Difference between revisions of "2002 AMC 12P Problems/Problem 25"
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<cmath>2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \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>\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>. The answer is <math>\boxed{(C)}</math>. | + | Dividing these equations tells us that <math>\tan(\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>. The 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 19:02, 3 October 2024
Problem
Let 
and 
be real numbers such that 
and 
Find 
Solution
Sum to product gives us
![\[2\sin(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \frac{\sqrt{2}}{2}\]](http://latex.artofproblemsolving.com/e/6/3/e6398be26614a0af045fd9dd4f6a7085f288c64c.png)
![\[2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \frac{\sqrt{6}}{2}\]](http://latex.artofproblemsolving.com/7/5/e/75e6b189eaefdbdc3e9a6af3f88c3faed5c7e73a.png)
Dividing these equations tells us that 
, so 
for an integer 
, so 
. The answer is 
.
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. 
