AIME 1996 · 第 10 题

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

Solution 5

Notice that $\sin 45^\circ = \cos 45^\circ$. Therefore, all terms in the numerator and the denominator could be multiplied by either $\sin 45^\circ$ or $\cos 45^\circ$.

$$\begin{aligned} \frac{\cos 96^\circ + \sin 96^\circ}{\cos 96^\circ - \sin 96^\circ} &= \frac{\cos 96^\circ \sin 45^\circ + \sin 96^\circ \cos 45^\circ}{\cos 96^\circ \cos 45^\circ - \sin 96^\circ \sin 45^\circ} \\[0.5em] &= \frac{\sin(96 + 45)^\circ}{\cos(96 + 45)^\circ} \\[0.5em] &= \tan 141^\circ \end{aligned}$$ Therefore, $\tan19x^\circ = \tan 141^\circ$. In other words, $19x = 141 + 180k$ for some integer $k$.

$$\begin{aligned} 141 + 180k &\equiv 0 \pmod{19} \\ 8 + 9k &\equiv 0 \pmod{19} \\ 9k &\equiv 11 \pmod{19} \\ 9k &\equiv 11, 30, 49, 68, 87, 106, 125, 144 \pmod{19} \\ \end{aligned}$$ Substituting $k$, the smallest value of $19x$ is $141 + 180 \cdot 16$, which is $\boxed{159}$.

~MaPhyCom