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Prove a formula for cosh (x+y)

Prove that

    \[ \cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y. \]


Proof. Using the definition of the hyperbolic cosine in terms of the exponential we have,

    \begin{align*}  \cosh(x+y) &= \frac{e^{x+y} + e^{-x-y}}{2} \\  &= \frac{2e^{x+y} + (e^{x-y} + e^{y-x}) - (e^{x-y} + e^{y-x}) + 2e^{-x-y}}{4} \\  &= \left( \frac{e^x+e^{-x}}{2} \right) \left( \frac{e^y + e^{-y}}{2} \right) + \left( \frac{e^x-e^{-x}}{2} \right) \left( \frac{e^y - e^{-y}}{2} \right) \\  &= \cosh x \cosh y + \sinh x \sinh y. \qquad \blacksquare \end{align*}

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