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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.

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  • $\begingroup$ Sorry, I mistypes -- I have corrected it $\endgroup$ – None Mar 19 '15 at 9:41
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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 $e^{\ln(z)}=z$

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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.

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