Question

How do you find the amplitude and period of `Arc Cot`?

Answer 1

`" "`
Amplitude of `color(red)(y=f(x)=tan^(-1)(x)=1`

Period of `color(red)(y=f(x)=tan^(-1)(x)=2pi`

Explanation:

`" "`
We are given the trigonometric inverse function:

`color(red)(y=f(x)=tan^(-1)(x)`

General form : `color(blue)(a(bx-c)+d`

In our problem,

`color(red)(a=1; b=1; c=0; d=0`

Amplitude:

`color(red)(Amplitude : |a|`

Hence, `color(blue)(|a|=1` ... Res.1

Period:

`color(red)(Period : (2pi)/(|b|`

Hence, `color(blue)((2pi)/|1|=2pi` ... Res.2

Hope it helps.

Answer 2

Period `pi`
Amplitude: `pi/2`

Explanation:

Seldom used is

`arctan x + arc cot x = pi/2`, on par with

`arcsinx + arccos x = pi/2`

Here

`y = arc cot x = pi/2 - arctan x in ( pi/2 - pi/2, pi/2 +pi/2 ) = ( 0, pi )`

This is an imposed constraint, oh the variation of y.

The period in y-sense is `pi`.

The axis is y = pi/2.

The amplitude is half-period, `pi/2`.

The asymptotes in x-directions are

`y = 0 and y = pi.` See graph, depicting all these aspects.
graph{(y -pi/2+arctan (x ))( y+0x)(y-pi/2+0x)(y-pi+0x)=0}

The wholesome graph from the inverse x = cot y:
graph{x sin y - cos y =0[ -20 20 -10 10]}