Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (doesn't work but gives the right answer))
Line 15: Line 15:
  
 
== Solution ==
 
== Solution ==
−
Suppose we substitute <math>\frac{a+b}{2} = x</math> and <math>\frac{a-b}{2} = y</math>. Sum to product gives us
+
Sum to product gives us
  
−
<cmath>2\sin{x}\cos{y} = \frac{\sqrt{2}}{2}</cmath>
+
<cmath>2\sin{\frac{a+b}{2}}\cos{\frac{a-b}{2}} = \frac{\sqrt{2}}{2}</cmath>
  
−
<cmath>2\cos{x}\cos{y} = \frac{\sqrt{6}}{2}.</cmath>
+
<cmath>2\cos{\frac{a+b}{2}}\cos{\frac{a-b}{2}} = \frac{\sqrt{6}}{2}.</cmath>
  
−
Dividing these equations tells us that <math>\tan{x} = \frac{1}{\sqrt{3}}</math>, so <math>x = \frac{\pi}{6} + \pi n</math> for an integer <math>n</math>. Note that <math>a+b = 2x</math>, so <math>\sin(a+b) = \sin{2x} = \sin (\frac{\pi}{3} + 2\pi n) = \frac{\sqrt{3}}{2}</math>, so our answer is <math>\boxed{(C)}</math>.
+
Dividing these equations tells us that <math>\frac{a+b}{2} = \frac{1}{\sqrt{3}}</math>, so <math>\frac{a+b}{2} = \frac{\pi}{6} + \pi n</math> for an integer <math>n</math>, so <math>\sin(a+b) = \sin (\frac{\pi}{3} + 2\pi n) = \frac{\sqrt{3}}{2}</math>, so our answer is <math>\boxed{(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 18:59, 3 October 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

Sum to product gives us

\[2\sin{\frac{a+b}{2}}\cos{\frac{a-b}{2}} = \frac{\sqrt{2}}{2}\]

\[2\cos{\frac{a+b}{2}}\cos{\frac{a-b}{2}} = \frac{\sqrt{6}}{2}.\]

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 $\sin(a+b) = \sin (\frac{\pi}{3} + 2\pi n) = \frac{\sqrt{3}}{2}$, so our answer is $\boxed{(C)}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12P_Problems/Problem_25&oldid=228904"