Question

Prove that sinh(x+y)=sinhxcoshy+sinhycoshx. I've tried it from both side. One way I just get ((e^x)(e^y)-(e^-x)(e^-y))/2 and the other way I just get everything to cancel out and I get 1/2..... Help XD

Answer 1

Hello,

$sh(x)= \dfrac{e^{x}-e^{-x}}{2}\\\\ ch(x)= \dfrac{e^{x}+e^{-x}}{2}\\\\ sh(x)ch(y)+ch(x)sh(y)= \dfrac{e^{x}-e^{-x}}{2}*\frac{e^{y}+e^{-y}}{2}+\frac{e^{x}+e^{-x}}{2}*\frac{e^{y}-e^{-y}}{2}$
$= \dfrac{e^{x+y}-e^{-x+y}+e^{x-y}-e^{-(x+y)}+e^{x+y}+e^{-x+y}-e^{-(x+y)}}{4}$
$=\dfrac{e^{x+y}-e^{-(x+y)}}{2}=sh(x+y)$