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

[richyw]

Homework Statement

Find $\sin^{-1}(-2)$ by writing [tex]sin w = -2[/itex] and using the rectangular representation of $\sin w$<br /> <br /> <h2>Homework Equations</h2><br /> <br /> Rectangular representation of $\sin w$<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> I think my biggest problem here is I have literally no idea what the "rectangular representation of $\sin w$ is. <br /> <br /> My textbook has this formula $sin(x)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)$. 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...[/tex]

 

[ehild]

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.

sin(sin^-1(w))=w=-2.


The complex Sine function is defined as sin(z)=(e^iz-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.

ehild

 

[richyw]

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!

 

[richyw]

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?

 

[ehild]

$$w=\sin^{-1}(-2)$$, so sin(w)=-2.

ehild

 

[richyw]

yes. that was in the initial question.

 

[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.

 

[SammyS]

yes. that was in the initial question.

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

 

[ehild]

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.

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

 

[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

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

 

[richyw]

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

 

[richyw]

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

 

[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

 

[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

 

[richyw]

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

 

[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 confusionehild