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Difference between revisions of "2012 AIME II Problems/Problem 9"

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== Solution ==
 
== Solution ==
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{{solution}}
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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>.
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 +
== The First Term ==
 +
 
 +
Using the identity <math>\sin 2\theta = 2\sin\theta\cos\theta</math>, we have:
 +
 
 +
<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 ==
 +
 
 +
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.
 +
Hungry for the widely-used identity <math>\sin^2\theta + \cos^2\theta = 1</math>, we cross multiply equation 1 by <math>\sin y</math> and multiply equation 2 by <math>\cos y</math>.
 +
 
 +
Equation 1 then becomes:
 +
 
 +
<math>\sin x = 3\sin y</math>.
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 +
Equation 2 then becomes:
 +
 
 +
<math>\cos x = \frac{1}{2} \cos y</math>
 +
 
 +
Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:
 +
 
 +
<math>1 = 9\sin^2 y + \frac{1}{4} \cos^2 y</math>
 +
 
 +
Applying the identity <math>\cos^2 y = 1 - \sin^2 y</math> (which is similar to <math>\sin^2\theta + \cos^2\theta = 1</math> but a bit different), we can change <math>1 = 9\sin^2 y + \frac{1}{4} \cos^2 y</math> into:
 +
 
 +
<math>1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y</math>
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Rearranging, we get <math>\frac{3}{4} = \frac{35}{4} \sin^2 y </math>.
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So, <math>\sin^2 y = \frac{3}{35}</math>.
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Squaring Equation 1 (leading to <math>\sin^2 x = 9\sin^2 y</math>), we can solve for <math>\sin^2 y</math>:
 +
 
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<math>\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}</math>
 +
 
 +
Using the identity <math>\cos 2\theta = 1 - 2\sin^2\theta</math>, we can solve for <math>\frac{\cos 2x}{\cos 2y}</math>.
 +
 
 +
<math>\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}</math>
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 +
<math>\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}</math>
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 +
Thus, <math>\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}</math>.
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== Now Back to the Solution! ==
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Finally, <math>\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}</math>.
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So, the answer is <math>49+58=\boxed{107}</math>.
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== See Also ==
 
== See Also ==
 
{{AIME box|year=2012|n=II|num-b=8|num-a=10}}
 
{{AIME box|year=2012|n=II|num-b=8|num-a=10}}

Revision as of 20:26, 3 April 2012

Problem 9

Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$. The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.


Solution

Examine the first term in the expression we want to evaluate, $\frac{\sin 2x}{\sin 2y}$, separately from the second term, $\frac{\cos 2x}{\cos 2y}$.

The First Term

Using the identity $\sin 2\theta = 2\sin\theta\cos\theta$, we have:

$\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}$

The Second Term

Let the equation $\frac{\sin x}{\sin y} = 3$ be equation 1, and let the equation $\frac{\cos x}{\cos y} = \frac12$ be equation 2. Hungry for the widely-used identity $\sin^2\theta + \cos^2\theta = 1$, we cross multiply equation 1 by $\sin y$ and multiply equation 2 by $\cos y$.

Equation 1 then becomes:

$\sin x = 3\sin y$.

Equation 2 then becomes:

$\cos x = \frac{1}{2} \cos y$

Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:

$1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$

Applying the identity $\cos^2 y = 1 - \sin^2 y$ (which is similar to $\sin^2\theta + \cos^2\theta = 1$ but a bit different), we can change $1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$ into:

$1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y$

Rearranging, we get $\frac{3}{4} = \frac{35}{4} \sin^2 y$.

So, $\sin^2 y = \frac{3}{35}$.

Squaring Equation 1 (leading to $\sin^2 x = 9\sin^2 y$), we can solve for $\sin^2 y$:

$\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}$

Using the identity $\cos 2\theta = 1 - 2\sin^2\theta$, we can solve for $\frac{\cos 2x}{\cos 2y}$.

$\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}$

$\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}$

Thus, $\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}$.

Now Back to the Solution!

Finally, $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}$.

So, the answer is $49+58=\boxed{107}$.

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
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