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

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== Problem 14 ==
 
== Problem 14 ==
(problem)
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14. For <math>\pi \le \theta < 2\pi</math>, let
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\begin{align*}
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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 \\ &\quad - \frac{1}{128} \cos 7\theta + \cdots
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\end{align*}
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and
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\begin{align*}
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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 \\
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&\quad +\frac{1}{128}\sin 7\theta + \cdots
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\end{align*}
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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>.
  
 
== Solution ==
 
== Solution ==

Revision as of 22:45, 16 March 2013

Problem 14

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 \\ &\quad - \frac{1}{128} \cos 7\theta + \cdots \end{align*} 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 \\ &\quad +\frac{1}{128}\sin 7\theta + \cdots \end{align*} 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)

See also

2013 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AIME Problems and Solutions