Question
Find the value of:
\[ \cos^{-1}(\cos \tfrac{2\pi}{3}) + \sin^{-1}(\sin \tfrac{2\pi}{3}) \]
Solution
We use principal value ranges:
- \( \cos^{-1}x \in [0, \pi] \)
- \( \sin^{-1}x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \)
Now,
\[ \cos^{-1}(\cos \tfrac{2\pi}{3}) = \tfrac{2\pi}{3} \]
Since \( \tfrac{2\pi}{3} \in [0, \pi] \)
Next,
\[ \sin^{-1}(\sin \tfrac{2\pi}{3}) \]
But \( \tfrac{2\pi}{3} \notin [-\tfrac{\pi}{2}, \tfrac{\pi}{2}] \), so we use identity:
\[ \sin^{-1}(\sin x) = \pi – x \quad \text{for } \tfrac{\pi}{2} < x < \tfrac{3\pi}{2} \]
Thus,
\[ \sin^{-1}(\sin \tfrac{2\pi}{3}) = \pi – \tfrac{2\pi}{3} = \tfrac{\pi}{3} \]
Therefore,
\[ \cos^{-1}(\cos \tfrac{2\pi}{3}) + \sin^{-1}(\sin \tfrac{2\pi}{3}) = \tfrac{2\pi}{3} + \tfrac{\pi}{3} \]
\[ = \pi \]
Final Answer:
\[ \boxed{\pi} \]
Key Concept
Always apply principal value ranges carefully when evaluating inverse trigonometric functions.