What is the value of \sin(2n\pi+\frac{5\pi}{6})\sin(2n\pi-\frac{5\pi}{6}), where n\in Z?
- A. -\frac{1}{4} ✓
- B. -\frac{3}{4}
- C. \frac{1}{4}
- D. \frac{3}{4}
Correct Answer: A. -\frac{1}{4}
Explanation
Since sine is periodic with period 2\pi, \sin(2n\pi + \theta) = \sin\theta. The expression simplifies to \sin(\frac{5\pi}{6}) \sin(-\frac{5\pi}{6}). Since \sin(\frac{5\pi}{6}) = \frac{1}{2} and \sin(-\frac{5\pi}{6}) = -\frac{1}{2}, their product is (\frac{1}{2})(-\frac{1}{2}) = -\frac{1}{4}.