Problem 63 - Algebra 2

Question

Graph each function in the interval from 0 to 2\(\pi .\) Describe any phase shift and vertical shift in the graph. $$ y=\csc 2 \theta-1 $$

Step-by-Step Solution

Answer

The graph of the function \( y=\csc 2 \theta-1 \) in the interval from 0 to 2\( \pi \) will look like a cosecant function with a period of \( \pi \) (completed in half the usual interval) and a downward shift of 1 unit.

Step 1: Understanding the Cosecant Function

The cosecant function, denoted as \( \csc \), is the reciprocal of the sine function. It means \( \csc \theta = 1/\sin \theta \). It has asymptotes (undefined values) where the sin function equals zero. The range of cosecant function is \( \csc \theta > 1 \) or \( \csc \theta < -1 \). Hence to plot it, first plot a sine function, then find the reciprocal at each given point to find the value of cosecant function.

Step 2: Identify the Phase Shift and Period

The general form of the sine and cosine functions include a phase shift. The general form is \( y=A \sin(Bx-C) \), where \( A \) is the amplitude, \( B \) determines the period, and \( C \) is the phase shift. In the given function \( y=\csc 2 \theta-1 \), there is no phase shift (value under parenthesis moves the graph left or right), but the period is changed due to the '2' in front of \( \theta \). The period is given by \( 2 \pi / B = \pi \), hence the graph will complete one full cycle in the interval [0, \( \pi \)].

Step 3: Identify the Vertical Shift

The '-1' at the end of the function indicates that there is a vertical shift downward by 1 unit. This means every point on the graph will be shifted down by 1 unit.

Step 4: Plot the Graph

To plot the graph correctly, we need to keep in mind the identified asymptotes, period, and shift. Begin by plotting a normal sine wave, label the asymptotes (where the sine function is 0), then plot the cosecant function, remembering to flip the sine wave over the x-axis. Lastly incorporate the vertical shift, moving all points down by 1 unit.

Key Concepts

Vertical Shift

When we talk about vertical shifts in a trigonometric function, we're essentially moving the entire graph up or down on the coordinate plane. For the function \( y = \csc 2\theta - 1 \), the "-1" refers to a vertical shift. This specific shift means that every point on the graph of \( \csc 2\theta \) is moved one unit downward.

• All peaks move down by 1.
• All troughs move down by 1.
• The asymptotes, points where the cosecant is undefined, remain in the same vertical lines, yet seen from a slightly lower starting point.

In simpler terms, if you were to draw the graph of \( \csc 2\theta \) normally and then just take the whole thing and shift it one unit down everywhere, you've executed a vertical shift. Remember, this operation doesn't change the shape, just the position of the graph on the y-axis.

Phase Shift

A phase shift in a trigonometric function refers to a horizontal shift along the x-axis. It dictates how the graph moves left or right. In the standard form \( y = A \sin(Bx - C) \), \( C \) is responsible for the phase shift. However, in our given function \( y = \csc 2\theta - 1 \), there is no parentheses attached to \( \theta \), meaning \( C = 0 \).
This tells us there is no phase shift. The graph of \( \csc 2\theta \) remains aligned in its usual start and end at the interval from 0 to \( 2\pi \) without horizontal movement. This means we don't slide the graph left or right at all, and the graph effectively adheres to its normal course from beginning to end.

Trigonometric Functions

Trigonometric functions are the building blocks for analyzing periodic phenomena. In this exercise, we focus on the cosecant function. Cosecant is the reciprocal of the sine function, expressed as \( \csc \theta = 1/\sin \theta \).
Here's what you should know about it:

• The range is \(\csc \theta > 1 \) or \(\csc \theta < -1\).
• It features vertical asymptotes wherever the sine function is zero because you can't divide by zero in math!
• The function graphed looks like arcs that open upwards at the peaks of \( \sin \theta \) and downwards at the troughs of \( \sin \theta \).

Remember, to graph \( \csc \theta \), you'd plot a regular \( \sin \theta \) curve first, marking where sine equals zero, which tells you where the asymptotes occur. Then plot the inverted arcs opening opposite to the direction of the sine peaks and troughs.

Period of Trigonometric Functions

The period of a trigonometric function tells you how often the pattern of the graph repeats itself. For sine and cosine, the standard period is \( 2\pi \). However, when you have the factor \( B \) as in \( y = \csc(B\theta) \), the period transforms to \( 2\pi/B \).
In our case with \( y = \csc 2\theta \), \( B = 2 \) signifies the graph's period becomes \( \pi \). This means the graph completes one full cycle in the interval from 0 to \( \pi \), and another from \( \pi \) to \( 2\pi \). So essentially, it cycles twice as fast compared to its original form.
Consequently, the change in period reflects a more frequent repetition of the cosecant 'arc' structures along the x-axis, which can be observed in the shorter interval for each repetitive pattern.