What is the value of \sin 10^{\circ}\cdot \sin 50^{\circ}+\sin 50^{\circ}\cdot \sin 250^{\circ}+\sin 250^{\circ}\cdot \sin 10^{\circ} equal to ?
- A. -\frac{1}{4}
- B. -\frac{3}{4} ✓
- C. \frac{3 \sin 10^{\circ}}{4}
- D. -\frac{3 \cos 10^{\circ}}{4}
Correct Answer: B. -\frac{3}{4}
Explanation
Substitute \sin 250^{\circ} = -\sin 70^{\circ}. The expression is \sin 10^{\circ}\sin 50^{\circ} - \sin 50^{\circ}\sin 70^{\circ} - \sin 70^{\circ}\sin 10^{\circ}. Multiply and divide by 2 to apply 2\sin A \sin B = \cos(A-B) - \cos(A+B). It simplifies to \frac{1}{2}(\cos 40^{\circ} - \cos 60^{\circ} - (\cos 20^{\circ} - \cos 120^{\circ}) - (\cos 60^{\circ} - \cos 80^{\circ})). Grouping terms and using \cos 40^{\circ} + \cos 80^{\circ} = \cos 20^{\circ}, everything cancels out leaving \frac{1}{2}(-\frac{1}{2} - \frac{1}{2} - \frac{1}{2}) = -\frac{3}{4}.