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ArcSin[2] {previously asked by a member}

412
2
1. 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.

2. Homework Equations

[tex]
e^{i \theta} = \cos{(\theta)} + i \sin{(\theta)}
[/tex]

3. The Attempt at a Solution

Let,

[tex]
ArcSin[2] = k
[/tex]

Then,

[tex]
Sin[k] = 2
[/tex]

Let,

[tex]
\lambda = \cos{(k)} + i \sin{(k)}
[/tex]

[tex]
\sin{(k)} = \frac{\lambda - \sqrt{1 - \sin(k)^2}}{i}
[/tex]

[tex]
\sin{(k)} = \frac{\lambda - \sqrt{1 - (2)^2}}{i}
[/tex]

[tex]
\sin{(k)} = \frac{\lambda - \sqrt{3}i}{i}
[/tex]

[tex]
2 = \frac{e^{ik}}{i} - \sqrt{3}
[/tex]

[tex]
e^{ik} = i(2 + \sqrt{3})
[/tex]

[tex]
k = \frac{1}{i} log_e(i(2 + \sqrt{3}))
[/tex]

[tex]
k = -i log_e(i(2 + \sqrt{3}))
[/tex]

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

thanks.
 

Answers and Replies

Gib Z
Homework Helper
3,344
4
Seems fine to me. When finding the inverse function of sine in exponential form, we get
[tex]\arcsin z = -i \log_e ( iz+ \sqrt{1-z^2})[/tex] and plugging z=2 there gets us the answer.
 
412
2
Seems fine to me. When finding the inverse function of sine in exponential form, we get
[tex]\arcsin z = -i \log_e ( iz+ \sqrt{1-z^2})[/tex] and plugging z=2 there gets us the answer.
thanks.. i didn't know there was a formula like this. :D
 

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