find the maximum value of `f(x) = (sin^(-1) (sin x))^(2) - sin^(-1) (sin x)`

Explanation

To find the maximum value of the function \( f(x) = (\sin^{-1}(\sin x))^2 - \sin^{-1}(\sin x) \), we will follow these steps:

### Step 1: Rewrite the Function
We start by rewriting the function:
\[ f(x) = (\sin^{-1}(\sin x))^2 - \sin^{-1}(\sin x) \]
Let \( y = \sin^{-1}(\sin x) \). Then, we can express \( f(x) \) in terms of \( y \):
\[ f(x) = y^2 - y \]

### Step 2: Complete the Square
Next, we complete the square for the quadratic expression:
\[ f(x) = y^2 - y = \left(y - \frac{1}{2}\right)^2 - \frac{1}{4} \]
This shows that the function \( f(x) \) can be expressed as a perfect square minus a constant.

### Step 3: Identify the Maximum Value
The term \( \left(y - \frac{1}{2}\right)^2 \) achieves its minimum value of \( 0 \) when \( y = \frac{1}{2} \). Therefore, the maximum value of \( f(x) \) occurs at:
\[ f(x)_{\text{max}} = 0 - \frac{1}{4} = -\frac{1}{4} \]

### Step 4: Determine the Corresponding \( x \) Value
Now, we need to find the value of \( x \) for which \( \sin^{-1}(\sin x) = \frac{1}{2} \). This occurs when:
\[ \sin x = \frac{1}{2} \]
The solutions for this equation are:
\[ x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2k\pi \quad \text{for any integer } k \]

### Conclusion
Thus, the maximum value of the function \( f(x) \) is:
\[ \boxed{-\frac{1}{4}} \]