Question

graph one period of each function in the space below the problem.

1. $y = 2\csc\theta + 1$
2. $y = -3\sec 4\theta$

Explanation:

Step1: Identify period of \(y=2\csc\theta +1\)

Period of \(\csc\theta\) is \(2\pi\). Vertical shift (1) and amplitude (2) don’t affect period. So period = \(2\pi\).

Step2: Identify period of \(y=-3\sec4\theta\)

Period of \(\sec\theta\) is \(2\pi\). For \(\sec(B\theta)\), period = \(\frac{2\pi}{|B|}\). Here \(B=4\), so period = \(\frac{2\pi}{4}=\frac{\pi}{2}\).

Answer:

1. Period of \(y=2\csc\theta +1\): \(2\pi\)
2. Period of \(y=-3\sec4\theta\): \(\frac{\pi}{2}\)

(Note: To graph one period:

1. For \(y=2\csc\theta +1\): Plot vertical asymptotes at \(\theta=k\pi\) (k integer), key points (e.g., \(\theta=\frac{\pi}{2}\), \(y=3\); \(\theta=\frac{3\pi}{2}\), \(y=-1\)) over \([0,2\pi]\) (excluding asymptotes).
2. For \(y=-3\sec4\theta\): Vertical asymptotes at \(4\theta=\frac{\pi}{2}+k\pi\) → \(\theta=\frac{\pi}{8}+\frac{k\pi}{4}\), key points (e.g., \(\theta=0\), \(y=-3\); \(\theta=\frac{\pi}{4}\), \(y=3\)) over \([0,\frac{\pi}{2}]\) (excluding asymptotes).)