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

<br /> e^{i \theta} = \cos{(\theta)} + i \sin{(\theta)}<br />

The Attempt at a Solution

Let,

<br /> ArcSin[2] = k<br />

Then,

<br /> Sin[k] = 2<br />

Let,

<br /> \lambda = \cos{(k)} + i \sin{(k)}<br />

<br /> \sin{(k)} = \frac{\lambda - \sqrt{1 - \sin(k)^2}}{i}<br />

<br /> \sin{(k)} = \frac{\lambda - \sqrt{1 - (2)^2}}{i}<br />

<br /> \sin{(k)} = \frac{\lambda - \sqrt{3}i}{i}<br />

<br /> 2 = \frac{e^{ik}}{i} - \sqrt{3}<br />

<br /> e^{ik} = i(2 + \sqrt{3})<br />

<br /> k = \frac{1}{i} log_e(i(2 + \sqrt{3}))<br />

<br /> k = -i log_e(i(2 + \sqrt{3}))<br />

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