Cosh x / sinhx in the form of e^x

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Homework Help Overview

The discussion revolves around the properties of hyperbolic functions, specifically the expressions for cosh and sinh in terms of exponential functions. Participants are examining the validity of the formula for cosh(2x) and its relationship to cosh(x) and sinh(x).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the correctness of the formula for cosh(2x) and discussing the implications of substituting different values into the exponential expressions. There is also a focus on clarifying the signs in the equations and the reliability of the sources of information.

Discussion Status

The discussion is ongoing, with some participants providing corrections and clarifications regarding the formulas. There is a mix of agreement and questioning about the validity of the expressions presented, indicating a productive exploration of the topic.

Contextual Notes

There are indications of potential typos or misunderstandings in the original statements about the formulas, which are being addressed by participants. The discussion reflects a need for careful attention to the definitions and properties of hyperbolic functions.

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Homework Statement


I was told that coshx = (e^x + e^-x) / 2 , why cosh2x = (e^2x - e^-2x) / 2 ?

Homework Equations

The Attempt at a Solution

 

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goldfish9776 said:

Homework Statement


I was told that coshx = (e^x + e^-x) / 2 , why cosh2x = (e^2x - e^-2x) / 2 ?

Homework Equations

The Attempt at a Solution


Because it is true for any value of x. Any number you can get with x you can get with 2x.
 
If cosh3x then e^x is substituted with e^3x ??
 
goldfish9776 said:
If cosh3x then e^x is substituted with e^3x ??
Yes. Although you really should use parentheses for the expression making up the exponent unless it's written as a superscript.

e^(3x) or e3x .
 
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goldfish9776 said:
I was told that coshx = (e^x + e^-x) / 2 , why cosh2x = (e^2x - e^-2x) / 2 ?
Hi goldfish9776:

I agree that the minus sign in
cosh 2x = (e2x - e-2x) / 2​
is peculiar.
(e2x - e-2x) / 2 = sinh 2x.​

Is the person who told you that
cosh 2x = (e2x - e-2x) / 2​
someone you would expect to be reliable?

Regards,
Buzz
 
@goldfish9776 ,

You have a typo in the OP. The correct statement is:

##\displaystyle \cosh(2x)=\frac{e^{2x}+e^{-2x}}{2} ##
 

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