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

(→‎Solution 3 (Only sine and cosine sum formulas))
(Adding a new solution.)
 
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<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}} \cos{45^{\circ}} + \sin{96^{\circ}} \sin{45^{\circ}} = \cos({96^{\circ}  + 45^{\circ}}) = \cos{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}} \cos{45^{\circ}} + \sin{96^{\circ}} \sin{45^{\circ}} = \cos({96^{\circ}  + 45^{\circ}}) = \cos{141^{\circ}}.</cmath>
 
This means <math>\theta = 141^{\circ},</math> so <math>19x = 141 + 180k</math> for some positive integer <math>k</math> (since the period of tangent is <math>180^{\circ}</math>), or <math>19 x \equiv 141 \pmod{180}</math>. Note that the inverse of <math>19</math> modulo <math>180</math> is itself as <math>19^2 \equiv 361 \equiv 1 \pmod {180}</math>, so multiplying this congruence by <math>19</math> on both sides gives <math>x \equiv 2679 \equiv 159 \pmod{180}.</math> For the smallest possible <math>x</math>, we take <math>x = \boxed{159}.</math>
 
This means <math>\theta = 141^{\circ},</math> so <math>19x = 141 + 180k</math> for some positive integer <math>k</math> (since the period of tangent is <math>180^{\circ}</math>), or <math>19 x \equiv 141 \pmod{180}</math>. Note that the inverse of <math>19</math> modulo <math>180</math> is itself as <math>19^2 \equiv 361 \equiv 1 \pmod {180}</math>, so multiplying this congruence by <math>19</math> on both sides gives <math>x \equiv 2679 \equiv 159 \pmod{180}.</math> For the smallest possible <math>x</math>, we take <math>x = \boxed{159}.</math>
 +
 +
== Solution 4 ==
 +
Multiplying the numerator and denominator of the right-hand side by
 +
<math>\cos(96^{\circ})+\sin(96^{\circ})</math>, we get
 +
 +
<math>{\tan(19x^{\circ})}
 +
={\frac{\cos(96^{\circ}) +\sin(96^{\circ})}{\cos(96^{\circ})-\sin(96^{\circ})}}\times{\frac{\cos(96^{\circ})+\sin(96^{\circ})}{\cos(96^{\circ})+\sin(96^{\circ})}} \\
 +
={\frac{(\cos(96^{\circ})+\sin(96^{\circ}))^2}{\cos^2(96^{\circ})-\sin^2(96^{\circ})}} \\
 +
={\frac{\cos^2(96^{\circ}) + 2\cos(96^{\circ})\sin(96^{\circ}) + \sin^2(96^{\circ})}{\cos(192^{\circ})}} \\
 +
={\frac{1+\sin(192^{\circ})}{\cos(192^{\circ})}}</math>
 +
 +
Using the fact that <math>\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}</math>, we get
 +
<math>\tan(19x^{\circ})=\frac{\sin(19x^{\circ})}{\cos(19x^{\circ})}=\frac{1+\sin(192^{\circ})}{\cos(192^{\circ})}</math>.
 +
 +
Cross-multiplying, we find that
 +
<math>\sin(19x^{\circ})\cos(192^{\circ})=\cos(19x^{\circ})+\cos(19x^{\circ})\sin(192^{\circ})</math>.
 +
 +
Rearranging the equation gives us
 +
<math>\cos(19x^{\circ})=\sin(19x^{\circ})\cos(192^{\circ})-\cos(19x^{\circ})\sin(192^{\circ})</math>
 +
which leads us to <math>\cos(19x^{\circ})=\sin(19x-192^{\circ})</math> by the sine difference formula.
 +
 +
Using the identity that
 +
<math>\cos(\theta)=\sin(90^{\circ}-\theta)</math>, we find that
 +
<math>\sin(90-19x^{\circ})=\sin(19x-192^{\circ})</math>.
 +
 +
Therefore, <math>90-19x \equiv 19x-192 \pmod{360}</math>, or <math>38x \equiv 282 \pmod{360}</math>.
 +
 +
We know that <math>38 \times 9=342</math> and <math>38 \times 10 \equiv 20 \pmod{360}</math> (by simple arithmetic).
 +
To "make" <math>282</math> we subtract <math>10</math> three times from <math>9</math>, giving us <math>-21</math>.
 +
 +
Finally, because <math>360|38 \times 180</math>, we can add <math>180</math> to get that
 +
<math>x=180-21=\boxed{159}</math> which is the final answer.
 +
 +
~primenumbersfun
 +
 +
  
 
== See Also ==
 
== See Also ==

Latest revision as of 20:59, 23 December 2024

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 $x = \boxed{159}.$

Solution 4

Multiplying the numerator and denominator of the right-hand side by $\cos(96^{\circ})+\sin(96^{\circ})$, we get

${\tan(19x^{\circ})} ={\frac{\cos(96^{\circ}) +\sin(96^{\circ})}{\cos(96^{\circ})-\sin(96^{\circ})}}\times{\frac{\cos(96^{\circ})+\sin(96^{\circ})}{\cos(96^{\circ})+\sin(96^{\circ})}} \\ ={\frac{(\cos(96^{\circ})+\sin(96^{\circ}))^2}{\cos^2(96^{\circ})-\sin^2(96^{\circ})}} \\ ={\frac{\cos^2(96^{\circ}) + 2\cos(96^{\circ})\sin(96^{\circ}) + \sin^2(96^{\circ})}{\cos(192^{\circ})}} \\ ={\frac{1+\sin(192^{\circ})}{\cos(192^{\circ})}}$

Using the fact that $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$, we get $\tan(19x^{\circ})=\frac{\sin(19x^{\circ})}{\cos(19x^{\circ})}=\frac{1+\sin(192^{\circ})}{\cos(192^{\circ})}$.

Cross-multiplying, we find that $\sin(19x^{\circ})\cos(192^{\circ})=\cos(19x^{\circ})+\cos(19x^{\circ})\sin(192^{\circ})$.

Rearranging the equation gives us $\cos(19x^{\circ})=\sin(19x^{\circ})\cos(192^{\circ})-\cos(19x^{\circ})\sin(192^{\circ})$ which leads us to $\cos(19x^{\circ})=\sin(19x-192^{\circ})$ by the sine difference formula.

Using the identity that $\cos(\theta)=\sin(90^{\circ}-\theta)$, we find that $\sin(90-19x^{\circ})=\sin(19x-192^{\circ})$.

Therefore, $90-19x \equiv 19x-192 \pmod{360}$, or $38x \equiv 282 \pmod{360}$.

We know that $38 \times 9=342$ and $38 \times 10 \equiv 20 \pmod{360}$ (by simple arithmetic). To "make" $282$ we subtract $10$ three times from $9$, giving us $-21$.

Finally, because $360|38 \times 180$, we can add $180$ to get that $x=180-21=\boxed{159}$ which is the final answer.

~primenumbersfun


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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