Problem 42 - Precalculus with Calculus Previews

Question

Find the period and the vertical asymptotes of the given function. Sketch at least one cycle of the graph. $$ y=\csc (4 x+\pi) $$

Step-by-Step Solution

Answer

Period: \( \frac{\pi}{2} \); Vertical asymptotes at \( x = \frac{(n-1)\pi}{4} \), where \( n \) is an integer.

Step 1: Understand the Cosecant Function

The function given is related to the sine function, since it is the cosecant function, which is defined as the reciprocal of the sine function. So, we have \( y = \csc(4x + \pi) = \frac{1}{\sin(4x + \pi)} \). This means that the function is undefined where sine is zero.

Step 2: Determine Where Sine is Zero

Determine the values of \( x \) where \( \sin(4x+\pi) = 0 \). This occurs when \( 4x + \pi = n\pi \) where \( n \) is an integer. Solving for \( x \), we get \( 4x = n\pi - \pi \) thus \( 4x = (n-1)\pi \) and \( x = \frac{(n-1)\pi}{4} \). These are the vertical asymptotes.

Step 3: Find the Period

The standard period of \( \csc(x) \) is \( 2\pi \). For \( \csc(4x+\pi) \), the input angle is scaled by a factor of 4, so the period becomes \( \frac{2\pi}{4} = \frac{\pi}{2} \). This is due to the horizontal stretch/compression.

Step 4: Sketching One Cycle of the Graph

To sketch one cycle, note the vertical asymptotes found in Step 2 and the period. Start with a vertical asymptote at \( x = \frac{-\pi}{4} \), then every \( \frac{\pi}{2} \) after this (so also at \( x = \frac{\pi}{4} \)). Between these asymptotes, the graph of \( \csc(4x+\pi) \) completes one cycle, with undefined values at the asymptotes and peaks/troughs between them.

Key Concepts

Cosecant Function

The cosecant function, denoted as \( \csc(x) \), is one of the fundamental trigonometric functions closely tied to its more familiar counterpart, the sine function. Specifically, it is the reciprocal of the sine function:

• \( \csc(x) = \frac{1}{\sin(x)} \)

This means that wherever the sine function equals zero, the cosecant function becomes undefined because division by zero is not possible in mathematics.
As we explore this function, it's essential to recognize that the graph of \( \csc(x) \) features discontinuities—those points where the function is not defined. These discontinuities are graphically represented as vertical asymptotes, lines that the graph approaches but never crosses.

Vertical Asymptotes

Vertical asymptotes are essential features in the graphing of certain functions, such as the cosecant function. They occur where the function is undefined. For \( y = \csc(4x + \pi) \), these asymptotes correspond to the zeros of \( \sin(4x + \pi) \).
To find them, solve the equation for when the sine equals zero. In this case, set \( 4x + \pi = n\pi \) where \( n \) is an integer.

• This simplifies to \( x = \frac{(n-1)\pi}{4} \)

Thus, vertical asymptotes for our specific function appear at these values of \( x \).
On the graph, these asymptotes show up as vertical lines where the graph of \( y = \csc(4x + \pi) \) reaches positively or negatively toward infinity.

Graphing Trigonometric Functions

Graphing trigonometric functions like \( y = \csc(4x + \pi) \) involves understanding the behaviors and patterns of these functions:

• Identify the vertical asymptotes where the function becomes infinite.
• Recognize the 'shape' or cycle of the graph between asymptotes.

In cosecant graphs, regions between asymptotes form what looks like segments of a hyperbola. From one vertical asymptote to the next, the graph will show an upward or downward curve depending on the defined points of the cycle.
To sketch one full cycle, first draw the vertical asymptotes as solid or dashed vertical lines, marking where the function is undefined. Then, show the upward and downward curves between these lines to complete the graph.

Period of Trigonometric Functions

The period of a trigonometric function refers to the length of one complete cycle before the function begins to repeat itself. For the basic \( \csc(x) \) function, one period is \( 2\pi \).
For scalar-modified functions like \( \csc(4x + \pi) \), adjustments in the period arise from the internal argument transformation. Here, the factor 4 compresses the period:

• The new period becomes: \( \frac{2\pi}{4} = \frac{\pi}{2} \)

Understanding the period helps in effective graph sketching, aiding in the placement of asymptotes, peaks, and troughs inherently dictated by this repeating cycle structure. Such insights ensure accurate representation of the function's behavior across the graph.