1999 AIME Problems/Problem 11
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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 
.
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 | ||
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