# How do you graph y=arccos(x/3)?

Jul 15, 2016

Instead of y-explicit form y = arccos (x/3), use the inverse x-explicit form x = 3 cos y. See the graphs and the explanation

#### Explanation:

I reiterate that the graphs of locally bijective y = f(x) and its inverse

$x = {f}^{- 1} y$ are one and the same.

The inverse of $y = a r c \cos \left(\frac{x}{3}\right)$ is

$x = 3 \cos y .$

For that matter, the same $x = 3 \cos y$ is the inverse of piecewise

bijective

$y = 2 k \pi \pm {\cos}^{- 1} \left(\frac{x}{3}\right) , y \in \left[2 \left(k - 1\right) \pi , 2 k \pi\right] , k = 0 , 1 , 2 , 3. . .$

and $y \in \left[2 k \pi , 2 \left(k - 1\right) \pi\right] , k = 0 , - 1 , - 2 , - 3. . .$

This waveform twines about y-axis. The amplitude (max $\left\mid x \right\mid$)

is 3 units and the period ( from the inverse x = 3 cos y) is $2 \pi$,

Data for making one half wave tor $y \in \left[0. \pi\right]$:

$\left(x , y\right) : \left(3 , 0\right) \left(\frac{3 \sqrt{3}}{2} , \frac{\pi}{6}\right) \left(\frac{3}{\sqrt{2}} , \frac{\pi}{4}\right) \left(\frac{3}{2} , \frac{\pi}{3}\right) \left(0 , \frac{\pi}{2}\right)$

$\left(- \frac{3 \sqrt{3}}{2} , \frac{7 \pi}{6}\right) \left(- \frac{3}{\sqrt{2}} , \frac{3 \pi}{4}\right) \left(- \frac{3}{2} , \frac{4 \pi}{3}\right) \left(- 3 , \pi\right)$

The other half wave is obtained by twining this beyond, for y in

[pi, 2pi]], around y-axis.

Observe that abs x <= 1 (amplitude).

Graph of $y = {\cos}^{- 1} \left(\frac{x}{3}\right)$, conventionally limited $y \in \left[0 , \pi\right]$:
graph{x-3 cos y = 0 [-3.14 3.14 0 3.14]}

Extended graph for $y = 2 k \pi \pm {\cos}^{- 1} \left(\frac{x}{3}\right) \in \left(- 25 , 25\right)$:
graph{x-3 cos y = 0 [ -50 50 -25 25]}

The graphs are on uniform scale.