# 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)?

Jul 8, 2018

The ( tangent-crossing-curve ) points of inflexion are
$\pm \left(- k \pi , k \pi\right) , k = 0 , 1 , 2 , 3 , \ldots$ 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 ${\left(\sin\right)}^{- 1} x$.

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

$= {\left(\sin\right)}^{- 1} \left(x + \left({\sin}^{- 1}\right) \left(x + \left({\sin}^{- 1}\right) \left(x + \ldots\right)\right)\right)$

$= {\left(\sin\right)}^{- 1} \left(x + y\right) , x + y \in \left[- 1 , 1\right]$.

$= k \pi + {\left(- 1\right)}^{k} {\sin}^{- 1} \left(x + y\right) , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$, in the

respective $y \in \left[k \pi - \frac{\pi}{2} , k \pi + \frac{\pi}{2}\right] .$

Inversely,

$x = {\left(- 1\right)}^{k} \sin \left(y - k \pi\right) - y , y \in \left[k \pi - \frac{\pi}{2} , k \pi + \frac{\pi}{2}\right]$

$= {\left(- 1\right)}^{k} \sin y \cos k \pi - y$

$= \sin y - y$.

It is easy to prove that

$\frac{{d}^{2} x}{\mathrm{dy}} ^ 2 = - \sin y = 0$ and

$\frac{{d}^{3} x}{{\mathrm{dy}}^{3}} = - \cos y \ne 0$, at $\pm \left(- k \pi , k \pi\right) , k = 0 , 1 , 2 , 3 , \ldots$.

So,

the points of inflexion (POI) are $\pm \left(- k \pi , k \pi\right) , k = 0 , 2 , 3. . .$

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 $x + y = \pm 1$ for the graph and

the POI $\pm \left(- k \pi , k \pi\right) , k = 0 , 1.$
Graph of y = (sin^(-1))_(fcs)( x; 1 )#, conventional $y \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$
graph{x-sin y+y=0[-3.14 3.14 -1.57 1.57]}

The graphs are on uniform scale.