The parameter, on which the value of the determinant `|(1,a,a^(2)),(cos(p-d)x,cospx,cos(p+d)x),(sin(p-d)x,sinpx,sin(p+d)x)|` does not depend, is

Options

A. a
B. p
C. d
D. x

Answer

To solve the problem of determining the parameter on which the value of the determinant

\[ D = \begin{vmatrix} 1 & a & a^2 \\ \cos(p-d)x & \cos(px) & \cos(p+d)x \\ \sin(p-d)x & \sin(px) & \sin(p+d)x \end{vmatrix} \]

does not depend, we will follow these steps:

### Step 1: Expand the Determinant
We will use the properties of determinants to expand \(D\). We can expand the determinant along the first row.

\[ D = 1 \cdot \begin{vmatrix} \cos(p-d)x & \cos(p+d)x \\ \sin(p-d)x & \sin(p+d)x \end{vmatrix} - a \cdot \begin{vmatrix} \cos(p-d)x & \cos(px) \\ \sin(p-d)x & \sin(px) \end{vmatrix} + a^2 \cdot \begin{vmatrix} \cos(px) & \cos(p+d)x \\ \sin(px) & \sin(p+d)x \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants
Now we will compute the 2x2 determinants.

1. For the first determinant:

\[ \begin{vmatrix} \cos(p-d)x & \cos(p+d)x \\ \sin(p-d)x & \sin(p+d)x \end{vmatrix} = \cos(p-d)x \sin(p+d)x - \sin(p-d)x \cos(p+d)x \]

Using the sine subtraction formula, this can be simplified to:

\[ \sin((p+d)x - (p-d)x) = \sin(2dx) \]

2. For the second determinant:

\[ \begin{vmatrix} \cos(p-d)x & \cos(px) \\ \sin(p-d)x & \sin(px) \end{vmatrix} = \cos(p-d)x \sin(px) - \sin(p-d)x \cos(px) \]

This can be simplified to:

\[ \sin(px - (p-d)x) = \sin(dx) \]

3. For the third determinant:

\[ \begin{vmatrix} \cos(px) & \cos(p+d)x \\ \sin(px) & \sin(p+d)x \end{vmatrix} = \cos(px) \sin(p+d)x - \sin(px) \cos(p+d)x \]

This can be simplified to:

\[ \sin((p+d)x - px) = \sin(dx) \]

### Step 3: Substitute Back into the Determinant
Now substituting these back into \(D\):

\[ D = 1 \cdot \sin(2dx) - a \cdot \sin(dx) + a^2 \cdot \sin(dx) \]

This simplifies to:

\[ D = \sin(2dx) + (a^2 - a) \sin(dx) \]

### Step 4: Analyze the Dependence on Parameters
From the expression for \(D\), we can see that it depends on \(d\) and \(x\), but it does not depend on \(p\).

### Conclusion
The parameter on which the value of the determinant does not depend is:

\[ \text{The parameter is } p. \]