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Ch. 4 - Graphs of the Circular Functions
Lial - Trigonometry 12th Edition
Lial12th EditionTrigonometryISBN: 9780136552161Not the one you use?Change textbook
All textbooksLial 12th EditionCh. 4 - Graphs of the Circular FunctionsProblem 4.24
Chapter 5, Problem 4.24

Graph each function over a one-period interval.
y = ½ sec x

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Identify the basic function: The function given is \( y = \frac{1}{2} \sec x \). The secant function, \( \sec x \), is the reciprocal of the cosine function, \( \cos x \).
Determine the period of the function: The period of \( \sec x \) is the same as \( \cos x \), which is \( 2\pi \). Therefore, the period of \( y = \frac{1}{2} \sec x \) is also \( 2\pi \).
Identify the vertical asymptotes: The secant function has vertical asymptotes where the cosine function is zero. For \( \cos x = 0 \), \( x = \frac{\pi}{2} + k\pi \), where \( k \) is an integer. Within one period \( [0, 2\pi] \), the vertical asymptotes are at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).
Determine the amplitude and transformations: The coefficient \( \frac{1}{2} \) affects the vertical stretch of the secant function. This means the graph of \( y = \frac{1}{2} \sec x \) will be half as tall as the standard \( \sec x \) graph.
Sketch the graph: Plot the vertical asymptotes at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \). Between these asymptotes, sketch the secant curve, which will have a minimum at \( x = 0 \) and a maximum at \( x = \pi \), with the curve approaching the asymptotes as \( x \) approaches \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Secant Function

The secant function, denoted as sec(x), is the reciprocal of the cosine function. It is defined as sec(x) = 1/cos(x). The secant function has a period of 2π, meaning it repeats its values every 2π units. It is important to note that sec(x) is undefined wherever cos(x) equals zero, which occurs at odd multiples of π/2.
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Graphs of Secant and Cosecant Functions

Graphing Trigonometric Functions

Graphing trigonometric functions involves plotting the values of the function over a specified interval. For the secant function, the graph will exhibit vertical asymptotes at points where the cosine function is zero. Understanding the behavior of the function, including its amplitude and period, is crucial for accurately representing it on a graph.
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Introduction to Trigonometric Functions

Amplitude and Vertical Stretch

Amplitude refers to the height of the wave from its midline to its peak. In the function y = ½ sec x, the coefficient ½ indicates a vertical stretch, meaning the graph of sec x is scaled down by a factor of 2. This affects the range of the function, making it oscillate between -½ and ½, while still maintaining the same periodicity.
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Related Practice
Textbook Question

An object in simple harmonic motion has position function s(t), in inches, from an equilibrium point, as follows, where t is time in seconds.

𝒮(t) = 5 cos 2t

What is the frequency?

Textbook Question

Graph each function over a two-period interval. Give the period and amplitude. See Examples 2–5.

y = -2 sin 2 πx

Textbook Question

Graph each function over the interval [-2π, 2π]. Give the amplitude. See Example 1.

y = -2 sin x

Textbook Question

Graph each function over a one-period interval.

y = 1 - (1/2) csc (x - 3π/4)

1
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Textbook Question

Graph each function over a one-period interval.

y = 2 + 3 sec (2x - π)

Textbook Question

Graph each function over a two-period interval. Give the period and amplitude. See Examples 2–5.

y = ½ cos π x

2