To analyze the function \( f(x) = \sin^{-1}(\sin x) \), we need to determine its behavior and periodicity. Let's break it down step by step.
### Step 1: Understand the Function
The function \( f(x) = \sin^{-1}(\sin x) \) combines the sine function and its inverse. The sine function oscillates between -1 and 1, and the inverse sine function \( \sin^{-1}(y) \) gives an angle whose sine is \( y \), with the range restricted to \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
### Step 2: Determine the Range of \( f(x) \)
Since \( \sin x \) can take any value between -1 and 1, the output of \( f(x) \) will be constrained to the range of \( \sin^{-1}(y) \), which is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
### Step 3: Analyze the Function Over Different Intervals
To understand how \( f(x) \) behaves, we can analyze it over specific intervals:
1. **For \( x \in [0, \frac{\pi}{2}] \)**:
- Here, \( \sin x \) is non-negative and increasing.
- Thus, \( f(x) = \sin^{-1}(\sin x) = x \).
2. **For \( x \in [\frac{\pi}{2}, \frac{3\pi}{2}] \)**:
- In this interval, \( \sin x \) takes negative values.
- Therefore, \( f(x) = \sin^{-1}(\sin x) = \pi - x \) (since \( \sin x \) is symmetric).
3. **For \( x \in [\frac{3\pi}{2}, 2\pi] \)**:
- Again, \( \sin x \) is non-negative and decreasing.
- Thus, \( f(x) = \sin^{-1}(\sin x) = x - 2\pi \).
### Step 4: Identify the Periodicity
From the analysis, we can see that:
- For \( x \in [0, 2\pi] \), the function \( f(x) \) behaves in a piecewise manner.
- The function repeats its values every \( 2\pi \) as \( \sin x \) is periodic with period \( 2\pi \).
### Conclusion
Thus, the function \( f(x) = \sin^{-1}(\sin x) \) is periodic with a period of \( 2\pi \).
### Final Answer
The function \( f(x) = \sin^{-1}(\sin x) \) is periodic with a period of \( 2\pi \).
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