Difference between revisions of "1984 AIME Problems/Problem 13"
Luimichael (talk | contribs) (→Solution) |
m (cleanup) |
||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Find the value of <math> | + | Find the value of <math>10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).</math> |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | + | === Solution 1 === | |
We know that <math>\tan(\arctan(x)) = x</math> so we can repeatedly apply the addition formula, <math>\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}</math>. Let <math>a = \arccot(3)</math>, <math>b=\arccot(7)</math>, <math>c=\arccot(13)</math>, and <math>d=\arccot(21)</math>. We have | We know that <math>\tan(\arctan(x)) = x</math> so we can repeatedly apply the addition formula, <math>\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}</math>. Let <math>a = \arccot(3)</math>, <math>b=\arccot(7)</math>, <math>c=\arccot(13)</math>, and <math>d=\arccot(21)</math>. We have | ||
Line 22: | Line 23: | ||
Thus our answer is <math>10\cdot\frac{3}{2}=15</math>. | Thus our answer is <math>10\cdot\frac{3}{2}=15</math>. | ||
− | + | === Solution 2 === | |
− | |||
− | |||
− | |||
− | |||
− | == | ||
Apply the formula <math>\cot^{-1}x + \cot^{-1} y = \cot^{-1}\left(\frac {xy-1}{x+y}\right)</math> repeatedly. | Apply the formula <math>\cot^{-1}x + \cot^{-1} y = \cot^{-1}\left(\frac {xy-1}{x+y}\right)</math> repeatedly. | ||
Line 33: | Line 29: | ||
== See also == | == See also == | ||
{{AIME box|year=1984|num-b=12|num-a=14}} | {{AIME box|year=1984|num-b=12|num-a=14}} | ||
− | + | ||
− | + | [[Category:Intermediate Trigonometry Problems]] | |
− |
Revision as of 11:56, 1 January 2008
Problem
Find the value of
Solution
Solution 1
We know that so we can repeatedly apply the addition formula, . Let $a = \arccot(3)$ (Error compiling LaTeX. Unknown error_msg), $b=\arccot(7)$ (Error compiling LaTeX. Unknown error_msg), $c=\arccot(13)$ (Error compiling LaTeX. Unknown error_msg), and $d=\arccot(21)$ (Error compiling LaTeX. Unknown error_msg). We have
,
So
and
,
so
.
Thus our answer is .
Solution 2
Apply the formula repeatedly.
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 |