Difference between revisions of "2012 AIME II Problems/Problem 9"
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Examine the first term in the expression we want to evaluate, <math>\frac{\sin 2x}{\sin 2y}</math>, separately from the second term, <math>\frac{\cos 2x}{\cos 2y}</math>. | Examine the first term in the expression we want to evaluate, <math>\frac{\sin 2x}{\sin 2y}</math>, separately from the second term, <math>\frac{\cos 2x}{\cos 2y}</math>. | ||
− | == The First Term == | + | === The First Term === |
Using the identity <math>\sin 2\theta = 2\sin\theta\cos\theta</math>, we have: | Using the identity <math>\sin 2\theta = 2\sin\theta\cos\theta</math>, we have: | ||
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<math>\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}</math> | <math>\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}</math> | ||
− | == The Second Term == | + | === The Second Term === |
Let the equation <math>\frac{\sin x}{\sin y} = 3</math> be equation 1, and let the equation <math>\frac{\cos x}{\cos y} = \frac12</math> be equation 2. | Let the equation <math>\frac{\sin x}{\sin y} = 3</math> be equation 1, and let the equation <math>\frac{\cos x}{\cos y} = \frac12</math> be equation 2. | ||
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Thus, <math>\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}</math>. | Thus, <math>\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}</math>. | ||
− | == Now Back to the Solution! == | + | === Now Back to the Solution! === |
Finally, <math>\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}</math>. | Finally, <math>\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}</math>. |
Revision as of 02:09, 4 April 2012
Contents
Problem 9
Let and
be real numbers such that
and
. The value of
can be expressed in the form
, where
and
are relatively prime positive integers. Find
.
Solution
Examine the first term in the expression we want to evaluate, , separately from the second term,
.
The First Term
Using the identity , we have:
The Second Term
Let the equation be equation 1, and let the equation
be equation 2.
Hungry for the widely-used identity
, we cross multiply equation 1 by
and multiply equation 2 by
.
Equation 1 then becomes:
.
Equation 2 then becomes:
Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:
Applying the identity (which is similar to
but a bit different), we can change
into:
Rearranging, we get .
So, .
Squaring Equation 1 (leading to ), we can solve for
:
Using the identity , we can solve for
.
Thus, .
Now Back to the Solution!
Finally, .
So, the answer is .
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |