ArcSin[2] {previously asked by a member}

[rohanprabhu]

Homework Statement

Find the value of ArcSin[2].

NOTE: This question was asked in: https://www.physicsforums.com/showthread.php?t=226670 I made a new thread since I wasn't sure about my solution and didn't want to confuse the OP or anybody else.

Homework Equations

$$e^{i \theta} = \cos{(\theta)} + i \sin{(\theta)}$$

The Attempt at a Solution

Let,

$$ArcSin[2] = k$$

Then,

$$Sin[k] = 2$$

Let,

$$\lambda = \cos{(k)} + i \sin{(k)}$$

$$\sin{(k)} = \frac{\lambda - \sqrt{1 - \sin(k)^2}}{i}$$

$$\sin{(k)} = \frac{\lambda - \sqrt{1 - (2)^2}}{i}$$

$$\sin{(k)} = \frac{\lambda - \sqrt{3}i}{i}$$

$$2 = \frac{e^{ik}}{i} - \sqrt{3}$$

$$e^{ik} = i(2 + \sqrt{3})$$

$$k = \frac{1}{i} log_e(i(2 + \sqrt{3}))$$

$$k = -i log_e(i(2 + \sqrt{3}))$$

My question is.. is this the right way to do it? Or.. all the assumptions that I've taken.. are they correct?

thanks.

 

[Gib Z]

Seems fine to me. When finding the inverse function of sine in exponential form, we get
$$\arcsin z = -i \log_e ( iz+ \sqrt{1-z^2})$$ and plugging z=2 there gets us the answer.

 

[rohanprabhu]

Seems fine to me. When finding the inverse function of sine in exponential form, we get
$$\arcsin z = -i \log_e ( iz+ \sqrt{1-z^2})$$ and plugging z=2 there gets us the answer.

thanks.. i didn't know there was a formula like this. :D