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Homework Help: I need Help, Half Angle, Emergency. Please Help

  1. Mar 14, 2010 #1
    1. The problem statement, all variables and given/known data
    How to verify this trigonometric equation? tanh z/2?
    verify that:

    tanh (z/2) = [sinh x + i sin y] / [cosh x + cos y]


    z is a complex number.

    How can i verify this equation?


    2. Relevant equations
    9957458708fa15c0a09a44ef7e3fbef8.png
    Inline3.gif

    # sin(x + i y) = sin(x) cosh(y) + i cos(x) sinh(y).
    # cos(x + i y) = cos(x) cosh(y) - i sin(x) sinh(y).
    # tan(x + i y) = (tan(x) + i tanh(y)) / (1 - i tan(x) tanh(y)).
    # cot(x + i y) = (cot(x) coth(y) - i) / (i cot(x) + coth(y)).




    3. The attempt at a solution

    i have tried to solve from right hand side but it didn't work, then i have simplified the equation tanh(z/2) = e^z-1 / e^z+1


    if you help me, i will be appreciate.
    thanx a lot.
     
  2. jcsd
  3. Mar 14, 2010 #2

    Dick

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    tanh(z/2)=(e^z-1)/(e^z+1) is an ok place to start. Now substitute z=x+iy and multiply numerator and denominator by the conjugate of (e^z+1). Multiply everything out. Where does that take you? What you eventually need is cosh(ix)=cos(x) and sinh(ix)=i*sin(x) for x real.
     
  4. Mar 14, 2010 #3
    thank u very much.

    ok i have substitued equation but after i don't know how can i simplify.
     
  5. Mar 14, 2010 #4

    Dick

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    What did you get? Did you multiply by the conjugate and collect terms? It works for me.
     
  6. Mar 14, 2010 #5
    yes i did sir.

    e^2(x+iy)-2e^(x+iy)+1 / e^2(x+iy)-1

    did u mean complex conjugate ? or normal conjugate?
     
  7. Mar 14, 2010 #6

    Dick

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    The conjugate of the denominator is (e^(x-iy)+1). It looks like you multiplied by (e^(x+iy)-1). Try the first way.
     
  8. Mar 14, 2010 #7

    Dick

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    Complex. BTW, I suggested that because the denominator of the expression you are supposed to get is real. See why that's a clue?
     
  9. Mar 14, 2010 #8
    ok result is

    e^2x + e^(x+iy) - e^(x-iy)-1 / e^2x + e^(x+iy) + e^(x-iy)
     
  10. Mar 14, 2010 #9
    than i don't know how can simplify this eq.

    sorry, i am taking your time.
     
  11. Mar 14, 2010 #10

    Dick

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    There's a 1 in the denominator too. Factor e^(x+iy) etc into e^x*e^(iy). Divide numerator and denominator by e^(x). Is anything looking familiar yet? Don't be sorry, just try a little harder.
     
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