If $x\sin\theta = y\sin\left(\theta+\frac{2\pi}{3}\right) = z\sin\left(\theta+\frac{4\pi}{3}\right)$, then $\sum xy=$
Step-by-Step Solution
Key Concept: Three sines equally spaced by $\frac{2\pi}{3}$ radians sum to zero due to symmetry in the unit circle.
Let each ratio $\frac{1}{x} = \frac{1}{y} = \frac{1}{z} = \frac{1}{k}$. Then $k\left(\sin\theta + \sin\left(\theta + \frac{2\pi}{3}\right) + \sin\left(\theta + \frac{4\pi}{3}\right)\right) = 0$. Using the product-to-sum identity with $\cos\frac{2\pi}{3} = -\frac{1}{2}$, this becomes $k\left(2\sin\left(\theta + \frac{2\pi}{3}\right)\cos\frac{2\pi}{3} + \sin\left(\theta + \frac{2\pi}{3}\right)\right) = 0$, which simplifies to $\sum xy = xy + yz + zx = 0$.
Correct Answer: 3