Difference between revisions of "2013 AIME I Problems/Problem 14"
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− | Then use the pythagorean identity to solve for <math>\sin\theta</math>, <math>\sin\theta = \boxed{036}</math> | + | Then use the pythagorean identity to solve for <math>\sin\theta</math>, <math>\sin\theta = -\frac{17}{19} \implies \boxed{036}</math> |
===Solution 3=== | ===Solution 3=== |
Revision as of 23:50, 26 February 2014
Problem 14
For , let
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) $P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots$ (Error compiling LaTeX. Unknown error_msg) $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
and
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) $Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots$ (Error compiling LaTeX. Unknown error_msg) $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
so that . Then
where
and
are relatively prime positive integers. Find
.
Solution
Solution 1
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg)
$\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
and
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg)
$\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Solving for P, Q we have
Square both side, and use polynomial rational root theorem to solve
The answer is
Solution 2
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg)
Use sum to product formulas to rewrite and
$\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg)
Therefore,
$\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg)
Using ,
$\end{align*}$ (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:
$\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Then use the pythagorean identity to solve for ,
Solution 3
Note that
Thus, the following identities follow immediately:
Consider, now, the sum . It follows fairly immediately that:
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:
Comparing real and imaginary parts, we find:
Squaring this equation and letting :
Clearing denominators and solving for gives sine as
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.