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ArcSin[2] {previously asked by a member}

  1. Apr 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the value of ArcSin[2].

    NOTE: This question was asked in: https://www.physicsforums.com/showthread.php?t=226670 I made a new thread since I wasn't sure about my solution and didn't want to confuse the OP or anybody else.

    2. Relevant equations

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

    3. The attempt at a solution

    Let,

    [tex]
    ArcSin[2] = k
    [/tex]

    Then,

    [tex]
    Sin[k] = 2
    [/tex]

    Let,

    [tex]
    \lambda = \cos{(k)} + i \sin{(k)}
    [/tex]

    [tex]
    \sin{(k)} = \frac{\lambda - \sqrt{1 - \sin(k)^2}}{i}
    [/tex]

    [tex]
    \sin{(k)} = \frac{\lambda - \sqrt{1 - (2)^2}}{i}
    [/tex]

    [tex]
    \sin{(k)} = \frac{\lambda - \sqrt{3}i}{i}
    [/tex]

    [tex]
    2 = \frac{e^{ik}}{i} - \sqrt{3}
    [/tex]

    [tex]
    e^{ik} = i(2 + \sqrt{3})
    [/tex]

    [tex]
    k = \frac{1}{i} log_e(i(2 + \sqrt{3}))
    [/tex]

    [tex]
    k = -i log_e(i(2 + \sqrt{3}))
    [/tex]

    My question is.. is this the right way to do it? Or.. all the assumptions that i've taken.. are they correct?

    thanks.
     
  2. jcsd
  3. Apr 5, 2008 #2

    Gib Z

    User Avatar
    Homework Helper

    Seems fine to me. When finding the inverse function of sine in exponential form, we get
    [tex]\arcsin z = -i \log_e ( iz+ \sqrt{1-z^2})[/tex] and plugging z=2 there gets us the answer.
     
  4. Apr 5, 2008 #3
    thanks.. i didn't know there was a formula like this. :D
     
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