Question

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

Answer 1

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=arc cos (x/3)` is

`x=3 cos y.`

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

bijective

`y = 2kpi +- cos^(-1)(x/3), y in [2(k-1)pi, 2kpi], k = 0, 1, 2, 3...`

and `y in [2kpi, 2(k-1)pi], k = 0, -1, -2, -3...`

This waveform twines about y-axis. The amplitude (max `abs x`)

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

Data for making one half wave tor `y in [0. pi]`:

`(x, y): (3, 0) ((3sqrt 3)/2, pi/6) (3/sqrt 2, pi/4) ((3)/2, pi/3) (0, pi/2)`

`(-(3sqrt 3)/2, (7pi)/6) (-3/sqrt 2, (3pi)/4) (-3/2, (4pi)/3) (-3, pi )`

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)(x/3)`, conventionally limited `y in [0, pi]`:
graph{x-3 cos y = 0 [-3.14 3.14 0 3.14]}

Extended graph for `y = 2kpi +- cos^(-1)(x/3) in (- 25, 25)`:
graph{x-3 cos y = 0 [ -50 50 -25 25]}

The graphs are on uniform scale.