If `f(x)=|(cosx,cosx,sinx),(-cosx,cosx,-sinx),(sinx,sinx,cosx)|`, then find its determinant

Answer

To find the determinant of the matrix given by \( f(x) = \begin{vmatrix} \cos x & \cos x & \sin x \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x \end{vmatrix} \), we will use the properties of determinants and some algebraic manipulations.

### Step-by-Step Solution:

1. **Write the Determinant**:
\[ D = \begin{vmatrix} \cos x & \cos x & \sin x \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x \end{vmatrix} \]

2. **Expand the Determinant**:
We can expand the determinant using the first row:
\[ D = \cos x \begin{vmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{vmatrix} - \cos x \begin{vmatrix} -\cos x & -\sin x \\ \sin x & \cos x \end{vmatrix} + \sin x \begin{vmatrix} -\cos x & \cos x \\ \sin x & \sin x \end{vmatrix} \]

3. **Calculate the 2x2 Determinants**:
- For the first determinant:
\[ \begin{vmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{vmatrix} = \cos^2 x + \sin^2 x = 1 \]
- For the second determinant:
\[ \begin{vmatrix} -\cos x & -\sin x \\ \sin x & \cos x \end{vmatrix} = -\cos x \cdot \cos x + \sin x \cdot \sin x = -\cos^2 x + \sin^2 x \]
- For the third determinant:
\[ \begin{vmatrix} -\cos x & \cos x \\ \sin x & \sin x \end{vmatrix} = -\cos x \cdot \sin x - \sin x \cdot \cos x = -2 \cos x \sin x \]

4. **Substituting Back**:
Now substituting these values back into the determinant:
\[ D = \cos x \cdot 1 - \cos x (-\cos^2 x + \sin^2 x) + \sin x (-2 \cos x \sin x) \]
Simplifying this gives:
\[ D = \cos x + \cos x (\sin^2 x - \cos^2 x) - 2 \sin^2 x \cos x \]

5. **Combine Like Terms**:
\[ D = \cos x + \cos x \sin^2 x - \cos^3 x - 2 \sin^2 x \cos x \]
\[ D = \cos x (1 + \sin^2 x - \cos^2 x - 2 \sin^2 x) \]
\[ D = \cos x (1 - \cos^2 x - \sin^2 x) \]
Since \( \sin^2 x + \cos^2 x = 1 \):
\[ D = \cos x (1 - 1) = 0 \]

### Final Answer:
The determinant \( D = 0 \).