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Chapter 5: Problem 48

Determine the amplitude, period, and phase shift of each function. Then graph one period of the function. $$\text { 48. } y=\frac{1}{2} \cos (2 x+\pi)$$

Short Answer

Expert verified
The function \(y=\frac{1}{2} \cos (2 x+\pi)\) has an amplitude of \(1/2\), a period of \(\pi\), and a phase shift of \(\pi\) units to the right. In other words, the function varies between \(1/2\) and \(-1/2\) as it fluctuates along the x-axis starting from \(x=\pi\) to \(x=2\pi\).

Step by step solution

01

Identify the amplitude

The amplitude of the cosine function is the absolute value of the coefficient of the cosine term. Here, the coefficient of the \(\cos\) term is \(1/2\). So, the amplitude A of the function \(y=\frac{1}{2} \cos (2 x+\pi)\) is \(|1/2|=1/2\).
02

Identify the period

The period of the cosine function is found by dividing \(2\pi\) by the absolute value of the coefficient of \(x\). Here, the coefficient of \(x\) is \(2\). So, the period T of the function \(y=\frac{1}{2} \cos (2 x+\pi)\) is \(2\pi/|2|=\pi\).
03

Identify the phase shift

The phase shift of the cosine function is the value by which the function is shifted horizontally from the origin. It's the value of \(C\) in our general expression \(y=A \cos(B(x-C))+D\). Here, it's \(-\pi\) (considering it's present as \(2x+\pi\) in the function), so the cosine function is shifted to the right by \(\pi\).
04

Graph the function

To graph the function \(y=\frac{1}{2} \cos (2 x+\pi)\), plot a point at the phase shift, \(C=\pi\). Then, move \(\pi\) to the right to complete a full cycle of the cosine function, keeping in mind that the maximum height of the curve is \(1/2\) and the minimum height is \(-1/2\). The curve fluctuates between these two heights as it moves rightward from \(x=\pi\), completing its first peak at \(\pi + \pi/4\), reaching back to zero at \(\pi + 2\pi/4\), achieving its lowest point at \(\pi + 3\pi/4\) and finally culminating back at \(1/2\) at \(x=2\pi\).

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Key Concepts

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

Understanding Amplitude
The amplitude of a trigonometric function like cosine or sine represents the height of its wave, defining how far the peaks and valleys of the graph extend from the middle line. In mathematical terms, the amplitude is the absolute value of the coefficient in front of the cosine or sine term.
For the function given, \(y=\frac{1}{2} \cos (2 x+\pi)\), the coefficient is \(\frac{1}{2}\). Therefore, the amplitude is calculated as \(|\frac{1}{2}|=\frac{1}{2}\).
  • The peaks of the function will reach a height of \(\frac{1}{2}\).
  • The valleys will descend to \(-\frac{1}{2}\).

Understanding amplitude helps to visually anticipate the wave's maximum and minimum from its center or horizontal axis, thus shaping how tall the graph appears.
Decoding the Period
The period of a trigonometric function determines the horizontal distance it takes to complete one full cycle. This is where the function returns to its starting point and begins repeating. The period is calculated using the formula \(\frac{2\pi}{|B|}\) where \(B\) is the coefficient of \(x\). Here, for the function \(y=\frac{1}{2} \cos (2x+\pi)\), \(B=2\). Therefore, the period is \(\frac{2\pi}{2}=\pi\).
  • The cycle starts repeating after \(\pi\) units along the x-axis.
  • This concept is key for graphing these functions, as it tells us where to expect each repetitive feature.

So, with a period of \(\pi\), this function's complete wave cycle fits twice in the "normal" \(2\pi\) span on the standard trigonometric circle.
Exploring Phase Shift
Phase shift indicates the horizontal translation of the graph of a trigonometric function. This tells how far the graph is shifted left or right from its original position. The phase shift is calculated from the expression \(B(x-C)\) and is \(-C/B\) where \(C\) is the constant term inside the parenthesis and it's added or subtracted from zero. For this function \(y=\frac{1}{2} \cos (2x+\pi)\), it's represented as \(2(x + \pi/2)\), so the phase shift is \(-\pi/2\).
  • The graph is shifted to the right by \(\pi/2\) units.
  • On the graph, you'd start plotting from \(\pi/2\) instead of zero.

Recognizing phase shift is crucial during graphing as it dictates the starting point of the trigonometric cycle on the graph.

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Most popular questions from this chapter

a. Graph the restricted secant function, \(y=\sec x,\) by restricting \(x\) to the intervals \(\left[0, \frac{\pi}{2}\right)\) and \(\left(\frac{\pi}{2}, \pi\right]\) b. Use the horizontal line test to explain why the restricted secant function has an inverse function. c. Use the graph of the restricted secant function to graph \(y=\sec ^{-1} x\).

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Determine the range of each of the following functions Then give a viewing rectangle, or window, that shows two periods of the function's graph. a. $$f(x)=3 \sin \left(x+\frac{\pi}{6}\right)-2$$ b. $$g(x)=\sin 3\left(x+\frac{\pi}{6}\right)-2$$

Your neighborhood movie theater has a 25 -foot-high screen located 8 feet above your eye level. If you sit too close to the screen, your viewing angle is too small, resulting in a distorted picture. By contrast, if you sit too far back, the image is quite small, diminishing the movie's visual impact. If you sit \(x\) feet back from the screen, your viewing angle, \(\theta,\) is given by $$\theta=\tan ^{-1} \frac{33}{x}-\tan ^{-1} \frac{8}{x}$$ (GRAPH CANNOT COPY) Find the viewing angle, in radians, at distances of 5 feet, 10 feet, 15 feet, 20 feet, and 25 feet.

Describe the restriction on the cosine function so that it has an inverse function.

Use a sketch to find the exact value of each expression. $$ \sin \left[\tan ^{-1}\left(-\frac{3}{4}\right)\right] $$
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