Graphing and Trigonometric Functions: Solutions

Here are some solutions to the Exercises to accompany the section Graphing and Trigonometric Functions.

Describing Trigonometric Graphs

Use our transformational graphing technique to graph the following trigonometric functions from the previous Exercise:

f(t) = 2sin(5(t - p/3)) - 4.

Solution

If we analyze f(t) = 2sin(5(t - p/3)) - 4:

Corresponding Algebraic formula	sin(t)	sin(5t)	sin(5(t - p/3))		2sin(5(t - p/3))	2sin(5(t - p/3)) - 4
Geometric Effect	Take the graph of the sin function	Shrink horizontally by a factor of 1/5	Shift right p/3 units		Stretch vertically by a factor of 2	Shift down 4 units
Numerical Effect		Divide by 5	Add p/3	Apply the sin function	Multiply by 2	Subtract 4
Numerical Results	Inputs			Outputs
	0	0	p/3	0	0	-4
	p/2	p/10	13p/30	1	2	-2
	p	p/5	8p/15	0	0	-4
	3p/2	3p/10	19p/30	-1	-2	-6
	2p	2p/5	11p/15	0	0	-4

we obtain the following table of values:

t	2sin(5(t - p/3)) - 4
p/3 13p/30 8p/15 19p/30 11p/15	-4 -2 -4 -6 -4

with the graph:

Notice that we have not drawn the axes to meet at the point (0, 0).

g(t) = 2 - 3cos(-4t - 5p).

Solution

If we analyze g(t) = 2 - 3cos(-4t - 5p):

Corresponding Algebraic formula	cos(t)	cos(t - 5p)	cos(-t - 5p)	cos(-4t - 5p)		3cos(-4t - 5p)	- 3cos(-4t - 5p)	2 - 3cos(-4t - 5p)
Geometric Effect	Take the graph of the cos function	Shift right 5p units	Flip horizontally	Shrink horizontally by a factor of 1/4		Stretch vertically by a factor of 3	Flip vertically	Shift up 2 units
Numerical Effect		Add 5p	Negate	Divide by 4	Apply the cos function	Multiply by 3	Negate	Add 2
Numerical Results	Inputs				Outputs
	0	5p	-5p	-5p/4	1	3	-3	-1
	p/2	11p/2	-11p/2	-11p/8	0	0	0	2
	p	6p	-6p	-3p/2	-1	-3	3	5
	3p/2	13p/2	-13p/2	-13p/8	0	0	0	2
	2p	7p	-7p	-7p/4	1	3	-3	-1

we obtain the following table of values:

t	2 - 3cos(-4t - 5p)
-5p/4 -11p/8 -3p/2 -13p/8 -7p/4	-1 2 5 2 -1

with the graph:

Again, we have not drawn the axes to meet at the point (0, 0).

h(t) = sin(t/2 - p)/3 + 1.

Solution

If we analyze h(t) = sin(t/2 - p)/3 + 1:

Corresponding Algebraic formula	sin(t)	sin(t - p)	sin(t/2 - p)		sin(t/2 - p)/3	sin(t/2 - p)/3 + 1
Geometric Effect	Take the graph of the sin function	Shift right p units	Stretch horizontally by a factor of 2		Shrink vertically by a factor of 1/3	Shift up 1 unit
Numerical Effect		Add p	Multiply by 2	Apply the sin function	Divide by 3	Add 1
Numerical Results	Inputs			Outputs
	0	p	2p	0	0	1
	p/2	3p/2	3p	1	1/3	4/3
	p	2p	4p	0	0	1
	3p/2	5p/2	5p	-1	-1/3	2/3
	2p	3p	6p	0	0	1

we obtain the following table of values:

t	sin(t/2 - p)/3 + 1
2p 3p 4p 5p 6p	1 4/3 1 2/3 1

with the graph:

Again, we have not drawn the axes to meet at the point (0, 0).

k(t) = -1 - 4cos(3(t/2 + p))/5.

Solution

It is quicker if first simplify this as k(t) = -1 - (4/5)cos((3/2)t + 3p)), so that our analysis would give:

Corresponding Algebraic formula	cos(t)	cos(t + 3p)	cos((3/2)t + 3p)		(4/5)cos((3/2)t + 3p)	-(4/5)cos((3/2)t + 3p)	-1 - (4/5)cos((3/2)t + 3p)
Geometric Effect	Take the graph of the cos function	Shift left 3p units	Shrink horizontally by a factor of 2/3		Shrink vertically by a factor of 4/5	Flip vertically	Shift down 1 units
Numerical Effect		Subtract 3p	Divide by 3/2	Apply the cos function	Multiply by 4/5	Negate	Add -1/ Subtract 1
Numerical Results	Inputs			Outputs
	0	-3p	-2p	1	4/5	-4/5	-9/5
	p/2	-5p/2	-5p/3	0	0	0	-1
	p	-2p	-4p/3	-1	-4/5	4/5	-1/5
	3p/2	-3p/2	-p	0	0	0	-1
	2p	-p	-2p/3	1	4/5	-4/5	-9/5

we obtain the following table of values:

t	-1 - (4/5)cos((3/2)t + 3p)
-2p -5p/3 -4p/3 -p -2p/3	-9/5 -1 -1/5 -1 -9/5

with the graph:

Again, we have not drawn the axes to meet at the point (0, 0).

