Which Graph or Graphs Appear to Show a Sinusoid

Which Graph or Graphs Appear to Show a Sinusoid.

two.2: Graphs of Sinusoidal Functions

  • Page ID
    7105
    • Grand Valley State University
      via
      ScholarWorks @Grand Valley Country University

    Focus Questions

    The following questions are meant to guide our study of the material in this department. After studying this section, nosotros should empathize the concepts mo- tivated past these questions and exist able to write precise, coherent answers to these questions.

    Allow \(A, B, C\), and \(D\) be constants with \(B > 0\) and consider the graph of \(f(t) = A\sin(B(t – C)) + D\) or \(f(t) = A\cos(B(t – C)) + D\).

    • How does the value of \(A\) affect the graph of \(f\)? How is A related to the amplitude of \(f\)?
    • How does the value of \(B\) bear on the graph of \(f\)? How is B related to the period of \(f\)?
    • How does the value of \(C\) affect the graph of \(f\)?
    • How does the value of \(D\) bear upon the graph of \(f\)?
    • What does a phase shift exercise to a sine or cosine graph? How exercise we recognize a stage shift from the equation of the sinusoid?
    • How do we accurately draw the graph of \(f(t) = A\sin(B(t – C)) + D\) or \(f(t) = A\cos(B(t – C)) + D\) without a figurer and how do nosotros correctly describe the effects of the constants \(A, B, C\), and \(D\) on the graph?

    Starting time Activity

    In this department, nosotros will study the graphs of functions whose equations are \(f(t) = A\sin(B(t – C)) + D\) and \(f(t) = A\cos(B(t – C)) + D\) where \(A, B, C\), and \(D\) are real numbers. These functions are called
    sinusoidal functions
    and their graphs are chosen
    sinusoidal waves. Nosotros will beginning focus on functions whose equations are \(y = \sin(Bt)\) and \(y = \cos(Bt)\). Now consummate Part 1 or Part 2 of this starting time activity.

    Part i – Using a Geogebra Applet

    To begin our exploration, nosotros volition use a Geogebra applet chosen
    Period of a Sinusoid. The web address for this applet is http://gvsu.edu/s/LY

    Afterward you open the applet, notice that there is an input box at the superlative of the screen where you can input a function. For now, get out this gear up at \(g(t) = \sin(t)\). The graph of the sine function should be displayed. The slider at the top can exist used to change the value of B . When this is done, the graph of \(y = A\sin(t)\) will exist displayed for the current value of B along with the graph of \(y = \sin(t)\).

    1. Utilise the slider to modify the value of B. Explicate in detail the deviation between the graph of \(y = 1000(t) = \sin(t)\) and \(y = f(t) = \sin(Bt)\) for a abiding \(B > 0\). Pay close attending to the graphs and make up one’s mind the period when \[(a)B = 2 \space (b) B = iii\space (c) B = 4 \space(d) B = 0.5\]. In particular, how does the period of \(y = \sin(Bt)\) appear to depend on \(B\)? Note: Consider doing 2 carve up cases: one when \(B > 1\) and the other when \(0< B <1\).
    2. At present click on the reset button in the upper correct corner of the screen. This volition reset the value of the B to its initial setting of \(B = 1\).
    3. Change the function to \(thou(t) = \cos(t)\)and repeat part (1) for the cosine role. Does changing the value of B bear on the graph of \(y = \cos(Bt)\) in the aforementioned way that changing the value for B affects the graph of \(t = \sin(Bt)\)?

    Part 2 – Using a Graphing Utility

    Make sure your graphing utility is set to radian manner. Note: Most graphing utilities require the use of \(x\) (or \(X\) ) equally the independent variable (input) for a function. Nosotros will use \(10\) for the independent variable when we discuss the use of a graphing utility.

    1. We will first examine the graph of \(y = \sin(Bx)\) for three dissimilar values of \(B\). Graph the iii functions: \[y = \sin(10) \space\infinite y = \sin(2x) \space\space y = \sin(4x)\] using the post-obit settings for the viewing window: \(0 \leq x \leq four\pi\) and \(-1.5 \leq y \leq 1.5\). If possible on your graphing utility, set it and then that the tickmarks on the ten-axis are space at \(\dfrac{\pi}{two}\) units. Examine these graphs closely and determine menstruation for each sinusoidal wave. In item, how does the flow of \(y = \sin(Bx)\) appear to depend on \(B\)?
    2. Clear the graphics screen. Nosotros will now examine the graph of \(y = \sin(Bx)\) for three different values of \(B\). Graph the following three functions: \[y = \sin(x) \space\space y = \sin(\dfrac{1}{2}x) \space\infinite y = \sin(\dfrac{i}{4}10)\] using the following settings for the viewing window: \(0 \leq x \leq \pi\) and \(-1.5 \leq y \leq 1.5\) If possible on your graphing utility, set it so that the tickmarks on the x-centrality are spaced at \(\dfrac{\pi}{2}\) units. Examine these graphs closely and make up one’s mind catamenia for each sinusoidal wave. In detail, how does the period of \(y = \sin(Bx)\) appear to depend on B?
    3. How does the graph of \(y = \sin(Bx)\) announced to be related to the graph of \(y = \sin(x)\). Note: Consider doing two separate cases: one when \(B > 1\) and the other when \(0 < B < 1\).

