/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Sketch the graph of the function... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 53

Sketch the graph of the function. (Include two full periods.) $$y=2-\sin \frac{2 \pi x}{3}$$

Short Answer

Expert verified
The graph of the function starts at point (0,2), then the next quarter of the period reaches its maximum, for the next quarter goes back to the middle amplitude, the third quarter reaches the lowest point and the final quarter brings it back up to the middle amplitude again. This pattern is repeated for another period to complete the sketch.

Step by step solution

01

Identify Basic Parameters

Start by identifying the basic parameters of the function: amplitude, wavelength, and phase shift. Here, the amplitude is given by 2, the wavelength is given by \( \frac{2\pi}{3} \), and there is no phase shift.
02

Setup the Axis

Set up the x and y-axes. The y-axis will range from -3 to 5 (i.e., -2 to 2 added to the amplitude 2), while the x-axis will depend on the period of the function. The period is \( \frac{2\pi}{3} \), so denote four critical points as 0, \( \frac{1}{4} \cdot \frac{2\pi}{3} \), \( \frac{1}{2} \cdot \frac{2\pi}{3} \), \( \frac{3}{4} \cdot \frac{2\pi}{3} \), and \( \frac{2\pi}{3} \) to complete one period.
03

Draw the Graph

For the sine function, at x=0 the value will be at the middle amplitude, in this case at 2. The next quarter of the period, the function reaches its maximum. For the next quarter, it goes back to the middle amplitude. The third quarter reaches the lowest point and the final quarter brings it back up to the middle amplitude again. Repeat this pattern for another period to complete the sketch.
04

Check the Graph

Check whether the drawn graph is consistent with the actual function values for a few points. Pay attention to hard-to-calculate points since sine function values are known for 0, \( \frac{1}{2}\pi \), \(\pi \), \( \frac{3}{2}\pi \), and \(2\pi \).

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Key Concepts

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

Graphing Trigonometric Functions
When working with trigonometric functions like sine, cosine, or tangent, graphing can provide a concrete visual representation of the cyclic patterns that these functions depict. Sine functions, in particular, have a wave-like form and repeat over intervals, showing periodic behavior. Visualizing this periodicity on a graph requires attention to specific parameters, such as the amplitude, period, and phase shift.

To graph a sine function, one begins by marking an x-axis (representing the angle or time) and a y-axis (representing the value of the sine function). You then identify several key points determined by the function's parameters, which guide you in sketching the smooth, wave-like form of the sine function. In the case of our exercise, for instance, the sine wave's critical points are marked at regular intervals along the x-axis, based on the function's period, dividing the wave into equal quarters. These aspects lead us to a detailed visualization of trigonometric functions.
Amplitude of Sine Function
The amplitude of a sine function causes the peak and trough of the wave to reach certain heights away from the middle line, or equilibrium. Technically, it's the measure of the wave's range from its highest point to the middle line of the function. With a standard sine function of the form y = A sin(Bx - C) + D, the amplitude is the absolute value of A.

For the function presented in the exercise, 测=2-蝉颈苍(2蟺虫/3), we interpret the amplitude indirectly. Rather than being a straightforward multiplier of the sine value, we start with an initial vertical shift of 2 units upward due to the '-sin' portion, which reflects the graph across the horizontal axis. In effect, it seems like the amplitude is 1 because the '-sin' term negates the height, but it is customary to speak about amplitude as a positive value. Thus, the amplitude is still 1, but the graph is inverted and shifted.
Period of Sine Function
The period of a sine function is the length it takes to complete one full cycle of the waveform, after which the pattern repeats itself. In a standard sine function y = A sin(Bx - C) + D, the period is 2蟺/叠. This understanding is crucial for graphing the sine function as it helps in spacing out the points correctly on the x-axis.

Taking our exercise as a reference, the function 测=2-蝉颈苍(2蟺虫/3) has a period calculated by 2蟺/(2蟺/3) = 3. This means it will complete one full cycle every 3 units along the x-axis. These cycles need to be accurately represented on the graph to allow for an accurate depiction of the function's behavior over time.
Phase Shift in Trigonometry
Phase shift in trigonometry refers to a horizontal shifting of the graph of a trigonometric function along the x-axis. It's caused by the C in the general formula y = A sin(Bx - C) + D, where the graph shifts to the right if C is positive and to the left if C is negative, by an amount of C/B units.

In the scenario of our exercise's function, 测=2-蝉颈苍(2蟺虫/3), there is no C term in the argument of the sine function, which means there is no phase shift here. However, understanding the concept of phase shift is important as it determines how much your sine curve needs to move horizontally from the standard position to represent the function accurately on the graph. Lack of phase shift means we start graphing from the origin, which greatly simplifies plotting the function.

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Most popular questions from this chapter

Define the inverse secant function by restricting the domain of the secant function to the intervals \([0, \pi / 2)\) and \((\pi / 2, \pi],\) and sketch the graph of the inverse trigonometric function.

Consider the functions \(f(x)=\sin x\) and \(f^{-1}(x)=\arcsin x.\) (a) Use a graphing utility to graph the composite functions \(f \circ f^{-1}\) and \(f^{-1} \circ f.\) (b) Explain why the graphs in part (a) are not the graph of the line \(y=x\). Why do the graphs of \(f \circ f^{-1}\) and \(f^{-1} \circ f\) differ?

Find two solutions of each equation. Give your answers in degrees \(\left(0^{\circ} \leq \theta<360^{\circ}\right)\) and in radians \((0 \leq \theta<2 \pi) .\) Do not use a calculator. (a) \(\sin \theta=\frac{\sqrt{3}}{2}\) (b) \(\sin \theta=-\frac{\sqrt{3}}{2}\)

For the simple harmonic motion described by the trigonometric function, find (a) the maximum displacement, (b) the frequency, (c) the value of \(d\) when \(t=5,\) and (d) the least positive value of \(t\) for which \(d=0 .\) Use a graphing utility to verify your results. $$d=\frac{1}{64} \sin 792 \pi t$$

A 20 -meter line is a tether for a helium-filled balloon. Because of a breeze, the line makes an angle of approximately \(85^{\circ}\) with the ground. (a) Draw a right triangle that gives a visual representation of the problem. Show the known quantities of the triangle and use a variable to indicate the height of the balloon. (b) Use a trigonometric function to write and solve an equation for the height of the balloon. (c) The breeze becomes stronger and the angle the line makes with the ground decreases. How does this affect the triangle you drew in part (a)? (d) Complete the table, which shows the heights (in meters) of the balloon for decreasing angle measures \(\theta\) $$\begin{array}{|l|l|l|l|l|} \hline \text { Angle, } \boldsymbol{\theta} & 80^{\circ} & 70^{\circ} & 60^{\circ} & 50^{\circ} \\ \hline \text { Height } & & & & \\ \hline \end{array}$$ $$\begin{array}{|l|l|l|l|l|} \hline \text { Angle, } \theta & 40^{\circ} & 30^{\circ} & 20^{\circ} & 10^{\circ} \\ \hline \text { Height } & & & & \\ \hline \end{array}$$ (e) As \(\theta\) approaches \(0^{\circ},\) how does this affect the height of the balloon? Draw a right triangle to explain your reasoning.
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