Prove a formula for sinh(x+y)

by RoRi

November 17, 2015

Prove that

$\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y.$

Proof. We use the definition of the hyperbolic sine in terms of the exponential,

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

One comment

February 22, 2024 at 6:21 am

Anonymous says:

why we put -y instead of y at this -e^-x-y

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Stumbling Robot

Prove a formula for sinh(x+y)

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