    The Menses of a Sinusoid

    When nosotros hash out an expression such as \(\sin(t)\) or \(\cos(t)\), we oft refer to the expression within the parentheses as the argument of the role. In the beginning action, we examined situations in which the argument was \(Bt\) for some number \(B\). We also saw that this number affects the menstruum of the sinusoid. If nosotros examined graphs close enough, we saw that the period of \(y = \sin(Bt)\) and \(y = \cos(Bt)\) is equal to \(\dfrac{2\pi}{B}\). The graphs in Effigy 2.xi illustrate this.

    Notice that the graph of \(y = \sin(2t)\) has i complete cycle over the interval \([0, \pi]\) and so its flow is \(\pi = \dfrac{ii\pi}{2}\). The graph of \(y = \sin(4t)\) has one complete cycle over the interval \([0, \dfrac{\pi}{two}]\) and so its period is \(\dfrac{\pi}{2} = \dfrac{2\pi}{4}\). In these two cases, nosotros had \(B > ane \) in \(y = \sin(Bt)\). Practice we get the same result when \(0 < B < 1\)? Figure 2.12 shows graphs for \(y = \sin(\dfrac{one}{2}t)\) and \(y = \sin(\dfrac{1}{iv}t)\)?

    Find that the graph of \(y = \sin(\dfrac{1}{2}t)\) has one complete wheel over the interval \([0, 4\pi]\)

    2.11.png

    Figure \(\PageIndex{1}\): Graphs of \(y = \sin(2t)\) and \(y = \sin(4t)\). The graph of \(y = \sin(t)\) is besides shown (dashes).

    2.12.png

    Effigy \(\PageIndex{2}\): Graphs of \(y = \sin(\dfrac{1}{2}t)\) and \(y = \sin(\dfrac{i}{four}t)\). The graph of \(y = \sin(t)\) is also shown (dashes).

    The period of \(y = \sin(\dfrac{ane}{two}t)\) is \(4\pi = \dfrac{2\pi}{\dfrac{1}{2}}\).

    The graph of \(y = \sin(\dfrac{ane}{four}t)\) has one-half of a complete bike over the interval \([0, 4\pi]\) and then the menstruation of \(y = \sin(\dfrac{1}{4}t)\) is \(eight\pi = \dfrac{2\pi}{\dfrac{1}{4}}\).
    A expert question now is, “Why are the periods of \(y = \sin(Bt)\) and \(y = \cos(Bt)\) equal to \(\dfrac{2\pi}{B}\)?” The idea is that when we multiply the independent variable \(t\) past a constant \(B\), it can modify the input nosotros need to get a specific output. For example, the input of \(t = 0\) in \(y = \sin(t)\) and \(y = \sin(Bt)\) yield the same output. To complete one menses in \(y = \sin(Bt)\) nosotros need to get through interval of length \(2\pi\) so that our input is \(2\pi\). All the same, in club for the statement \((Bt)\) in \(y = \sin(Bt)\) to be \(two\pi\), nosotros need \(Bt = ii\pi\) and if we solve this for \(t\), we get \(t = \dfrac{2\pi}{B}\). So the function given by \(y = \sin(Bt)\) (or \(y = \cos(Bt)\)) will complete one complete bike when \(t\) varies from \(t = 0\) to \(t = \dfrac{2\pi}{B}\), and hence, the period is \(\dfrac{2\pi}{B}\). Notice that if we utilise \(y = A\sin(Bt)\) or \(y = A\cos(Bt)\), the value of A only affects the aamplitude of the sinusoid and does non affect the period.

    If \(A\) is a existent number and \(B\) is a positive real number, then the period of the functions given by \(y = A\sin(Bt)\)and \(y = A\cos(Bt)\) is \(\dfrac{ii\pi}{B}\).

