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Basic Trig Identity question

  1. Feb 25, 2012 #1
    I need to show that sin i*theta= i* sinh(theta).
    where sinh(theta) = .5[e^theta - e^(-theta)]
    and cos(theta) = .5[e^theta + e^(-theta)]
    and e^(i*theta) = cos(theta) + isin(theta)



    if I start with the formula sinh(theta) = .5[e^theta - e^(-theta)]
    and plug in e^(i*theta) = cos(theta) + isin(theta)
    I get

    sinh(theta) = .5*[{cos(theta) + isin(theta)}+e^i - {cos(-theta) + isin(-theta)} + e^i]

    since cos(-theta) = cos(theta) and sin(-theta) = -sin(theta)

    sinh(theta) = .5*2*i*sin(theta)
    or
    sinh(theta) = i*sin(theta)

    now how do I go from here to
    sin(i*theta) = i*sinh(theta)

    I know I am almost there I just need a little last nudge.
    Thanks
    Stephen
     
    Last edited: Feb 25, 2012
  2. jcsd
  3. Feb 25, 2012 #2
    Are you sure it isnt..

    [tex]sinh(i \theta) = isin\theta[/tex]??
     
  4. Feb 25, 2012 #3
    nope..

    sin(i*theta) = i*sinh(theta)

    Attached is the question as given by the professor
     

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  5. Feb 25, 2012 #4
    Okay, fair enough.

    This isn't true. Try finding [itex] \sin(i\theta)[/itex] first with Euler's formula, and then compare the result to what you were given for [itex]\sinh(\theta)[/itex].
     
  6. Feb 25, 2012 #5
    How would I find \sin(i\theta)

    from Eulers formula
    e^itheta = cos(theta) +isin(theta)

    so if theta = itheta then would
    e^itheta = e^-theta = cos(itheta) +i*sin(itheta)
    so
    sin(i*theta) = (e^-theta -cos(itheta))/i

    how does this help???
     
  7. Feb 25, 2012 #6
    Alright, I'm going to show you a little bit of wizardry. We know that

    [tex]e^{i\theta}=\cos(\theta) + i \sin(\theta) \; \; \; [1][/tex]

    Now let [itex]\theta = -\theta[/itex] so that:

    [tex]e^{i(-\theta)}=\cos(-\theta) + i \sin(-\theta)[/tex]

    Using the facts that cosine is even and sine is odd (which you already knew):

    [tex]e^{i(-\theta)}=\cos(\theta) - i \sin(\theta) \; \; \; [2][/tex]

    What happens if you subtract [2] from [1]?
     
  8. Feb 25, 2012 #7

    SammyS

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    I don't see how you get
    sinh(θ) = .5*[{cos(θ) + i sin(θ)}+ei - {cos(-θ) + i sin(-θ)} + ei]
    ...​

    That's equivalent to [itex]\sinh(\theta)=.5\{e^{i\theta}+e^{i}-e^{-i\theta}+e^{i} \}[/itex] which is definitely not true.

    BTW: It is true that sin(iθ) = i sinh(θ) .
     
    Last edited: Feb 25, 2012
  9. Feb 25, 2012 #8
    ok..Screwdriver

    I did that and got
    .5(e^(itheta)-e^(-itheta) = isin(theta)
    so
    sinh(theta) = i sin(theta)

    now how do I get from here to
    sin(i*theta) = i*sinh(theta)
     
  10. Feb 25, 2012 #9
    You've determined that [itex]\sin(\theta)=\frac{1}{2i}(e^{i\theta}-e^{i\theta})[/itex]. Here is where you should sub in [itex]\theta = i\theta[/itex].
     
  11. Feb 26, 2012 #10
    ok if you plug in i*theta into sin(theta) = 1/i *sinh(\theta) you get sin(i*theta) = 1/i * sinh(i*theta)

    How do you go from this to sin(i*theta) = i*sinh(theta)
     
  12. Feb 26, 2012 #11

    SammyS

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    That should be [itex]\sin(\theta)=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})\,.[/itex] What you had was equivalent to zero.
     
  13. Feb 26, 2012 #12
    Oops, thanks for catching that typo :redface:

    You should be subbing in [itex]i\theta[/itex] into [itex]\sin(\theta)=\frac{1}{2i}(e^{i\theta} - e^{i(-\theta)})[/itex]. In other words, simplify the following:

    [tex]\sin(i\theta)=\frac{1}{2i}(e^{i(i\theta)} - e^{i(-i\theta)})[/tex]

    And then compare it to the definition of [itex]\sinh(\theta)[/itex] in terms of exponential functions (given in the question.)
     
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