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Solving simultaneous trigonometric equation

  1. Aug 12, 2009 #1
    how to solve
    eqn1 = sin((alpha)*a)*cosh((alpha)*b)=sin(alpha)*a
    eqn2 = sinh((alpha)*b)*cos((alpha)*a) = b*sin(alpha)
    alpha = 0 to 2pi....

    Find a & b...

    Plz help me with any method that would help solving this...
  2. jcsd
  3. Aug 12, 2009 #2
    I can get it reduced to:

    [tex]\sin(\alpha x) = x \sin(\alpha)[/tex]

    [tex]\sin(\alpha y) = y \sin(\alpha)[/tex]

    [tex]x = a + (b/i) \quad \& \quad y= a - (b/i)[/tex]

    beyond that, I am a bit stuck. Anyone else?
    Last edited: Aug 12, 2009
  4. Aug 13, 2009 #3
    What is that "h" in front of cos in the first equation and sin in the second equation?

    Anyway, the solution is:

    eq1 - eq2

    Now use the sum difference formula

    [tex]sin(u \pm v)=sinucosv \pm cosusinv[/tex]

    and you got

    [tex]sin(\alpha*a - \alpha*b) = (a-b)sin\alpha[/tex]




  5. Aug 13, 2009 #4

    Thank for ur reply,

    Reply 1:
    I will look in to it... thanx...

    Reply 2:
    'h' means hyperbolic function.... and the result u have showed is from where i started to get these equations.... these are transcedental equations having many roots so that eqn formed at the end that you got is not solvable(that is i dont know to solve).... if you know any method plz reply...

    Thank you.
  6. Aug 13, 2009 #5
    Aaah... I see now...
    Could you possibly provide me the original problem?

  7. Aug 13, 2009 #6
    sinZ = C1*Z
    Z = (\alpha)*(\lambda)
    C1 =(+/-)[Sin(\alpha)] / (\alpha)

    hence we get,

    Sin((\alpha)*(\lambda)) = \lambda *sin(\alpha)

    \lambda = a+ib

    hence eqn1 and eqn2...
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