Back to Exercises.

Fill-in-the-blanks, putting either sin or cos into the second blank, to create a formula for a trigonometric function, then use our transformational graphing technique to graph it.

y = __·__(__x + __) + __

Repeat this Exercise as often as necessary until you are confident in your ability to plot trigonometric graphs.

Solution: Use the Multigraph Utility in XFunctions to make a plot to check your work:; Check that the points that you computed on your graph are truly on the graph.

Back to Exercises.

Determine the amplitude, baseline, and wavelength in each of the following functions.

Solution

From the graph, we can see that that the amplitude is 1 - -1 = 2, the baseline shift is down 1, and the wavelength is 4p/3 - p/3 = p.
f(t) = 2sin(5(t - p/3)) - 4.

Solution

Using the theorem in the text, we can see that the amplitude is 2, the baseline shift is -4 (i.e., down 4), and the wavelength is 2p/5. This agrees with the graph from the previous Exercise.
g(t) = 2 - 3cos(-4t - 5p).

Solution

Before applying the theorem in the text, we should rewrite this as g(t) = -3cos(-4(t - 5p/(-4)) + 2 = -3cos(-4(t + 5p/4)) + 2, by switching the order of terms and factoring. Now it is clear that the amplitude is 3, the baseline shift is up 2, and the wavelength is 2p/4 = p/2. This agrees with the graph from the previous Exercise.
h(t) = sin(t/2 - p)/3 + 1. Hint: Remember to use proper order of operations in order to interpret this expression correctly.

Solution

Before applying the theorem in the text, we should rewrite this as h(t) = (1/3)sin((1/2)(t - 2p)) + 1. Now it is clear that the amplitude is 1/ 3, the baseline shift is up 1, and the wavelength is 2p/(1/2) = 4p. This agrees with the graph from the previous Exercise.
k(t) = -1 - 4cos(2(t/3 + p))/5.

Solution

Before applying the theorem in the text, we should rewrite this as k(t) = (-4/5)cos((2/3)(t + 3p)) - 1. Now it is clear that the amplitude is -4/5, the baseline shift is down 1, and the wavelength is 2p/(2/3) = 3p. This agrees with the graph from the previous Exercise.

Back to Exercises.

Determine a trigonometric function which applies to each of the following situations.

f is given by the following table of values:

t f(t)

0
5
10
15
20 6
3
6
9
6

Solution

We can see that this has a period of 20. It oscillates between a maximum of 9 and a minimum of 3 from a baseline of 6 = (9 + 3)/2, so that its amplitude is 3 (we can calculate this as (9 - 3)/2 or 9 - 6 or 6 - 3). Since it starts at the baseline and goes down, it must be a -sin function, so that f(t) = -3sin((2p/20)t) + 6.
g(t) has the following graph:

Solution

We can see that this has a period of 5p. It oscillates between a maximum of 2 and a minimum of -6 from a baseline of -2, so that its amplitude is 4. Since it starts at the maximum and goes down, it must be a cos function, so that g(t) = 4cos((2p/5p)t) - 2 = 4cos((2/5)t) - 2.
A weight is hanging on a large spring, originally at a height of 8 in. from the top of the table. If you pull down on the spring, until it is only 4 in. off the table and let go, it will begin to oscillate from that minimum height of 4 in. to a maximum height of 12 in. If it makes one bounce in 2 seconds, give the equation for h(t) = height above the table after letting go.

Solution

We are told that the pendulum has a period of 2 sec. Note: It is more common to give the frequency of 30 oscillations/min., which is another way of saying (1 oscillation)/(2 sec.). We are also told that the weight bounces between a maximum of 12 in. and a minimum of 4 in. from a baseline of 8 in., so that its amplitude is 4. Since it starts at the minimum and goes up, it must be a -cos function, so that h(t) = -4cos((2p/2)t) + 8 = -4cos(pt) + 8.
Have your partner construct a problem similar to one of parts a) - c). Have him/her describe this situation to you, and see if you can write down a correct equation.

Solution

Your partner should correct your work.

Back to Exercises.

Graphing the Advanced Trigonometric Functions

Explain why the amplitude of each of the advanced trigonometric functions (i.e., tan, cot, sec, and csc) is infinite.

Solution: We have defined the amplitude as the magnitude of the maximum displacement from the baseline. However, all of these graphs go to ±¥, that is, they get infinitely far away from their respective baselines. Therefore, we can reasonably say that their amplitude is ¥. It is also common to simply say that the amplitude is undefined for these trigonometric functions.

Back to Exercises.

Use our transformational graphing technique to graph the following trigonometric functions. Determine the baseline shift and wavelength in each case:

f(t) = 3tan(2(t - p/4)) - 1.

