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Help with substitution of cosh(A),etc, into equation!

  1. Mar 27, 2006 #1
    Hi there,

    I was given this equation,
    cosh(A+B)=cos(A).cosh(B) + sinh(A).sinh(B)
    then I was told to write down the definition of cosh(A+B) which I did.

    But then i'm told to substitue the defintions of cosh(A), sinh(A), cosh(B) and sinh(B) into the right hand side of the equation, and to simplify....now I don't understand what it means by that? Is it doing the same way as the definition for cosh(A+B) (where you replace the e^t with A...and so on??) I'm not quite sure what it wants out of that one??
     
  2. jcsd
  3. Mar 27, 2006 #2

    benorin

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    Definition [tex]\mbox{cosh}(t)=\frac{e^t+e^{-t}}{2}[/tex] hence [tex]\mbox{cosh}(A+B)=\frac{e^{A+B}+e^{-(A+B)}}{2}[/tex] do likwise for the righhand side, simplify to get what I put.

    edit: good excuse.
     
    Last edited: Mar 27, 2006
  4. Mar 27, 2006 #3

    J77

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    If by the definitions, you mean (or have been taught), eg.

    [tex]\cosh(x)=\frac{e^x+e^{-x}}{2}[/tex]

    then I'd say go through the working and show the result.

    edit: was using that as an excuse for trying of the tex thingy :-)
     
  5. Mar 27, 2006 #4

    arildno

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    First of all, you've not been given an equation you are asked to SOLVE; you've been given an IDENTITY you are asked to PROVE.

    Remember that an identity is true for any choice of argument, whereas an equation is only true for some particular choices of argument (i.e, members of the solution set of the equation)

    Let us start with our definitions of the hyperbolic functions, and see how this develops:
    [tex]\cosh(A)\cosh(B}+\sinh(A)\sinh(B)=(\frac{e^{A}+e^{-A}}{2})*(\frac{e^{B}+e^{-B}}{2})+(\frac{e^{A}-e^{-A}}{2})*(\frac{e^{B}-e^{-B}}{2})=[/tex]
    [tex]\frac{1}{4}(e^{A+B}+e^{-(A+B)}+e^{A-B}+e^{B-A})+\frac{1}{4}(e^{A+B}+e^{-(A+B)}-e^{A-B}-e^{B-A})[/tex]
    Can you simplify this further?
     
  6. Mar 27, 2006 #5
    Benorin, J77- that's what i've already been told.
    arildno, that's what I got (the last line) when i did cosh(A+B)=1/2(e^t+e^-t), I think that was as far as I simplified it, unless I could do something with the two 1/4...

    Sorry if this has been made obvious...but I don't understand what it means by to substitute the definitions of cosh(A)...does it only want the squares^ to be A then? Same for sinh(A)? Because in my notes, I'm only given the definition for things like cosh(A+B) and sinh(A+B).
    But seeing how what you just wrote, I already have...maybe I've done the problem all in one?? lol (I hope so haha) Just that some of my minus and positive signs are a bit different...
     
  7. Mar 27, 2006 #6

    arildno

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    Well, sum my two brackets together; what do you get?
     
  8. Mar 27, 2006 #7
    Are you supposed to get something like 1/2(2e^A+B-e^-A-B) ?? Well I think I did the right thing...i did the same for cosh(B) and the others...because it says to prove that the right hand side of the equation is the same as cosh(A+B) and I think I've done that...if I haven't done it properly...well then that'll teach me for doing this all the day before it's due! Now I'm going to run to the room and hand in this small piece of assesssment! I think I have proved it, well at least shown the idea and attempt to do so.
    Thank you for you valuable help arildno!
     
  9. Mar 28, 2006 #8

    J77

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    It's just hard graft from there then, or did you want the answer? :wink:
     
  10. Mar 28, 2006 #9

    arildno

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    Yes, you're done! :smile:
     
  11. Mar 29, 2006 #10
    I'm just worried a little bit about my positive negative signs, but i made sure my final answer looked like what I was proving:D
     
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