Period of the function `f(x) = sin((pi x)/(2)) cos((pi x)/(2))` is

Answer

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

### Step 1: Use the Product-to-Sum Identity
We start by using the product-to-sum identity for sine and cosine. The identity states that:
\[ 2 \sin A \cos B = \sin(A + B) + \sin(A - B) \]
In our case, we can rewrite \( f(x) \) as:
\[ f(x) = \sin\left(\frac{\pi x}{2}\right) \cos\left(\frac{\pi x}{2}\right) = \frac{1}{2} \cdot 2 \sin\left(\frac{\pi x}{2}\right) \cos\left(\frac{\pi x}{2}\right) \]
This simplifies to:
\[ f(x) = \frac{1}{2} \left( \sin\left(\frac{\pi x}{2} + \frac{\pi x}{2}\right) + \sin\left(\frac{\pi x}{2} - \frac{\pi x}{2}\right) \right) \]
However, we can also directly use the double angle identity:
\[ \sin A \cos A = \frac{1}{2} \sin(2A) \]
Thus, we can write:
\[ f(x) = \frac{1}{2} \sin\left(\pi x\right) \]

### Step 2: Determine the Period of the Sine Function
The standard sine function \( \sin(kx) \) has a period given by:
\[ T = \frac{2\pi}{k} \]
In our case, we have:
\[ f(x) = \frac{1}{2} \sin(\pi x) \]
Here, \( k = \pi \).

### Step 3: Calculate the Period
Using the formula for the period:
\[ T = \frac{2\pi}{\pi} = 2 \]

### Conclusion
Thus, the period of the function \( f(x) = \sin\left(\frac{\pi x}{2}\right) \cos\left(\frac{\pi x}{2}\right) \) is:
\[ \boxed{2} \]