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

(Solution 3 (Only sine and cosine sum formulas))
(Solution 3 (Only sine and cosine sum formulas))
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== Solution 3 (Only sine and cosine sum formulas) ==
 
== 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>
 
<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  
 
from the sine sum formula. For the denominator, from the cosine sum formula, we have  

Revision as of 15:45, 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 $x = \boxed{159}.$

See Also

1996 AIME (ProblemsAnswer KeyResources)
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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