lecture notes – trigonometric functions

IntroductionTrigonometric Functions

Trigonometric ModelsModel Properties

Calculus for the Life Sciences

Lecture Notes – Trigonometric Functions

Joseph M. Mahaffy,〈[email protected]〉

Department of Mathematics and StatisticsDynamical Systems Group

Computational Sciences Research Center

San Diego State UniversitySan Diego, CA 92182-7720

http://www-rohan.sdsu.edu/∼jmahaffy

Spring 2017

Joseph M. Mahaffy, 〈[email protected]〉Lecture Notes – Trigonometric Functions —(1/68)

http://www-rohan.sdsu.edu/~jmahaffy



Outline

1 IntroductionAnnual Temperature CyclesSan Diego and Chicago

2 Trigonometric FunctionsBasic Trig FunctionsRadian MeasureProperties of Sine and CosineIdentities

3 Trigonometric ModelsVertical Shift and AmplitudeFrequency and PeriodPhase ShiftExamples

4 Model PropertiesPhase Shift of Half a PeriodEquivalent Sine and Cosine ModelsReturn to Annual Temperature VariationOther Examples




Annual Temperature CyclesSan Diego and Chicago

Introduction — Trigonometric Functions

Introduction — Trigonometric Functions

Natural physical cycles

Daily cycle of lightAnnual cycle of the seasons

Many phenomena in biology appear in cycles

Circadian rhythmsHormonal fluctuationsPredator-prey cycles

Oscillations are often modeled using trigonometric functions





Annual Temperature Cycles

Annual Temperature Cycles

Weather reports give the average temperature for a day

There are seasonal differences in the average daily temperature

Higher averages occur in the summerLower averages occur in the winter

Long term averages

Compare background noise from annual variationGlobal warming if there is a clear increase in annualaverage over the long termThis is not a simple trig function model





Modeling Annual Temperature Cycles

Modeling Annual Temperature Cycles

What mathematical tools can help predict the annualtemperature cycles?

Polynomials and exponentials do not exhibit the periodicbehavior

Trigonometric functions exhibit periodicity

Fit any specific periodManage amplitude of variationShift the maximum or minimum for a data set





Average Temperatures for San Diego and Chicago 1

Average Temperatures for San Diego and Chicago: Table ofthe monthly average high and low temperatures for San Diego andChicago

Month Jan Feb Mar Apr May JunSan Diego 66/49 67/51 66/53 68/56 69/59 72/62Chicago 29/13 34/17 46/29 59/39 70/48 80/58

Month Jul Aug Sep Oct Nov DecSan Diego 76/66 78/68 77/66 75/61 70/54 66/49Chicago 84/63 82/62 75/54 63/42 48/32 34/19






Graph of Temperature for San Diego and Chicago with best fittingtrigonometric functions (cosine)

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan20

30

40

50

60

70

80

90

Month

Ave

rage

Tem

pera

ture

, o F

Temperatures for San Diego and Chicago

San DiegoChicago






Models of Annual Temperature Cycles for San Diego andChicago

The two graphs have similarities and differences

Same seasonal period as expectedSeasonal variation or amplitude of oscillation for Chicago ismuch greater than San DiegoOverall average temperature for San Diego is greater thanthe average for Chicago

Overlying models use cosine functions (could equally fit withsine functions)




Basic Trig FunctionsRadian MeasureProperties of Sine and CosineIdentities

Trigonometric Functions 1

Trigonometric Functions are often called circular functions

Let (x, y) be a point on a circle of radius r centered at the origin

Define the angle θ between the ray connecting the point to theorigin and the x-axis

