Question:
Find the principal value of:
\[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) \]
Concept:
The principal value range of \( \sin^{-1}x \) is:
\[ -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \]
—Solution:
Step 1: Recall standard value
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Step 2: Apply negative sign
Since sine is negative, the angle lies in the principal range:
\[ \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \]
Step 3: Check range
\[ -\frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]
So, it is the principal value.
—Final Answer:
\[ \boxed{-\frac{\pi}{3}} \]