TITLE: Weierstrass $ \tanh \frac{\theta}{2} $ substitution confusion.
QUESTION [6 upvotes]: I'm already familiar with the trigonometric version of this substitution $ t = \tan \frac{\theta}{2} $ and it's geometrical derivation involving the unit circle found here. However, I'm not sure how the hyperbolic equivalents (shown below) are derived.
$$ t = \tanh \dfrac{\theta}{2} = \dfrac{\sinh \theta}{\cosh \theta + 1} = \dfrac{\cosh \theta - 1}{\sinh \theta} $$
$$ \cosh \theta = \dfrac{1+t^2}{1-t^2}, \ \sinh \theta = \dfrac{2t}{1-t^2}$$
At the bottom of the page, it refers to projecting the point $ (\cosh \theta, \sinh \theta)$ which is found on the right branch of a hyperbola onto the y-axis from the center $ (-1, 0) $ but I'm unsure of what this means.
Can anybody provide a geometrical take on deriving these?

REPLY [9 votes]: Here is the hyperbola $(\cosh(\theta),\sinh(\theta))$ and the line from $(-1,0)$ to a point on the hyperbola projected on the $y$-axis. The coordinate of the point is
$$
\left(0,\frac{\sinh(\theta)}{1+\cosh(\theta)}\right)=\left(0,\tanh\left(\frac\theta2\right)\right)
$$
$\hspace{6mm}$
If $t=\dfrac{\sinh(\theta)}{1+\cosh(\theta)}$, then $1-t^2=\dfrac{2}{1+\cosh(\theta)}$. Therefore,
$$
\begin{align}
\sinh(\theta)&=\dfrac{2t}{1-t^2}\\
\cosh(\theta)&=\dfrac{1+t^2}{1-t^2}
\end{align}
$$