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Find arcsine(-2) using the rectangular representation of sin w

  1. Feb 28, 2014 #1
    1. The problem statement, all variables and given/known data

    Find [itex]\sin^{-1}(-2)[/itex] by writing [tex]sin w = -2[/itex] and using the rectangular representation of [itex]\sin w[/itex]

    2. Relevant equations

    Rectangular representation of [itex]\sin w[/itex]

    3. The attempt at a solution

    I think my biggest problem here is I have literally no idea what the "rectangular representation of [itex]\sin w[/itex] is.

    My textbook has this formula [itex] sin(x)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)[/itex]. Other than that, I am sure where to start here really. I can get the answer using the formula for arcsin, but not from starting with sin...
  2. jcsd
  3. Mar 1, 2014 #2


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    The rectangular representation means z=x+iy for the complex number z.

    sin-1(w) is a complex number, sin-1(w)=z=x+iy.


    The complex Sine function is defined as sin(z)=(eiz-e-iz)/(2i) = sin(x) cosh(y) +i cos(x) sinh(y).

    So you have the equation sin(x) cosh(y) +i cos(x) sinh(y)=-2. Solve for x, y.

  4. Mar 1, 2014 #3
    thanks. I thought that I needed to use that formula, but I was unaware that it was a definition! I tried doing this but I will try again!
  5. Mar 2, 2014 #4
    nope. I'm still incredibly lost here. I don't even know what w is. That is the definition for sin(z). I am not looking for sin(z). I am looking for sin(w). If sin(w)=sin(x)cosh(y)+icos(x)sinh(y), then I just end up going in circles.

    my textbook says in order to define arcsin(z), we write w=arcsin(z) when z=sin(w). That's pretty much useless to me. What is w?
  6. Mar 2, 2014 #5


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    [tex]w=\sin^{-1}(-2)[/tex], so sin(w)=-2.

  7. Mar 2, 2014 #6
    yes. that was in the initial question.
  8. Mar 2, 2014 #7
    what is the rectangular representaion of sin(w) though? no matter what I do I just keep ending up back at arcsin(-2), which is where I started.
  9. Mar 2, 2014 #8


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    If it had said, " sin(z) = -2 , find z." could you find z ?
  10. Mar 2, 2014 #9


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    w=x+iy ( x is the real component and y is the imaginary component of the complex number w) .

    sin(w)=sin(x) cosh(y) +i cos(x) sinh(y)=-2.

    What are x and y? Compare the real parts and the imaginary parts of both sides of the equation. Show your work.

  11. Mar 2, 2014 #10
    really? that is the definition of w? how does that make it any different from z then?

    anyways I am getting stuck even trying to solve that.

    I have

    sin(x)cosh(y)+i cos(x)sinh(y)=-2



    the second equation says that either cos(x)=0 or sinh(y)=0, if sinh(y)=0 then y=0, but then in the first equation sin(x)=-2 which can't hapen. So then I have

    x = π (n - 1/2)

    plugging into the first equation is then

  12. Mar 2, 2014 #11
    uh, so I guess[tex]y=\cosh^{-1}2[/tex][tex]x=\pi\left(n-\frac{1}{2}\right)[/tex]
  13. Mar 2, 2014 #12
  14. Mar 2, 2014 #13


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    Almost correct. Check if x=pi/2 is possible? What has to be the sign of sin(x)? Can cosh(y) negative for real y?

    Last edited: Mar 2, 2014
  15. Mar 2, 2014 #14
    no. it's not I don't think because I would want cosh(y)=2, so I would need to have sin(x) = -1, so
    x = π(2n-1/2) right?

    if I do this I think I can hack together something that gives me the correct answer. However I am still uncomfortable with how we can just say w = x+iy . I don't get that at all. To me it seems like saying w = x + iy and then saying sin(w)=z is like saying sin(z)=z
  16. Mar 2, 2014 #15
    oh wait, I see now why that is not a problem. This all makes sense to me now. Thank you for helping me.
  17. Mar 2, 2014 #16


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    You are welcome :smile:

    z is just an arbitrary complex number. An arbitrary complex number z can be written with it real part (x) and imaginary part (y) as z=x+iy. We have a special complex number now: w=arcsin(-2) which also has a real part and an imaginary part. We could have denoted them something else , u, v instead of x and y to avoid confusion

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