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Chapter 4: Problem 42

Sketch the graph of the function. (Include two full periods.) $$y=4 \cos x$$

Short Answer

Expert verified
To graph \(y = 4 \cos x\), one simply needs to note that the graph will oscillate between 4 and -4. The function reaches 4 at \(x = 0\) and \(x = 2\pi\), 0 at \(x = \pi/2\) and \(x = 3\pi/2\), and -4 at \(x = \pi\). Upon replicating these points, an appropriate graph of the function over two full periods can be sketched.

Step by step solution

01

Identification of the Amplitude, Period, Phase Shift and Vertical Shift

From the function \(y = 4 \cos x\), it's clear that the amplitude is 4 (the maximum and minimum value will be 4 and -4, respectively). The period is \(2\pi\) since there's no variable value in front of \(x\) that might change the period. There's no phase shift and no vertical shift since there's no addition or subtraction inside or outside the cosine function.
02

Creating a Full Period

Now that the characteristics have been identified, a full period can be drawn. The full period of the cosine function goes from 0 to \(2\pi\). At \(x = 0\), \(\cos 0 = 1\), but due to the amplitude, the function will start at 4, so plot (0,4). At \(x = \pi/2\), \(\cos \pi/2 = 0\), so plot \((\pi/2, 0)\). At \(x = \pi\), \(\cos \pi = -1\), but due to the amplitude, the function will be at -4, so plot \((\pi,-4)\). At \(x = 3\pi/2\), \(\cos 3\pi/2 = 0\), so plot \((3\pi/2, 0)\). Finally, at \(x = 2\pi\), \(\cos 2\pi = 1\), so, getting back to the beginning, plot \( (2\pi,4)\).
03

Creation of the Second Full Period

Creating another full period is straightforward, just repeat the process from Step 2. Mirror the points of the first period from \(2\pi\) to \(4\pi\). Plot \((2\pi + \pi/2, 0)\), \((2\pi + \pi, -4)\), \((2\pi + 3\pi/2, 0)\) and \((2\pi + 2\pi, 4)\). With these plotted points, it will be easier to draw the graph.

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Key Concepts

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

Cosine Function
The cosine function, often represented as \( \cos x \), is a fundamental trigonometric function. It's known for its wave-like pattern and is periodic in nature. This means it repeats its values in regular intervals.
The standard cosine function oscillates between -1 and 1, taking values along the y-axis for each x-value. It is symmetrical around the y-axis, making it an even function.
  • Starts at maximum value at \( x=0 \)
  • Dips to zero at \( x=\pi/2 \) and \( x=3\pi/2 \)
  • Reaches minimum at \( x=\pi \)
Understanding cosine鈥檚 behavior is key when graphing functions like \( y = 4 \cos x \). Here, it has been stretched vertically to a new amplitude, which affects its maximum and minimum values.
Amplitude
Amplitude describes the height of the wave measured from the equilibrium position to the peak. It's half the distance between the maximum and minimum values of the function.
In the function \( y = 4 \cos x \), the amplitude is 4.
  • Maximum value is 4
  • Minimum value is -4
Amplitude doesn't affect the horizontal position of the graph, only the vertical stretch or compression.
This results in a tall wave that reaches up to 4 and down to -4 instead of the usual 1 and -1 in a basic cosine function.
Period of Trigonometric Functions
The period is the length of one complete cycle of the wave. For cosine, this is the distance needed to return to a starting point.
For \( y = \cos x \), the default period is \( 2\pi \). Since there are no coefficients affecting \( x \), the period remains \( 2\pi \) in \( y = 4 \cos x \) as well.
  • The full cycle from peak to peak is \( 2\pi \)
  • Key points at \( 0, \pi/2, \pi, 3\pi/2, 2\pi \)
Period is crucial for accurately plotting the function over intervals. In this problem, two periods from 0 to \( 4\pi \) showcase how the wave repeats identically.
Phase Shift
A phase shift occurs when the graph of the function is moved horizontally. It's determined by inside additions or subtractions to the variable.
In \( y = 4 \cos x \), there is no phase shift because there's no horizontal translation term like \( \cos(x + c) \).
  • No changes in starting point
  • Graph begins at \( x=0 \)
Understanding phase shift helps recognize where the graph begins. Even when no phase shift is present, identifying this assures you plot the accurate starting position. Here, without a shift, the cycle starts directly from its peak at (0,4).

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