The formula for sinh x in exponential form is,
[tex]\sinh x=\frac{e^x-e^{-x}}{2}[/tex]
Substitute x + y for x in formula to determine the eponential expression for sinh (x + y).
[tex]\sinh (x+y)=\frac{e^{x+y}-e^{-(x+y)}}{2}[/tex]
Simplify the exponential expression of sinh (x + y) to prove the identity.
[tex]\begin{gathered} \frac{e^{x+y}-e^{-(x+y)}}{2}=\frac{2e^{x+y}-2e^{-x-y}}{4} \\ =\frac{2e^{x+y}+(e^{x-y}-e^{y-x})-(e^{x-y}-e^{y-x})-2e^{-x-y}}{4} \\ =\frac{(e^{x+y}+e^{x-y}-e^{-x-y}-e^{y-x})+(e^{x+y}-e^{x-y}+e^{-x-y}-e^{y-x})}{4} \\ =\frac{e^{x+y}+e^{x-y}-e^{-x-y}-e^{y-x}}{4}+\frac{e^{x+y}-e^{x-y}+e^{-x-y}-e^{y-x}}{4} \\ =\frac{e^x(e^y+e^{-y})-e^{-x}(e^{-y}+e^y)}{4}+\frac{e^x(e^y-e^{-y})+e^{-x}(e^{-y}-e^y)}{4} \\ =(\frac{e^y+e^{-y}}{2})(\frac{e^x-e^{-x}}{2})+(\frac{e^y-e^{-y}}{2})(\frac{e^x+e^{-x}}{2}) \\ =\cosh y\sinh x+\sinh y\cosh x \end{gathered}[/tex]
Hence it is proved that,
sinh (x + y) = sinh x cosh y + sinh y cosh x
