Relationship between $\sin(x)$ and $\sinh(x)$ 
Given that $\tan(y) =\sinh(x)$ show that $\sin(y) = \pm \tanh(x) $.

I  know:
$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$
$\tanh\theta=\dfrac{\sinh\theta}{\cosh\theta}$
Also,
$\tanh\theta = \dfrac{e^x-e^{-x}}{e^x+e^{-x}}$
$\sinh\theta=\dfrac{e^x-e^{-x}}{2}$
$\cosh\theta=\dfrac{e^x+e^{-x}}{2}$
I'm struggling to link the two together, please assume little to no knowledge of calculus.
 A: with $$\frac{\sin(y)}{\cos(y)}$$ we get
$$\frac{\sin(y)}{\pm\sqrt{1-\sin^2(y)}}=\sinh(x)$$ square this equation and solve the equation for $\sin(y)$
with Algebra we get
$$\sin^2(y)=\frac{\sinh^2(x)}{1+\sinh^2(x)}$$
can you finish?
after squaring the given equation we obtain
$$\frac{\sin^2(y)}{1-\sin^2(y)}=\sinh^2(x)$$ and then
$$\sin^2(y)=\sinh^2(x)(1-\sin^2(y))$$
expanding
$$\sin^2(y)=\sinh^2(x)-\sinh^2(x)\sin^2(y)$$
this gives
$$\sin^2(y)+\sin^2(y)\sinh^2(x)=\sinh^2(x)$$
or
$$\sin^2(y)=\frac{\sinh^2(x)}{1+\sinh^2(x)}$$
and note that $$-\sinh^2(x)+\cosh^2(x)=1$$
A: First of all notice that to keep consistency  $$\tanh\theta = \dfrac{e^x-e^{-x}}{e^x+e^{-x}}$$
Should have been $$\tanh(x) = \dfrac{e^x-e^{-x}}{e^x+e^{-x}}$$
Same with your other two functions.
There are many similarities and differences between hyperbolic functions and trig functions.
For example trig functions are periodic but hyperbolic functions are not periodic.
$sin(x)$ and $cos(x)$ are bounded but $sinh(x)$ and $cosh(x)$ are not bounded. 
The identities $$ cos^2(x) + sin ^2(x) =1$$ 
turn into $$ cosh^2(x) - sinh ^2(x) =1$$ 
and 
$$\cosh(x)=\dfrac{e^x+e^{-x}}{2}$$
turns into
$$\cos(x)=\dfrac{e^ix+e^{-ix}}{2}$$
You will learn more about their infinite series and derivatives  in  your calculus courses.
A: We know that,
$$\sinh x = \frac{e^{x} - e^{-x}}{2}$$
Going by the definition of $e^{ix}$,
$$
\begin{aligned}
e^{ix} = \cos x + i\sin x \\
e^{-ix} = \cos x - i\sin x
\end{aligned}
$$
Putting $\pmb{ix}$ in place of $\pmb{x}$,
$$
\left.
\begin{aligned}
e^{i(ix)} = \cos (ix) + i\sin (ix) \\
e^{-i(ix)} = \cos (ix) - i\sin (ix)
\end{aligned}
\right\} \tag{1}\label{1}
$$
Now we know,
$$i^2 = -1 \tag{2}\label{2}$$
$\eqref{2}\; in\; \eqref{1}$,
$$e^{-x} = \cos ix + i\sin ix \tag{3}\label{3}$$
$$e^{x} = \cos (ix) - i\sin (ix) \tag{4}\label{4}$$
Finally, subtract $\eqref{3}\; from\; \eqref{4}$ and divide by two, to get in  the form of $\pmb{\sinh x}$,
$$\frac{e^{x} - e^{-x}}{2} =\sinh x=\frac{-i\sin (ix) - i\sin (ix)}{2}$$
$$\bbox[5px,border:2px solid red]
{
\sinh x=-i\sin(ix)
}
$$
Since, $\frac{1}{i}=-i$, we can also have,
$$\bbox[5px,border:2px solid red]
{
\sinh x=\frac{sin(ix)}{i}
}
$$
I know this isn't strictly the relation between $\sinh x$ and $\pmb{\sin x}$, in the sense that this involves complex part.
Hope this helps someone who needs this relation between the two functions.
