hyperbolic-functionstrigonometry
As far as my understanding goes, trigonometry is the math of right triangles. Sine is the opposite side over the hypotenuse, cosine is the adjacent side over the hypotenuse, etc. The unit circle definition comes as a result of this fact.
What connection do the the hyperbolic trig functions have to the actual trig functions?
We have $$a\,\sin\theta+b\,\cos\theta=\sqrt{a^2+b^2}\,\sin(\theta+\alpha)$$ where $\alpha$ is the unique angle such that $\cos\alpha=a/\sqrt{a^2+b^2}$ and $\sin\alpha=b/\sqrt{a^2+b^2}$, in case $a^2+b^2 >0$.
Note that $\alpha$ is the angle of the vector $(a,b)$, measured from the positive half of $x$-axis, and $r:=\sqrt{a^2+b^2}$ is its length.
It means that, $a\,\sin\theta+b\,\cos\theta$ is the $y$ coordinate of the rotation of $(\cos\theta,\,\sin\theta)$ by $\alpha$, multiplied by $r$.
In your example $a=b=1$ so $r=\sqrt2$ and $\alpha=\pi/4$. Then the rotated and stretched vector will have angle $\pi/4+\pi/4=\pi/2$ and length $\sqrt2$. Its $y$ coordinate is indeed $\sqrt2$.
The meaning of the parameter $a$ in $\sinh(a)$ and in the case $\cosh(a)$ is the same as in the case of $\sin(a)$ and $\cos(a)$. Take a look at the figure below.
In both cases the parameter equals the half area of the red region.
Best Answer
$\cos(ix)=\cosh(x)$ and $\sin(ix)=i\sinh(x)$. Beyond that, most of the hyperbolic trig identities have a similar form compared to their "circle" trig counterparts. For instance,
$$\sinh(x+y)=\sinh x\cosh y+\cosh x+\sinh y$$ $$\cosh^2x-\sinh^2=1 $$
Also, as noted in the comments and on Wikipedia, there is a connection to the hyperbola $x^2-y^2=1$ analogous to the unit circle.