Difference between revisions of "1996 AIME Problems/Problem 10"

Revision as of 14:45, 3 August 2022

Problem

Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$ .

Solution

$\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{186^{\circ}}+\sin{96^{\circ}}}{\sin{186^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{(141^{\circ}+45^{\circ})}+\sin{(141^{\circ}-45^{\circ})}}{\sin{(141^{\circ}+45^{\circ})}-\sin{(141^{\circ}-45^{\circ})}} =$ $\dfrac{2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}$ .

The period of the tangent function is $180^\circ$ , and the tangent function is one-to-one over each period of its domain.

Thus, $19x \equiv 141 \pmod{180}$ .

Since $19^2 \equiv 361 \equiv 1 \pmod{180}$ , multiplying both sides by $19$ yields $x \equiv 141 \cdot 19 \equiv (140+1)(18+1) \equiv 0+140+18+1 \equiv 159 \pmod{180}$ .

Therefore, the smallest positive solution is $x = \boxed{159}$ .

Solution 2

$\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \dfrac{1 + \tan{96^{\circ}}}{1-\tan{96^{\circ}}}$ which is the same as $\dfrac{\tan{45^{\circ}} + \tan{96^{\circ}}}{1-\tan{45^{\circ}}\tan{96^{\circ}}} = \tan{141{^\circ}}$ .

So $19x = 141 +180n$ , for some integer $n$ . Multiplying by $19$ gives $x \equiv 141 \cdot 19 \equiv 2679 \equiv 159 \pmod{180}$ . The smallest positive solution of this is $x = \boxed{159}$

Solution 3 (Only sine and cosine sum formulas)

It seems reasonable to assume that $\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \tan{\theta}$ for some angle $\theta$ . This means $\dfrac{\alpha (\cos{96^{\circ}}+\sin{96^{\circ}})}{\alpha (\cos{96^{\circ}}-\sin{96^{\circ}})} = \frac{\sin{\theta}}{\cos{\theta}}$ for some constant $\alpha$ . We can set $\alpha (\cos{96^{\circ}}+\sin{96^{\circ}}) = \sin{\theta}$ .Note that if we have $\alpha$ equal to both the sine and cosine of an angle, we can use the sine sum formula (and the cosine sum formula on the denominator). So, since $\sin{45^{\circ}} = \cos{45^{\circ}} = \tfrac{\sqrt{2}}{2}$ , if $\alpha = \tfrac{\sqrt{2}}{2}$ we have $\alpha (\cos{96^{\circ}}+\sin{96^{\circ}} = \cos{96^{\circ}} \frac{\sqrt{2}}{2} + \sin{96^{\circ}} \frac{\sqrt{2}}{2} = \cos{96^{\circ}} \sin{45^{\circ}} + \sin{96^{\circ}} \cos{45^{\circ}} = \sin{45^{\circ} + 96^{\circ}} = \sin{141^{\circ}}$ from the sine sum formula. For the denominator, from the cosine sum formula, we have $\alpha (\cos{96^{\circ}} - \sin{96^{\circ}}) = \cos{96^{\circ}} \frac{\sqrt{2}}{2} + \sin{96^{\circ}} \frac{\sqrt{2}}{2} = \cos{96^{\circ}} \cos{45^{\circ}} + \sin{96^{\circ}} \sin{45^{\circ}} = \cos{96^{\circ} + 45^{\circ}} = \cos{141^{\circ}}.$ This means $\theta = 141^{\circ},$ so $19x = 141 + 180k$ for some positive integer $k$ (since the period of tangent is $180^{\circ}$ ), or $19 x \equiv 141 \pmod{180}$ . Note that the inverse of $19$ modulo $180$ is itself as $19^2 \equiv 361 \equiv 1 \pmod {180}$ , so multiplying this congruence by $19$ on both sides gives $x \equiv 2679 \equiv 159 \pmod{180}.$ For the smallest possible $x$ , we take $x = \boxed{159}.$

@@ Line 21: / Line 21: @@
 == Solution 3 (Only sine and cosine sum formulas) ==
-It seems reasonable to assume that <math>\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \tan{\theta}</math> for some angle <math>\theta</math>. This means <cmath>\dfrac{\alpha (\cos{96^{\circ}}+\sin{96^{\circ}})}{\alpha (\cos{96^{\circ}}-\sin{96^{\circ}})} = \frac{\sin{\theta}}{\cos{\theta}}</cmath> for some constant <math>\alpha</math>. We can set <math>\alpha (\cos{96^{\circ}}+\sin{96^{\circ}}) = \sin{\theta}</math>.Note that if we have <math>\alpha</math> equal to both the sine and cosine of an angle, we can use the sine sum formula (and the cosine sum formula on the denominator). So, since <math>\sin{45^{\circ}} = \cos{45^{\circ}} = \tfrac{\sqrt{2}}{2}</math>, if <math>\alpha = \frac{\sqrt{2}}{2}</math> we have
+It seems reasonable to assume that <math>\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \tan{\theta}</math> for some angle <math>\theta</math>. This means <cmath>\dfrac{\alpha (\cos{96^{\circ}}+\sin{96^{\circ}})}{\alpha (\cos{96^{\circ}}-\sin{96^{\circ}})} = \frac{\sin{\theta}}{\cos{\theta}}</cmath> for some constant <math>\alpha</math>. We can set <math>\alpha (\cos{96^{\circ}}+\sin{96^{\circ}}) = \sin{\theta}</math>.Note that if we have <math>\alpha</math> equal to both the sine and cosine of an angle, we can use the sine sum formula (and the cosine sum formula on the denominator). So, since <math>\sin{45^{\circ}} = \cos{45^{\circ}} = \tfrac{\sqrt{2}}{2}</math>, if <math>\alpha = \tfrac{\sqrt{2}}{2}</math> we have
 <cmath>\alpha (\cos{96^{\circ}}+\sin{96^{\circ}} = \cos{96^{\circ}} \frac{\sqrt{2}}{2} + \sin{96^{\circ}} \frac{\sqrt{2}}{2} = \cos{96^{\circ}} \sin{45^{\circ}} + \sin{96^{\circ}} \cos{45^{\circ}} = \sin{45^{\circ} + 96^{\circ}} = \sin{141^{\circ}}</cmath>
 from the sine sum formula. For the denominator, from the cosine sum formula, we have

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

Page

Toolbox

Search