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Difference between revisions of "2013 AIME I Problems/Problem 14"

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For <math>\pi \le \theta < 2\pi</math>, let
 
For <math>\pi \le \theta < 2\pi</math>, let
  
−
<math>\begin{align*}</math>
+
<math>\begin{align*}
−
<math>P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots</math>
+
P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots
−
<math>\end{align*}</math>
+
\end{align*}</math>
  
 
and
 
and
  
−
<math>\begin{align*}</math>
+
<math>\begin{align*}
−
<math>Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots</math>
+
Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots
−
<math>\end{align*}</math>
+
\end{align*}</math>
  
 
so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.

Revision as of 17:53, 10 March 2015

Problem 14

For $\pi \le \theta < 2\pi$, let

$\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}$ (Error compiling LaTeX. Unknown error_msg)

and

$\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}$ (Error compiling LaTeX. Unknown error_msg)

so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$. Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Solution 1

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) $P \sin\theta\ + Q \cos\theta\ = \cos\theta\ - \frac{1}{2}\ P$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg) and $\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) $P \cos\theta\ + Q \sin\theta\ = -2(Q-1)$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Solving for P, Q we have


$\frac{P}{Q} = \frac{\cos\theta }{2 + \sin\theta } = \frac{2\sqrt2}{7}$

Square both side, and use polynomial rational root theorem to solve $\sin\theta$

$\sin\theta = -\frac{17}{19}$

The answer is $\boxed{036}$

Solution 2

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) Use sum to product formulas to rewrite $P$ and $Q$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

$P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ...$

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) Therefore, $P \sin \theta - Q \cos \theta = -2P$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) Using $\frac{P}{Q} = \frac{2\sqrt2}{7}$, $Q = \frac{7}{2\sqrt2} P$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) Plug in to the previous equation and cancel out the "P" terms to get: $\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Then use the pythagorean identity to solve for $\sin\theta$, $\sin\theta = -\frac{17}{19} \implies \boxed{036}$

Solution 3

Note that \[e^{i\theta}=\cos(\theta)+i\sin(\theta)\]

Thus, the following identities follow immediately: \[ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)\] \[i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)\] \[i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)\]

Consider, now, the sum $Q+iP$. It follows fairly immediately that:

\[Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}\] \[Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\]

This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:

\[Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)\] \[Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}\]

Comparing real and imaginary parts, we find: \[\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}\]

Squaring this equation and letting $\sin^2(\theta)=x$:

$\frac{P^2}{Q^2}=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}$

Clearing denominators and solving for $x$ gives sine as $x=-\frac{17}{19}$.

$017+019=\boxed{036}$

See also

2013 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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