Problem 19
Question
The graphs of which two trigonometric functions have an asymptote at \(x=\frac{\pi}{2} ?\)
Step-by-Step Solution
Verified Answer
The tangent and secant functions have asymptotes at \(x=\frac{\pi}{2}\).
1Step 1: Understanding the Problem
We need to identify which trigonometric functions have an asymptote at \(x=\frac{\pi}{2}\). An asymptote is a line that the graph of a function approaches but never touches, indicating the function is undefined or infinite at those points.
2Step 2: Trigonometric Functions Overview
Focus on trigonometric functions that potentially have vertical asymptotes. The common trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant.
3Step 3: Examine Tangent Function
The tangent function, \(\tan(x)\), is defined as \(\frac{\sin(x)}{\cos(x)}\). The function is undefined where \(\cos(x) = 0\). \(\cos(x) = 0\) at \(x = \frac{\pi}{2} + k\pi\), where \(k\) is an integer. Hence, \(\tan(x)\) has an asymptote at \(x = \frac{\pi}{2}\).
4Step 4: Examine Secant Function
The secant function, \(\sec(x)\), is defined as \(\frac{1}{\cos(x)}\). It is also undefined where \(\cos(x) = 0\). Thus, \(\sec(x)\) has an asymptote at \(x = \frac{\pi}{2} + k\pi\), indicating an asymptote at \(x = \frac{\pi}{2}\).
5Step 5: Identify Functions with Asymptote
We identified that both \(\tan(x)\) and \(\sec(x)\) have an asymptote at \(x = \frac{\pi}{2}\), due to \(\cos(x)\) being zero at this point.
Key Concepts
Tangent Function AsymptotesSecant Function AsymptotesTrigonometric Functions
Tangent Function Asymptotes
The tangent function, written as \( \tan(x) \), is one of the six main trigonometric functions used in mathematics. This function exhibits a periodic pattern and has vertical asymptotes, which are lines that the graph approaches but never touches. These occur where the tangent function becomes undefined, specifically where \( \cos(x) = 0 \).
Vertical asymptotes in \( \tan(x) \) occur at \( x = \frac{\pi}{2} + k\pi \) for any integer \( k \). This is because the tangent function is \( \tan(x) = \frac{\sin(x)}{\cos(x)} \), meaning it is undefined wherever the cosine of \( x \) is zero. At these points, \( \tan(x) \) tends towards infinity.
The behavior of the tangent function around its vertical asymptotes is crucial for sketching its graph and understanding how it behaves in different intervals. It rises sharply towards positive infinity on one side of the asymptote and falls steeply into negative infinity on the other.
Vertical asymptotes in \( \tan(x) \) occur at \( x = \frac{\pi}{2} + k\pi \) for any integer \( k \). This is because the tangent function is \( \tan(x) = \frac{\sin(x)}{\cos(x)} \), meaning it is undefined wherever the cosine of \( x \) is zero. At these points, \( \tan(x) \) tends towards infinity.
The behavior of the tangent function around its vertical asymptotes is crucial for sketching its graph and understanding how it behaves in different intervals. It rises sharply towards positive infinity on one side of the asymptote and falls steeply into negative infinity on the other.
Secant Function Asymptotes
Among trigonometric functions, the secant function, denoted as \( \sec(x) \), is closely linked to the cosine function. It is defined as the reciprocal of the cosine function: \( \sec(x) = \frac{1}{\cos(x)} \). Due to this relationship, the secant function has vertical asymptotes at the same values of \( x \) where \( \cos(x) = 0 \).
So, just like the tangent function, \( \sec(x) \) has vertical asymptotes where \( x = \frac{\pi}{2} + k\pi \). At these points, the value of \( \cos(x) \) is zero, making \( \sec(x) \) undefined and causing it to shoot off to infinity.
This means the graph of \( \sec(x) \) will display sharp discontinuities at these asymptotic points, moving towards positive or negative infinity. Such characteristics are vital in understanding how this function behaves over its period.
So, just like the tangent function, \( \sec(x) \) has vertical asymptotes where \( x = \frac{\pi}{2} + k\pi \). At these points, the value of \( \cos(x) \) is zero, making \( \sec(x) \) undefined and causing it to shoot off to infinity.
This means the graph of \( \sec(x) \) will display sharp discontinuities at these asymptotic points, moving towards positive or negative infinity. Such characteristics are vital in understanding how this function behaves over its period.
Trigonometric Functions
Trigonometric functions are vital concepts in the study of mathematics, particularly in fields involving waves, oscillations, and various engineering applications. They include six main functions: sine, cosine, tangent, cotangent, secant, and cosecant.
Each of these functions has unique properties and characteristics. While some maintain continuous graphs without interruptions, others like the tangent and secant functions, have vertical asymptotes. These asymptotes represent values of \( x \) where the function remains undefined, and the graph approaches infinity.
Recognizing these asymptotic behaviors is crucial for anyone studying trigonometric functions. It helps in sketching accurate graphs and understanding how the functions interact over their cycles. The periodic nature of these functions means patterns repeat after specific intervals, highlighted by these points of discontinuity within the cycles.
Each of these functions has unique properties and characteristics. While some maintain continuous graphs without interruptions, others like the tangent and secant functions, have vertical asymptotes. These asymptotes represent values of \( x \) where the function remains undefined, and the graph approaches infinity.
Recognizing these asymptotic behaviors is crucial for anyone studying trigonometric functions. It helps in sketching accurate graphs and understanding how the functions interact over their cycles. The periodic nature of these functions means patterns repeat after specific intervals, highlighted by these points of discontinuity within the cycles.
Other exercises in this chapter
Problem 18
Find the period of each function. \(y=\cos 0.75 x\)
View solution Problem 19
In \(15-26,\) find each exact value in radians, expressing each answer in terms of \(\pi\) \(y=\arcsin \left(-\frac{\sqrt{3}}{2}\right)\)
View solution Problem 19
Find the phase shift of each function. \(y=\cos \left(x+\frac{\pi}{2}\right)\)
View solution Problem 20
In \(15-26,\) find each exact value in radians, expressing each answer in terms of \(\pi\) \(y=\arccos \frac{1}{2}\)
View solution