(x,y)

r

θ

X

Y





Trigonometric Functions

Trig Functions – 6 basic Trigonometric functions

(x,y)

r

θ

X

Y

sin(θ) =y

rcos(θ) =

x

rtan(θ) =

y

x

csc(θ) =r

ysec(θ) =

r

xcot(θ) =

x

y

We will concentrate almost exclusively on the sine and cosine





Radian Measure

Radian Measure

Trigonometry courses start with degrees measuring an angle

This is not the appropriate unit to use in Calculus

The radian measure of the angle uses the unit circle

The distance around the perimeter of the unit circle is 2π

The radian measure of the angle θ is simply the distance alongthe circumference of the unit circle

A 45◦ angle (18the distance around the unit circle) becomes

π

4radians

90◦ and 180◦ angles convert to π

2and π radians, respectively

Conversions

1◦ = π

180= 0.01745 radians or 1 radian = 180

◦

π= 57.296◦





Sine and Cosine 1

Sine and Cosine: The unit circle has r = 1, so the trig functionssine and cosine satisfy

cos(θ) = x and sin(θ) = y

The formula for cosine (cos) gives the x value of the angle, θ,(measured in radians)

The formula for sine (sin) gives the y value of the angle, θ

The tangent function (tan) gives the slope of the line (y/x)





Sine and Cosine 2

Graph of sin(θ) and cos(θ) for angles θ ∈ [−2π, 2π]

−2π −π 0 π 2π

−0.5

0

0.5

θ (radians)

Sine and Cosine in Radians

sin(θ)cos(θ)





Sine and Cosine 3

Sine and Cosine - Periodicity and Bounded

Notice the 2π periodicity: The functions repeat the samepattern every 2π radians

Consider a point moving around a circleAfter 2π radians, the point returns to the same position(circular function)

Note: Both the sine and cosine functions are boundedbetween −1 and 1





Sine and Cosine 6

Table of Some Important Values of Trig Functions

x sin(x) cos(x)0 0 1π

6

1

2

√3

2

π

4

√2

2

√2

2

π

3

√3

2

1

2π

21 0

π 0 −13π

2−1 0

2π 0 1





Properties of Sine and Cosine 1

Properties of Cosine

Periodic with period 2π and bounded by −1 and 1

Cosine is an even function

Maximum at x = 0, cos(0) = 1

By periodicity, other maxima at xn = 2nπ with cos(2nπ) = 1 (nany integer)

Minimum at x = π, cos(π) = −1

By periodicity, other minima at xn = (2n+ 1)π withcos(xn) = −1 (n any integer)

Zeroes of cosine separated by π with cos(xn) = 0 whenxn = π

2+ nπ (n any integer)





Properties of Sine and Cosine 2

Properties of Sine

Periodic with period 2π and bounded by −1 and 1

Sine is an odd function

Maximum at x = π

2, sin

(

π

2

)

= 1

By periodicity, other maxima at xn = π

2+ 2nπ with sin(xn) = 1

(n any integer)

Minimum at x = 3π

2, sin

(

3π

2

)

= −1

By periodicity, other minima at xn = 3π

2+ 2nπ with

sin(xn) = −1 (n any integer)

Zeroes of sine separated by π with sin(xn) = 0 when xn = nπ(n any integer)





Some Identities for Sine and Cosine

Some Identities for Cosine and Sine

cos2(x) + sin2(x) = 1 for all values of x (Pythagorean Theorem)

Adding and Subtracting angles for cosine

cos(x + y) = cos(x) cos(y)− sin(x) sin(y)

cos(x − y) = cos(x) cos(y) + sin(x) sin(y)

Adding and Subtracting angles for sine

sin(x+ y) = sin(x) cos(y) + cos(x) sin(y)

sin(x− y) = sin(x) cos(y)− cos(x) sin(y)





Example of Shifts 1

Example of Shifts for Sine and Cosine:

Use the trigonometric identities to show

cos(x) = sin(

x+ π

2

)

sin(x) = cos(

x− π

2

)

This first example shows the cosine is the same as the sine functionshifted to the left by π

2(a quarter period)

This second example shows the sine is the same as the cosine functionshifted to the right by π

2(a quarter period)





Example of Shifts 2

Solution: Use the additive identity for sine

sin(

x+π

2

)

