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Trigonometric Inverse

  1. Jul 27, 2011 #1
    1. The problem statement, all variables and given/known data
    I understand that y = sin -1 x. However, why is it that when one writes x = sin y, one leaves out the -1


    2. Relevant equations



    3. The attempt at a solution
    I know that when you invert a trig function, the dependent variable essentially switches from y to x and the independent variable switches from x to y. However, why do you leave out the -1 when you write the inverse sin as x = sin y ?
     
  2. jcsd
  3. Jul 27, 2011 #2

    lanedance

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    not sure if I understand the question, but if you start from
    [tex] y = sin^{-1} x [/tex]

    take the sin of both sides
    [tex] sin(y) = sin(sin^{-1} x) = x [/tex]
     
  4. Jul 27, 2011 #3
    I'm sorry I wasn't very clear. That answers my question though! I had no idea you could take the sin of the sin-1(x) and it would cancel them out. Thank you! :smile:
     
  5. Jul 27, 2011 #4

    SteamKing

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    Well, that is how inverse functions work.
     
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