Question

The value of \(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right)\) is equal to

• A. $\frac{\pi}{3}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{6}$
• D. $\frac{\pi}{2}$
• E. $\pi$

Hint:

Always check principal value ranges before evaluating inverse trigonometric functions.

Answer 1

Answer: (E)

Concept: Use principal value ranges: \[ \sin^{-1}x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], \quad \cos^{-1}x \in [0, \pi] \]

Step 1: Evaluate $\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
We know: \[ \cos\theta = -\frac{\sqrt{3}}{2} \] Reference angle: \[ \frac{\pi}{6} \] Since cosine is negative in second quadrant and $\cos^{-1}x \in [0,\pi]$: \[ \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]

Step 2: Evaluate $\sin^{-1}\left(\frac{1}{2}\right)$
\[ \sin\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6} \] Since $\sin^{-1}x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$: \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \]

Step 3: Add both values
\[ \frac{5\pi}{6} + \frac{\pi}{6} = \pi \] Final Conclusion:
Option (E)