If `f(x)` is a real-valued function defined as `f(x)=In (1-sinx),` then the graph of `f(x)` is

Options

A. symmetric about the line `x =pi`
B. symmetric about the y-axis
C. symmetric and the line ` x=(pi)/(2)`
D. symmetric about the origin

Explanation

To determine the symmetry of the function \( f(x) = \ln(1 - \sin x) \), we will analyze the function step by step.

### Step 1: Check for symmetry about the line \( x = \pi \)

To check if the function is symmetric about the line \( x = \pi \), we need to evaluate \( f(\pi - x) \) and \( f(\pi + x) \).

1. Calculate \( f(\pi - x) \):
\[ f(\pi - x) = \ln(1 - \sin(\pi - x)) \]
Since \( \sin(\pi - x) = \sin x \), we have:
\[ f(\pi - x) = \ln(1 - \sin x) \]

2. Calculate \( f(\pi + x) \):
\[ f(\pi + x) = \ln(1 - \sin(\pi + x)) \]
Since \( \sin(\pi + x) = -\sin x \), we have:
\[ f(\pi + x) = \ln(1 - (-\sin x)) = \ln(1 + \sin x) \]

Since \( f(\pi - x) \neq f(\pi + x) \), the function is **not symmetric about the line \( x = \pi \)**.

### Step 2: Check for symmetry about the y-axis

To check for symmetry about the y-axis, we need to evaluate \( f(-x) \) and see if it equals \( f(x) \).

1. Calculate \( f(-x) \):
\[ f(-x) = \ln(1 - \sin(-x)) \]
Since \( \sin(-x) = -\sin x \), we have:
\[ f(-x) = \ln(1 + \sin x) \]

Since \( f(-x) \neq f(x) \), the function is **not symmetric about the y-axis**.

### Step 3: Check for symmetry about the line \( x = \frac{\pi}{2} \)

To check for symmetry about the line \( x = \frac{\pi}{2} \), we need to evaluate \( f\left(\frac{\pi}{2} - x\right) \) and \( f\left(\frac{\pi}{2} + x\right) \).

1. Calculate \( f\left(\frac{\pi}{2} - x\right) \):
\[ f\left(\frac{\pi}{2} - x\right) = \ln\left(1 - \sin\left(\frac{\pi}{2} - x\right)\right) \]
Since \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \), we have:
\[ f\left(\frac{\pi}{2} - x\right) = \ln(1 - \cos x) \]

2. Calculate \( f\left(\frac{\pi}{2} + x\right) \):
\[ f\left(\frac{\pi}{2} + x\right) = \ln\left(1 - \sin\left(\frac{\pi}{2} + x\right)\right) \]
Since \( \sin\left(\frac{\pi}{2} + x\right) = \cos x \), we have:
\[ f\left(\frac{\pi}{2} + x\right) = \ln(1 - \cos x) \]

Since \( f\left(\frac{\pi}{2} - x\right) = f\left(\frac{\pi}{2} + x\right) \), the function is **symmetric about the line \( x = \frac{\pi}{2} \)**.

### Conclusion

The graph of the function \( f(x) = \ln(1 - \sin x) \) is symmetric about the line \( x = \frac{\pi}{2} \).

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