Answer
$\begin{align*}
\displaystyle \sinh^{2}x&=(\displaystyle \frac{e^{x}-e^{-x}}{2})^{2}\\\\
& =\displaystyle \frac{e^{2x}-2+e^{-2x}}{4}\\\\
& =\displaystyle \frac{-2+(e^{2x}+e^{-2x})}{4} \tag{ $\frac{a+b}{4}=\frac{\frac{a+b}{2}}{2} $ } \\\\
&=\displaystyle \frac{\frac{-2}{2}+\frac{e^{2x}+e^{-2x}}{2}}{2} \\\\
&=\displaystyle \frac{-1+\cosh 2x}{2} \end{align*}$
Work Step by Step
Definitions of the Hyperbolic Functions
$\displaystyle \sinh x=\frac{e^{x}-e^{-x}}{2} \qquad \mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}\ x=\displaystyle \frac{1}{\sinh x}=\frac{2}{e^{x}-e^{-x}},\quad x\neq 0$
$\displaystyle \cosh x=\frac{e^{x}+e^{-x}}{2} \qquad \mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\ x=\displaystyle \frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}$
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$\begin{align*}
\displaystyle \sinh^{2}x&=(\displaystyle \frac{e^{x}-e^{-x}}{2})^{2}\\\\
& =\displaystyle \frac{e^{2x}-2+e^{-2x}}{4}\\\\
& =\displaystyle \frac{-2+(e^{2x}+e^{-2x})}{4} \tag{ $\frac{a+b}{4}=\frac{\frac{a+b}{2}}{2} $ } \\\\
&=\displaystyle \frac{\frac{-2}{2}+\frac{e^{2x}+e^{-2x}}{2}}{2} \\\\
&=\displaystyle \frac{-1+\cosh 2x}{2} \end{align*}$
So, $\displaystyle \qquad \sinh^{2}x=\frac{-1+\cosh 2x}{2}$.