    Practise \(\PageIndex{i}\)

    1. Determine the amplitude and period of the following sinusoidal functions.: \[(a) y = three\cos(\dfrac{1}{three}t)\] \[(b) y = -ii\sin(\dfrac{\pi}{ii}t)\]
    2. The graph below is a graph of a sinusoidal office. Make up one’s mind an equation for this role.

    pc2.png

    Reply

    1. (a) For \(y = 3\cos(\dfrac{i}{iii}t)\), the amplitude is \(3\) and the period is \(\dfrac{2\pi}{\dfrac{ane}{iii}} = vi\pi\)

    (b) For \(y = -two\sin(\dfrac{\pi}{two}t)\),the amplitude is \(2\) and the menstruum is \(\dfrac{ii\pi}{\dfrac{pi}{two}} = four\).

    ii. From the graph, the amplitude is \(2.v\) and the flow is \(2\). Using a cosine function, nosotros accept \(A = ii.v\) and \(\dfrac{2\pi}{B} = 2\). Solving for \(B\) gives \(B = \pi\). And then an equation is \(y = ii.v\cos(\pi t)\).

    Phase Shift

    Nosotros will at present investigate the effect of subtracting a abiding from the argument (independent variable) of a round function. That is, we volition investigate what effect the value of a real number \(C\) has the graph of \(y = \sin(t – C)\) and \(y = \cos(t – C)\).

    Activity 2.10 (The Graph of \(y = \sin(t – C)\))

    Complete Part 1 or Part 2 of this activity.

    Part ane – Using a Geogebra Applet

    Nosotros will utilize a Geogebra applet called
    Sinusoid – Phase Shift. The web address for this applet is http://gvsu.edu/south/Mu

    After you open up the applet, observe that in that location is an input box at the top of the screen where yous tin input a function. For now, leave this fix at \(one thousand(t) = \sin(t)\). The graph of the sine function should be displayed. The slider at the top can be used to modify the value of C. When this is done, the graph of \(y = A\sin(t – C)\) will be displayed for the current value of C along with the graph of \(y = \sin(t)\).

    1. Utilise the slider to alter the value of \(C\). Explain in detail the difference between the graph of \(y = g(t) = \sin(t)\) and \(y = f(t) = \sin(t – C)\) for a constant C. Pay close attention to the graphs and determine the horizontal shift when \[(a) C=ane\space (b) C=ii\space(c) C=3\infinite (d) C=-1\infinite (e) C=-2\space (f) C=-3.\] In item, describe the departure between the graph of \(y = \sin(t – C)\) and the graph of \(y = \sin(t)\)? Note: Consider doing two separate cases: one when \(C >0\) and the other when \(C < 0\).
    2. Now click on the reset button in the upper right corner of the screen. This will reset the value of the C to its initial setting of \(C = 0\).
    3. Change the function to \(g(t) = \cos(t)\) and repeat part (ane) for the cosine function. Does changing the value of C affect the graph of \(y = \cos(t – C)\) affect the sinusoidal moving ridge in the same way that changing the value for C affects the graph of \(y = \sin(t – C)\)?

    Part 2 – Using a Graphing Utility

    Brand certain your graphing utility is set to radian mode.

    1. We volition first examine the graph of \(y = \sin(t – C)\) for two different values of \(C\). Graph the three functions: \[y = \sin(ten) \infinite y = \sin(x – one) \infinite y = \sin(x – ii)\] using the following settings for the viewing window: \(0 \leq x \leq 4\pi\) and \(-ane.5 \leq y \leq one.5\). Examine these graphs closely and describe the difference between the graph of \(y = \sin(x – C)\) and the graph of \(y = \sin(10)\) for these values of C .
    2. Clear the graphics screen. We will now examine the graph of \(y = \sin(x – C)\) for two dissimilar values of C. Graph the following three functions: \[y = \sin(ten)\] \[y = \sin(x + i) = \sin(x-(-i))\] \[y = \sin(x)\] \[y = \sin(10 + 2) = \sin(x-(-two))\] using the following settings for the viewing window: \(0 \leq x \leq \pi\) and \(-ane.5 \leq y \leq 1.5\). Examine these graphs closely and describe the difference between the graph of \(y = \sin(t – C)\) and the graph of \(y = \sin(t)\) for these values of C.
    3. Depict the difference between the graph of \(y = \sin(x – C)\) and the graph of \(y = \sin(x)\)? Note: Consider doing two separate cases: one when \(C > 0\) and the other when \(C < 0\).

    Past exploring the graphs in Activity 2.10, we should observe that when \(C > 0\), the graph of \(y = \sin(t – C)\) is the graph of \(y = \sin(t)\) horizontally translated to the right by C units. In a like mode, the graph of \(y = \cos(t – C)\) is the graph of \(y = \cos(t)\) horizontally translated to the right by C units. When working with a sinusoidal graph, such a horizontal translation is called a
    stage shift. This is illustrated in Figure ii.13 , which shows the graphs of \(y = \sin(t)\) and \(y = \sin(t – \dfrac{\pi}{2})\). For reference, the graph of \(y = \sin(t)\) is also shown.

