Rearrangeing Inverse Hyperbolic functions

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SUMMARY

The discussion centers on rearranging the equation for hyperbolic functions, specifically the identity cosh²(y) - sinh²(y) = 1. The user initially misarranged the equation as cosh²(y) = 1 + sinh²(y), while the correct form is cosh²(y) = sinh²(y) + 1. Both expressions are mathematically equivalent, as confirmed through definitions of hyperbolic sine and cosine. The definitions provided clarify the relationship between the two functions, reinforcing that both rearrangements yield the same result.

PREREQUISITES
  • Understanding of hyperbolic functions, specifically sinh and cosh.
  • Familiarity with algebraic manipulation of equations.
  • Basic knowledge of exponential functions and their properties.
  • Ability to differentiate inverse hyperbolic functions.
NEXT STEPS
  • Study the definitions and properties of hyperbolic functions in detail.
  • Practice rearranging hyperbolic identities and verifying their equivalence.
  • Learn about the differentiation of inverse hyperbolic functions.
  • Explore applications of hyperbolic functions in calculus and physics.
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Students studying calculus, mathematicians focusing on hyperbolic functions, and educators teaching algebraic manipulation of mathematical identities.

Observables
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Hi,

My brain is not working today. So can someone please tell me what I am doing wrong.

(^2 = squared)

coshy^2 - sinhy^2 = 1, how do I rearrange this for coshy^2

I keep getting: coshy^2 = 1 + Sinhy^2

The book that I'm looking at has it this way: coshy^2 = Sinhy^2 + 1

Thanks

Obs
 
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Time for brain overhaul. Both expressions evaluate the same.
 
Observables said:
Hi,

My brain is not working today. So can someone please tell me what I am doing wrong.

(^2 = squared)

coshy^2 - sinhy^2 = 1, how do I rearrange this for coshy^2

I keep getting: coshy^2 = 1 + Sinhy^2

The book that I'm looking at has it this way: coshy^2 = Sinhy^2 + 1

Thanks

Obs

Hey observables and welcome to the forums.

The easiest way to find if the book is right (ie coshy^2 - sinhy^2 = 1) is to just use the definition of sinhy and coshy.

sinhy = 1/2 (e^x - e^-x)
coshy = 1/2(e^x + e^-x)

so sinhy^2 = 1/4 [e^2x - 2 + e^-2x)
coshy^2 = 1/4(e^2x + 2 - e^-2x)

coshy^2 - sinhy^2 = 1/2 - (-1/2) = 1
 
SteamKing said:
Time for brain overhaul. Both expressions evaluate the same.

Hi,

I realized that, but I couldn't figure out how the book was doing the rearranging. I've been differentiating a few Inverse Hyperbolic functions, and I thought the reason that the book rearranged differently to me was that somewhere down the line it would become clear why it was done.

Obs.
 
chiro said:
Hey observables and welcome to the forums.

The easiest way to find if the book is right (ie coshy^2 - sinhy^2 = 1) is to just use the definition of sinhy and coshy.

sinhy = 1/2 (e^x - e^-x)
coshy = 1/2(e^x + e^-x)

so sinhy^2 = 1/4 [e^2x - 2 + e^-2x)
coshy^2 = 1/4(e^2x + 2 - e^-2x)

coshy^2 - sinhy^2 = 1/2 - (-1/2) = 1

Hi Chiro,

Thanks for the info.

Obs.
 
Gezzz breakfast time must be confusing at your place. Trying to decide whether to have "eggs and toast" or "toast and eggs". It must keep you occupied for hours.
 

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