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Proving a hyperbolic identity

  1. Nov 25, 2008 #1
    \equiv1. The problem statement, all variables and given/known data
    Hi, I've been given a hyperbolic identity to prove:

    [tex]2sinhAsinhB \equiv Cosh(A+B) - Cosh(A-B)[/tex]


    2. Relevant equations

    [tex]Cos(A\pm B) \equiv CosACosB \mp SinASinB[/tex]

    3. The attempt at a solution

    I have Cosh(A+B) and - Cosh(A-B) so i can kind of see that there will be two lots of SinhASinhB and the CoshACoshB will cancel, but how do I prove it? I mean how do I know that [tex]Cosh(A\pm B) \equiv CoshACoshB \mp SinhASinhB[/tex]

    Thanks :)
     
  2. jcsd
  3. Nov 25, 2008 #2

    tiny-tim

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    Danger!

    (have a ± :wink:)

    Warning, warning, thomas49th!

    Cosh(A±B) = coshAcoshB ± sinhAsinhB (the opposite sign to cos).

    (this is because, from Euler's formula … cosx = coshix, isinx = sinhix, so i2sinAsinB = sinhAsinhB :wink:)
     
  4. Nov 25, 2008 #3
    Errr sorry I don't follow :S. I've only just started hyperbolics today and havn't used an imaginary numbers with them yet?

    Thanks :)
     
  5. Nov 25, 2008 #4

    gabbagabbahey

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    I assume that you have been taught the definitions of these functions:

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

    That is all you need to prove this identity....What are [itex]\sinh A[/itex] and [itex]\sinh B[/itex]? What does that make the LHS of your identity? What are [itex]\cosh(A+B)[/itex] and [itex]\cosh(A-B)[/itex]? What does that make the RHS of your identity? Can you show that the two expressions are equivalent? If so, then you prove the identity.
     
  6. Nov 25, 2008 #5

    tiny-tim

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    oops!

    Hi thomas49th! :smile:
    Oops! :redface:

    In that case, all you need to know is that "hyperbolic" trig functions cosh sinh tanh sech coth and cosech work almost the same as ordinary trig functions (for example, sinh(2x) = 2sinhx coshx), but occasionally you get a + instead of a minus (or vice versa) :rolleyes: … I think only when you have two sinh's.

    But, to be on the safe side, use gabbagabbahey's :smile: method!
     
  7. Nov 25, 2008 #6
    using the identities i got

    [tex]\frac{1}{2} e^{2x} - e^{-2x}[/tex]
    which is equilivlent to cosh2x. but where next?

    Thanks ;)
     
  8. Nov 25, 2008 #7

    gabbagabbahey

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    I think you'd better show me your work :wink:
     
  9. Nov 25, 2008 #8
    [tex]2(\frac{e^{2x} - e^{-2x}}{2})(\frac{e^{2x} + e^{-2x}}{2})[/tex]
    one 2 cancels so you get a half overall. the difference of two squares acts nicely leaving us with:
    [tex]
    \frac{1}{2} e^{2x} - e^{-2x}
    [/tex]
    and i was looking back over the notes in class and I saw that we identified cosh2x as that

    Right?
    Thanks :)
     
  10. Nov 25, 2008 #9

    tiny-tim

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    oh i see … you're proving 2 sinhx coshx = sinh 2x (not cosh 2x! :rolleyes: … cosh is the positive one :wink:)

    but what about the original problem, with A and B? :smile:
     
  11. Nov 25, 2008 #10
     
  12. Nov 25, 2008 #11

    gabbagabbahey

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    Well, if [tex]
    \sinh(x)=\frac{e^x-e^{-x}}{2}
    [/tex]....then [tex]
    \sinh(A)=\frac{e^A-e^{-A}}{2}
    [/tex]....so [itex]\sinh(B)=[/itex]___? And so [itex]
    2\sinh(A)\sinh(B)=[/itex]___?
     
  13. Nov 25, 2008 #12
    [tex]

    \sinh(B)=\frac{e^B-e^{-B}}{2}

    [/tex]

    [itex]

    2\sinh(A)\sinh(B)= \frac{1}{2} [ e^{AB} - 2e^{-AB} + e^{AB}]

    [/itex]

    [tex]

    e^{AB} - e^{-AB}

    [/tex]

    I'm almost there?
    Thanks :)
     
  14. Nov 25, 2008 #13

    gabbagabbahey

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    Hmmm... [tex]e^Ae^B=e^{A+B} \neq e^{AB}[/tex] :wink:
     
  15. Nov 25, 2008 #14
    [tex]


    e^{A+B} - e^{A-B}


    [/tex]

    Notice how there is an A+B and A-B from JUST like in cos(A+B)

    now how do I show that cos(A+B) = [tex]e^{A+B} [/tex]
     
  16. Nov 25, 2008 #15

    gabbagabbahey

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    aren't you missing a couple of terms in that expression? :wink:
     
  17. Nov 25, 2008 #16
    arrrg it was just starting to look nice:

    stick
    [tex] -e^{-A+B} -e^{-A-B}[/tex]

    I've goto go to bed... knackard sorry. it's almost midnight ere in merry old england

    i'd read any other message people post on here in the morning. thanks for everything!
    :)
     
  18. Nov 25, 2008 #17

    gabbagabbahey

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    Looks good, except your missing a factor of 1/2, and one of your signs is incorrect..... you should have:

    [tex]2\sinh (A) \sinh (B)=\frac{e^{A+B}-e^{A-B}-e^{-A+B}+e^{-A-B}}{2}[/tex]

    You also know the definition of cosh: [tex]\cosh(x)=\frac{e^x+e^{-x}}{2}[/tex]....so [itex]\cosh (A+B)=[/itex]___? And [itex]\cosh (A-B)=[/itex]__? And so [itex]\cosh (A+B)-\cosh (A-B)=[/itex]___?
     
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