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Chapter 10: Problem 5

Fill in the following table for each finction and plot the graph from these points. $$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|c}x & -\frac{\pi}{2} & -\frac{\pi}{3} & -\frac{\pi}{4} & -\frac{\pi}{6} & 0 & \frac{\pi}{6} & \frac{\pi}{4} & \frac{\pi}{3} & \frac{\pi}{2} & \frac{2 \pi}{3} & \frac{3 \pi}{4} & \frac{5 \pi}{6} & \pi \\\\\hline y & & & & & & & & & & & & &\end{array}$$ $$y=\sec x$$

Short Answer

Expert verified
The table is filled and the graph is plotted with asymptotes at \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \).

Step by step solution

01

Understanding the Function

The function is given as \( y = \sec x \). The secant function is defined as \( \sec x = \frac{1}{\cos x} \). We need to find the secant of various angles listed as x-values and fill them into a table.
02

Calculating \( y \) for each \( x \)-value

We'll calculate \( y \) for each \( x \) value provided:1. \( x = -\frac{\pi}{2} \): \( \cos(-\frac{\pi}{2}) = 0 \), \( \sec(-\frac{\pi}{2}) = \frac{1}{0} = undefined \)2. \( x = -\frac{\pi}{3} \): \( \cos(-\frac{\pi}{3}) = \frac{1}{2} \), \( \sec(-\frac{\pi}{3}) = 2 \)3. \( x = -\frac{\pi}{4} \): \( \cos(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \), \( \sec(-\frac{\pi}{4}) = \sqrt{2} \)4. \( x = -\frac{\pi}{6} \): \( \cos(-\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \), \( \sec(-\frac{\pi}{6}) \approx 1.15 \)5. \( x = 0 \): \( \cos(0) = 1 \), \( \sec(0) = 1 \)6. \( x = \frac{\pi}{6} \): \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \), \( \sec(\frac{\pi}{6}) \approx 1.15 \)7. \( x = \frac{\pi}{4} \): \( \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \), \( \sec(\frac{\pi}{4}) = \sqrt{2} \)8. \( x = \frac{\pi}{3} \): \( \cos(\frac{\pi}{3}) = \frac{1}{2} \), \( \sec(\frac{\pi}{3}) = 2 \)9. \( x = \frac{\pi}{2} \): \( \cos(\frac{\pi}{2}) = 0 \), \( \sec(\frac{\pi}{2}) = \frac{1}{0} = undefined \)10. \( x = \frac{2\pi}{3} \): \( \cos(\frac{2\pi}{3}) = -\frac{1}{2} \), \( \sec(\frac{2\pi}{3}) = -2 \)11. \( x = \frac{3\pi}{4} \): \( \cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} \), \( \sec(\frac{3\pi}{4}) = -\sqrt{2} \)12. \( x = \frac{5\pi}{6} \): \( \cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} \), \( \sec(\frac{5\pi}{6}) \approx -1.15 \)13. \( x = \pi \): \( \cos(\pi) = -1 \), \( \sec(\pi) = -1 \)
03

Filling the Table

Let's fill in the table with the calculated \( y \) values:\[\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|c}x & -\frac{\pi}{2} & -\frac{\pi}{3} & -\frac{\pi}{4} & -\frac{\pi}{6} & 0 & \frac{\pi}{6} & \frac{\pi}{4} & \frac{\pi}{3} & \frac{\pi}{2} & \frac{2\pi}{3} & \frac{3\pi}{4} & \frac{5\pi}{6} & \pi \hliney & \text{undefined} & 2 & \sqrt{2} & \approx 1.15 & 1 & \approx 1.15 & \sqrt{2} & 2 & \text{undefined} & -2 & -\sqrt{2} & \approx -1.15 & -1 \\end{array}\]
04

