How to prove the identity cosh(x+y)=cosh(x)cosh(y)+sinh(x)sinh(y)
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How to prove the identity cosh(x+y)=cosh(x)cosh(y)+sinh(x)sinh(y)

[From: ] [author: ] [Date: 12-11-01] [Hit: ]
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You could go back to basics and argue from the definitions:

cosh (x+y) = {e^(x+y)+e^(-x-y)}/2.
cosh (x) = {e^x+e^(-x)}/2, cosh(y) = {e^y+e^(-y)}/2 ==> cosh(x)cosh(y) = {e^(x+y) + e^(y-x) + e^(x-y)+e^(-x-y)}/4.
sinh (x) = {e^x-e^(-x)}/2, sinh(y) = {e^y-e^(y)}/2 ==> sinh(x)sinh(y) = {e^(x+y) - e^(-x+y)-e^(y-x)+e^(-x-y)}/4.

Their sum is easily seen to be {e^(x+y)+e^(-x-y)}/2.

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use definitions and simplify the right side
cosh x = (e^x + e^-x)/2 and
sinh x = (e^x - e^-x)/2
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