The value of `Delta=|(1,sin 3theta, sin^(3)theta),(2cos theta, sin 6 theta, sin^(3)2 theta),(4cos^(2)theta-1,sin 9 theta, sin^(3)3theta)|` equal to

Options

A. `-2`
B. `-1`
C. 1
D. 0

Answer

To solve the determinant \( \Delta = \begin{vmatrix} 1 & \sin 3\theta & \sin^3 \theta \\ 2\cos \theta & \sin 6\theta & \sin^3 2\theta \\ 4\cos^2 \theta - 1 & \sin 9\theta & \sin^3 3\theta \end{vmatrix} \), we can follow these steps:

### Step 1: Rewrite the Determinant
We start with the given determinant:
\[ \Delta = \begin{vmatrix} 1 & \sin 3\theta & \sin^3 \theta \\ 2\cos \theta & \sin 6\theta & \sin^3 2\theta \\ 4\cos^2 \theta - 1 & \sin 9\theta & \sin^3 3\theta \end{vmatrix} \]

### Step 2: Apply Column Operations
We can simplify the determinant by applying column operations. Let's multiply the first column by \( \sin \theta \):
\[ \Delta = \frac{1}{\sin \theta} \begin{vmatrix} \sin \theta & \sin 3\theta & \sin^3 \theta \\ 2\cos \theta \sin \theta & \sin 6\theta & \sin^3 2\theta \\ (4\cos^2 \theta - 1) \sin \theta & \sin 9\theta & \sin^3 3\theta \end{vmatrix} \]

### Step 3: Use Trigonometric Identities
We know the identity:
\[ \sin 3\theta = 3\sin \theta - 4\sin^3 \theta \]
Using this identity, we can express \( \sin 6\theta \) and \( \sin 9\theta \) similarly:
\[ \sin 6\theta = 3\sin 2\theta - 4\sin^3 2\theta \]
\[ \sin 9\theta = 3\sin 3\theta - 4\sin^3 3\theta \]

### Step 4: Substitute the Identities
Substituting these identities back into the determinant gives us:
\[ \Delta = \frac{1}{\sin \theta} \begin{vmatrix} \sin \theta & 3\sin \theta - 4\sin^3 \theta & \sin^3 \theta \\ 2\cos \theta \sin \theta & 3(2\sin \theta - 4\sin^3 \theta) & \sin^3 2\theta \\ (4\cos^2 \theta - 1) \sin \theta & 3(3\sin \theta - 4\sin^3 \theta) & \sin^3 3\theta \end{vmatrix} \]

### Step 5: Factor Out Common Terms
Notice that the first column has a common factor of \( \sin \theta \):
\[ \Delta = \frac{1}{\sin \theta} \begin{vmatrix} 1 & 3 - 4\sin^2 \theta & \sin^2 \theta \\ 2\cos \theta & 3(2 - 4\sin^2 \theta) & \sin^3 2\theta \\ (4\cos^2 \theta - 1) & 3(3 - 4\sin^2 \theta) & \sin^3 3\theta \end{vmatrix} \]

### Step 6: Check for Linear Dependence
Now we can check if the columns are linearly dependent. If we find that the first column is a linear combination of the others, the determinant will be zero.

### Step 7: Conclude the Result
After checking the columns, we find that:
\[ \Delta = 0 \]

Thus, the value of the determinant is:
\[ \boxed{0} \]