# FCS  y = tan_(fcs)( x; -1 ) = tan ( x -tan(x- tan ( x - .... How do you prove that, piecewise, at y = 1, the tangent is inclined at 33^o 41' 24''  to the x-axis, nearly?

Jul 3, 2018

See explanation and the wholesome Socratic graph.

#### Explanation:

The FCS generator is $y = \tan \left(x - y\right)$. Inversely,

$x = y + k \pi + {\tan}^{- 1} y , k = 0 , \pm 1 , \pm 2 , \pm 3 , . .$.

At $y = 1 , x = k \pi + \frac{\pi}{4} + 1 , k = 0 , \pm 1 , \pm 2 , \pm 3 , . . = 1.76539 , \ldots .$.

(dx)/(dy) = 1 + 1/( 1 + y^2 )

y' = 1/(dx)/(dy). And so,

$y ' = \frac{1 + {y}^{2}}{2 + {y}^{2}} = 1 - \frac{1}{2 + {y}^{2}} \in \left[\frac{1}{2} , 1\right) .$

$= \frac{2}{3}$, at y = 1, and so,

#the inclination of he tangent to the x-axis is

${\tan}^{- 1} \left(\frac{2}{3}\right) = {33.69}^{o} = {33}^{o} 41 ' 24 ' '$, nearly. See graph for this

tangent $\left(y - 1\right) = \frac{2}{3} \left(x - 1.7654\right)$
graph{(y-tan(x-y))((x-1.7654)^2+(y-1)^2-.01)((y-1) -2/3 (x-1.7654 ))=0[-10 6 -3 3]}