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.
Best Answer
\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 $159$.