Question

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

Answer 1

See explanation and graphs.

Explanation:

`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 ]`.