Chapter 1: Problem 66
For $$f(x)=A \sin \left(\frac{2 \pi}{B}(x-C)\right)+D$$, identify \(A, B, C,\) and \(D\) for the sine functions and sketch their graphs. $$y=\frac{1}{2} \sin (\pi x-\pi)+\frac{1}{2}$$
Short Answer
Expert verified
The parameters are: \( A = \frac{1}{2} \), \( B = 2 \), \( C = 1 \), \( D = \frac{1}{2} \).
Step by step solution
01
Identify Amplitude (A)
The given function is in the form of \( y = A \sin \left(\frac{2\pi}{B}(x - C)\right) + D \). Compare the given function \( y = \frac{1}{2} \sin (\pi x - \pi) + \frac{1}{2} \) to the general form. The amplitude \( A \) is the coefficient of \( \sin \), which is \( \frac{1}{2} \).
02
Determine Period (B)
Identify the period \( B \) by analyzing the coefficient inside the sine function. The expression \( \pi x - \pi \) can be rewritten as \( \pi(x - 1) \). Thus, \( \frac{2\pi}{B} = \pi \), leading to \( B = 2 \).
03
Shift the Function Horizontally (C)
The horizontal shift \( C \) is identified from the expression \((x-C)\). Since we rewrite \( \pi x - \pi \) as \( \pi(x - 1) \), we conclude that \( C = 1 \).
04
Compute the Vertical Shift (D)
The vertical shift \( D \) corresponds to the constant added outside the sine function. Here, it is \( \frac{1}{2} \).
05
Sketch the Graph
Using the identified values: Amplitude \( A = \frac{1}{2} \), Period \( B = 2 \), Horizontal Shift \( C = 1 \), and Vertical Shift \( D = \frac{1}{2} \), sketch the sine wave. Begin the wave at \( x = 1 \), with a maximum at \( y = 1 \) and a minimum at \( y = 0 \). Complete one full cycle from \( x = 1 \) to \( x = 3 \) since the period is 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Amplitude
The amplitude of a trigonometric function, such as a sine or cosine wave, tells us how tall the waves are. It represents the maximum displacement from the center or the "midline" of the wave. This is a crucial feature because it reflects the wave's strength or intensity.
In the general sinusoidal function, given by \( y = A \sin\left(\frac{2\pi}{B}(x - C)\right) + D \), the amplitude is represented by \( A \). To determine the amplitude, look directly at the coefficient of the sine function. In our specific example:
When you look at the graph, the amplitude directly affects the height of the wave rather than its width or position.
In the general sinusoidal function, given by \( y = A \sin\left(\frac{2\pi}{B}(x - C)\right) + D \), the amplitude is represented by \( A \). To determine the amplitude, look directly at the coefficient of the sine function. In our specific example:
- The given function is \( y = \frac{1}{2} \sin(\pi x - \pi) + \frac{1}{2} \).
- Here, \( A = \frac{1}{2} \).
When you look at the graph, the amplitude directly affects the height of the wave rather than its width or position.
Period
The period of a trigonometric function tells us how long it takes for the wave to repeat itself. In other words, it's the distance along the x-axis that covers one complete cycle of the wave.
For the general function \( y = A \sin\left(\frac{2\pi}{B}(x - C)\right) + D \), the period can be calculated using the formula \( \frac{2\pi}{B} \). In our example function, we have the expression inside the sine:
Understanding the period is essential because it helps in predicting the wave's characteristics over a certain interval.
For the general function \( y = A \sin\left(\frac{2\pi}{B}(x - C)\right) + D \), the period can be calculated using the formula \( \frac{2\pi}{B} \). In our example function, we have the expression inside the sine:
- The initial expression is \( \pi x - \pi \), which can be rewritten as \( \pi(x - 1) \).
- Here, you're solving for \( B \) by equating \( \frac{2\pi}{B} = \pi \).
- Solving gives \( B = 2 \).
Understanding the period is essential because it helps in predicting the wave's characteristics over a certain interval.
Phase Shift
The phase shift of a trigonometric function shows how the graph of the function is translated horizontally along the x-axis. It's essentially where the function starts its wave cycle.
In the general form \( y = A \sin\left(\frac{2\pi}{B}(x - C)\right) + D \), the term \( C \) determines the phase shift. The function shifts to the right if \( C > 0 \) and to the left if \( C < 0 \). Let's break it down for our example:
Recognizing and calculating phase shift is key to accurately sketching and interpreting trigonometric wave graphs.
In the general form \( y = A \sin\left(\frac{2\pi}{B}(x - C)\right) + D \), the term \( C \) determines the phase shift. The function shifts to the right if \( C > 0 \) and to the left if \( C < 0 \). Let's break it down for our example:
- The function component inside the sine is \( x - 1 \) after rewriting \( \pi x - \pi = \pi(x - 1) \).
- This indicates a phase shift of \( C = 1 \).
Recognizing and calculating phase shift is key to accurately sketching and interpreting trigonometric wave graphs.