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Trig expansion

  1. Dec 25, 2008 #1
    is there a way to expand ArcSin[x+y] into sin cos Arcsin and/or Arccos?
    it looks like a Tan curve, but then so does Arcsin[x+y]-Tan[x+y]
    thanks
     
  2. jcsd
  3. Dec 25, 2008 #2
    I doubt very much that such a way exists. The enquiry is a formula for ArcSin in the complex domain looks like
    ArcSin z=-i Ln(z+\sqrt{z^2-1}).

    For example, if one use an equality Sin(z)=(e^{iz}-e^{-iz})/2i, we will easy derive an expansion for Sin(x+y). I don't see how we can obtain an expansion for ArcSin(x+y), using the expression for ArcSin that I wrote above.

    I'm sorry for my English. I don't even know whether you have understood me.
     
  4. Dec 25, 2008 #3

    HallsofIvy

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    You English is excellent GoodMax and I agree that there probabaly is no good way to rewrite arcsin(x+y). There exist formulas for sin(x+y) or cos(x+y) precisely because we can treat them as exponentials and ex+y= exey. But arcsin would involve a logarithm and there is no good formula for ln(x+ y).
     
  5. Dec 25, 2008 #4
    You can get sin(x+y)=sin(x)cos(y)+sin(y)cos(x) simply by looking on the unit circle, did you forget this HallsofIvy?
     
  6. Dec 25, 2008 #5
    It's possible precisely because the unit circle is a curve specified by the equation z=e^{i\phi}.

    By the way, how can I use LaTeX at this forum?
     
    Last edited: Dec 25, 2008
  7. Dec 25, 2008 #6
    I am not so sure. When Arcsin[a+b] is plotted, it is almost like there is only one curve and by shifting it in a straight line, one can obtain the full surface.
    since the same thing is true of other "expandable" graphs like Sin[a+b] i think that Arcsin[a+b] could be expanded, at least in the domain where it is real.
    if i am unclear, see pdf.
    the last example one where i do not think that there is a formula
     

    Attached Files:

  8. Dec 25, 2008 #7
    Explain to me, please, what does "the full surface" mean? And why do you base one's conclusions upon similiraty of the functions' graphs?

    For some obscure reason I can't download the pdf-file. Could you send it me via e-mail (maximt@gorodok.net)?
     
    Last edited: Dec 25, 2008
  9. Dec 25, 2008 #8
    You would do so by inserting [ tex] type stuff here [/ tex] (without the spaces).

    So for example if you wanted to write integral of f(x) dx from a to b, as you know the TeX for that is \int_{a}^{b} f(x) dx : [tex] \int_{a}^{b} f(x) dx [/tex]. If you ever want to see anyone's code for their expression, just hit the quote button :) Enjoy
     
  10. Dec 25, 2008 #9
    why i think there is a formula:

    If one rotates the plot to view it from a specific angle, it looks like a curve (2-D). to find the equation of such a curve, would it be possible to find the equation of the plane that we are viewing from (this is where every point has co-ordinates (x,y,z) and x=y and -oo<z<oo i believe) and solve the system of equations.

    would this work?
    what would be the equation of such a plane?
     
  11. Dec 26, 2008 #10
    There should be a Fourier expansion for arcsin(x) in sines (it's an odd function), but there's not even a closed form on the coefficients because the integrals are messy. Here's a quick dirty approximation:

    .696*sin(pi*x) - .390*sin(2*pi*x) + .273*sin(3*pi*x) - .211*sin(4*pi*x) + .172*sin(5*pi*x) - ...

    EDIT: correct me if i'm wrong but I believe this also proves that there is no finite representation for arcsin in sines/cosines.
     
    Last edited: Dec 26, 2008
  12. Dec 26, 2008 #11

    Gib Z

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    True, it merely depends on your preferred definition of the Trigonometric functions, to which would be the easiest method of proof. In "higher" mathematics I think they define sin and cos in terms of exponentials quite often, so the exponential method of proof would seem quite natural.

    And of course, the proof from the unit circle takes somewhat longer than the exponential proof. =]
     
  13. Dec 27, 2008 #12
    does anyone think what i suggested above is possible?
    or is that just a dead end.
     
  14. Dec 28, 2008 #13

    Gib Z

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    Dead End: Read posts 2 and 3 to see why.
     
  15. Dec 28, 2008 #14
    Here is your expansion:

    [tex] \arcsin(x) = \sum_{n=0}^{\infty} a_{n} \sin(x)[/tex]

    Where

    [tex] a_{n} = \int_{-1}^{1} \arcsin(x) \sin(n \pi x) \ dx [/tex]

    [tex] n = 1, 2, 3, ...[/tex]

    What I am really saying, then, is that there is no closed form for arcsin when expressed as sines/cosines. Because 1, sin(n*pi*x), and cos(n*pi*x) form a complete basis, this expansion is unique, i.e., there cannot be a simpler expansion in terms of sines, cosines, and constants.
     
    Last edited: Dec 28, 2008
  16. Dec 28, 2008 #15
    1) does that imply there there is no expansion for anything like arcsin[x+y], arcsin[xy]?
    2) is there something for arcsin[x]+arcsin[y], in the same way that
    arctan[x]+arctan[y] = arctan[(x+y)/(1-xy)]?
    3) is there a way to expand arcsin[x+y] into arcsins and arccos that only have one variable in them?
     
  17. Dec 28, 2008 #16
    It looks like the Fourier coefficients can be computed exactly in terms of Bessel functions.
     
  18. Dec 28, 2008 #17

    Well, why not draw a right triangle? Make one of the legs (a) a length of x, the other leg (b) a length of 1. Then the hypothenuse (c) has a length of sqrt[1+x^2]. The angle between b and c can then be expressed as arctan(x) and as arcsin[x/sqrt(1+x^2)].

    So, it looks like you got a relation between the arctan function and the arcsin function! Also you got a nice addition formula for arctan........
     
  19. Dec 28, 2008 #18
    still don't see it
    the addition thing only works for arctan+arctan, not arcsin+arctan
    i get the feeling that i am missing something obvious.
     
  20. Dec 29, 2008 #19
    Why not compute sin[arxsin(x)+arcsin(y)] by using cos(x) = sqrt[1-sin^2(x)] if x is between minus and plus pi/2.
     
  21. Dec 29, 2008 #20
    so that leaves me with
    Sqrt[1 - x^2] y + x Sqrt[1 - y^2] = Sin[ArcSin[x] + ArcSin[y]]
    meaning that ArcSin[Sqrt[1 - x^2] y + x Sqrt[1 - y^2]] = ArcSin[x] + ArcSin[y]
    True, or did i make a mistake somewhere

    still don't see why there cant be anything for Arcsin[x+y] though.
     
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