Prove that sinh(x+y)=sinxcoshy+coshxsinhy

Prove that $\text{sinh}\left(x+y\right)=\mathrm{sin}x\text{cosh}y+\text{cosh}x\text{sinh}y$
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krolaniaN

$\text{sinh}\left(A\right)=\frac{eˣ-e⁻ˣ}{2}$ by definition when A=x, so, if A=x+y:
$\text{sinh}\left(x+y\right)=\frac{{e}^{x+y}-\left({e}^{-\left(x+y\right)}\right)}{2}.$
$\text{cosh}\left(x\right)=\frac{eˣ+e⁻ˣ}{2}$ by definition.
$\text{sinh}\left(x\right)\text{cosh}\left(y\right)=\left({e}^{x}-{e}^{-x}\right)\frac{{e}^{y}+{e}^{-y}}{4}=\frac{{e}^{x+y}+{e}^{x-y}-{e}^{y-x}-{e}^{-\left(x+y\right)}}{4}.$
$\text{cosh}\left(x\right)\text{sinh}\left(y\right)=\left({e}^{x}+{e}^{-x}\right)\frac{{e}^{y}-{e}^{-y}}{4}=\frac{{e}^{x+y}-{e}^{x-y}+{e}^{y-x}-{e}^{-\left(x+y\right)}}{4}.$
When we add these last two equations we get $\mathrm{sin}h\left(x\right)\mathrm{cos}h\left(y\right)+\mathrm{cos}h\left(x\right)\mathrm{sin}h\left(y\right)=$
$2\frac{{e}^{x+y}-{e}^{-\left(x+y\right)}}{4}=\frac{{e}^{x+y}-{e}^{-\left(x+y\right)}}{2}=\text{sinh}\left(x+y\right)$