Question

Defining the wholesome inverse operator `(sin)^(-1)` by `Y = (sin)^(-1)(X) = k pi + (-1)^k sin^(-1)X, k= 0, +-1, +-2, +-3,...`, how do you find the points of inflexion of the FCS `y = ((sin)^(-1))_(fcs) (x; 1) = (sin)^(-1)(x+y)?`

Answer 1

The ( tangent-crossing-curve ) points of inflexion are
`+- (-k pi, k pi ), k = 0, 1, 2, 3, ...` See the graph, for these plots. Please do not edit my answer. You could give another answer.

Explanation:

Currently, there seems to be no provision in Calculators, for the

piecewise ( one x - many answer ) output, for the wholesome

inversion `(sin)^(-1)x`.

`y = ((sin)^(-1))_(fcs)( x; 1 )`

`= ( sin )^(-1)( x + (sin^(-1))( x +(sin^(-1))( x + ...)))`

`= (sin)^(-1)( x + y ), x + y in [ -1, 1 ]`.

`= k pi + (-1)^ksin^(-1)( x + y ), k = 0, +-1, +-2, +-3, ...`, in the

respective `y in [ k pi - pi/2, kpi + pi/2].`

Inversely,

`x = (-1)^k sin ( y - k pi ) - y , y in [ k pi - pi/2, kpi + pi/2]`

`= (-1)^k sin y cos k pi - y`

`= sin y - y`.

It is easy to prove that

`(d^2x)/dy^2 = - sin y = 0` and

`(d^3x)/(dy^3) = - cos y ne 0`, at `+-(-kpi, kpi), k = 0, 1, 2, 3, ...`.

So,

the points of inflexion (POI) are `+-(-kpi, kpi), k = 0, 2, 3. ..`

The graph of x = sin y - y includes the ( y-restricted ) inverse of

`y = ((sin)^(-1))_(fcs)( x; 1 )`.

Graph of #y = ((sin)^(-1))_(fcs)( x; 1 ), wth POI plots.
graph{(x-sin y+y)(x+y)(x^2+y^2 -.01)((x+3.14)^2+(y-3.14)^2-.01) ((x-3.14)^2+(y+3.14)^2-.01)=0[-10 10 -5 5]}

graph{(x-sin y+y)(x+y)(x^2+y^2 -.01)((x+3.14)^2+(y-3.14)^2-.01) ((x-3.14)^2+(y+3.14)^2-.01)((x+y)^2-1)=0[-20 20 -10 10]}
This graph includes the alignment of the points of inflexion along

x + y = 0, the limits `x + y = +-1` for the graph and

the POI `+-(-kpi, kpi), k = 0, 1.`
Graph of `y = (sin^(-1))_(fcs)( x; 1 )`, conventional `y in [-pi/2, pi/2]`
graph{x-sin y+y=0[-3.14 3.14 -1.57 1.57]}

The graphs are on uniform scale.