= sin(x) cos(π

2

)

+ cos(x) sin(π

2

)

Since cos(

π

2

)

= 0 and sin(

π

2

)

= 1, sin(

x+ π

2

)

= cos(x)

Similarly,

cos(

x−π

2

)

= cos(x) cos(π

2

)

+ sin(x) sin(π

2

)

Again cos(

π

2

)

= 0 and sin(

π

2

)

= 1, so cos(

x− π

2

)

= sin(x)




Vertical Shift and AmplitudeFrequency and PeriodPhase ShiftExamples

Model of Predator 1

G. F. Gauss (The Struggle for Existence) studied the system of the predatorDidinium nasutum and its prey Paramecium caudatum

Below is a graph of his data and a trigonometric model

0 2 4 6 8 10 12 14 16 18t, days

0

5

10

15

20

25

30

Did

iniu

m n

asu

tum





Model of Predator 2

Didinium Model shown is

D(t) = 14 + 12.5 cos

(

2π

8.5(t− 6)

)

or

D(t) = 14 + 12.5 sin

(

2π

8.5(t− 3.875)

)

1 How do we go from the Gauss data to the models above?

2 What do the various numbers in the models represent?

3 How do we find the numbers?

4 What methods can be employed to improve the model? (Lab)





Model of Predator 3

Observations:

1 The population fluctuates about an average

2 The population appears to have a regular periodic cycle

3 The degree of fluctuation or amplitude of variation is fairlyconsistent

4 The maximum population is shifted from t = 0

These observations can be converted into the 4 key parameters of themodel.





Trigonometric Models

Trigonometric Models are appropriate when data follows a simpleoscillatory behavior (as seen in the example above)

The Cosine Model

y(t) = A+B cos(ω(t− φ))

The Sine Model

y(t) = A+B sin(ω(t− φ))

Each model has Four Parameters





Vertical Shift and Amplitude

Trigonometric Model Parameters: For the cosine model


The model parameter A is the vertical shift, which isassociated with the average height of the model

The model parameter B gives the amplitude, whichmeasures the distance from the average, A, to themaximum (or minimum) of the model

There are similar parameters for the sine model





Frequency and Period



The model parameter ω is the frequency, which gives thenumber of periods of the model that occur as t varies over2π radians

The period is given by T =2π

ω

There are similar parameters for the sine model





Phase Shift



The model parameter φ is the phase shift, which shiftsour models to the left or right

This gives a right horizontal shift for positive φ

If the period is denoted T = 2πω , then the principle phase

shift satisfies φ ∈ [0, T )

By periodicity of the model, if φ is any phase shift

φ1 = φ+ nT = φ+ 2nπω , n an integer

is a phase shift for an equivalent model

There is a similar parameter for the sine modelJoseph M. Mahaffy, 〈[email protected]〉

Lecture Notes – Trigonometric Functions —(27/68)




Model Parameters

Trigonometric Model Parameters: For the cosine and sine models



The vertical shift parameter A is unique

The amplitude parameter B is unique in magnitude but thesign can be chosen by the modeler

The frequency parameter ω is unique in magnitude but thesign can be chosen by the modeler

By periodicity, phase shift has infinitely many choices

One often selects the unique principle phase shift satisfying0 ≤ φ < T





Parameters for Didinium Model

Overview: Consider the cosine model for Didinium

D(t) = 14 + 12.5 cos

(

2π

8.5(t − 6)

)

Best fit would use the Sum of Square Errors (SSE), but narrow peaks indicatethat the best model is not quite sinusoidal, which skews the parameters

1 Observe the average between the high and low is approximated by 14, sothe vertical shift, A = 14.

2 A reasonable approximation to the distance between the average and thehigh or low data is 12.5, so the amplitude, B = 12.5.

3 The distance between peaks (or troughs) is about 8.5, the period, T = 8.5,which gives ω = 2π

8.5= 0.73930.

