How do you sketch one cycle of $y=-cotx$ ?

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Answer 1 · 2018-08-09T06:23:33

See explanation and graphs.

$y = - cot x = - cos x/sin x,$

$x ne$ ( asymptotic ) zeros of the denominator

$kpi, k = 0, +-1, +-2, +-3, ...$ .

The period = period of the reciprocal $( - tan x ) = pi$

= the space between consecutive asymptotes, $x = kpi$ .

The amplitude is $1/2 pi$ .

See graph, depicting all these aspects.
graph{(y sin x + cos x)( x +.0001y) ( x- pi +.0001y)=0[ 0 pi -pi/4 pi/4]}

Here, 1-cycle graph is precisely given by the inverse

$x = - arctan ( 1/y ), x in [ - pi/2, pi/2 ]$ .

graph{x + arctan(1/y)=0}.

This phenomenon is attributed to the constraint on the range of arctan values as $[ - pi/2, pi/2 ]$ .

How do you sketch one cycle of y=-cotxy=-cotx?