(-\frac{1}{2}\right) \).

Hint:

Always check the principal value range before giving answers for inverse trigonometric functions.

Answer 1

Concept: The principal value range of the inverse sine function is: \[ -\frac{\pi}{2} \le \sin^{-1}(x) \le \frac{\pi}{2} \] We need the angle within this interval whose sine equals \( -\frac{1}{2} \).
Step 1: Recall standard values \[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Step 2: Use odd property of sine Since sine is an odd function: \[ \sin^{-1}(-x) = -\sin^{-1}(x) \] So, \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\sin^{-1}\left(\frac{1}{2}\right) \]
Step 3: Substitute known value \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] \[ \therefore \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \] Final Answer: \[ \boxed{-\frac{\pi}{6}} \] Explanation: The angle must lie in the principal range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), so the negative acute angle is chosen.