Difference between revisions of "1996 AIME Problems/Problem 10"
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== Solution 2 == | == Solution 2 == | ||
<math>\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \dfrac{1 + \tan{96^{\circ}}}{1-\tan{96^{\circ}}}</math> | <math>\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \dfrac{1 + \tan{96^{\circ}}}{1-\tan{96^{\circ}}}</math> | ||
− | which is the same as <math>\dfrac{\tan{45^{\circ}} + \tan{96^{\circ}}}{1-\tan{45^{\circ}}\tan{96^{\circ}}}</math> | + | which is the same as <math>\dfrac{\tan{45^{\circ}} + \tan{96^{\circ}}}{1-\tan{45^{\circ}}\tan{96^{\circ}}}</math> i.e. <math>\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 20:11, 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
which is the same as i.e.
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.