4 A first maximum of this cosine model occurs at t = 6. Since cosine is at amaximum when its argument is zero, the phase shift, φ = 6.





Improved Didinium Model

We saw that a cosine model reasonably fit the Gauss data.

D(t) = 14 + 12.5 cos

(

2π

8.5(t− 6)

)

D2(t) = 11.47 + 12.56 cos(0.7423(t − 6.064)) + 4.013 cos(1.485(t − 1.686))

Below is a graph of the Didinium Model and a simple variation, which adds oneadditional cosine term with double the frequency (examined in Lab later)

0 2 4 6 8 10 12 14 16 18t, days

0

5

10

15

20

25

30

Did

iniu

m n

asu

tum

Model 1Model 2Data





Example: Sine Function 1

Example 1: Consider the model

y(x) = 3 sin(2x)− 2

Skip Example

Find the vertical shift, amplitude, and period

Sketch a graph

Determine all maxima and minima for x ∈ [0, 2π]






Solution: Fory(x) = 3 sin(2x)− 2

The vertical shift is A = −2

The amplitude is B = 3, so solution oscillates with−5 ≤ y(x) ≤ 1

The frequency is ω = 2, so the period, T , satisfies

T =2π

ω=

2π

2= π

No phase shift, so divide the period into 4 even parts,x = 0, π

4, π

2, 3π

4, π






Graphing: Steps for graphing y(x) = 3 sin(2x)− 2.

1 With no phase shift, start at x = 0. Create a line along theaverage value y = −2, which extends the length of one period.

2 The amplitude of 3 means creating parallel lines above andbelow y = −2 at y = 1 and y = −5, then complete the rectanglefor one period.

3 Finally draw parallel vertical lines along each quarter of theperiod, x = 0, π/4, π/2, 3π/4, π.

4 This drawing is on next slide.






Graphing (cont): For y(x) = 3 sin(2x)− 2, previous slide describes image on theleft, while the image on the right shows the easy 5 function evaluations,y(0), y(π/4), y(π/2), y(3π/4), y(π).

0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2π−6

−5

−4

−3

−2

−1

0

1

2

x

y

0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2πx

-6

-5

-4

-3

-2

-1

0

1

2

y






Graphing (cont): For y(x) = 3 sin(2x) − 2, smoothly connect the points on theprevious graph. Graph on the right shows the periodic extension over anotherperiod.

0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2πx

-6

-5

-4

-3

-2

-1

0

1

2

y

0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2πx

-6

-5

-4

-3

-2

-1

0

1

2

y






From the graph ofy(x) = 3 sin(2x)− 2,

it is easy to obtain the maxima and minima.

By periodicity, the maxima are separated by π, and they occur at

(π

4, 1)

and

(

5π

4, 1

)

By periodicity, the minima are separated by π, and they occur at

(

3π

4,−5

)

and

(

7π

4,−5

)





Example: Cosine Function 1


y(x) = 4− 3 cos(π

4(x− 3)

)

Skip Example

Find the vertical shift, amplitude, and period

Sketch a graph

Determine all maxima and minima for x ∈ [0, 16]






Solution: Fory(x) = 4− 3 cos

(π

4(x− 3)

)

The vertical shift is A = 4

The amplitude is B = 3, so solution oscillates with 1 ≤ y(x) ≤ 7

The frequency is ω = π

4, so the period, T , satisfies

T =2π

ω= 8

The phase shift is φ = 3, so starting at x = 3 add the periodT = 8 and divide x ∈ [3, 11] into 4 even parts, x = 3, 5, 7, 9, 11






Graphing: Steps for graphing y(x) = 4− 3 cos(

π

4(x− 3)

)

.

1 With phase shift, φ = 3, start at x = 3. Create a line along theaverage value y = 4, which extends the length of one period.

2 The amplitude of 3 means creating parallel lines above andbelow y = 4 at y = 1 and y = 7, then complete the rectangle forone period.

