Difference between revisions of "1984 AIME Problems/Problem 13"
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=== Solution 2 === | === Solution 2 === | ||
− | Apply the formula <math>\cot^{-1}x + \cot^{-1} y = \cot^{-1}\left(\frac {xy-1}{x+y}\right)</math> repeatedly. Using it twice on the inside, the desired sum becomes <math>\cot (\cot^{-1}2+\cot^{-1}8)</math> | + | Apply the formula <math>\cot^{-1}x + \cot^{-1} y = \cot^{-1}\left(\frac {xy-1}{x+y}\right)</math> repeatedly. Using it twice on the inside, the desired sum becomes <math>\cot (\cot^{-1}2+\cot^{-1}8)</math>. This sum can then be tackled by taking the cotangent of both sides of the inverse cotangent addition formula shown at the beginning. |
== See also == | == See also == |
Revision as of 00:08, 18 August 2012
Problem
Find the value of
Solution
Solution 1
We know that so we can repeatedly apply the addition formula,
. Let
,
,
, and
. We have
,
So
and
,
so
.
Thus our answer is .
Solution 2
Apply the formula repeatedly. Using it twice on the inside, the desired sum becomes
. This sum can then be tackled by taking the cotangent of both sides of the inverse cotangent addition formula shown at the beginning.
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |