How do you prove $Cosh(A+B) = CoshACoshB+SinhBSinhA$ ?

Question

18187 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-03T19:27:59

See below

Use the definition of Hyperbolic Functions for cosh x
$cosh x =(e^x + e^-x)/2$ and $sinh x=(e^x-e^-x)/2$

$cosh(A+B)=coshAcoshB+sinhAsinhB$

Right Side: $=[(e^A+e^-A)/2 xx (e^B+e^-B)/2]+[(e^A-e^-A)/2 xx (e^B-e^-B)/2]$

$=(e^(A+B)+e^(A-B)+e^(B-A)+e^(-(A+B)))/4 + (e^(A+B)-e^(A-B)-e^(B-A)+e^(-(A+B)))/4$

$=(e^(A+B)+cancel(e^(A-B))+cancel(e^(B-A))+e^(-(A+B))+ e^(A+B)-cancel(e^(A-B))-cancel(e^(B-A))+e^(-(A+B)))/4$

$=(2e^(A+B)+2e^(-(A+B)))/4$

$=(2(e^(A+B)+e^(-(A+B))))/4$

$=(e^(A+B)+e^(-(A+B)))/2$

$=cosh(A+B)$

$=$ right side

How do you prove Cosh(A+B) = CoshA*CoshB+SinhB*SinhACosh(A+B) = CoshA*CoshB+SinhB*SinhA?