    2.13.png

    Figure \(\PageIndex{three}\): Graphs of \(y = \sin(t – 1)\) and \(y = \sin(t – \dfrac{\pi}{two})\). The graph of \(y = \sin(t)\) is too shown (dashes).page112image3410956576

    Then, why are we seeing this stage shift? The reason is that the graph of \(y = \sin(t)\) will become through i complete bike over the interval divers by \(0 \leq t \leq 2\pi\). Similarly, the graph of \(y = \sin(t – C)\) will go through one consummate cycle over the interval divers by \(0 \leq t – C \leq two\pi\). Solving for \(t\), we come across that \(C \leq t \leq 2\pi + C\). Then we see that this bicycle for \(y = \sin(t)\) has been shifted by C units.

    This statement besides works when \(C < 0\) and when we use the cosine function instead of the sine function. Figure 2.14 illustrates this with \(y = \cos(t – (-one))\) and \(y = \cos(t – (-\dfrac{\pi}{2}))\). Notice that we can rewrite these two equations every bit follows: \[y = \cos(t – (-1))\] \[y = \cos(t + 1))\] \[y = \cos(t – (-\dfrac{\pi}{2}))\] \[y = \cos(t + \dfrac{\pi}{2})\]

    We summarize the results for stage shift as follows:

    2.14.png

    Figure \(\PageIndex{4}\): Graphs of \(y = \cos(t + ane))\) and \(y = \cos(t + \dfrac{\pi}{2})\). The graph of \[y = \cos(t)\] is also shown (dashes).

    For \(y = \sin(t – C)\) and \(y = \cos(t – C)\), where C is any nonzero real number:

    • The graph of \(y = \sin(t)\) (or \(y = \cos(t)\)) is shifted horizontally \(|C|\) units. This is called the
      phase shift
      of the sinusoid.
    • If \(C > 0\), the graph of \(y = \sin(t)\) (or \(y = \cos(t)\)) is shifted horizontally C units to the right. That is, in that location is a phase shift of C units to the correct.
    • If \(C < 0\), the graph of \(y = \sin(t)\) (or \(y = \cos(t)\)) is shifted horizontally C units to the left. That is, in that location is a stage shift of C units to the left.

    Exercise \(\PageIndex{two}\)

    1. Determine the aamplitude and phase shift of the post-obit sinusoidal functions. \[(a) y = iii.2(\sin(t – \dfrac{\pi}{three})) \] \[(b) y = 4\cos(t + \dfrac{\pi}{vi})\]
    2. The graph below is a graph of a sinusoidal part (a) Determine an equation for this role. (b) Determine a 2d equation for this part.

    pc 2.11 2.png

    Reply

    ane. (a) For \(y = three.2(\sin(t – \dfrac{\pi}{3}))\), the amplitude is \(iii.ii\) and the phase shift is \(\dfrac{\pi}{three}\)
    (b) For \(y = 4\cos(t + \dfrac{\pi}{6})\), notice that \(y = 4\cos(t – (-\dfrac{\pi}{6}))\). Then the aamplitude is \(4\) and the stage shift is \(-\dfrac{\pi}{vi}\).

    2. There are several possible equations for this sinusoid. Some of these equations are:

    \[y = iii\sin(t + \dfrac{3\pi}{iv})\]

    \[y = -iii\sin(t – \dfrac{\pi}{4})\]

    \[y = 3\cos(t + \dfrac{\pi}{4})\]

    \[y = -3\cos(t – \dfrac{3\pi}{four})\]

    A graphing utility tin can be used to verify that whatever of these equations produce the given graph.

    Vertical Shift

    We accept one more transformation of a sinusoid to explore, the and so-chosen vertical shift. This is 1 by adding a constant to the equation for a sinusoid and is explored in the following activity.

    Activity 2.12 (The Vertical Shift of a Sinusoid)

    Complete Role i or Role 2 of this activity.

    Office i – Using a Geogebra Applet

    Nosotros will at present investigate the issue of adding a constant to a sinusoidal function. That is, we will investigate what effect the value of a real number \(D\) has the graph of \(y = A\sin(B(t – C)) + D\) and \(y = A\cos(B(t – C)) + D\). Consummate Part i or Part 2 of this activity. Nosotros will use a Geogebra applet called
    Exploring a Sinusoid. The spider web address for this applet is http://gvsu.edu/s/Lx

    After you open the applet, discover that there is an input box at the top of the screen where you can input a function. For now, leave this ready at \(g(t) = \sin(t)\). The graph of the sine function should be displayed. There are 4 sliders at the top that can be used to change the values of \(A, B, C\), and \(D\).

    i. Get out the values \(A = i, B = 1\) and \(C = 0\) set. Utilize the slider for D to change the value of C. When this is done, the graph of \(y = \sin(t) + D\) volition be displayed for the electric current value of D forth with the graph of \(y = \sin(t)\).

