2013 AIME I Problems/Problem 14
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 | ||
