Taking a Break From Number Theory

I came up with a formula that has to do with extending the arcsine function beyond limited input. My interest was sparked when I noticed that there is a simple pattern among the commonly known values for sine:

sin(π/6) = sqrt(1)/2 = 1/2
sin(π/4) = sqrt(2)/2
sin(π/3) = sqrt(3)/2
sin(π/2) = sqrt(4)/2 = 1

As the angles increase, the value for sin(Θ) increases as sqrt(n)/2, for n = 1,2,3,4. I then asked: For what angle will the sine of the angle be sqrt(5)/2?

Clearly, the angle must be imaginary: the range of sine for real input is [-1,1]. Here's how I went about solving it. (The solution I give here will give the angle for any given n, not just 5- though that is how I started out.)

I started with the identity

sin(Θ) = [e^(iΘ) - e^(-iΘ)]/2i

(If you are not familiar with this, it is easily derived with the knowledge that
e^iΘ = cosΘ + isinΘ- I'll leave the proof of sin(Θ) = [e^(iΘ) - e^(-iΘ)]/2i up to you.)

Continuing,

i*Sqrt(n) = e^iΘ - e^-iΘ, since sin(Θ) = Sqrt(n)/2

e^2iΘ - i*Sqrt(n)*e^iΘ -1 = 0

This is quadratic in e^iΘ, so

e^iΘ = {i*Sqrt(n) +- Sqrt(4-n)}/2

Θ = -i*log({iSqrt(n)+/- Sqrt(4-n)}/2)

This says that

arcsin[Sqrt(n)/2] = -i*log({iSqrt(n)+/- Sqrt(4-n)}/2)

Or, substituting n = 4k^2,

arcsin[k] = -i*log(ik+-Sqrt(1-k^2))

This is essentially an analytic continuation of the arcsin function, whose domain is usually only [-1,1]; this continuation defines arcsin(k) for any and all k, including complex values. You may notice that for values of k on [-1,1], the Sqrt(1-k^2) part is real, and within the log we are left with a familiar quantity. For example, take k = 1/2. Then

arcsin[1/2] = -i*log(i/2 + Sqrt(3)/4)

Since i/2 + Sqrt(3)/4 = e^(iπ/6),

arcsin[1/2] = -i*iπ/6

arcsin[1/2] = π/6, which is true.

For my original example of k = Sqrt(5)/2, the angle Θ is -i*log(iφ), where φ is phi, the golden ratio, so

sin(-ilogiφ) = sqrt(5)/2, an elegant result.

Beyond just arcsin, I also created analytic continuations for arccos and arctan, both derived in almost the exactly same manner.

arccos[k] = -i*log(k +- Sqrt(k^2-1)), derived from cos Θ = (e^iΘ + e^-iΘ)/2

arctan[k] = -i/2*log{(1+ik)/(1-ik)}, derived from tan Θ = (e^iΘ-e^-iΘ)/(ie^iΘ+ie^-iΘ)

Happy New Year!