Difference between revisions of "1999 AIME Problems/Problem 11"
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Let <math>s = \sum_{k=1}^{35}\sin 5k = \sin 5 + \sin 10 + \ldots + \sin 175</math>. 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 [[trigonometric identity|identity]] <math>\sin a \sin b = \frac 12(\cos (a-b) - \cos (a+b))</math>, we can rewrite <math>s</math> as | Let <math>s = \sum_{k=1}^{35}\sin 5k = \sin 5 + \sin 10 + \ldots + \sin 175</math>. 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 [[trigonometric identity|identity]] <math>\sin a \sin b = \frac 12(\cos (a-b) - \cos (a+b))</math>, we can rewrite <math>s</math> as | ||
− | <cmath>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)) | + | <cmath> |
− | + | \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))\\ | ||
+ | &= \frac{0.5(\cos 0 - \cos 10 + \cos 5 - \cos 15 + \cos 10 \ldots + \cos 165 - \cos 175+ \cos 170 - \cos 180)}{\sin 5}\end{align*}</cmath> | ||
− | This telescopes to < | + | 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>. |
==Alternate Solution== | ==Alternate Solution== |
Revision as of 11:22, 12 October 2020
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.