What is \sin 12^{\circ}\sin 48^{\circ} equal to?
- A. \frac{\sqrt{5}-1}{4}
- B. \frac{\sqrt{5}+1}{4}
- C. \frac{\sqrt{5}-1}{8} ✓
- D. \frac{\sqrt{5}+1}{8}
Correct Answer: C. \frac{\sqrt{5}-1}{8}
Explanation
Using the product-to-sum formula \sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)], we have \sin 12^{\circ}\sin 48^{\circ} = \frac{1}{2}[\cos(48^{\circ}-12^{\circ}) - \cos(48^{\circ}+12^{\circ})] = \frac{1}{2}[\cos 36^{\circ} - \cos 60^{\circ}]. Substituting the known values \cos 36^{\circ} = \frac{\sqrt{5}+1}{4} and \cos 60^{\circ} = \frac{1}{2}, we get \frac{1}{2}[\frac{\sqrt{5}+1}{4} - \frac{2}{4}] = \frac{\sqrt{5}-1}{8}.