If `theta in [(pi)/(2),3(pi)/(2)] then sin^(-1)(sin theta)` equals

Options

A. `theta`
B. `pi - theta`
C. `2pi - theta`
D. `-pi + theta`

Explanation

To solve the problem \( \sin^{-1}(\sin \theta) \) for \( \theta \) in the interval \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \), we can follow these steps:

### Step 1: Understand the Range of \( \sin^{-1} \)
The function \( \sin^{-1}(x) \) (or arcsin) is defined for \( x \) in the range of \([-1, 1]\) and its output is restricted to the interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

### Step 2: Identify the Interval of \( \theta \)
Given that \( \theta \) is in the interval \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \), we need to find a way to express \( \theta \) such that it falls within the range of \( \sin^{-1} \).

### Step 3: Adjust \( \theta \) to Fit the Range
To bring \( \theta \) within the range of \( \sin^{-1} \), we can subtract \( \pi \) from \( \theta \):
\[ \theta - \pi \]
This transformation will shift \( \theta \) into the range of \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

### Step 4: Calculate the New Values
Now, let's compute the new values:
- For \( \theta = \frac{\pi}{2} \):
\[ \frac{\pi}{2} - \pi = -\frac{\pi}{2} \]
- For \( \theta = \frac{3\pi}{2} \):
\[ \frac{3\pi}{2} - \pi = \frac{\pi}{2} \]
Thus, \( \theta - \pi \) will range from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

### Step 5: Apply the Property of Inverse Functions
Using the property of inverse functions:
\[ \sin^{-1}(\sin x) = x \quad \text{if } x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]
we can write:
\[ \sin^{-1}(\sin \theta) = \theta - \pi \]

### Step 6: Final Result
Thus, the final answer is:
\[ \sin^{-1}(\sin \theta) = \theta - \pi \]

### Conclusion
Since we need to express the result in terms of \( \theta \), we can rewrite it as:
\[ \sin^{-1}(\sin \theta) = \pi - \theta \]