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Is arcsin sin rotated by 90°?

  1. Jun 25, 2010 #1

    zxh

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    Sry, noob, but i didn't find this anywhere.
     
  2. jcsd
  3. Jun 25, 2010 #2
    No. Try Google.
     
  4. Jun 25, 2010 #3

    Mark44

    Staff: Mentor

    The two functions are inverses of one another. In general, the graphs of y = f(x) and x = f-1(y) are identical, but the graphs of y = f(x) and y = f-1(x) are reflections of each other in the line y = x.

    The situation is a little more complicated with y = sin(x) and y = arcsin(x) = sin-1(x) since the graph of the sine function isn't one-to-one (making the inverse not a function). The usual way around this is to restrict the domain of the sine function, defining y = Sin(x) = sin(x), with x restricted to the interval -pi/2 <= x <= pi/2.
     
  5. Jun 25, 2010 #4

    zxh

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    So the title would be true for -sin, viewed as a curve?
     
  6. Jun 25, 2010 #5

    Mark44

    Staff: Mentor

    Are you asking whether y = arcsin(x) is the rotation by 90 deg of y = -sin(x)? If that's the question, then no.

    If that isn't the question, then what are you asking?
     
  7. Jun 25, 2010 #6
    No it's only a segment of the curve, but the graph of y=arcsin(x) would fit over y=-sin(x) if rotated 90° either way about the origin.
     
    Last edited: Jun 25, 2010
  8. Jun 25, 2010 #7

    zxh

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    Thanks, that's what i was looking for. I'm not too concerned about range definitions.
    I came to this looking for a trig definition of a (half) circle (not the pythagorean Sqrt(r-x^2)).
    At first i was wondering why Cos(Sin(x)) (given that the 2 functions for a circle in a parametric plot are sinx and cosx) didn't work but it turns out it's
    Cos(ArcSin(x)).
     
  9. Jun 25, 2010 #8

    disregardthat

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    Science Advisor

    You can show this algebraically. If you are familiar with complex numbers, this is easy.
    1) multiply x+i (-sin x) with e^(i*pi/2) to rotate it by 90 degrees.
    2) reflect x+i(-sin x) over the curve y=x to invert it.
     
  10. Jun 25, 2010 #9

    zxh

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    thanks, good one.
     
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