\documentclass{../style}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\euler

\begin{gather*}
	\therefore \cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}
\end{gather*}

\begin{gather*}
	\text{let} \quad \cos(\theta) = x \\
	2x = e^{i\theta} + e^{-i\theta} \\
	2xe^{i\theta} = (e^{i\theta})^2 + 1 \\ 
	(e^{i\theta})^2 + (-2x)e^{i\theta} + 1 = 0
\end{gather*}

\begin{gather*}
	e^{i\theta} = \frac{-(-2x) \pm \sqrt{(-2x)^2 - 4}}{2} = x \pm \sqrt{x^2 - 1} \\
	i\theta = \ln(x \pm \sqrt{x^2 - 1}) \\
	\theta = -i\ln(x \pm \sqrt{x^2 - 1})
\end{gather*}

\begin{gather*}
	\therefore \arccos(\theta) = -i\ln(\theta \pm \sqrt{\theta^2 - 1})
\end{gather*}
\end{document}