Obviously no regular angle, no real number, has a sine of 2. We're getting in the realm of complex numbers.
Euler's Formula is valid for complex numbers too:
e^{iz} = cos z + i sin z
We want to know about e^{i(-z)} = e^{-iz}. If z is real that would be the conjugate, but it's not for complex z. But even for complex z we want cosine to be even, cos z = cos(-z), and sine to be odd, sin z = -sin(-z). So
e^{-iz} = e^{i(-z)} = cos(-z) + i sin(-z) = cos z - i sin z
Subtracting,
e^{iz} - e^{-iz} = 2 i sin z
sin z = 1/{2i}(e^{iz} - e^{-iz})
From here out we'll use that essentially as the definition of sin z.
It's actually easier to get a relatively nice closed form for the inverse sine of a real number greater than one than of a regular sine.
We want to solve
sin z = a for real a>1.
1/{2i}(e^{iz} - e^{-iz}) = a
Let y=e^{iz}. Then e^{-iz}=1/y.
y - 1/y = 2i a
y^2 - 2i a y - 1 = 0
I call it the Shakespeare Quadratic Formula (2b or -2b): x^2-2bx+c has zeros x=b pm sqrt{b^2-c}.
y = ia pm sqrt{(-ia)^2+1} = ia pm sqrt{1-a^2}
Since a>1 the radicand is negative. Let's make that explicit.
y = ia pm sqrt{ (a^2-1) (-1) } = i (a pm sqrt{a^2-1} )
e^{iz} = i (a pm sqrt{a^2-1} )
Since i=e^{i pi/2} and e^{2pi k i} = 1 quad integer k,
e^{iz} = e^{i pi /2 } e^{2pi k i} e^{ln (a pm sqrt{a^2-1} ) }
iz = i pi/2 + 2pi k i + ln ( a pm sqrt{a^2-1} )
z = (pi/2 + 2 pi k) - i ln ( a pm sqrt{a^2-1} )
That's the general solution to sin z = a for real a>1. There are a few things to note about it. Let's look at the two imaginary parts first.
(a + sqrt{a^2 - 1} ) (a - sqrt{a^2 - 1 }) = a^2 - (a^2-1) =1
a - sqrt{a^2 - 1 } = 1/{ a + sqrt{a^2 - 1} }
ln (a - sqrt{a^2 - 1 } ) = -ln ( a + sqrt{a^2 - 1} )
z = (pi/2 + 2 pi k) pm i ln ( a - sqrt{a^2-1} )
The two imaginary parts are negations of each other! That means the zs come in complex conjugate pairs.
Considering k=0, when we add the two roots together we'll get pi. In other words those two roots, even though complex, are supplementary angles who share the same sine, just like real angles. Just like real angles, we can add or subtract 2pi as many times as we like and get another number with the same sine.
OK, let's plug in a=2 for the big finish.
arcsin 2 = (pi/2 + 2pi k ) pm i ln(2 - sqrt{3}) quad integer k