    (a)Use the slider to change the value of D. Explain in detail the departure between the graph of \(y = g(t) = \sin(t)\) and \(y = f(t) = \sin(t) + D\) for a abiding D. Pay close attending to the graphs and determine the vertical shift when \[i. D = 1.\] \[ii. D = 2.\] \[three. D = 3.\] \[iv. D = -ane.\] \[v. D = -ii.\] \[half dozen. D = -3.\] In item, draw the difference between the graph of \(y = \sin(t) + D\) and the graph of \(y = \sin(t)\)? Note: Consider doing ii divide cases: i when \(D > 0\) and the other when \(D < 0\).

    (b)At present click on the reset push in the upper right corner of the screen. This will reset the value of the C to its initial setting of \(C = 0\).

    (c)Alter the function to \(g(t) = \cos(t)\) and repeat function (1) for the cosine office. Does irresolute the value of D affect the graph of \(y = \cos(t) + D\) affect the sinusoidal wave in the same way that changing the value for D affects the graph of \(y = \sin(t) + D\) ?

    2. Now change the value of A to 0.five, the value of B to 2, and the value of C to 0.five. The graph of \(g(t) = \cos(t)\) volition still be displayed simply nosotros will now have \(f(t) = 0.5\cos(2(t – 0.five)) + D\). Does changing the value of D affect the graph of \(y = 0.5\cos(2(t – 0.5)) + D\) affect the sinusoidal moving ridge in the same way that changing the value for D affects the graph of \(y = \cos(t)\)?

    Function two – Using a Graphing Utility

    Make sure your graphing utility is set to radian fashion.

    1. We will first examine the graph of \(y = \cos(t) + D\) for iv different values of \(D\). Graph the five functions: \[y = \cos(x)\] \[y = \cos(x) + 1\] \[y = \cos(x) – i\] \[y = \cos(x) + two\] \[y = \cos(x) – two\] using the following settings for the viewing window: \(0 \leq 10 \leq two\pi\) and \(-3 \leq y \leq 3\). Examine these graphs closely and describe the difference between the graph of \(y = \cos(x) + D\) and the graph of \(y = \cos(x)\) for these values of D.
    2. Articulate the graphics screen. We volition now examine the graph of \(y = 0.v\cos(2(x – 0.v))\) for two different values of \(D\). Graph the following three functions: \[y = 0.5\cos(2(x – 0.5))\] \[y = 0.5\cos(2(x – 0.5)) + 2\] \[y = 0.5\cos(2(x – 0.five)) – 2\] using the following settings for the viewing window: \(0 \leq x \leq two\pi\) and \(-3 \leq y \leq three\). Examine these graphs closely and describe the difference between the graph of \(y = 0.5\cos(2(x – 0.v)) + D\) and the graph of \(y = 0.5\cos(2(x – 0.5))\) for these values of \(D\).

    Past exploring the graphs in Activity two.12, we should notice that the graph of something like \(y = A\sin(B(t – C)) + D\) is the graph of \(y = A\sin(B(t – C))\) shifted upwards \(D\) units when \(D > 0\) and shifted downwards \(|D|\) units when \(D < 0\). When working with a sinusoidal graph, such a vertical translation is called a vertical shift. This is illustrated in Figure 2.15 for a situation in which \(D > 0\).

    2.15.png

    Figure \(\PageIndex{5}\): Graphs of \(y = A\sin(B(t – C))\) (dashes) and \(y = A\sin(B(t – C)) + D\) (solid) for \(D > 0\).

    The \(y\)-axis is not shown in Effigy \(\PageIndex{5}\) because this shows a general graph with a stage shift.

    What we have done in Activeness two.12 is to start with a graph such as \(y = A\sin(B(t – C))\) and added a constant to the dependent variable to go \(y = A\sin(B(t – C)) + D\). Then when t stays the same, we are adding \(D\) to the the dependent variable. The effect is to translate the unabridged graph upward by \(D\) units if \(D > 0\) and down by \(|D|\) units if \(D < 0\).

    Amplitude, Period, Stage Shift, and Vertical Shift

    The post-obit is a summary of the work we have done in this department dealing with aamplitude, menstruum, phase shift, and vertical shift for a sinusoidal function.

    Let \(A, B, C\), and \(D\) exist nonzero existent numbers with \(B > 0\). For \(y = A\sin(B(t – C)) + D\) and \(y = A\cos(B(t – C)) + D\)

    The
    amplitude
    of the sinusoidal graph is \(|A|\).

    • If \(|A| > one\), then there is a vertical stretch of the pure sinusoid by a gene of \(|A|\).
    • If \(|A| > 1\), and so there is a vertical contraction of the pure sinusoid past a factor of \(|A|\).

    The
    menstruum
    of the sinusoidal graph is \(two\pi\).

    • When \(B > 1\), there is a horizontal compression of the graphs.
    • When \(0 < B < i\), at that place is a horizontal stretches of the graph.

