The determinant
`|(cos(theta+phi),-sin(theta+phi),cos2phi),(sintheta,costheta,sinphi),(-costheta,sintheta,cosphi)|` is

Options

A. 0
B. independent of `theta`
C. independent of `phi`
D. independent of `theta` & `phi` both

Answer

To solve the determinant

\[ D = \begin{vmatrix} \cos(\theta + \phi) & -\sin(\theta + \phi) & \cos(2\phi) \\ \sin(\theta) & \cos(\theta) & \sin(\phi) \\ -\cos(\theta) & \sin(\theta) & \cos(\phi) \end{vmatrix} \]

we will use properties of determinants and trigonometric identities.

### Step 1: Expand the determinant using the first row.

Using the first row to expand the determinant, we have:

\[ D = \cos(\theta + \phi) \begin{vmatrix} \cos(\theta) & \sin(\phi) \\ \sin(\theta) & \cos(\phi) \end{vmatrix} - (-\sin(\theta + \phi)) \begin{vmatrix} \sin(\theta) & \sin(\phi) \\ -\cos(\theta) & \cos(\phi) \end{vmatrix} + \cos(2\phi) \begin{vmatrix} \sin(\theta) & \cos(\theta) \\ -\cos(\theta) & \sin(\theta) \end{vmatrix} \]

### Step 2: Calculate the 2x2 determinants.

1. For the first determinant:

\[ \begin{vmatrix} \cos(\theta) & \sin(\phi) \\ \sin(\theta) & \cos(\phi) \end{vmatrix} = \cos(\theta) \cos(\phi) - \sin(\theta) \sin(\phi) = \cos(\theta + \phi) \]

2. For the second determinant:

\[ \begin{vmatrix} \sin(\theta) & \sin(\phi) \\ -\cos(\theta) & \cos(\phi) \end{vmatrix} = \sin(\theta) \cos(\phi) - (-\cos(\theta) \sin(\phi)) = \sin(\theta) \cos(\phi) + \cos(\theta) \sin(\phi) = \sin(\theta + \phi) \]

3. For the third determinant:

\[ \begin{vmatrix} \sin(\theta) & \cos(\theta) \\ -\cos(\theta) & \sin(\theta) \end{vmatrix} = \sin^2(\theta) + \cos^2(\theta) = 1 \]

### Step 3: Substitute back into the determinant expression.

Substituting these results back into the determinant expression gives:

\[ D = \cos(\theta + \phi) \cdot \cos(\theta + \phi) + \sin(\theta + \phi) \cdot \sin(\theta + \phi) + \cos(2\phi) \cdot 1 \]

This simplifies to:

\[ D = \cos^2(\theta + \phi) + \sin^2(\theta + \phi) + \cos(2\phi) \]

### Step 4: Use the Pythagorean identity.

Using the identity \(\cos^2(x) + \sin^2(x) = 1\):

\[ D = 1 + \cos(2\phi) \]

### Step 5: Use the double angle formula for cosine.

Using the identity \(\cos(2\phi) = 2\cos^2(\phi) - 1\):

\[ D = 1 + (2\cos^2(\phi) - 1) = 2\cos^2(\phi) \]

### Final Result:

Thus, the value of the determinant is

\[ D = 2\cos^2(\phi) \]