2002 AMC 12P Problems/Problem 25

Revision as of 19:00, 3 October 2024 by Alexanderruan (talk | contribs) (Solution)

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

\[2\sin(\frac{a+b}{2}}\cos{\frac{a-b}{2}) = \frac{\sqrt{2}}{2}\] (Error compiling LaTeX. Unknown error_msg)
\[2\cos(\frac{a+b}{2}}\cos{\frac{a-b}{2}) = \frac{\sqrt{6}}{2}.\] (Error compiling LaTeX. Unknown error_msg)

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 $\boxed{(C)}$.

See also

2002 AMC 12P (ProblemsAnswer KeyResources)
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