    The
    stage shift
    of the sinusoidal graph is \(|C|\).

    • If \(C > 0\), there is a horizontal shift of \(C\) units to the right.
    • If \(C < 0\), at that place is a horizontal shift of \(|C|\) units to the left.

    The
    vertical shift
    of the sinusoidal graph is \(|D|\).

    • If \(D > 0\), the vertical shift is \(D\) units up.
    • If \(D < 0\), the vertical shift is \(|D|\) units down.

    Example \(\PageIndex{i}\): The Graph of a Sinusoid

    This example volition illustrate how to use the characteristics of a sinusoid and volition serve as an introduction to the more general discussion that follows. The graph of \(y = 3\sin(4(t – \dfrac{\pi}{eight})) + 2\) will expect similar the post-obit: Notice that the axes accept non yet been drawn. We want to state the coordinates of the points \(P, Q, R, Due south\), and \(T\). There are several choices but we will make the point \(P\) every bit close to the origin every bit possible. Post-obit are the important characteristics of this sinusoid:

    2.16.png

    Figure \(\PageIndex{vi}\): Graph of a Sinusoid

    • The stage shift is \(\dfrac{\pi}{8}\).
    • The amplitude is \(3\).
    • The period is \(\dfrac{2\pi}{4} = \dfrac{\pi}{two}\)
    • The vertical shift is \(2\).

    Then for the graph in Figure \(\PageIndex{half-dozen}\), we can brand the post-obit conclusions.

    • Since the vertical shift is \(2\), the horizontal line that is the center line of the sinusoid is \(y = 2\).
    • Since the phase shift is \(\dfrac{\pi}{viii}\) and this is a sine office, the coordinates of \(P\) are \((\dfrac{\pi}{8}, 2)\).
    • Since the menstruum is \(\dfrac{\pi}{2}\), the \(t\)-coordinate of \(R\) is \(\dfrac{\pi}{8} + \dfrac{1}{ii}(\dfrac{\pi}{2}) = \dfrac{3\pi}{viii}\). So the coordinates of \(R\) are \((\dfrac{3\pi}{8}, two)\).
    • Since the period is \(\dfrac{\pi}{2}\), the \(t\)-coordinate of \(T\) is \(\dfrac{\pi}{8} + \dfrac{\pi}{2} = \dfrac{5\pi}{eight}\). And then the coordinates of \(R\) are \((\dfrac{v\pi}{viii}, 2)\).
    • Since the menstruation is \(\dfrac{\pi}{2}\), the \(t\)-coordinate of \(Q\) is \(\dfrac{\pi}{eight} + \dfrac{1}{iv}(\dfrac{\pi}{two}) = \dfrac{\pi}{4}\).Also, since the amplitude is \(3\), the \(y\)-coordinate of \(Q\) is \(2 + iii = five\). And so the coordinates of \(Q\) are \((\dfrac{\pi}{4}, 5)\).
    • Since the period is \(\dfrac{\pi}{ii}\), the \(t\)-coordinate of \(S\) is \(\dfrac{\pi}{viii} + \dfrac{3}{4}(\dfrac{\pi}{2}) = \dfrac{\pi}{2}\).Besides, since the amplitude is \(3\), the \(y\)-coordinate of \(S\) is \(2 – 3 = -1\). Then the coordinates of \(S\) are \((\dfrac{\pi}{2}, -ane)\).page119image3414898048
      page119image3414898336
      page119image3414898624
      page119image3414898912
      page119image3414890736
      page119image3414891024
      page119image3414891312

    We can verify all of these results past using a graphing utility to draw the graph of \(y = 3\sin(4(t – \dfrac{\pi}{viii})) + 2\) using \(0 \leq t \leq \dfrac{5\pi}{8}\) and \(-two\ leq y \leq vi\). If the utility allows, set the t-scale to one-quarter of a period, which is \(\dfrac{\pi}{eight}\).

    Important Notes nigh Sinusoids

    • For \(y = A\sin(B(t – C)) + D\) and \(y = A\cos(B(t – C)) + D\), the amplitude, the period, and the vertical shift volition be the same.
    • The graph for both functions will await like that shown in Figure \(\PageIndex{7}\). The deviation betwixt \(y = A\sin(B(t – C)) + D\) and \(y = A\cos(B(t – C)) + D\) will be the phase shift.
    • The horizontal line shown is not the t-axis. It is the horizontal line \(y = D\), which we frequently phone call the center line for the sinusoid.

