Period of the function
`f(x)=sin((x)/(2))cos [(x)/(2)]-cos((x)/(2))sin[(x)/(2)]`, where [.] denotes the greatest integer function, is _________.

Explanation

To find the period of the function \( f(x) = \sin\left(\frac{x}{2}\right) \cos\left(\left\lfloor \frac{x}{2} \right\rfloor\right) - \cos\left(\frac{x}{2}\right) \sin\left(\left\lfloor \frac{x}{2} \right\rfloor\right) \), we will follow these steps:

### Step 1: Simplify the Function
We can use the sine subtraction formula:
\[ \sin A \cos B - \cos A \sin B = \sin(A - B) \]
Here, let \( A = \frac{x}{2} \) and \( B = \left\lfloor \frac{x}{2} \right\rfloor \). Thus, we can rewrite the function as:
\[ f(x) = \sin\left(\frac{x}{2} - \left\lfloor \frac{x}{2} \right\rfloor\right) \]

### Step 2: Understand the Argument of the Sine Function
The term \( \frac{x}{2} - \left\lfloor \frac{x}{2} \right\rfloor \) represents the fractional part of \( \frac{x}{2} \). The fractional part function, denoted as \( \{x\} \), is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
Thus, we have:
\[ f(x) = \sin\left(\left\{\frac{x}{2}\right\}\right) \]

### Step 3: Determine the Period of the Fractional Part
The fractional part function \( \{x\} \) has a period of 1. Therefore, the function \( \left\{\frac{x}{2}\right\} \) will have a period of:
\[ \text{Period of } \left\{\frac{x}{2}\right\} = 2 \]
This is because if \( x \) increases by 2, \( \frac{x}{2} \) increases by 1, which resets the fractional part.

### Step 4: Find the Period of \( f(x) \)
Since \( f(x) = \sin\left(\left\{\frac{x}{2}\right\}\right) \) and the argument \( \left\{\frac{x}{2}\right\} \) has a period of 2, it follows that:
\[ \text{Period of } f(x) = 2 \]

### Conclusion
Thus, the period of the function \( f(x) \) is:
\[ \boxed{2} \]