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Difference between revisions of "2002 AMC 12P Problems/Problem 25"

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get <math>\sin{a}\cos{a}  + \sin{b} \cos{b} = 0 </math> as <math> \sin^{2}{a}+\cos^{2}{a} = 1</math> and <math> \sin^{2}{b}+\cos^{2}{b} = 1</math>.  
 
get <math>\sin{a}\cos{a}  + \sin{b} \cos{b} = 0 </math> as <math> \sin^{2}{a}+\cos^{2}{a} = 1</math> and <math> \sin^{2}{b}+\cos^{2}{b} = 1</math>.  
  
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So <math>\sin{(a+b)} = \frac{\sqrt{3}}{2} \Longrightarrow \boxed{\mathrm{C}}</math>
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So <math>\sin{(a+b)} = \frac{\sqrt{3}}{2} \Longrightarrow \boxed{\textbf{(C) }\frac{\sqrt{3}}{2}}</math>
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Comment: This problem is pretty much identical to [[2007 AMC 12A Problems/Problem 17|2007 AMC 12A Problem 17]] except with different numbers.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=24|after=Last question}}
 
{{AMC12 box|year=2002|ab=P|num-b=24|after=Last question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:53, 15 July 2024

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

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.

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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