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Prove that cosh x + sinh x = ex

Prove that

    \[ \cosh x + \sinh x = e^x. \]


Proof. We use the definitions of hyperoblic sine and cosine to compute,

    \begin{align*}  \cosh x + \sinh x &= \frac{e^x+e^{-x}}{2} + \frac{e^x - e^{-x}}{2} \\  &= \frac{e^x + e^{-x} + e^x -e^{-x}}{2} \\  &= \frac{2e^x}{2} \\  & = e^x. \qquad \blacksquare \end{align*}

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