2002 AMC 12P Problems/Problem 25

Problem

Let $a$ and $b$ be real numbers such that $\sin{a} + \sin{b} = \frac{\sqrt{2}}{2}$ and $\cos {a} + \cos {b} = \frac{\sqrt{6}}{2}.$ Find $\sin{(a+b)}.$

$\text{(A) }\frac{1}{2} \qquad \text{(B) }\frac{\sqrt{2}}{2} \qquad \text{(C) }\frac{\sqrt{3}}{2} \qquad \text{(D) }\frac{\sqrt{6}}{2} \qquad \text{(E) }1$

Solution

Sum to product gives us

$2\sin(\frac{a+b}{2})\cos)\frac{a-b}{2}) = \frac{\sqrt{2}}{2}$

$2\cos(\frac{a+b}{2})\cos)\frac{a-b}{2}) = \frac{\sqrt{6}}{2}$

Dividing these equations tells us that $\frac{a+b}{2} = \frac{1}{\sqrt{3}}$ , so $\frac{a+b}{2} = \frac{\pi}{6} + \pi n$ for an integer $n$ , so $\sin(a+b) = \sin (\frac{\pi}{3} + 2\pi n) = \frac{\sqrt{3}}{2}$ . The answer is $\boxed{(C)}$ .

2002 AMC 12P (Problems • Answer Key • Resources)
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2002 AMC 12P Problems/Problem 25

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