Difference between revisions of "2002 AMC 12P Problems/Problem 25"
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<cmath>2\cos{x}\cos{y} = \frac{\sqrt{6}}{2}.</cmath> | <cmath>2\cos{x}\cos{y} = \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{(B)}</math>. | + | 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{(B)}</math>. |
== Solution 2 (doesn't work but gives the right answer) == | == Solution 2 (doesn't work but gives the right answer) == |
Revision as of 01:54, 28 September 2024
Problem
Let and be real numbers such that and Find
Solution
Suppose we substitute and . Sum to product gives us
Dividing these equations tells us that , so for an integer . Note that , so , so our answer is .
Solution 2 (doesn't work but gives the right answer)
Given We multiply both sides of the syetem, , then we get . i.e. .
We must get the sum of the first part of the equation, then we calculate , we will get as and .
So
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) | |
Preceded by Problem 24 |
Followed by Last question |
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All AMC 12 Problems and Solutions |
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