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

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== Solution ==
 
== Solution ==
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If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
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Given <math>\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} </math>
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We multiply both sides of the syetem, <math>\textcircled{1} \times \textcircled{2}</math>, then we get <math> (\sin{a}\cos{a}  + \sin{b} \cos{b}  )+( \sin{a}\cos{b}  + \sin{b} \cos{a} )= \frac{\sqrt{3}}{2}</math>. i.e.
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<math>(\sin{a}\cos{a}  + \sin{b} \cos{b}  )+\sin{(a+b)}= \frac{\sqrt{3}}{2}</math>.
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We must get the sum of the first part of the equation, then we calculate <math>\textcircled{1}^2+\textcircled{2}^2 </math>, we will
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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>.  
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So <math>\sin{(a+b)} = \frac{\sqrt{3}}{2} \Longrightarrow \boxed{\mathrm{C}}</math>
  
 
== 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}}

Revision as of 14:22, 21 January 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 $\sin{(a+b)} = \frac{\sqrt{3}}{2} \Longrightarrow \boxed{\mathrm{C}}$

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last question
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All AMC 12 Problems and Solutions

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