Trigonometric Functions

Precalculus Project

  Objectives

To study the graphs of y = a + b sin(cx + d) for various values of a, b, c and d.  

To study period, amplitude, phase shift and vertical  and horizontal shifts.

To solve word problems using trigonometric functions.

Use graphical capabilities of Maple  to graph trigonometric functions.

  Solved Problem 1:

(a)  Draw the graphs of sin(x) for two periods.

(b)  Draw the graphs of sin(x) and sin (2x)  on the same set of axes.

(c) Draw the graphs of sin(x) and -2 + sin (x) for two periods on the same set of axes.

(d)  Draw the graphs of sin(x) and sin (x + Pi /3) for two periods on the same set of axes.

(e)  Draw the graphs of sin(x) and - sin (x) for two periods on the same set of axes.

(f)  Draw the graphs of sin(x) and 4 sin (x) for two periods on the same set of axes.

(g)  Draw the graphs of sin(x) and 1 + 2sin (3x + 4) on the same set of axes.

 Solution:

  Graph of sin (x)

 First define the sine function and graph it.

 

>    f:= x -> sin (x);

>    with(plots);

>    plot(f(x),x=-2*Pi..2*Pi);

The period of the function sin x is 2 Pi .  (2 Pi  ~ 6.28)  That means the function starts repeating

every 2 Pi  interval.  In other words sin (x + 2 Pi ) = sin x,  and 2 Pi  is the smallest such number

for which you get one complete cycle of sin x.   Two cycles have been plotted in the graph.

The maximum value of the graph is 1.  The maximum value is called the amplitude of sin(x).

The domain of sin(x) is all reals.  The range of the function is [-1, 1].  

  Graph of sin(x) and sin(2x)

Define the function.

>    g := x -> sin(2*x);

>    plot({f(x),g(x)},x=-2*Pi..2*Pi);

Observe that  the period of the function sin 2x is Pi .  The above graph shows four cycles

of sin(2x) and only two cycles of sin(x) in the same interval [ - 2*Pi  , 2*Pi  ].   The amplitude

of sin(2x) is 1.

 

To find this manually use the fact that sin(2x + 2 Pi ) =  sin (2x)  

                                                            sin (2(x + Pi )) = sin(2x)

  In other words Pi   is the smallest interval for which you get a comple cycle of sin(2x).  

  Graph of sin(x) and -2 + sin(x)

>    g:= x-> -2 + sin(x);

>    plot({f(x), g(x)}, x = -2*Pi..2*Pi);

 Note that the period of the two functions is the same, 2 Pi  .  But adding -2, the graph sin(x)

has a vertical shift  -2 to get the graph of -2 + sin(x).    

The amplitude is 1.

  Graph of sin(x) and sin(x + 2*Pi/3 )

>    g:=x->sin(x + Pi/3);

>    plot({f(x),g(x)},x=-2*Pi..2*Pi);

 The period of sin(x + Pi/3  ) is 2 Pi .  The amplitude is 1.  The graph is obtained from sin(x)

by a horizontal shift of - Pi/3   .                     

Graph of sin(x) and -sin(x)

>    g := x -> -sin(x);

>    plot({f(x),g(x)},x=-2*Pi..2*Pi);

Observe that changing the sign of the function sin x gives a reflection in the x-axis.

  Graphs of sin(x) and 4sin(x)

>    g := x -> 4*sin(x);

>    plot({f(x),g(x)},x=-2*Pi..2*Pi);

The period is 2 Pi   and the amplitude of 4sin(x) is 4.  

 Observe that multiplying sin x by a constant >1 increases the amplitude

by the constant.  (Vertical stretch by the constant).  Similarly multilying sin x

by a positive constant < 1 will result in shrinking the amplitude.

  Graphs of sin(x) and 1 + 2*sin(3*x + 4)

>    g := x -> 1 + 2*sin(3*x + 4);

>    plot({f(x),g(x)},x=-2*Pi..2*Pi);

Observe that the graph of y = sin x is changed into the graph of g by a vertical shift of 1, reducing the period to Pi /3, taking a horizontal shift of 4, and changing the amplitude to 2.  In the same interval [-2 Pi , 2 Pi ], there are two cycles of sin x and six cycles of g(x) .

  Solved Problem 2:

 Blood Pressure:  The function P = 100 - 20 cos (5 Pi  t/3) approximates the blood pressure P in mm of mercury at time t in seconds for a person at rest.

   (a) Find the period of the function.

   (b) Find the number of heartbeats per minute.

   (c) Use a graphing utility to graph the pressure function.

Solution:

  (a) Period

The period of this function is found by considering the fact cos (5 Pi t/3) = cos (5 Pi  t/3 + 2 Pi ).  

 After taking out 5 Pi /3 common from the two terms of the right argument we get   

          cos(5 Pi  t/3) = cos (5 Pi /3(t + 6/5)).

  The period is 6/5.

(b) Number of Heartbeats

Since the period is 6/5 seconds, the number of heartbeats per second is 5/6.  

         For a minute the number of heartbeats is (5/6)(60) = 50.

  (c) Graph of blood pressure function

>    f:= t -> 100 - 20*cos(5*Pi/3* t);

>    plot(f(t),t=-2*Pi..2*Pi);

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ASSIGNMENT

   

  Problems:

Sketch by hand the two functions in each exercise on the same coordinate plane and verify using Maple:  

 1.    (a) f(x) = sin x                  (b) g(x) = sin ( x/3 )

 2.   (a) f(x) = sin x                  (b) g(x) = cos (x - Pi/2 )

 3.    (a) f(x) = - sin (2 Pi  x)      (b) g(x) = 10 cos ( Pi/6   x)

 4.  Fuel consumption:  The daily consumption C (in gallons) of diesel fuel on a farm is modeled by

                                 C = 30.3 + 21.6 sin ( 2*Pi*t/365    + 10.9).

    where t is the time in days, with t = 1 corresponding to January 1.

(a) What is the period of the model?  Is it what you expected?  Explain.

(b) What is the average daily fuel consumption?  Which term of the model did you use?  Explain.

(c) Use Maple to graph the model. Use the graph to approximate the time of the year when consumption

exceeds 40 gallons per day.

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MSEIP Grant #P120AA010031:  "Four Colleges: Calculus + Enhancements", 2001-2004