If `theta, phi epsilon R`, then the determinant
`Delta=|(cos theta, - sin theta, 1),(sin theta, cos theta, 1),(cos(theta+phi),-sin(theta+phi),0)|`
lies in the interval

Options

A. `[-sqrt(2),sqrt(2)]`
B. `[-1,1]`
C. `[-sqrt(2),1]`
D. `[-1,sqrt(2)]`

Answer

To solve the determinant

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

we will use the properties of determinants and some trigonometric identities.

### Step 1: Expand the determinant

Using the determinant expansion along the third column, we have:

\[ \Delta = 1 \cdot \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} + 1 \cdot \begin{vmatrix} \cos \theta & 1 \\ \sin \theta & 1 \end{vmatrix} \cdot (-1) + 0 \]

### Step 2: Calculate the first 2x2 determinant

The first determinant is:

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

### Step 3: Calculate the second 2x2 determinant

The second determinant is:

\[ \begin{vmatrix} \cos \theta & 1 \\ \sin \theta & 1 \end{vmatrix} = \cos \theta \cdot 1 - \sin \theta \cdot 1 = \cos \theta - \sin \theta \]

### Step 4: Substitute back into the determinant

Now substituting back into our expression for \(\Delta\):

\[ \Delta = 1 - (\cos \theta - \sin \theta) = 1 - \cos \theta + \sin \theta \]

### Step 5: Simplify the expression

Thus, we have:

\[ \Delta = \sin \theta - \cos \theta + 1 \]

### Step 6: Find the range of \(\Delta\)

To find the range of \(\Delta\), we can express it as:

\[ \Delta = 1 + \sin \theta - \cos \theta \]

Using the identity \( \sin \theta - \cos \theta = \sqrt{2} \sin\left(\theta - \frac{\pi}{4}\right) \), we can rewrite:

\[ \Delta = 1 + \sqrt{2} \sin\left(\theta - \frac{\pi}{4}\right) \]

The range of \(\sin\left(\theta - \frac{\pi}{4}\right)\) is \([-1, 1]\). Therefore, the range of \( \sqrt{2} \sin\left(\theta - \frac{\pi}{4}\right) \) is \([- \sqrt{2}, \sqrt{2}]\).

Thus, the range of \(\Delta\) becomes:

\[ 1 - \sqrt{2} \leq \Delta \leq 1 + \sqrt{2} \]

### Final Result

The determinant \(\Delta\) lies in the interval:

\[ [1 - \sqrt{2}, 1 + \sqrt{2}] \]