Evaluate \( \sin^{-1}(\sin \frac{17\pi}{8}) \)
Step-by-Step Solution
We need to evaluate:
\[ \sin^{-1}\left(\sin \frac{17\pi}{8}\right) \]
Step 1: Use periodic property
\[ \frac{17\pi}{8} = 2\pi + \frac{\pi}{8} \]
\[ \sin\left(\frac{17\pi}{8}\right) = \sin\left(2\pi + \frac{\pi}{8}\right) = \sin\left(\frac{\pi}{8}\right) \]
Step 2: Apply inverse sine
\[ \sin^{-1}\left(\sin \frac{\pi}{8}\right) \]
The principal value range of \( \sin^{-1}x \) is:
\[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]
Since \( \frac{\pi}{8} \) lies in this interval, we get:
\[ \sin^{-1}\left(\sin \frac{17\pi}{8}\right) = \frac{\pi}{8} \]
Final Answer
\[ \boxed{\frac{\pi}{8}} \]