Question
Find the value of:
\[ \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) \]
Solution
Using standard values:
\[ \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \]
(since \( \cos \frac{\pi}{3} = \frac{1}{2} \) and lies in principal range)
Also,
\[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \]
(since \( \sin \frac{\pi}{6} = \frac{1}{2} \))
Therefore,
\[ \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} + 2 \cdot \frac{\pi}{6} \]
\[ = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} \]
Final Answer:
\[ \boxed{\frac{2\pi}{3}} \]
Key Concept
Use standard inverse trigonometric values within principal ranges for quick evaluation.