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

To solve the given problem, we need to determine the value of the determinant:

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

We shall proceed with the step-by-step calculation:

1. Use the trigonometric identities. We know:
   • \(\sin(a+b) = \sin a \cos b + \cos a \sin b\)
   • \(\cos(a+b) = \cos a \cos b - \sin a \sin b\)
2. For the second row:
   • \(\sin\left(\theta + \frac{2\pi}{3}\right) = \sin\theta \cdot \cos\frac{2\pi}{3} + \cos\theta \cdot \sin\frac{2\pi}{3}\)
   • \(\cos\left(\theta + \frac{2\pi}{3}\right) = \cos\theta \cdot \cos\frac{2\pi}{3} - \sin\theta \cdot \sin\frac{2\pi}{3}\)
   • \(\sin\left(2\theta + \frac{4\pi}{3}\right) = \sin 2\theta \cdot \cos\frac{4\pi}{3} + \cos 2\theta \cdot \sin\frac{4\pi}{3}\)
3. Similarly for the third row:
   • \(\sin\left(\theta - \frac{2\pi}{3}\right) = \sin\theta \cdot \cos\left(-\frac{2\pi}{3}\right) + \cos\theta \cdot \sin\left(-\frac{2\pi}{3}\right)\)
   • \(\cos\left(\theta - \frac{2\pi}{3}\right) = \cos\theta \cdot \cos\left(-\frac{2\pi}{3}\right) - \sin\theta \cdot \sin\left(-\frac{2\pi}{3}\right)\)
   • \(\sin\left(2\theta - \frac{4\pi}{3}\right) = \sin 2\theta \cdot \cos\left(-\frac{4\pi}{3}\right) + \cos 2\theta \cdot \sin\left(-\frac{4\pi}{3}\right)\)
4. We know the values of the trigonometric functions for these standard angles:
   • \(\cos\frac{2\pi}{3} = -\frac{1}{2}, \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}\)
   • \(\cos(-\frac{2\pi}{3}) = -\frac{1}{2}, \sin(-\frac{2\pi}{3}) = -\frac{\sqrt{3}}{2}\)
   • \(\cos\frac{4\pi}{3} = -\frac{1}{2}, \sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2}\)
   • \(\cos(-\frac{4\pi}{3}) = -\frac{1}{2}, \sin(-\frac{4\pi}{3}) = \frac{\sqrt{3}}{2}\)
5. Substitute the identities back into the determinant and simplify row operations for mathematical clarity and convenience.
6. Observe that the determinant of this particular trigonometric matrix equals zero by properties of periodicity and linear dependence of angles used.

Thus, the final value of the given determinant is 0.