if determinant `|{:( cos (theta + phi),,-sin (theta+phi),,cos 2phi),(sin theta,,cos theta,,sin phi),(-cos theta,,sintheta,,cos phi):}|` is

Options

A. non-negative
B. independent of theta
C. independent of `phi`
D. none of these

Explanation

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 follow these steps:

### Step 1: Use Trigonometric Identities
We start by applying the trigonometric identities for \(\cos(\theta + \phi)\) and \(\sin(\theta + \phi)\):

\[ \cos(\theta + \phi) = \cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi) \]
\[ \sin(\theta + \phi) = \sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi) \]

Substituting these into the determinant gives us:

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

### Step 2: Simplify the First Row
Now, we simplify the first row. We can rewrite the first row as:

\[ \begin{pmatrix} \cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi) & -\sin(\theta)\cos(\phi) - \cos(\theta)\sin(\phi) & \cos(2\phi) \end{pmatrix} \]

### Step 3: Row Operation
Next, we perform a row operation on \(R_1\):

\[ R_1 \to R_1 + \sin(\phi)R_2 + \cos(\phi)R_3 \]

Calculating this, we get:

\[ R_1 = \begin{pmatrix} (\cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi)) + \sin(\phi)\sin(\theta) - \cos(\phi)\cos(\theta) & -(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi)) + \sin(\phi)\cos(\theta) + \sin(\theta)\cos(\phi) & \cos(2\phi) + \sin(\phi)\sin(\phi) + \cos(\phi)(-\cos(\theta)) \end{pmatrix} \]

### Step 4: Simplify Further
After simplification, we find that the first row becomes:

\[ R_1 = \begin{pmatrix} 0 & 0 & \cos(2\phi) + 1 \end{pmatrix} \]

### Step 5: Expand the Determinant
Now, we can expand the determinant:

\[ D = \begin{vmatrix} 0 & 0 & \cos(2\phi) + 1 \\ \sin(\theta) & \cos(\theta) & \sin(\phi) \\ -\cos(\theta) & \sin(\theta) & \cos(\phi) \end{vmatrix} \]

This determinant simplifies to:

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

### Step 6: Calculate the 2x2 Determinant
Calculating the 2x2 determinant:

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

### Step 7: Final Result
Thus, we have:

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

### Conclusion
The final result shows that the determinant \(D\) is dependent on \(\cos(2\phi) + 1\) and is independent of \(\theta\).