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

The solution states to use $\sin(\theta) = \cos(90-\theta)$ and simplify the fraction to $-\cot{51}$, then use some number theory to finish it off.


My approach:

We can use difference of squares on the RHS. \begin{align} &\frac{(\cos{96°} + \sin{96°})(\cos{96°} - \sin{96°})}{(\cos{96°} - \sin{96°})^2} = \\ &\qquad\frac{\cos^2{96°}-\sin^2{96°}}{\cos^2{96°}+\sin^2{96°}-2\cos{96°}\sin{96°}} = \frac{\cos{192°}}{1-\sin{192°}}. \end{align}

However, finding the value for this is hard. I did note the resemblance of the half-angle tangent formula. It states that for any angle $\theta$, $$\tan{\frac{\theta}{2}} = \frac{\sin{\theta}}{1+\cos{\theta}} = \frac{1-\cos{\theta}}{\sin{\theta}}.$$

My question is, can $\frac{\cos{192°}}{1-\sin{192°}}$ be used in any way to relate to the half-angle tangent formula? An added bonus is that we want to find $\tan{19x}$, and having a tangent formula only helps. However, I was unable to find a relation.

Problem from 1996 AIME Problem 10. The official solution is linked here.

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  • $\begingroup$ Maybe you can use $96=4 \cdot 19 +1$? (Or not, just glanced at the key!) $\endgroup$ – copper.hat Sep 7 '20 at 1:26
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    $\begingroup$ Substitute $\sin 2x=\frac{2\tan x}{1+\tan^2x}$ and $\cos 2x=\frac{1-\tan^2x}{1+\tan^2x}$. $\endgroup$ – SarGe Sep 7 '20 at 8:23
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\begin{align} \frac{\cos(192°)}{1 - \sin(192°)} & = \frac{\sin(90° - 192°)}{1 - \cos(90° - 192°)} \\ & = \frac{\sin258°}{1 - \cos258°} \\ & = \frac{1}{\frac{1 - \cos258°}{\sin258°}} \\ & = \frac{1}{\tan(\frac{258°}{2})} \\ & = \frac{1}{\tan129°} \\\\ & = \tan(270° - 129°) \\\\ & = \tan141° \end{align}

It suffices to find smallest positive x such that $$ 19x \equiv 141 \pmod {180}$$ notice that $19^2 \equiv 1 \pmod {180}$, thus \begin{align} x &\equiv 19 \cdot 141 \\ &\equiv 159\pmod{180} \end{align} The smallest positive $x$ is thus $159$.

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  • $\begingroup$ $141$ is not an integer multiple of $19$ $\endgroup$ – enzotib Sep 8 '20 at 13:09
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A mush simpler manipulation, in my opinion, using addition formulae $$ \frac{\cos y+\sin y}{\cos y-\sin y}=\frac{\sqrt{2}\sin(y+45^\circ)}{\sqrt{2}\cos(y+45^\circ)}=\tan(y+45^\circ) $$ then, for $y=96^\circ,$ $$ 19^\circ x=141^\circ+180^\circ k,\quad k\in\mathbb{Z} $$

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