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2002 AMC 12P Problems/Problem 25

Revision as of 01:54, 28 September 2024 by Alexanderruan (talk | contribs)

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 $\boxed{(B)}$.

Solution 2 (doesn't work but gives the right answer)

Given $\begin{cases}\sin{a} + \sin{b} = \frac{\sqrt{2}}{2} \dots \textcircled{1}\\ \cos {a} + \cos {b} = \frac{\sqrt{6}}{2} \dots \textcircled{2}\end{cases}$ We multiply both sides of the syetem, $\textcircled{1} \times \textcircled{2}$, then we get $(\sin{a}\cos{a} + \sin{b} \cos{b} )+( \sin{a}\cos{b} + \sin{b} \cos{a} )= \frac{\sqrt{3}}{2}$. i.e. $(\sin{a}\cos{a} + \sin{b} \cos{b} )+\sin{(a+b)}= \frac{\sqrt{3}}{2}$.

We must get the sum of the first part of the equation, then we calculate $\textcircled{1}^2+\textcircled{2}^2$, we will get $\sin{a}\cos{a} + \sin{b} \cos{b} = 0$ as $\sin^{2}{a}+\cos^{2}{a} = 1$ and $\sin^{2}{b}+\cos^{2}{b} = 1$.

So $\sin{(a+b)} = \frac{\sqrt{3}}{2} \Longrightarrow \boxed{\textbf{(C) }\frac{\sqrt{3}}{2}}$

Comment: This problem is pretty much identical to 2007 AMC 12A Problem 17 except with different numbers.

Note: This solution is wrong since equation 1 square plus equation 2 squared gives sin a sin b and cos a cos b.

See also

2002 AMC 12P (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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