What do you do if you have to evaluate arcsin2?

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The evaluation of arcsin(2) is not possible since 2 is outside the domain of the sine function, which is restricted to values between -1 and 1. Consequently, there is no real solution for sin(x) = 2. In the complex domain, solutions exist, expressed as x = -i ln(2 ± i√3) + 2kπ for any integer k. Attempting to use a series expansion for arcsin(2) would lead to divergence. A Pade approximant might yield a numerical approximation, but it would not represent a valid solution.
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what do you do if you have to evaluate arcsin2?
 
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fizzzzzzzzzzzy said:
what do you do if you have to evaluate arcsin2?

Nothing, since 2 is not in the domain.
 
-1\le sin(x) \le 1

There is no x such that sin(x)= 2 and so arcsin 2 is not defined.
 
Perhaps get the series for it and sub in 2, see what happens :S
 
fizzzzzzzzzzzy said:
what do you do if you have to evaluate arcsin2?

It comes down to solving the equation

\sin x =2

which has no solution in \mathbb{R}, and in \mathbb{C} has the solution

x=-i\ln\left(2\pm i\sqrt{3}\right)+2k\pi \ , \forall k\in\mathbb{Z}
 
Gib Z said:
Perhaps get the series for it and sub in 2, see what happens :S

Which would diverge.
A Pade approximant might give you some silly number, though. :smile:
 

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