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

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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 $\sin^2 x$ :

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

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 $49+58=\boxed{107}$ .

Solution 2

As mentioned above, the first term is clearly $\frac{3}{2}.$ For the second term, we first wish to find $\frac{\cos 2x}{\cos 2y} =\frac{2\cos^2 x - 1}{2 \cos^2y -1}.$ Now we first square the first equation getting $\frac{\sin^2x}{\sin^2y} =\frac{1-\cos^2x}{1 - \cos^2y} =9.$ Squaring the second equation yields $\frac{\cos^2x}{\cos^2y} =\frac{1}{4}.$ Let $\cos^2x = a$ and $\cos^2y = b.$ We have the system of equations $ $\begin{aligned} 1 - a & = 9 - 9 b \\ 4 a & = b \end{aligned}$ $ (Error compiling LaTeX. Unknown error_msg) Multiplying the first equation by $4$ yields $4-4a = 36 - 36b$ and so $4-b =36 - 36b \implies b =\frac{32}{35}.$ We then find $a =\frac{8}{35}.$ Therefore the second fraction ends up being $\frac{\frac{64}{35}-1}{\frac{16}{35}-1} = -\frac{19}{29}$ so that means our desired sum is $\boxed{\frac{49}{58}}$ so the desired sum is $\boxed{107}.$

@@ Line 58: / Line 58: @@
 === Solution 2===
-As mentioned above, the first term is clearly <math>\displaystyle\frac{3}{2}.</math> For the second term, we first wish to find <math>\displaystyle\frac{\cos 2x}{\cos 2y} = \displaystyle\frac{2\cos^2 x - 1}{2 \cos^2y -1}.</math> Now we first square the first equation getting <math>\displaystyle\frac{\sin^2x}{\sin^2y} = \displaystyle\frac{1-\cos^2x}{1 - \cos^2y} =9.</math> Squaring the second equation yields <math>\displaystyle\frac{\cos^2x}{\cos^2y} = \displaystyle\frac{1}{4}.</math> Let <math>\cos^2x = a</math> and <math>\cos^2y = b.</math> We have the system of equations
+As mentioned above, the first term is clearly <math>\frac{3}{2}.</math> For the second term, we first wish to find <math>\frac{\cos 2x}{\cos 2y} =\frac{2\cos^2 x - 1}{2 \cos^2y -1}.</math> Now we first square the first equation getting <math>\frac{\sin^2x}{\sin^2y} =\frac{1-\cos^2x}{1 - \cos^2y} =9.</math> Squaring the second equation yields <math>\frac{\cos^2x}{\cos^2y} =\frac{1}{4}.</math> Let <math>\cos^2x = a</math> and <math>\cos^2y = b.</math> We have the system of equations
+<math>
 \begin{align*}
 -a &= 9-9b \
 a &= b \
 \end{align*}
+</math>
-Multiplying the first equation by <math>4</math> yields <math>4-4a = 36 - 36b</math> and so <math>4-b =36 - 36b \implies b = \displaystyle\frac{32}{35}.</math> We then find <math>a = \displaystyle\frac{8}{35}.</math> Therefore the second fraction ends up being <math>\displaystyle\frac{\displaystyle\frac{64}{35}-1}{\displaystyle\frac{16}{35}-1} = -\displaystyle\frac{19}{29}</math> so that means our desired sum is <math>\boxed{\frac{49}{58}}</math> so the desired sum is <math>\boxed{107}.</math>
+Multiplying the first equation by <math>4</math> yields <math>4-4a = 36 - 36b</math> and so <math>4-b =36 - 36b \implies b =\frac{32}{35}.</math> We then find <math>a =\frac{8}{35}.</math> Therefore the second fraction ends up being <math>\frac{\frac{64}{35}-1}{\frac{16}{35}-1} = -\frac{19}{29}</math> so that means our desired sum is <math>\boxed{\frac{49}{58}}</math> so the desired sum is <math>\boxed{107}.</math>
 == See Also ==
 {{AIME box|year=2012|n=II|num-b=8|num-a=10}}
 {{MAA Notice}}

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