3 Finally draw parallel vertical lines along each quarter of theperiod, x = 3, 5, 7, 9, 11.

4 This drawing is on next slide.






Graphing (cont): For y(x) = 4− 3 cos(

π

4(x− 3)

)

, previous slide describes imageon the left.The image on the right shows the easy 5 function evaluations,y(3), y(5), y(7), y(9), y(11).

0 2 4 6 8 10 12 14 160

1

2

3

4

5

6

7

8

x

y

0 2 4 6 8 10 12 14 160

1

2

3

4

5

6

7

8

x

y






Graphing (cont): For y(x) = 4− 3 cos(

π

4(x− 3)

)

, smoothly connect the pointson the previous graph.Graph on the right shows the periodic extension over another period.

0 2 4 6 8 10 12 14 160

1

2

3

4

5

6

7

8

x

y

0 2 4 6 8 10 12 14 160

1

2

3

4

5

6

7

8

x

y






From the graph of

y(x) = 4− 3 cos(π

4(x− 3)

)

,

it is easy to obtain the maxima and minima.

By periodicity, the maxima are separated by 8, and they occur at

(7, 7) and (15, 7)

By periodicity, the minima are separated by 8, and they occur at

(3, 1) and (11, 1)






0 2 4 6 8 10 12 14 160

1

2

3

4

5

6

7

8

x

y

There is a maximum at (7, 7)for the function:

y(x) = 4− 3 cos(

π

4(x− 3)

)

.

If we start at (7, 7),using a cosine model:

y(x) = 4 + 3 cos(

π

4(x− φ)

)

.

Since the cosine function

has a maximum with itsargument being zero, it follows that φ = 7, which is half a periodshift, so

y(x) = 4 + 3 cos(π

4(x− 7)

)

.




Phase Shift of Half a PeriodEquivalent Sine and Cosine ModelsReturn to Annual Temperature VariationOther Examples

Phase Shift in Models

Phase Shift of Half a Period

A phase shift of half a period creates an equivalent sine orcosine model with the sign of the amplitude reversed

Models Matching Data

Phase shifts are important matching data in periodicmodels

The cosine model is easiest to match, since themaximum of the cosine function occurs when theargument is zero

The maximum of the sine model occurs when theargument is π

2





Example: Cosine Model with Phase Shift 1


y(x) = 4 + 6 cos(

1

2(x− π)

)

, x ∈ [0, 8π]

Skip Example

Find the vertical shift, amplitude, period, and phase shift

Sketch a graph

Determine all maxima and minima for x ∈ [0, 4π]

Find the equivalent sine model






Solution: For the model

y(x) = 4 + 6 cos(

1

2(x− π)

)

The vertical shift is A = 4

The amplitude is B = 6, so y(x) oscillates between −2 and 10

The frequency is ω = 1

2

The period, T , satisfies

T = 2π

ω= 4π

The phase shift is φ = π, which means the cosine model isshifted horizontally x = π units to the right

Since cosine has a maximum with argument zero, a maximumwill occur at x = π





Example: Cosine Function with Phase Shift 3

Graphing (cont): For y(x) = 4+6 cos(

12(x− π)

)

, we use the vertical shift (4),amplitude (6), and phase shift (π) to create the appropriate rectangular box.The image on the right shows the rectangular box with the 5 easy functionevaluations, y(π), y(2π), y(3π), y(4π), y(5π).

0 π 2π 3π 4π 5π 6π 7π 8π−4

−2

0

2

4

6

8

10

12

x

y

0 π 2π 3π 4π 5π 6π 7π 8π−4

−2

0

2

4

6

8

10

12

x

y





Example: Cosine Function with Phase Shift 4

Graphing (cont): For y(x) = 4 + 6 cos(

12(x− π)

)

, smoothly connect the pointson the previous graph.Graph on the right shows the periodic extension over another period.

