Hyperbolic analog to sinx = cos(pi/2 - x)

Question

Is there a hyperbolic analog to that? I understand the circular trigonometric relation comes from analysing a right-angled triangle in Euclidean space, but how can we visualise the analog in hyperbolic space?

Answer 1

Yes,but not exactly like that.

$$\sinh x = e^x - \cosh x$$

Answer 2

$\displaystyle-\,\mathrm{i}\sinh\left(\mathrm{i}x\right) = \cosh\left(\mathrm{i}\left[{\pi \over 2} - x\right]\right)$. Set $\displaystyle z = \,\mathrm{i}x \implies \bbox[15px,#ffe,border:1px dotted navy]{ \sinh\left(z\right) =\,\mathrm{i}\cosh\left(z - {\pi \over 2}\,\mathrm{i}\right)}$