If `Delta=|(1,a,a^2),(cos(n-1)x,cosnx,cos(n+1)x),(sin(n-1)x,sinnx,sin(n+1)x)|` then `Delta` is

Options

A. independent of x
B. independent of a
C. independent of n
D. none of these

Answer

To solve the determinant \( \Delta = \begin{vmatrix} 1 & a & a^2 \\ \cos((n-1)x) & \cos(nx) & \cos((n+1)x) \\ \sin((n-1)x) & \sin(nx) & \sin((n+1)x) \end{vmatrix} \), we will expand it step by step.

### Step 1: Expand the Determinant
We will expand the determinant along the first row. The formula for the determinant of a 3x3 matrix is given by:

\[ \Delta = a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix} \]

In our case, \( a_{11} = 1 \), \( a_{12} = a \), \( a_{13} = a^2 \).

Calculating the minors:

1. For \( a_{11} = 1 \):
\[ \begin{vmatrix} \cos(nx) & \cos((n+1)x) \\ \sin(nx) & \sin((n+1)x) \end{vmatrix} = \cos(nx)\sin((n+1)x) - \sin(nx)\cos((n+1)x) = \sin((n+1)x - nx) = \sin(x) \]

2. For \( a_{12} = a \):
\[ \begin{vmatrix} \cos((n-1)x) & \cos((n+1)x) \\ \sin((n-1)x) & \sin((n+1)x) \end{vmatrix} = \cos((n-1)x)\sin((n+1)x) - \sin((n-1)x)\cos((n+1)x) = \sin((n+1)x - (n-1)x) = \sin(2x) \]

3. For \( a_{13} = a^2 \):
\[ \begin{vmatrix} \cos((n-1)x) & \cos(nx) \\ \sin((n-1)x) & \sin(nx) \end{vmatrix} = \cos((n-1)x)\sin(nx) - \sin((n-1)x)\cos(nx) = \sin(nx - (n-1)x) = \sin(x) \]

### Step 2: Substitute Back into the Determinant
Now substituting back into the determinant:

\[ \Delta = 1 \cdot \sin(x) - a \cdot \sin(2x) + a^2 \cdot \sin(x) \]

### Step 3: Simplify the Expression
Combining the terms gives:

\[ \Delta = \sin(x) + a^2 \sin(x) - a \sin(2x) = \sin(x)(1 + a^2) - a \sin(2x) \]

### Final Result
Thus, the value of the determinant \( \Delta \) is:

\[ \Delta = \sin(x)(1 + a^2) - a \sin(2x) \]