0 π 2π 3π 4π 5π 6π 7π 8π−4

−2

0

2

4

6

8

10

12

x

y

0 π 2π 3π 4π 5π 6π 7π 8π−4

−2

0

2

4

6

8

10

12

x

y






Solution (cont): From the graph of

y(x) = 4 + 6 cos(

1

2(x− π)

)

,

there are clearly maxima at x = π and 5π, so for x ∈ [0, 4π], themaximum is (π, 10), (which agrees with the phase shift).

It is easily seen that the minima occur at x = 3π and 7π, so forx ∈ [0, 4π], the minimum is (3π,−2).






Solution (cont): The appropriate sine model has the same verticalshift, A, amplitude, B, and frequency, ω,

y(x) = 4 + 6 sin(

1

2(x− ψ)

)

We must find the appropriate phase shift, ψ

The maximum of the sine function occurs when its argument is π

2

Since the maximum occurs at x = π, it follows that

1

2(π − ψ) =

π

2or ψ = 0

The equivalent sine model is

y(x) = 4 + 6 sin(

x

2

)

= 4 + 6 cos(

1

2(x− π)

)





Equivalent Sine and Cosine Models

Phase Shift for Equivalent Sine and Cosine Models

Suppose that the sine and cosine models are equivalent, so

sin(ω(x− φ1)) = cos(ω(x− φ2)).

The relationship between the phase shifts, φ1 and φ2 satisfies:

φ1 = φ2 −π

2ω,

which is a quarter period shift

Note: Remember that the phase shift is not uniqueIt can vary by integer multiples of the period, T = 2π

ω





Return to Annual Temperature Model 1

Annual Temperature Model: Started section with data andgraphs of average monthly temperatures for Chicago and SanDiego

Fit data to cosine model for temperature, T ,

T (m) = A+B cos(ω(m− φ))

where m is in months

Find best model parameters, A, B, ω, and φ

The frequency, ω, is constrained by a period of 12 months

It follows that

12ω = 2π or ω = π6 = 0.5236






Annual Temperature Model:

T (m) = A+B cos(ω(m− φ))

Choose A to be the average annual temperature

Average for San Diego is A = 64.29Average for Chicago is A = 49.17

Perform least squares best fit to data for B and φ

For San Diego, obtain B = 7.29 and φ = 6.74For Chicago, obtain B = 25.51 and φ = 6.15






Annual Temperature Model for San Diego:

T (m) = 64.29 + 7.29 cos(0.5236(m − 6.74))

Annual Temperature Model for Chicago:

T (m) = 49.17 + 25.51 cos(0.5236(m − 6.15))

The amplitude of models

Temperature in San Diego only varies ±7.29◦F, giving it a“Mediterranean” climateTemperature in Chicago varies ±25.51◦F, indicating coldwinters and hot summers






Annual Temperature Model for San Diego:

T (m) = 64.29 + 7.29 cos(0.5236(m − 6.74))

Annual Temperature Model for Chicago:

T (m) = 49.17 + 25.51 cos(0.5236(m − 6.15))

The phase shift for the models

For San Diego, the phase shift of φ = 6.74, so the maximumtemperature occurs at 6.74 months (late July)For Chicago, the phase shift of φ = 6.15, so the maximumtemperature occurs at 6.15 months (early July)






Convert Cosine Model to Sine Model:

T (m) = A+B sin(ω(m− φ2))

Formula showsφ2 = φ− π

2ω

where φ is from the cosine modelFor San Diego, φ2 = 3.74For Chicago, φ2 = 3.15

Sine Model for San Diego:

T (m) = 64.29 + 7.29 sin(0.5236(m − 3.74))

Sine Model for Chicago:

T (m) = 49.17 + 25.51 sin(0.5236(m − 3.15))





Population Model with Phase Shift 1

Population Model: Suppose population data show a 10 yearperiodic behavior with a maximum population of 26 (thousand)at t = 2 and a minimum population of 14 (thousand) at t = 7

