Difference between revisions of "2013 AIME I Problems/Problem 14"
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== Problem 14 == | == Problem 14 == | ||
| − | + | 14. For <math>\pi \le \theta < 2\pi</math>, 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 <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 21:45, 16 March 2013
Problem 14
14. For 
, 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 
. Then 
where 
and 
are relatively prime positive integers. Find 
.
Solution
(solution)
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 | ||
