We will show that for any real element x, y the trigonometric formula (difference identity) 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{aligned} \sinh (x-y) & =\frac{e^{x+y}-e^{-(x-y)}}{2} \\ & =\frac{e^{x+y}-e^{-x+y}}{2} \\ & =\frac{2 e^{x-y}-2 e^{-x+y}}{4} \\ & =\frac{2 e^{x-y}-2 e^{-x+y}+\left(e^{x+y}-e^{-x-y}\right)-\left(e^{x+y}-e^{-x-y}\right)}{4} \\ & =\frac{2 e^{x-y}+\left(e^{x+y}-e^{-x-y}\right)-\left(e^{x+y}-e^{-x-y}\right)-2 e^{-x+y}}{4} \\ & =\frac{e^{x-y}+e^{x-y}+\left(e^{x+y}-e^{-x-y}\right)-\left(e^{x+y}-e^{-x-y}\right)- e^{-x+y}- e^{-x+y}}{4} \\ & =\frac{e^{x-y}- e^{-x+y}+\left(e^{x+y}-e^{-x-y}\right)+e^{x-y}-\left(e^{x+y}-e^{-x-y}\right) - 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) \\ & =\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{aligned}

## Easy Proof/Demonstration

We start from the right hand side of the equality:

\begin{aligned} \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{2 e^{x-y}-2 e^{-x+y}}{4} \\ & =\frac{e^{x+y}-e^{-x+y}}{2} \\ & =\frac{e^{x+y}-e^{-(x-y)}}{2} \\ & =\sinh (x-y) \end{aligned}