    2.17.png

    Figure \(\PageIndex{seven}\): Graph of a Sinusoid

    And then to use the results about sinusoids and Effigy \(\PageIndex{7}\), we have:

    1. The amplitude, which we will phone call
    amp, is equal to the lengths of the segments \(QV\) and \(WS\).

    1. The period, which we will call
      per, is equal to \(\dfrac{2\pi}{B}\). In addition, the length of the segments \(PV,VR,RW\), and \(WT\) are equal to \(\dfrac{1}{iv} per\).
    2. For \(y = A\sin(B(t – C)) + D\), we tin can apply the betoken \(P\) for the phase shift. And then the \(t\)-coordinate of the point \(P\) is \(C\) and \(P\) has coordinates \((C, D)\). We tin can decide the coordinates of the other points as need by using the amplitude and period. For example:

      The signal \(Q\) has coordinates \((C + \dfrac{ane}{4} per, D + amp)\)

      The point \(R\) has coordinates \((C + \dfrac{1}{2} per, D)\)

      The indicate \(South\) has coordinates \((C + \dfrac{3}{iv} per, D – amp)\)

      The betoken \(T\) has coordinates \((C + per, D)\)

    3. For \(y = A\cos(B(t – C)) + D\), we can use the betoken \(Q\) for the stage shift. And so the \(t\)-coordinate of the indicate \(Q\) is \(C\) and \(Q\) has coordinates \((C, D + amp)\). Nosotros can make up one’s mind the coordinates of the other points as demand by using the amplitude and menstruum. For example:

      The point \(P\) has coordinates \((C – \dfrac{1}{4} per, D)\)

      The point \(R\) has coordinates \((C + \dfrac{1}{4} per, D)\)

    4. The point \(South\) has coordinates \((C + \dfrac{ane}{two} per, D – amp)\)

      The bespeak \(T\) has coordinates \((C + \dfrac{3}{4} per, D)\)

    Please note that information technology is not necessary to try to recollect all of the facts in items (3) and (4). What nosotros should call up is how to use the concepts of 1-quarter of a menstruation and the amplitude illustrated in items (3) and (iv). This will be done in the side by side ii progress checks, which in reality are guided examples.

    Do \(\PageIndex{3}\)

    The characteristics of a sinusoid can be helpful in setting an appropriate viewing window when producing a useable graph of a sinusoid on a graphing utility. This is especially truthful when the catamenia is small or large. For case, consider the sinusoidal function \[y = 6.3\cos(50\pi(t + 0.01)) + 2\]

    one. What is the aamplitude?

    2. What is the phase shift?

    3. What is the period?

    4. What is the vertical shift?

    5. Use this data to determine coordinates for the bespeak \(Q\) in the post-obit diagram.

    pc 2.14 5.png

    6. Now determine the coordinates of points \(P, R, S\), and \(T\).

    7. Utilize this information and a graphing utility to draw a graph of (slightly more than) one menstruation of this sinusoid that shows the points \(P, Q\), and \(T\).

    Answer
    1. The aamplitude is \(half-dozen.3\).
    2. The period is \(\dfrac{ii\pi}{l\pi} = \dfrac{1}{25}\)
    3. We write \(y = six.3\cos(50\pi(t – (-0.01))) + 2\) and see that the phase shift is \(-0.01\) or \(0.01\) units to the left.
    4. The vertical shift is \(2\).
    5. Because nosotros are using a cosine and the phase shift is \(-0.01\), we can use \(-0.01\) every bit the \(t\)-coordinate of \(Q\). The \(y\)-coordinate will exist the vertical shift plus the aamplitude. So the \(y\)-coordinate is \(eight.3\). Point \(Q\) has coordinates \((-0.01, 8.3)\)
    6. Nosotros now utilize the fact that the horizontal altitude between \(P\) and \(Q\) is ane-quarter of a period. Since the period is \(\dfrac{ane}{25} = 0.04\), we run across that one-quarter of a period is \(0.01\). The point \(P\) also lies on the centre line, which is \(y = 2\). And so the coordinates of \(P\) are \((-0.02, two)\).

    We at present use the fact that the horizontal distance between \(Q\) and \(R\) is one-quarter of a menstruum. The point \(R\) is on the heart line of the sinusoid and then \(R\) has coordinates \((0, 2)\).

    The betoken \(Southward\) is a depression point on the sinusoid. And then its \(y\)-coordinate will exist \(D\) minus the amplitude, which is \(two – 6.3 = -4.3\). Using the fact that the horizontal distance from \(R\) to \(S\) is one-quarter of a period, the coordinates of \(S\) are \((0.01, -4.3)\). Since the point \(T\) is on the middle line and the horizontal distance from \(S\) to \(T\) is ane-quarter of a period, the coordinates of \(T\) are \((0.03, two)\).

    7. We will use a viewing window that is one-quarter of a period to the left of \(P\) and one-quarter of a period to the correct of \(T\). So we will use \(-0.03

    leq t \leq 0.03\). Since the maximum value is \(8.3\) and the minimum value is \(-4.iii\), nosotros volition utilize \(-v \leq y \leq 9\).

