Difference between revisions of "1996 AIME Problems/Problem 10"
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== Solution == | == Solution == | ||
+ | <math>\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} =</math> <math>\dfrac{\sin{186^{\circ}}+\sin{96^{\circ}}}{\sin{186^{\circ}}-\sin{96^{\circ}}} =</math> <math>\dfrac{\sin{(141^{\circ}+45^{\circ})}+\sin{(141^{\circ}-45^{\circ})}}{\sin{(141^{\circ}+45^{\circ})}-\sin{(141^{\circ}-45^{\circ})}} =</math> <math>\dfrac{2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}</math>. | ||
− | + | The period of the [[Trigonometry#Tangent|tangent]] function is <math>180^\circ</math>, and the tangent function is [[one-to-one]] over each period of its domain. | |
− | |||
+ | Thus, <math>19x \equiv 141 \pmod{180}</math>. | ||
+ | |||
+ | Since <math>19^2 \equiv 361 \equiv 1 \pmod{180}</math>, multiplying both sides by <math>19</math> yields <math>x \equiv 141 \cdot 19 \equiv (140+1)(18+1) \equiv 0+140+18+1 \equiv 159 \pmod{180}</math>. | ||
+ | |||
+ | Therefore, the smallest positive solution is <math>x = \boxed{159}</math>. | ||
+ | == 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> | ||
+ | which is the same as <math>\dfrac{\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>. | ||
+ | Multiplying by <math>19</math> gives <math>x \equiv 141 \cdot 19 \equiv 2679 \equiv 159 \pmod{180}</math>. | ||
+ | The smallest positive solution of this is <math>x = \boxed{159}</math> | ||
+ | |||
+ | == See Also == | ||
+ | |||
{{AIME box|year=1996|num-b=9|num-a=11}} | {{AIME box|year=1996|num-b=9|num-a=11}} | ||
+ | |||
+ | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 00:53, 28 May 2020
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 .
So , for some integer . Multiplying by gives . The smallest positive solution of this is
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 |
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