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Trig problem

  1. Apr 13, 2010 #1
    im generating potentials, however the coulomb is proving tricky. stumbling block is finding a way to write

    A coth(2x) + B cosech(2x)= (A+B)/2 tanh(x)+(A_B)/2 coth(x)

    i am at the RHS and need to get the tanh to come out which im told can be done.
    the a's and b's might not be quite right but the trig functions are what im after
     
  2. jcsd
  3. Apr 13, 2010 #2
    1. The problem statement, all variables and given/known data

    something like
    A coth(2x) + B cosech(2x)= (A+B)/2 tanh(x)+(A_B)/2 coth(x)
    the a's and b's may not be right but the trig is what's important

    2. Relevant equations

    anything

    3. The attempt at a solution
     
  4. Apr 13, 2010 #3

    Mark44

    Staff: Mentor

    What do you mean "something like?" If you need to prove some statement, give us exactly the statement that needs to be proved.

    What equations are relevant to this problem?

    What have you tried? Before anyone can give you any help, you need to show some effort at working the problem.
     
  5. Apr 13, 2010 #4

    Cyosis

    User Avatar
    Homework Helper

    Use the identities [itex]\cosh(2x)=\sinh^2(x)+cosh^2(x)[/itex], [itex]\sinh(2x)=2\cosh(x)\sinh(x)[/itex] and [itex]\cosh^2 (2x)-\sinh^2(2x)=1[/itex].
     
  6. Apr 13, 2010 #5
    the exact statement is

    A coth(2ax) + B cosech(2ax)= (A+B)/2 tanh(ax)+(A-B)/2 coth(ax)

    i have tried using the double angle formulas but seem to be going in circles
     
  7. Apr 13, 2010 #6
    am i better off going from the RHS to LHS?
     
  8. Apr 13, 2010 #7

    Mark44

    Staff: Mentor

    Show us what you've tried.
     
  9. Apr 13, 2010 #8
    need to show

    (1) A coth(2ax) - B cosech(2ax)= (A+B)/2 tanh(ax)+(A-B)/2 coth(ax)

    starting from the RHS

    (2) A/2 (1+tanh^2(ax))/tanh(ax) - B/2 cosech(ax)sech(ax)
    (3) A/2 coth(ax) + A/2 tanh(ax) - B/2 cosech(ax)sech(ax)

    from here i just starting going in circles
    from line 2 i could have instead gone to

    A/2 coth(ax) (2-sech^2(ax))- B/2 cosech(ax)sech(ax)
    A coth(ax) -A/2 coth(ax)sech^2(ax)- B/2 cosech(ax)sech(ax)
    A coth(ax) -A/2 (cosech(ax) / sech(ax)) sech^2(ax)- B/2 cosech(ax)sech(ax)
    A coth(ax) -A/2 cosech(ax) sech(ax)- B/2 cosech(ax)sech(ax)
    A coth(ax) -(A+B)/2 cosech(ax) sech(ax)

    i don't see any other openings
     
  10. Apr 13, 2010 #9

    Mark44

    Staff: Mentor

    How did you get to (2) from the right side of (1)? 1 + tanh2(ax) = sech2(ax). If I were doing this, I'd start on the left side and see if I could make it look like the right side.
    I don't know if this would work, but this is what I would try. Starting from (3) I would add (B/2)tanh(ax) - (B/2)coth(ax) and then subtract it off, with the possibility that (-B/2)tanh(ax) + (B/2)coth(ax) -(B/2)cosech(ax)sech(ax) adds up to zero. If so, that what remains is what you want to end up with.
     
    Last edited: Apr 13, 2010
  11. Apr 13, 2010 #10
    sorry, i meant i started from the LHS
     
  12. Apr 13, 2010 #11
    I should also point out i don't actually know if it possible to solve this, the paper im reading claims it is but im not convinced.
     
  13. Apr 13, 2010 #12

    Cyosis

    User Avatar
    Homework Helper

    Start on the left hand side and write the coth and cosech in terms of cosh/sinh. Now apply the three identities from post 4 everywhere you can. And don't forget to show your attempt if you get stuck.
     
  14. Apr 13, 2010 #13
    solved it thanks, i was being a fool
     
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