Find the principal value (in radians) of $displaystylesi{n}^{-1}!\left(-\frac{3}{5}\right)$.

Step-by-step solution

1. Recall the range of $si{n}^{-1}$ is $[-\frac{\pi }{2},\frac{\pi }{2}]$.
2. Thus $si{n}^{-1}!\left(-\frac{3}{5}\right)=-,si{n}^{-1}!\left(\frac{3}{5}\right)$.
3. Compute $si{n}^{-1}!\left(\frac{3}{5}\right)\approx 0.6435$ rad using a calculator.
4. Hence the principal value is $-0.6435$ rad.
5. (Rounded to four decimal places.)

Concept tested: principal value of arcsine for negative arguments

Published 7 October 2026 · How this solution was prepared and checked