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

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Answer 1 · 2016-07-15T01:56:15

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

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.

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