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

2 Answers
Aug 4, 2018

#" "#
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.

Aug 4, 2018

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]}