Difference between revisions of "1999 AIME Problems/Problem 11"
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\begin{align*} | \begin{align*} | ||
s \cdot \sin 5 = \sum_{k=1}^{35} \sin 5k \sin 5 &= \sum_{k=1}^{35} \frac{1}{2}(\cos (5k - 5)- \cos (5k + 5))\\ | s \cdot \sin 5 = \sum_{k=1}^{35} \sin 5k \sin 5 &= \sum_{k=1}^{35} \frac{1}{2}(\cos (5k - 5)- \cos (5k + 5))\\ | ||
| − | &= | + | &= 0.5(\cos 0 - \cos 10 + \cos 5 - \cos 15 + \cos 10 \ldots + \cos 165 - \cos 175+ \cos 170 - \cos 180) |
| + | \end{align*}</cmath> | ||
This telescopes to <cmath>s = \frac{\cos 0 + \cos 5 - \cos 175 - \cos 180}{2 \sin 5} = \frac{1 + \cos 5}{\sin 5}.</cmath> Manipulating this to use the identity <math>\tan x = \frac{1 - \cos 2x}{\sin 2x}</math>, we get <cmath>s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},</cmath> and our answer is <math>\boxed{177}</math>. | This telescopes to <cmath>s = \frac{\cos 0 + \cos 5 - \cos 175 - \cos 180}{2 \sin 5} = \frac{1 + \cos 5}{\sin 5}.</cmath> Manipulating this to use the identity <math>\tan x = \frac{1 - \cos 2x}{\sin 2x}</math>, we get <cmath>s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},</cmath> and our answer is <math>\boxed{177}</math>. | ||
Revision as of 17:00, 25 September 2023
Problem
Given that 
where angles are measured in degrees, and 
and 
are relatively prime positive integers that satisfy 
find 
Solution
Let 
. We could try to manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to telescope the sum. Using the identity 
, we can rewrite 
as
This telescopes to ![\[s = \frac{\cos 0 + \cos 5 - \cos 175 - \cos 180}{2 \sin 5} = \frac{1 + \cos 5}{\sin 5}.\]](http://latex.artofproblemsolving.com/6/8/2/68204002e17fa96b4bd85331949cf679f8ca45e8.png)
Manipulating this to use the identity 
, we get ![\[s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},\]](http://latex.artofproblemsolving.com/2/a/7/2a7fb83217837fd611ebaa7fe87686c1620c75e1.png)
and our answer is 
.
Alternate Solution
We note that 
. We thus have that
![\begin{align*} \sum_{k = 1}^{35} \sin 5k &= \sum_{k = 1}^{35} \mbox{Im } e^{5ki}\\ &= \mbox{Im } \sum_{k = 1}^{35} e^{5ki}\\ &= \mbox{Im } \frac{e^{5i}(1 - e^{180i})}{1 - e^{5i}}\\ &= \mbox{Im } \frac{2\cos5 + 2i \sin 5}{(1 - \cos 5) - i \sin 5}\\ &= \mbox{Im } \frac{(2 \cos 5 + 2i \sin 5)[(1 - \cos 5) + i \sin 5]}{(1 - \cos 5)^2 + \sin^2 5}\\ &= \frac{2 \sin 5}{2 - 2 \cos 5}\\ &= \frac{\sin 5}{1 - \cos 5}\\ &= \frac{\sin 175}{1 + \cos 175} \\ &= \tan \frac{175}{2}.\\ \end{align*}](http://latex.artofproblemsolving.com/6/4/2/6422d893048a9c0d41bfec17a4ca96f6ab10fbc5.png)
The desired answer is thus 
.
- Only if
is in radians, which it is not. However, the solution is still viable, so keep reading.
is in radians, which it is not. However, the solution is still viable, so keep reading.See also
| 1999 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
