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Homework Help: How can you find two different values for sin^-1(0.750)?

  1. Oct 13, 2007 #1
    I am working on a statics question and one of the angles is sin^-1(0.750) which I calculate as 48.6 degrees though the book gives 131.41 degrees

    If I calculate sin(48.6) it equals 0.750
    If I calculate sin(131.41) it equals 0.750

    Clearly there is some relation or rule that I am unaware of. How do I ensure that I find the proper angle?
  2. jcsd
  3. Oct 13, 2007 #2


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    Just draw the unit circle and joggle a bit then all becomes clear.

    sin^1(0.75) means "what angle gives y = 0.75" if you do the unit circle. You'll see that there is two angles in the range 0 - 360 degree that gives you that value.

    Which is correct depends on what the problem is, what physical situation it is and so on.
  4. Oct 13, 2007 #3


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    sin x = a has infinitely many solutions for x.

    If you're working on a problem and expect a unique answer, then you'll have to find some other constraint on x.
  5. Oct 13, 2007 #4
    Ok, I figured it out. Because of the direction of the vector it is not 0 degrees + 48.6 degrees but 180 degrees - 48.6 degrees which equals 131.4 degrees
  6. Oct 13, 2007 #5


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    You realize that you said nothing at all about vectors in your original post. And even now, we do not know WHY the direction is "180 degrees- 48.6 degrees"!
  7. Oct 13, 2007 #6


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    Your calculator cannot help you with interpreting the output in relation with the problem you are working on. The inverse trig functions on a calculator will only give output in the ranges

    arcsin or arctan: -90º to +90º (not quite 90º, of course, for arctan)

    arccos: 0º to 180º .

    In order to define these as functions, it is necessary to have restricted ranges on
    y=f(x) and the calculator is programmed to follow these definitions (this is why the inverse trig functions are sometimes identified as Sin^{-1}, Cos^{-1}, etc., rather than using lower-case).

    As you note that the positive sine value is associated with two angles (in the first and second quadrants), it is important to have a picture in mind for the physical situation in order to decide which angle is appropriate to the solution. (I particularly warn engineering students about this all the time: you have to know what the result should look like because the calculator can't.)
  8. Oct 13, 2007 #7
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