Suppose
`f(a,b,x,y)=|(1,x,x^(2)),(cos((a-b)y),cos(ay),cos((a+b)y)),(sin((a-b)y),sin(ay),sin(a+b)y)|`
then `f(pi,(pi)/2,sqrt(3),0)=`_______

Answer

To solve the problem, we need to evaluate the determinant defined by the function \( f(a, b, x, y) \) at the specific values \( a = \pi \), \( b = \frac{\pi}{2} \), \( x = \sqrt{3} \), and \( y = 0 \).

The function is given as:

\[ f(a, b, x, y) = \begin{vmatrix} 1 & x & x^2 \\ \cos((a-b)y) & \cos(ay) & \cos((a+b)y) \\ \sin((a-b)y) & \sin(ay) & \sin((a+b)y) \end{vmatrix} \]

### Step 1: Substitute the values into the determinant

We substitute \( a = \pi \), \( b = \frac{\pi}{2} \), \( x = \sqrt{3} \), and \( y = 0 \):

\[ f\left(\pi, \frac{\pi}{2}, \sqrt{3}, 0\right) = \begin{vmatrix} 1 & \sqrt{3} & (\sqrt{3})^2 \\ \cos\left((\pi - \frac{\pi}{2}) \cdot 0\right) & \cos(\pi \cdot 0) & \cos\left((\pi + \frac{\pi}{2}) \cdot 0\right) \\ \sin\left((\pi - \frac{\pi}{2}) \cdot 0\right) & \sin(\pi \cdot 0) & \sin\left((\pi + \frac{\pi}{2}) \cdot 0\right) \end{vmatrix} \]

### Step 2: Simplify the expressions

Calculating the values:

- \( (\sqrt{3})^2 = 3 \)
- \( (a-b) = \pi - \frac{\pi}{2} = \frac{\pi}{2} \)
- \( (a+b) = \pi + \frac{\pi}{2} = \frac{3\pi}{2} \)

Now substituting these into the determinant:

\[ = \begin{vmatrix} 1 & \sqrt{3} & 3 \\ \cos(0) & \cos(0) & \cos(0) \\ \sin(0) & \sin(0) & \sin(0) \end{vmatrix} \]

### Step 3: Evaluate the trigonometric functions

Since \( \cos(0) = 1 \) and \( \sin(0) = 0 \), we have:

\[ = \begin{vmatrix} 1 & \sqrt{3} & 3 \\ 1 & 1 & 1 \\ 0 & 0 & 0 \end{vmatrix} \]

### Step 4: Evaluate the determinant

The determinant of a matrix with a row of zeros is always zero. Therefore:

\[ = 0 \]

### Final Answer

Thus, the value of \( f\left(\pi, \frac{\pi}{2}, \sqrt{3}, 0\right) \) is:

\[ \boxed{0} \]