`sin^(-1)(sinx)=x` if

Options

A. `x in R`
B. `x in [-1,1]`
C. `x in [-pi/2,pi/2]`
D. `x in [-pi,pi]`

Answer

To solve the equation \( \sin^{-1}(\sin x) = x \), we need to determine the conditions under which this equation holds true.

### Step-by-Step Solution:

1. **Understanding the Functions**:
The function \( \sin^{-1}(y) \) is the inverse sine function, which gives the angle whose sine is \( y \). The output of \( \sin^{-1}(y) \) is restricted to the range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

2. **Domain of \( \sin^{-1} \)**:
The domain of \( \sin^{-1}(y) \) is \( y \in [-1, 1] \). Therefore, for \( \sin^{-1}(\sin x) \) to be defined, \( \sin x \) must also lie within this range, which it does since the sine function always outputs values between -1 and 1.

3. **Principal Values**:
The equation \( \sin^{-1}(\sin x) = x \) holds true when \( x \) is within the range of the inverse sine function, which is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

4. **General Case**:
If \( x \) is outside the range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we can express \( x \) in terms of its reference angle. For any angle \( x \), we can write:
\[ x = n\pi + (-1)^n \theta \]
where \( n \) is an integer and \( \theta \) is the reference angle such that \( \theta = \sin^{-1}(\sin x) \).

5. **Conclusion**:
Thus, the equation \( \sin^{-1}(\sin x) = x \) holds true if:
\[ x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] + 2n\pi \quad \text{for any integer } n \]

### Final Condition:
The condition for \( \sin^{-1}(\sin x) = x \) is that \( x \) must be of the form:
\[ x = n\pi + (-1)^n \theta \quad \text{where } \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]