How do you graph $y = csc (π - x)$ $y=csc(pi-x)$ and include two full periods?

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Answer 1 · 2018-07-28T15:40:03

See graph and details.

$y = csc (π - x) = \frac{1}{sin (π - x)} = \frac{1}{sin x} \notin (- 1, 1)$ $y = csc (pi - x ) = 1/sin ( pi - x ) = 1/sin x notin ( - 1, 1 )$

And, asymptotic x = any zero of the denominator $sin x$ $sin x$

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

The period = period of $sin x = 2 pi$ .

Graph for two periods,

$x in ( - 2 pi, 2 pi )$ , with asymptotes $x = +- pi$ and

y not-to-be-in $( -1, 1 )$ :
graph{(y sin x - 1)(y^2-1)(x)(x^2-(pi)^2)(x^2-4(pi)^2)=0[-10 10 -5 5]}

How do you graph y=csc(pi-x)y=csc(π−x)y=csc(pi-x) and include two full periods?