Solution

If we analyze f(t) = 3tan(2(t - p/4)) - 1:

Corresponding Algebraic formula	tan(t)	tan(2t)	tan(2(t - p/4))		3tan(2(t - p/4))	3tan(2(t - p/4)) - 1
Geometric Effect	Take the graph of the tan function	Shrink horizontally by a factor of 1/2	Shift right p/4 units		Stretch vertically by a factor of 3	Shift down 1 unit
Numerical Effect		Divide by 2	Add p/4	Apply the tan function	Multiply by 3	Subtract 1
Numerical Results	Inputs			Outputs
	0	0	p/4	0	0	-1
	p/4	p/8	3p/8	1	3	2
	p/2	p/4	p/2	undefined	undefined	undefined
	3p/4	3p/8	5p/8	-1	-3	-4
	p	p/2	3p/4	0	0	-1

we obtain the following table of values:

t	3tan(2(t - p/4)) - 1
p/4 3p/8 p/2 5p/8 3p/4	-1 2 undefined -4 -1

with the graph:

We can see that the baseline of the graph is shifted down 1, while the wavelength is shrunk to p/2 from p.

g(t) = -3cot(2(t - p/4)) - 1.

Solution

If we analyze g(t) = -3cot(2(t - p/4)) - 1:

Corresponding Algebraic formula	cot(t)	cot(2t)	cot(2(t - p/4))		3cot(2(t - p/4))	-3cot(2(t - p/4))	-3cot(2(t - p/4)) - 1
Geometric Effect	Take the graph of the cot function	Shrink horizontally by a factor of 1/2	Shift right p/4 units		Stretch vertically by a factor of 3	Flip vertically	Shift down 1 unit
Numerical Effect		Divide by 2	Add p/4	Apply the cot function	Multiply by 3	Negate	Subtract 1
Numerical Results	Inputs			Outputs
	0	0	p/4	undefined	undefined	undefined	undefined
	p/4	p/8	3p/8	1	3	-3	-4
	p/2	p/4	p/2	0	0	0	-1
	3p/4	3p/8	5p/8	-1	-3	3	2
	p	p/2	3p/4	undefined	undefined	undefined	undefined

we obtain the following table of values:

t	-3cot(2(t - p/4)) - 1
p/4 3p/8 p/2 5p/8 3p/4	undefined -4 -1 2 undefined

with the graph:

We can see that the baseline of the graph is shifted down 1, while the wavelength is again shrunk to p/2 from p.

h(t) = -2sec(3t - p) - 4.

Solution

If we analyze h(t) = -2sec(3t - p) - 4:

Corresponding Algebraic formula	sec(t)	sec(t - p)	sec(3t - p)		2sec(3t - p)	-2sec(3t - p)	-2sec(3t - p) - 4
Geometric Effect	Take the graph of the sec function	Shift right p units	Shrink horizontally by a factor of 1/3		Stretch vertically by a factor of 2	Flip vertically	Shift down 4 units
Numerical Effect		Add p	Divide by 3	Apply the sec function	Multiply by 2	Negate	Subtract 4
Numerical Results	Inputs			Outputs
	0	p	p/3	1	2	-2	-6
	p/2	3p/2	p/2	undefined	undefined	undefined	undefined
	p	2p	2p/3	-1	-2	2	-2
	3p/2	5p/2	5p/6	undefined	undefined	undefined	undefined
	2p	3p	p	1	2	-2	-6

we obtain the following table of values:

t	-2sec(3t - p) - 4
p/3 p/2 2p/3 5p/6 p	-6 undefined -2 undefined -6

with the graph:

We can see that the baseline of the graph is shifted down 4, while the wavelength is 2p/3.

k(t) = csc(2(t + p/3))/5 + 1.

Solution

If we analyze k(t) = csc(2(t + p/3))/5 + 1:

Corresponding Algebraic formula	csc(t)	csc(2t)	csc(2(t + p/3))		csc(2(t + p/3))/5	csc(2(t + p/3))/5 + 1
Geometric Effect	Take the graph of the csc function	Shrink horizontally by a factor of 1/2	Shift left p/3 units		Shrink vertically by a factor of 1/5	Shift up 1 unit
Numerical Effect		Divide by 2	Subtract p/3	Apply the csc function	Divide by 5	Add 1
Numerical Results	Inputs			Outputs
	0	0	-p/3	undefined	undefined	undefined
	p/2	p/4	-p/12	1	1/5	6/5
	p	p/2	p/6	undefined	undefined	undefined
	3p/2	3p/4	5p/12	-1	-1/5	4/5
	2p	p	2p/3	undefined	undefined	undefined

we obtain the following table of values:

t	csc(2(t + p/3))/5 + 1
-p/3 -p/12 p/6 5p/12 2p/3	undefined 6/5 undefined 4/5 undefined

with the graph:

We can see that the baseline of the graph is shifted up 1, while the wavelength is p.

Back to Exercises.

Fill-in-the-blanks, putting either tan, cot, sec, or csc into the second blank, to create a formula for a trigonometric function, then use our transformational graphing technique to graph it.

y = __·__(__x + __) + __

Repeat this Exercise as often as necessary until you are confident in your ability to plot trigonometric graphs.

Solution: Use the Multigraph Utility in XFunctions to make a plot to check your work:; Check that the points that you computed on your graph are truly on the graph.

Back to Exercises.