Difference between revisions of "1999 AIME Problems/Problem 11"
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Mathischess (talk | contribs) (Geo series formula was wrong) |
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\sum_{k = 1}^{35} \sin 5k &= \sum_{k = 1}^{35} \mbox{Im } e^{5ki}\ | \sum_{k = 1}^{35} \sin 5k &= \sum_{k = 1}^{35} \mbox{Im } e^{5ki}\ | ||
&= \mbox{Im } \sum_{k = 1}^{35} e^{5ki}\ | &= \mbox{Im } \sum_{k = 1}^{35} e^{5ki}\ | ||
− | &= \mbox{Im } \frac{e^{5i}(1 - e^{ | + | &= \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\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}\ | &= \mbox{Im } \frac{(2 \cos 5 + 2i \sin 5)[(1 - \cos 5) + i \sin 5]}{(1 - \cos 5)^2 + \sin^2 5}\ |
Revision as of 22:03, 8 October 2022
Contents
[hide]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 Manipulating this to use the identity , we get and our answer is .
Alternate Solution
We note that . We thus have that 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.