Trigonometry addition formula sinh(x + y) = sinh x cosh y +cosh x sinh y

We will show that for any real element x, y the trigonometric formula sinh(x + y) = sinh(x)cosh(y) +cosh(x)sinh(y)

Difficult Proof/Demonstration

We start from the left hand side of the equality:

$$\begin{array}{rl}\sinh (x+y)& =\frac{e^{x+y}-e^{-(x+y)}}{2}\\ & =\frac{e^{x+y}-e^{-x-y}}{2}\\ & =\frac{2e^{x+y}-2e^{-x-y}}{4}\\ & =\frac{2e^{x+y}-2e^{-x-y}+(e^{x-y}-e^{-x+y})-(e^{x-y}-e^{-x+y})}{4}\\ & =\frac{2e^{x+y}+(e^{x-y}-e^{-x+y})-(e^{x-y}-e^{-x+y})-2e^{-x-y}}{4}\\ & =\frac{e^{x+y}+e^{x+y}+(e^{x-y}-e^{-x+y})-(e^{x-y}-e^{-x+y})-e^{-x-y}-e^{-x-y}}{4}\\ & =\frac{e^{x+y}-e^{-x-y}+(e^{x-y}-e^{-x+y})+e^{x+y}-(e^{x-y}-e^{-x+y})-e^{-x-y}}{4}\\ & =\left(\frac{e^{x+y}+e^{x-y}-e^{-x+y}-e^{-x-y}}{4}\right)+\left(\frac{e^{x+y}-e^{x-y}+e^{-x+y}-e^{-x-y}}{4}\right)\\ & =\left(\frac{e^x-e^{-x}}{2}\right)\left(\frac{e^y+e^{-y}}{2}\right)+\left(\frac{e^x+e^{-x}}{2}\right)\left(\frac{e^y-e^{-y}}{2}\right)\\ & =\sinh x\cosh y+\cosh x\sinh y\end{array}$$

Easy Proof/Demonstration

We start from the right hand side of the equality:

$$\begin{array}{rl}\sinh x\cosh y+\cosh x\sinh y& =\left(\frac{e^x-e^{-x}}{2}\right)\left(\frac{e^y+e^{-y}}{2}\right)+\left(\frac{e^x+e^{-x}}{2}\right)\left(\frac{e^y-e^{-y}}{2}\right)\\ & =\left(\frac{e^{x+y}+e^{x-y}-e^{-x+y}-e^{-x-y}}{4}\right)+\left(\frac{e^{x+y}-e^{x-y}+e^{-x+y}-e^{-x-y}}{4}\right)\\ & =\frac{2e^{x+y}-2e^{-x-y}}{4}\\ & =\frac{e^{x+y}-e^{-x-y}}{2}\\ & =\frac{e^{x+y}-e^{-(x+y)}}{2}\\ & =\sinh (x+y)\end{array}$$