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

Aug 9, 2018

See explanation and graphs.

#### Explanation:

$y = - \cot x = - \cos \frac{x}{\sin} x ,$

$x \ne$ ( asymptotic ) zeros of the denominator

$k \pi , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$.

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

= the space between consecutive asymptotes, $x = k \pi$.

The amplitude is $\frac{1}{2} \pi$.

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

$x = - \arctan \left(\frac{1}{y}\right) , x \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.

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

This phenomenon is attributed to the constraint on the range of arctan values as $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.