The function `f(x)=sin^(-1)(sinx)` , is

Options

A. periodic with period `2pi`
B. periodic with period `pi`
C. periodic with period `(pi)/(2)`
D. non-periodic

Explanation

To determine the nature of the function \( f(x) = \sin^{-1}(\sin x) \), we will analyze its periodicity step by step.

### Step 1: Understanding the Function
The function \( f(x) = \sin^{-1}(\sin x) \) involves the sine function and its inverse. The sine function is periodic with a period of \( 2\pi \).

### Step 2: Finding \( f(x + 2\pi) \)
To check for periodicity, we will evaluate \( f(x + 2\pi) \):
\[ f(x + 2\pi) = \sin^{-1}(\sin(x + 2\pi)) \]
Using the property of the sine function:
\[ \sin(x + 2\pi) = \sin x \]
Thus,
\[ f(x + 2\pi) = \sin^{-1}(\sin x) \]
This simplifies to:
\[ f(x + 2\pi) = f(x) \]

### Step 3: Conclusion on Periodicity
Since we have shown that \( f(x + 2\pi) = f(x) \) for all \( x \in \mathbb{R} \), we conclude that the function \( f(x) \) is periodic with a period of \( 2\pi \).

### Final Answer
The function \( f(x) = \sin^{-1}(\sin x) \) is periodic with a period of \( 2\pi \).

---