#(sin )^(-1)# is the piecewise-wholesome inverse sine operator. The FCS y = #(sin )^(-1)(x+(sin )^(-1)(x+(sin )^(-1)(x+...)))#. How do you find the amplitude and the period, of this FCS wave?

Definition of piecewise-wholesome #(sin)^(-1)X#:

#Y = (sin)^(-1 )X = kpi + (-1)^k sin^(-1) X,#

#k = 0, +-1, +-2, +-3, ..., Y in [ k pi-pi/2, k pi + pi/2 ]#.

https://socratic.org/questions/on-what-interval-is-the-identity-sin-1-sin-x-x-valid#639442

The FCS wave, with axis #x + y = 0 # and demarcation, marked by

# x + y = +- 1#, is created by

#x = sin y - y#. See uniform-scale graph. Observe the points of

inflexion aligned in x + y = 0 and crests and troughs aligned in

#x + y = +- 1#.
graph{(x-sin y + y)(x+y)((x+y)^2-1)=0}

Graph of just one wave from #y = sin^(-1)( x + y )# instead:

graph{(y-arcsin (x+y))(x+y)((x+y)^2-1)=0} .

The points of the meet with the axis are

#+- ( k pi, - k pi ), k = 0, +-1, +-2, +-3, ...#

The amplitude = half-width between # x + y = +- 1# is #1/sqrt2#.

Period = 2 ( distance between two consecutive axial points)

#= 2sqrt2 pi#.

Plots revealing a crust, a trough, period length and width.

graph{(x-sin y + y)(x-y)((x+y)^2-1)((x-3.14)^2+(y+3.14)^2-.01)((x+3.14)^2+(y-3.14)^2-.01)((x+0.57)^2+(y-1.57)^2 -.01)((x-0.57)^2+(y+1.57)^2 -.01)((x-0.5)^2 +(y-0.5)^2-.01)((x+0.5)^2 +(y+0.5)^2-.01)=0[-7 7 -3.5 3.5]}

Also see

https://socratic.org/questions/defining-the-wholesome-inverse-operator-sin-1-by-y-sin-1-x-k-pi-1-k-sin-1-x-y-in#63919