If `sin^(-1) theta = sin^(-1)(sin5)` then `theta` is:

Options

A. `sin(5-pi)`
B. `sin(2pi -5)`
C. `sin(5-2pi)`
D. None of these

Explanation

To solve the equation \( \sin^{-1} \theta = \sin^{-1}(\sin 5) \), we need to determine the value of \( \theta \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( \sin^{-1}(x) \) (also known as arcsin) gives the angle whose sine is \( x \). The range of \( \sin^{-1}(x) \) is limited to \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

2. **Finding the Value of \( \sin^{-1}(\sin 5) \)**:
Since \( 5 \) is outside the range of \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), we need to find an equivalent angle within this range.

The sine function is periodic with a period of \( 2\pi \). To find an equivalent angle, we can subtract \( 2\pi \) from \( 5 \) until the angle falls within the range of \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

\[ 5 - 2\pi \approx 5 - 6.2832 \approx -1.2832 \]

Since \( -1.2832 \) is within the range of \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), we can use this value.

3. **Setting Up the Equation**:
Now we have:
\[ \sin^{-1}(\sin 5) = 5 - 2\pi \]
Therefore, we can rewrite the original equation:
\[ \sin^{-1} \theta = 5 - 2\pi \]

4. **Finding \( \theta \)**:
To find \( \theta \), we take the sine of both sides:
\[ \theta = \sin(5 - 2\pi) \]
Since \( \sin \) is periodic with a period of \( 2\pi \):
\[ \sin(5 - 2\pi) = \sin(5) \]
Thus, we conclude:
\[ \theta = \sin(5) \]

### Final Answer:
\[ \theta = \sin(5) \]