Question

the trigonometric function ( y=csc x ) has period
( x=\frac{pi}{2}+2 n pi ) ( ( n ) is an integer)
( x=n pi ) ( ( n ) is an integer)
( x=\frac{pi}{2}+n pi ) ( ( n ) is an integer)
( x=\frac{3 pi}{2}+2 n pi ) ( ( n ) is an integer)
( x=(2 n + 1) pi ) ( ( n ) is an integer)
sketch a graph of this function on the interval ( (-pi, pi) ).

Explanation:

Step1: Recall the period formula for \(y = \csc x\)

The general form of the cosecant function is \(y = A\csc(Bx - C)+D\). For \(y=\csc x\), \(B = 1\). The period of \(y=\csc x\) is given by the formula \(T=\frac{2\pi}{|B|}\). Substituting \(B = 1\) into the formula, we get \(T = 2\pi\).

Step2: Recall the asymptotes of \(y=\csc x\)

We know that \(\csc x=\frac{1}{\sin x}\). The function \(y = \sin x\) has zeros at \(x=n\pi\), \(n\in\mathbb{Z}\). Since \(\csc x\) is undefined when \(\sin x=0\), the vertical asymptotes of \(y=\csc x\) occur at \(x = n\pi\), \(n\in\mathbb{Z}\).

Answer:

The period is \(2\pi\) and the asymptotes are \(x = n\pi\) ( \(n\) is an integer)