Question
Find the value of:
\[ \sin^{-1}(\cos \tfrac{\pi}{9}) \]
Solution
Use identity:
\[ \cos \theta = \sin\left(\frac{\pi}{2} – \theta\right) \]
So,
\[ \sin^{-1}(\cos \tfrac{\pi}{9}) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} – \frac{\pi}{9}\right)\right) \]
\[ = \sin^{-1}\left(\sin \tfrac{7\pi}{18}\right) \]
Now check principal range of \( \sin^{-1}x \):
\[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]
Since \( \tfrac{7\pi}{18} \in \left(0, \frac{\pi}{2}\right) \),
\[ \sin^{-1}(\sin \tfrac{7\pi}{18}) = \tfrac{7\pi}{18} \]
Final Answer:
\[ \boxed{\tfrac{7\pi}{18}} \]
Key Concept
Convert cosine into sine using complementary angle identity, then apply principal value rule.