Proving the Cosh and Sinh Identity

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The discussion revolves around verifying the identity exp(2x)sinh(y)² + exp(-2x) = exp(2x)cosh(y)² - 2sinh(2x). Participants suggest using the definitions of sinh and cosh in terms of exponentials to prove the identity. There is a clarification that the relation cosh² - sinh² = 1 is the correct approach, as opposed to the incorrect assumption that cosh² + sinh² = 1. The need for a proper mathematical approach is emphasized, with hints provided to guide the proof. Overall, the focus is on finding a correct method to demonstrate the identity for an exercise.
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Is it true that:

exp(2x)sinh(y)2 + exp(-2x) = exp(2x)cosh(y)2-2sinh(2x)

I need this to be correct for an exercise but I don't know how to show it. I tried using something like cosh2+sinh2=1, but it didn't work.
 
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Perhaps use the definition of sinh and cosh to express everything as exponentials?
 
aaaa202 said:
I tried using something like cosh2+sinh2=1, but it didn't work.

In general cosh2+sinh2 ≠1 :wink:
 
aaaa202 said:
Is it true that:

exp(2x)sinh(y)2 + exp(-2x) = exp(2x)cosh(y)2-2sinh(2x)

I need this to be correct for an exercise but I don't know how to show it. I tried using something like cosh2+sinh2=1, but it didn't work.

It might also help if you use the correct relation. cosh^2-sinh^2=1.
 
I thought ;) would propel him there. ;)
 
epenguin said:
I thought ;) would propel him there. ;)

Yeah, I didn't see your hint until after I posted. Sorry.
 
No matter. :smile:
 

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