Trying to prove a hyperbolic identity

[MrGinsu]

Hello,

I'm ok with turning formula cranks but proofs are another story. I am confused on how to provide a proof for:

\(\displaystyle sinh(x+y) = sinh(x)cosh(y) + cosh(x)sinh(y)\)

Here's what I have tried so far:

\(\displaystyle d/dx[sinh(x)+sinh(y)] = d/dx[sinh(x)cosh(y)] = d/dx[cosh(x)sinh(y)]\)

\(\displaystyle cosh(x) + cosh(y) = cosh(x)cosh(y) + sinh(x)sinh(y) + sinh(x)sinh(y) + cosh(x)cosh(y)\)

\(\displaystyle cosh(x)cosh(y) = 2(cosh(x)cosh(y) + sinh(x)sinh(y))\)

Now, I'm not certain how this is getting the right side to equal anything close to the left side of the identity. Can anyone post a tip? Please don't provide the answer. I just need some help on the best way to tackle this proof.

Thanks.

 

[soroban]

Hello, MrGinsu!

Taking derivatives is not the way to prove identities.

Have you considered using the definitions of those hyperbolic functions?

\(\displaystyle \;\;\text{sinh} x \:= \:\frac{e^x\,-\,e^{-x}}{2}\;\;\;\text{cosh} x\:=\:\frac{e^x\,+\,e^{-x}}{2}\;\;\;\text{sinh }y\:=\:\frac{e^y\,-\,e^{-y}}{2}\;\;\;\text{cosh {y\:=\:\frac{e^y\,+\,e^{-y}}{2}\)

Plug them into the right side . . . and simplify it to resemble the left side.

 

[MrGinsu]

Hello, MrGinsu!

Taking derivatives is not the way to prove identities.

Have you considered using the definitions of those hyperbolic functions?

\(\displaystyle \;\;\text{sinh} x \:= \:\frac{e^x\,-\,e^{-x}}{2}\;\;\;\text{cosh} x\:=\:\frac{e^x\,+\,e^{-x}}{2}\;\;\;\text{sinh }y\:=\:\frac{e^y\,-\,e^{-y}}{2}\;\;\;\text{cosh {y\:=\:\frac{e^y\,+\,e^{-y}}{2}\)

Plug them into the right side . . . and simplify it to resemble the left side.

Yahoo! That go me moving in the right direction. I started to use the definitions but then thought that I should use implicit differentiation to prove it.

Thanks for the tip.