Expanding ArcSin[x+y] into Sin/Cos/ArcSin/ArcCos

[soandos]

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

 

[GoodMax]

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.

 

[HallsofIvy]

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 e^x+y= e^xe^y. But arcsin would involve a logarithm and there is no good formula for ln(x+ y).

 

[MathematicalPhysicist]

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 e^x+y= e^xe^y. But arcsin would involve a logarithm and there is no good formula for ln(x+ y).

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?

 

[GoodMax]

You can get sin(x+y)=sin(x)cos(y)+sin(y)cos(x) simply by looking on the unit circle

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?

 

[soandos]

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

 

[GoodMax]

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)?

 

[NoMoreExams]

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?

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 : \int_{a}^{b} f(x) dx. If you ever want to see anyone's code for their expression, just hit the quote button :) Enjoy

 

[soandos]

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?

 

[king vitamin]

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.

 

[Gib Z]

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?

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. =]

 

[soandos]

does anyone think what i suggested above is possible?
or is that just a dead end.

 

[Gib Z]

Dead End: Read posts 2 and 3 to see why.

 

[king vitamin]

does anyone think what i suggested above is possible?
or is that just a dead end.

Here is your expansion:

\arcsin(x) = \sum_{n=0}^{\infty} a_{n} \sin(x)

Where

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

n = 1, 2, 3, ...

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.

 

[soandos]

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?

 

[Count Iblis]

Here is your expansion:

\arcsin(x) = \sum_{n=0}^{\infty} a_{n} \sin(x)

Where

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

n = 1, 2, 3, ...

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.

It looks like the Fourier coefficients can be computed exactly in terms of Bessel functions.

 

[Count Iblis]

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?

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...

 

[soandos]

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.

 

[Count Iblis]

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.

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.

 

[soandos]

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 can't be anything for Arcsin[x+y] though.