ABC is a triangle such that angle C=60^{\circ}, then what is \frac{\cos A+\cos B}{\cos(\frac{A-B}{2})} equal to ?
- A. 2
- B. \sqrt{2}
- C. 1 ✓
- D. \frac{1}{\sqrt{2}}
Correct Answer: C. 1
Explanation
Using the sum-to-product formula, \cos A + \cos B = 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2}). The given expression simplifies to 2\cos(\frac{A+B}{2}). Since A+B+C = 180^{\circ} and C = 60^{\circ}, we have A+B = 120^{\circ}. Thus, \frac{A+B}{2} = 60^{\circ}. Evaluating the expression gives 2\cos 60^{\circ} = 2(\frac{1}{2}) = 1.