Evaluate: sin-1(−1/2) + 2cos-1(−√3/2)
Solution:
Using principal values:
\[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \]
(Since range of sin-1(x) is \([- \pi/2, \pi/2]\))
\[ \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6} \]
(Since range of cos-1(x) is \([0, \pi]\))
Substitute values:
\[ \sin^{-1}\left(-\frac{1}{2}\right) + 2\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{6} + 2 \times \frac{5\pi}{6} \]
\[ = -\frac{\pi}{6} + \frac{10\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2} \]
Final Answer:
Value = \[ \frac{3\pi}{2} \]