Difference between revisions of "1996 AIME Problems/Problem 10"
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Therefore, the smallest positive solution is <math>x = \boxed{159}</math>. | Therefore, the smallest positive solution is <math>x = \boxed{159}</math>. | ||
== Solution 2 == | == Solution 2 == | ||
| − | <math>\dfrac{\cos{96^\circ}}+\sin{96^{\circ}}}{\cos96^{\circ}}-\sin{96^{\circ}}}= | + | <math>\dfrac{\cos{96^\circ}}+\sin{96^{\circ}}}{\cos96^{\circ}}-\sin{96^{\circ}}} = \dfrac{1+\tan{96^{\circ}}}{1-\tan{96^{\circ}}} = \drac{\tan{45^{\circ}}+\tan{96^{\circ}}}{1-\tan{45^{\circ}}\tan{96^{\circ}}} = \tan{141^{\circ}}</math> |
So <math>19x = 141 +180n</math>, for some integer <math>n</math>. | So <math>19x = 141 +180n</math>, for some integer <math>n</math>. | ||
Revision as of 19:01, 17 August 2015
Contents
Problem
Find the smallest positive integer solution to 
.
Solution
.
The period of the tangent function is 
, and the tangent function is one-to-one over each period of its domain.
Thus, 
.
Since 
, multiplying both sides by 
yields 
.
Therefore, the smallest positive solution is 
.
Solution 2
$\dfrac{\cos{96^\circ}}+\sin{96^{\circ}}}{\cos96^{\circ}}-\sin{96^{\circ}}} = \dfrac{1+\tan{96^{\circ}}}{1-\tan{96^{\circ}}} = \drac{\tan{45^{\circ}}+\tan{96^{\circ}}}{1-\tan{45^{\circ}}\tan{96^{\circ}}} = \tan{141^{\circ}}$ (Error compiling LaTeX. Unknown error_msg)
So 
, for some integer 
.
Then 
, from which 
after some manipulation. The smallest suitable value of is therefore 16 from which
See also
| 1996 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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. 
