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

##### 2 Answers
Aug 4, 2018

$\text{ }$
Amplitude of color(red)(y=f(x)=tan^(-1)(x)=1

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

#### Explanation:

$\text{ }$
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.

Aug 4, 2018

Period $\pi$
Amplitude: $\frac{\pi}{2}$

#### Explanation:

Seldom used is

$\arctan x + a r c \cot x = \frac{\pi}{2}$, on par with

$\arcsin x + \arccos x = \frac{\pi}{2}$

Here

$y = a r c \cot x = \frac{\pi}{2} - \arctan x \in \left(\frac{\pi}{2} - \frac{\pi}{2} , \frac{\pi}{2} + \frac{\pi}{2}\right) = \left(0 , \pi\right)$

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, $\frac{\pi}{2}$.

The asymptotes in x-directions are

$y = 0 \mathmr{and} y = \pi .$ 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]}