Question

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

• A. \(3\sin\theta\)
• B. \(\sin^3\theta\)
• C. 0
• D. None of these

Hint:

If one row is a linear combination of others, the determinant is zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Check for linear dependence between rows.

Step 2: Detailed Explanation:
Note that \(\sin(\theta+2\pi/3) = \sin\theta\cos(2\pi/3) + \cos\theta\sin(2\pi/3) = -\frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta\). Similarly \(\sin(\theta-2\pi/3) = -\frac{1}{2}\sin\theta - \frac{\sqrt{3}}{2}\cos\theta\). So these are linear combinations of \(\sin\theta\) and \(\cos\theta\). The same holds for the cosine terms. The third column: \(\sin(2\theta+4\pi/3) = \sin(2\theta)\cos(4\pi/3) + \cos(2\theta)\sin(4\pi/3) = -\frac{1}{2}\sin2\theta - \frac{\sqrt{3}}{2}\cos2\theta\). The three rows are linear combinations of the same basis functions. It can be shown that the sum of the second and third rows is a multiple of the first row, making the determinant zero. Alternatively, add row 2 and row 3: Row2+Row3 = \((-\sin\theta, -\cos\theta, -\sin2\theta) = -Row1\). So rows are linearly dependent. Hence determinant = 0.

Step 3: Final Answer:
0, which corresponds to option (C).