Evaluate: sin-1(−√3/2) + cos-1(√3/2)
Solution:
Using principal values:
\[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \]
(Since range of sin-1(x) is \([- \pi/2, \pi/2]\))
\[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \]
(Since range of cos-1(x) is \([0, \pi]\))
Substitute values:
\[ -\frac{\pi}{3} + \frac{\pi}{6} = -\frac{2\pi}{6} + \frac{\pi}{6} = -\frac{\pi}{6} \]
Final Answer:
Value = \[ -\frac{\pi}{6} \]