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

Suppose we substitute $\frac{a+b}{2} = x$ and $\frac{a-b}{2} = y$ . Sum to product gives us

$2\sin{x}\cos{y} = \frac{\sqrt{2}}{2}$

$2\cos{x}\cos{y} = \frac{\sqrt{6}}{2}.$

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

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2002 AMC 12P Problems/Problem 25

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