If f(x) =`|{:(cos (x+alpha),cos(x+beta),cos(x+gamma)),(sin (x+alpha),sin(x+beta),sin(x+gamma)),(sin(beta+gamma),sin(gamma+alpha),sin(alpha+beta)):}|` then `f(theta)-2f(phi)+f(psi)` is equal to

Options

A. (A)0
B. (B)`alpha-beta`
C. (C)`alpha+beta+gamma`
D. (D)`alpha+beta+-gamma`

Answer

To solve the problem, we need to evaluate the expression \( f(\theta) - 2f(\phi) + f(\psi) \) given the function:

\[ f(x) = \begin{vmatrix} \cos(x + \alpha) & \cos(x + \beta) & \cos(x + \gamma) \\ \sin(x + \alpha) & \sin(x + \beta) & \sin(x + \gamma) \\ \sin(\beta + \gamma) & \sin(\gamma + \alpha) & \sin(\alpha + \beta) \end{vmatrix} \]

### Step 1: Evaluate \( f(x) \)

We can start by evaluating the determinant \( f(x) \).

Using properties of determinants, we can simplify the determinant by performing row operations. We can subtract the third row from the first two rows to simplify our calculations.

\[ f(x) = \begin{vmatrix} \cos(x + \alpha) - \sin(\beta + \gamma) & \cos(x + \beta) - \sin(\gamma + \alpha) & \cos(x + \gamma) - \sin(\alpha + \beta) \\ \sin(x + \alpha) - \sin(\beta + \gamma) & \sin(x + \beta) - \sin(\gamma + \alpha) & \sin(x + \gamma) - \sin(\alpha + \beta) \\ \sin(\beta + \gamma) & \sin(\gamma + \alpha) & \sin(\alpha + \beta) \end{vmatrix} \]

### Step 2: Analyze the determinant

Notice that the first two rows now contain terms that can be further simplified. The determinant can be evaluated using the sine and cosine addition formulas. However, we can also observe that the first two rows are linear combinations of sine and cosine functions.

### Step 3: Find \( f(\theta) - 2f(\phi) + f(\psi) \)

Now we need to evaluate:

\[ f(\theta) - 2f(\phi) + f(\psi) \]

By substituting \( \theta, \phi, \psi \) into the function \( f(x) \), we can express this as:

\[ f(\theta) = \begin{vmatrix} \cos(\theta + \alpha) & \cos(\theta + \beta) & \cos(\theta + \gamma) \\ \sin(\theta + \alpha) & \sin(\theta + \beta) & \sin(\theta + \gamma) \\ \sin(\beta + \gamma) & \sin(\gamma + \alpha) & \sin(\alpha + \beta) \end{vmatrix} \]

\[ f(\phi) = \begin{vmatrix} \cos(\phi + \alpha) & \cos(\phi + \beta) & \cos(\phi + \gamma) \\ \sin(\phi + \alpha) & \sin(\phi + \beta) & \sin(\phi + \gamma) \\ \sin(\beta + \gamma) & \sin(\gamma + \alpha) & \sin(\alpha + \beta) \end{vmatrix} \]

\[ f(\psi) = \begin{vmatrix} \cos(\psi + \alpha) & \cos(\psi + \beta) & \cos(\psi + \gamma) \\ \sin(\psi + \alpha) & \sin(\psi + \beta) & \sin(\psi + \gamma) \\ \sin(\beta + \gamma) & \sin(\gamma + \alpha) & \sin(\alpha + \beta) \end{vmatrix} \]

### Step 4: Use properties of determinants

Using the properties of determinants and the fact that the sine and cosine functions are periodic, we can conclude that \( f(x) \) is independent of the variable \( x \) when evaluated at specific angles. Thus, the expression simplifies to:

\[ f(\theta) - 2f(\phi) + f(\psi) = 0 \]

### Final Answer

Thus, the final answer is:

\[ \boxed{0} \]