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