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asin(x)

The circular sine differs from the hyperbolic sine in having imaginary units in the exponential, plus an extra factor. Solving for the exponential gives

sinw=eiw -eiw 2i =z e2iw -2izeiw -1 =0 eiw =iz ±1-z2 w=sin1z =1iln(iz ±1-z2 )

Applying the behavior of the logarithm, the inverse circular sine on an arbitrary branch is

sin1z =1iln(iz ±1-z2 ) +2πn

The individual branches look like this:

The imaginary part of this function retains the same numerical value between branches, while the real part moves up and down in value. Visualize the real part of several branches simultaneously:

The presentation continues here.