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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?

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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 $$ \left(0,\frac{\sinh(\theta)}{1+\cosh(\theta)}\right)=\left(0,\tanh\left(\frac\theta2\right)\right) $$

$\hspace{6mm}$enter image description here

If $t=\dfrac{\sinh(\theta)}{1+\cosh(\theta)}$, then $1-t^2=\dfrac{2}{1+\cosh(\theta)}$. Therefore, $$ \begin{align} \sinh(\theta)&=\dfrac{2t}{1-t^2}\\ \cosh(\theta)&=\dfrac{1+t^2}{1-t^2} \end{align} $$

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  • $\begingroup$ Thanks so much for your help :) $\endgroup$ – Khallil Mar 8 '14 at 16:24

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