Question
Find the principal value of:
\[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) \]
Solution
The principal value range of \( \sin^{-1}x \) is:
\[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]
We know:
\[ \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \]
So,
\[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \]
(since \( -\frac{\pi}{3} \) lies in the principal range)
Final Answer:
\[ \boxed{-\frac{\pi}{3}} \]
Key Concept
Always select the angle within the principal value range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).