How to turn sin(arcsinh(x)) into algebraic form? How can I turn $\sin({\sinh^{-1}{x}})$ into explicit algebraic form ? I've tried to plug in $\sinh^{-1}{x}$ into sine's exponential form $\frac{e^{ix} - e^{-ix}}{2i}$, but then I cannot think of any ways to simplify that.
 A: Using this, 
We need $\sin\left[\ln(x+\sqrt{x^2+1})\right]$
Now $e^{i\left[\ln(x+\sqrt{x^2+1})\right]}=\left[e^{\left(\ln(x+\sqrt{x^2+1})\right)}\right]^i$
and $e^{\ln(z)}=z$
A: This is not possible.
Impossibility proofs are always very hard; therefore hear the following arguments:
If there were a "simple algebraic relation" $\Phi(x,y)=0$ between the variables $x:=\sinh t$ and  $y:=\sin t$,  valid for all $t\in{\mathbb R}$, then this "algebraic relation"  would have to mimic somehow that for all $y\in[{-1},1]$ there are an infinity of values $x_k$, all of them satisfying $\Phi(x_k,y)=0$. An "algebraic" expression $\Phi(x,y)$ would not be ale to accomplish this.
Another hint in this direction is the following: If such a $\Phi$ existed in your example then a similar $\Psi(u,p)$ would exist for the much simpler example relating the variables $u:=\cos t+i\sin t=e^{it}$ and $p:=e^t$ valid for all real $t$. Such a relation would empower you to calculate trigonometric functions using a table of logarithms.
