The value of the determinant
`|(sintheta,costheta,sin2theta),(sin(theta+(2pi)/3),cos(theta+(2pi)/3),sin(2theta+(4pi)/3)),(sin(theta-(2pi)/3),cos(theta-(2pi)/3),sin(2theta-(4pi)/3))|` is

Options

A. `sin theta`
B. `cos theta`
C. 0
D. none of these

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} \sin \theta & \cos \theta & \sin 2\theta \\ \sin\left(\theta + \frac{2\pi}{3}\right) & \cos\left(\theta + \frac{2\pi}{3}\right) & \sin\left(2\theta + \frac{4\pi}{3}\right) \\ \sin\left(\theta - \frac{2\pi}{3}\right) & \cos\left(\theta - \frac{2\pi}{3}\right) & \sin\left(2\theta - \frac{4\pi}{3}\right) \end{vmatrix} \]

we can use properties of determinants and trigonometric identities.

### Step 1: Use properties of determinants
We can perform row operations on the determinant. Specifically, we will add the second row and the third row together and replace the third row with the result.

\[ R_3 \to R_2 + R_3 \]

This gives us the new determinant:

\[ D = \begin{vmatrix} \sin \theta & \cos \theta & \sin 2\theta \\ \sin\left(\theta + \frac{2\pi}{3}\right) & \cos\left(\theta + \frac{2\pi}{3}\right) & \sin\left(2\theta + \frac{4\pi}{3}\right) \\ \sin\left(\theta + \frac{2\pi}{3}\right) + \sin\left(\theta - \frac{2\pi}{3}\right) & \cos\left(\theta + \frac{2\pi}{3}\right) + \cos\left(\theta - \frac{2\pi}{3}\right) & \sin\left(2\theta + \frac{4\pi}{3}\right) + \sin\left(2\theta - \frac{4\pi}{3}\right) \end{vmatrix} \]

### Step 2: Simplify using trigonometric identities
Using the identities for sine and cosine:

1. \(\sin a + \sin b = 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right)\)
2. \(\cos a + \cos b = 2 \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right)\)

We can simplify the elements of the new third row:

- For sine:
\[ \sin\left(\theta + \frac{2\pi}{3}\right) + \sin\left(\theta - \frac{2\pi}{3}\right) = 2 \sin\left(\theta\right) \cos\left(\frac{2\pi}{3}\right) = -\sin\theta \]

- For cosine:
\[ \cos\left(\theta + \frac{2\pi}{3}\right) + \cos\left(\theta - \frac{2\pi}{3}\right) = 2 \cos\left(\theta\right) \cos\left(\frac{2\pi}{3}\right) = -\cos\theta \]

- For sine (again):
\[ \sin\left(2\theta + \frac{4\pi}{3}\right) + \sin\left(2\theta - \frac{4\pi}{3}\right) = 2 \sin\left(2\theta\right) \cos\left(\frac{4\pi}{3}\right) = -\sin 2\theta \]

### Step 3: Substitute back into the determinant
Now substituting these back into the determinant:

\[ D = \begin{vmatrix} \sin \theta & \cos \theta & \sin 2\theta \\ \sin\left(\theta + \frac{2\pi}{3}\right) & \cos\left(\theta + \frac{2\pi}{3}\right) & \sin\left(2\theta + \frac{4\pi}{3}\right) \\ -\sin \theta & -\cos \theta & -\sin 2\theta \end{vmatrix} \]

### Step 4: Factor out -1 from the third row
Factoring out -1 from the third row gives:

\[ D = -1 \cdot \begin{vmatrix} \sin \theta & \cos \theta & \sin 2\theta \\ \sin\left(\theta + \frac{2\pi}{3}\right) & \cos\left(\theta + \frac{2\pi}{3}\right) & \sin\left(2\theta + \frac{4\pi}{3}\right) \\ \sin \theta & \cos \theta & \sin 2\theta \end{vmatrix} \]

### Step 5: Notice that two rows are identical
Since the third row is now identical to the first row, the determinant evaluates to zero:

\[ D = 0 \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{0} \]