Plotting the Graph

Using the table, plot the points on a graph where the x-axis represents \( x \), and the y-axis represents \( y = \sec x \). Note that at \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \), the graph will have vertical asymptotes since the secant function is undefined at these points. Plot the remaining points and connect them smoothly to show the overall trend of the secant function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Secant Function
The secant function, denoted as \( y = \sec x \), is a fundamental trigonometric function. It is defined as the reciprocal of the cosine function, meaning \( \sec x = \frac{1}{\cos x} \). This function gives the ratio of the length of the hypotenuse to the adjacent side in a right-angled triangle.
  • Secant is undefined where cosine is zero because division by zero is not allowed in mathematics.
  • This function is unique as it does not cycle through the same values, unlike sine or cosine, because it involves reciprocation.
  • It is periodic, repeating every \(2\pi\), which we'll explore later.
Understanding the secant function deepens comprehension of other trigonometric functions as they all interrelate.
Graphing Trigonometric Functions
Graphing trigonometric functions involves plotting them on a coordinate plane. The x-values typically represent angles in radians, while the y-values represent the function values.
To graph \( y = \sec x \), follow these steps:
  • First, compute the secant values for multiple x-values to visualize trends.
  • Use these computed values to plot points on the graph.
  • Pay attention to undefined points where cosine equals zero, as these result in vertical asymptotes.
  • Finally, smoothly connect the points to illustrate the periodic nature and overall shape of the secant function.
By following this process, you can interpret and understand the behavior and structure of trigonometric functions more intuitively.
Periodicity in Trigonometry
Periodicity refers to the repeating nature of trigonometric functions over regular intervals.For \( y = \sec x \), this interval, known as the period, is \(2\pi\). This means the behavior of the function, including values and asymptotes, repeats itself after each interval of \(2\pi\).
Essential points about periodicity include:
  • Patterns in the function's values help quickly predict future values based on known points.
  • The graph patterns, including peaks and asymptotes, recur every \(2\pi\), making it predictable.
  • Periodicity helps in solving equations and in applications such as signal processing and physics.
Understanding periodicity allows students to recognize these common patterns across different trigonometric functions.
Asymptotes in Trigonometry
In trigonometry, an asymptote is a line that a graph approaches but never touches. For the secant function, vertical asymptotes occur where \( \cos x = 0 \) because \( \sec x = \frac{1}{\cos x} \) becomes undefined.
These asymptotes demonstrate some important concepts:
  • Asymptotes occur at \( x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \ldots \) where each represents an undefined point on the secant graph.
  • The graph seems to "shoot off" to infinity as it nears these points, creating a vertical line that defines the function's non-existence there.
  • Understanding where asymptotes appear helps avoid undefined calculations and helps accurately sketch trigonometric graphs.
Grasping the concept of asymptotes is crucial when dealing with trigonometric functions, helping students navigate complex problems with confidence.

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Most popular questions from this chapter

Solve the given problems. Find the function and graph it for a function of the form \(y=a \sin x\) that passes through \((3 \pi / 2,-2)\)

In Exercises \(37-40\), use the trace feature of a calculator to find the value of a for either \(y=a \sin x\) or \(y=a \cos x,\) whichever is correct. such that the given point is on the graph. The amplitude of each function is \(2.50 .\) Thereby determine the function. (All points are located such that the \(x\) value is between \(-\pi\) and \(\pi .\) ) (0.67,-1.55)

Sketch the appropriate curves. A calculator may be used. The intensity \(I\) of an alarm (in \(\mathrm{d} B-\) decibel) signal is given by \(I=40+50 \sin t-20 \cos 2 t,\) where \(t\) is measured in seconds. Display two cycles of \(I\) as a function of \(t\) on a calculator.

Sketch the required curves. The angular displacement \(\theta\) of a motorboat propeller blade in terms of the initial displacement \(\theta_{0}\) is \(\theta=\theta_{0} \cos \omega t .\) If \(\omega=360 \mathrm{rad} / \mathrm{s}\) and \(\theta_{0}=0.10,\) draw two cycles for the resulting equation.

Sketch the graphs of the given functions. Check each using a calculator. (These are the same functions as in Exercises \(3-22 .)\) $$y=15 \cos \frac{\pi x}{10}$$
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