Assume a model of the form


Skip Example

Find the constants A, B, ω, and φ with B > 0, ω > 0,andφ ∈ [0, 10)

Since φ is not unique, find values of φ with φ ∈ [−10, 0)and φ ∈ [10, 20)

Sketch a graph

Find the equivalent cosine model






Solution: Compute the various parameters

The vertical shift satisfies

A =26 + 14

2= 20

The amplitude satisfies

B = 26− 20 = 6

Since the period is T = 10 years, the frequency, ω,satisfies

ω = 2π10 = π

5






Solution (cont): Compute the phase shift

The maximum of 26 occurs at t = 2, so the model satisfies:

y(2) = 26 = 20 + 6 sin(

π5 (2− φ)

)

Clearlysin

(

π5 (2− φ)

)

= 1

The sine function is at its maximum when its argument isπ2 , so

π5 (2− φ) = π

2

2− φ = 52

φ = −12






Solution (cont): Continuing, the phase shift was

φ = −12

This value of φ is not in the interval [0, 10)

The periodicity, T = 10, of the model is also reflected inthe phase shift, φ

φ = −12 + 10n, n an integer

φ = ...− 10.5,−0.5, 9.5, 19.5, ...

The principle phase shift is φ = 9.5






Solution (cont): The sine model is

−10 −5 0 5 1010

15

20

25

30

t (years)

y

y(t) = 20 + 6 sin(π(t − 9.5)/5)






Solution (cont): The cosine model has the form

y(t) = 20 + 6 cos(

π5 (t− φ2)

)

,

The vertical shift, amplitude, and frequency match the sinemodel

The maximum of the cosine function occurs when itsargument is zero, so

π5 (2− φ2) = 0,

φ2 = 2.

The cosine model satisfies

y(t) = 20 + 6 cos(

π5 (t− 2)

)





Body Temperature 1

Circadian Rhythms:

Humans, like many organisms, undergo circadianrhythms for many of their bodily functions

Circadian rhythms are the daily fluctuations that aredriven by the light/dark cycle of the Earth

Seems to affect the pineal gland in the head

This temperature normally varies a few tenths of a degreein each individual with distinct regularity

The body is usually at its hottest around 10 or 11 AM andat its coolest in the late evening, which helps encouragesleep





Body Temperature 2

Body Temperature Model: Suppose that measurements ona particular individual show

A high body temperature of 37.1◦C at 10 am

A low body temperature of 36.7◦C at 10 pm

Assume body temperature T and a model of the form

T (t) = A+B cos(ω(t− φ))

Find the constants A, B, ω, and φ with B > 0, ω > 0,andφ ∈ [0, 24)

Graph the model

Find the equivalent sine model





Body Temperature 3

Solution: Compute the various parameters

The vertical shift satisfies

A =37.1 + 36.7

2= 36.9

The amplitude satisfies

B = 37.1 − 36.9 = 0.2

Since the period is P = 24 hours, the frequency, ω,satisfies

ω =2π

24=

π

12





Body Temperature 4

Solution (cont): Compute the phase shift

The maximum of 37.1◦C occur at t = 10 am

The cosine function has its maximum when its argument is0 (or any integer multiple of 2π)

The appropriate phase shift solves

ω(10− φ) = 0 or φ = 10





Body Temperature 5

Solution (cont): The cosine model is

T (t) = 36.9 + 0.2 cos(

π12 (t− 10)

)

0 5 10 15 20

36.75

36.8

36.85

36.9

36.95

37

37.05

t (hours)

Tem

pera

ture

(o C)

Body Temperature





Body Temperature 6

Solution (cont): The sine model for body temperature is

T (t) = 36.9 + 0.2 sin(

π12(t− φ2)

)

The vertical shift, amplitude, and frequency match thecosine model

From our formula above

φ2 = 10−π

2ω= 10− 6 = 4

The sine model satisfies

T (t) = 36.9 + 0.2 sin(

π12(t− 4)

)


lecture notes – trigonometric functions

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