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

1. Feb 28, 2014

### richyw

1. The problem statement, all variables and given/known data

ehild

6. Mar 2, 2014

### richyw

yes. that was in the initial question.

7. Mar 2, 2014

### richyw

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.

8. Mar 2, 2014

### SammyS

Staff Emeritus
If it had said, " sin(z) = -2 , find z." could you find z ?

9. Mar 2, 2014

### ehild

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.

ehild

10. Mar 2, 2014

### richyw

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

so

sin(x)cosh(y)=-2
cos(x)sinh(y)=0

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

Cos(x)=0
x = π (n - 1/2)

plugging into the first equation is then

$$\sin(\pi(n-1/2))\cosh(y)=-2$$
$$(-1)^n\cosh(y)=-2$$

11. Mar 2, 2014

### richyw

uh, so I guess$$y=\cosh^{-1}2$$$$x=\pi\left(n-\frac{1}{2}\right)$$

12. Mar 2, 2014

### richyw

$$y=\ln\left(2+\sqrt{3}\right)$$

13. Mar 2, 2014

### ehild

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?

ehild

Last edited: Mar 2, 2014
14. Mar 2, 2014

### richyw

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

15. Mar 2, 2014

### richyw

oh wait, I see now why that is not a problem. This all makes sense to me now. Thank you for helping me.

16. Mar 2, 2014

### ehild

You are welcome

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

ehild