We will show that for any real element x, y the trigonometric 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:

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

Easy Proof/Demonstration

We start from the right hand side of the equality:

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