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

Difficult Proof/Demonstration

We start from the left hand side of the equality:

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

Easy Proof/Demonstration

We start from the right hand side of the equality:

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