    Practice \(\PageIndex{iv}\)

    We will determine two equations for the sinusoid shown in Figure \(\PageIndex{8}\).

    1. Make up one’s mind the coordinates of the points \(Q\) and \(R\). The vertical distance between these ii points is equal to two times the amplitude. Use the y-coordinates of these points to determine two times the amplitude then the amplitude.
    2. The center line for the sinusoid is one-half-mode between the high point \(Q\) and the low bespeak \(R\). Employ the y-coordinates of \(Q\) and \(R\) to determine the center line \(y = D\). This volition exist the vertical shift.

    2.18.png

    Figure \(\PageIndex{8}\): The Graph of a Sinusoid

    1. The horizontal altitude between \(Q\) and \(R\) is equal to 1-half of a period. Use the t-coordinates of \(Q\) and \(R\) to decide the length of one-half of a period and hence, the period. Use this to determine the value of \(B\).
    2. Nosotros will now discover an equation of the form \(y = A\cos(B(t – C)) + D\). We still need the stage shift \(C\). Utilize the point \(Q\) to decide the phase shift and hence, the value of \(C\). We now have values for \(A, B, C\), and \(D\) for the equation \(y = A\cos(B(t – C)) + D\).
    3. To determine an equation of the form \(y = A\sin(B(t – C’)) + D\)., nosotros volition apply the signal \(P\) to determine the phase shift \(C’\). (A different symbol was used considering \(C’\) will be different than \(C\) in part (iv).) Discover that the y-coordinate of \(P\) is 4 and then \(P\) lies on the center line. We can use the fact that the horizontal altitude betwixt P and \(Q\) is equal to ane-quarter of a period. Decide the t coordinate of \(P\), which will exist equal to \(C’\). Now write the equation \(y = A\sin(B(t – C’)) + D\). using the values of \(A, B, C’\), and \(D\) that we accept determined.

    We can check the equations we found in parts (4) and (5) by graphing these equations using a graphing utility.

    Answer
    pc 2.15.png

    1. The coordinates of \(Q\) are \((\dfrac{seven\pi}{12}, vii)\) and the coordinates of \(R\) are \((\dfrac{13\pi}{12}, ane)\). So two times the aamplitude is \(seven – i = half-dozen\) and the amplitude is \(3\).

    2. We add the amplitude to the lowest \(y\)-value to determine \(D\). This gives \(D = 1 + 3 = 4\) and the centre line is \(y = 4\).

    3. The horizontal distance between \(Q\) and \(R\) is \(\dfrac{13\pi}{12} – \dfrac{7\pi}{12} = \dfrac{six\pi}{12}\). So we run into that half of a period is \(\dfrac{\pi}{2}\) and the period is \(\pi\). So \(B = \dfrac{2\pi}{\pi} = ii\).

    four. For \(y = A\cos(B(t – c)) + D\), nosotros can use the signal \(Q\) to determine a stage shift of \(\dfrac{vii\pi}{12}\). So an equation for this sinusoid is \[y = iii\cos(2(t – \dfrac{seven\pi}{12})) + 4.\]

    5. The point \(P\) is on the middle line and then the horizontal altitude between \(P\) and \(Q\) is one-quarter of a catamenia. So this horizontal distance is \(\dfrac{\pi}{iv}\) and the \(t\)-coordinate of \(P\) is \[\dfrac{7\pi}{12} – \dfrac{\pi}{4} = \dfrac{four\pi}{12} = \dfrac{\pi}{iii}.\]
    This can be the phase shift for \(y = A\sin(B(t – C’)) + D\). So another equation for this sinusoid is \[y = three\sin(2(t – \dfrac{\pi}{3})) + 4\]

    Summary

    In this section, we studied the following important concepts and ideas: For a sinusoidal function of the form \(f(t) = A\sin(B(t – C)) + D\) or \(f(t) = A\cos(B(t – C)) + D\) where \(A, B, C\),and \(D\) are real numbers with \(B > 0\):

    • The value of \(|A|\) is the amplitude of the sinusoidal part.
    • The value of \(B\) determines the period of the sinusoidal office. the period is equal to \(2\pi\).
    • The value of \(C\) is the phase shift (horizontal shift) of the sinusoidal function. The graph is shifted to the right if \(C > 0\) and shifted to the left if \(C < 0\).
    • The value of \(D\) is the vertical shift of the sinusoid. The horizontal line \(y = D\) is the so-called
      center line
      for the graph of the sinusoidal office.
    • The important notes about sinusoids on page 103.

    Which Graph or Graphs Appear to Show a Sinusoid

    Source: https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Trigonometry_(Sundstrom_and_Schlicker)/02%3A_Graphs_of_the_Trigonometric_Functions/2.02%3A_Graphs_of_Sinusoidal_Functions