Periodic and trigonometric functions.

We will concentrate our attention on periodic functions, although most of thematerial will consist on trigonometric functions. There is a good reason for this:periodic functions are approximated by trigonometric functions and so it is essential,from the point of view of constructing models, to understand well the properties oftrigonometric functions.Since trigonometry has been a subject of enquiry for thousands of years, we will

also present some of its traditional applications.Definition. Let f : R R be a function. We will say that f is periodic of period p 0

if fx p fx,x R.Observation. If f is periodic of period p 0, then the graph of f "repeats itself " on

0,p, p, 2p, 2p, 3p, etc.A graph of a periodic function is shown below.

52.50-2.5-5

10

8

6

4

x

y

x

y

There are many periodic phenomena in nature, which is induced by he day-nightcycle, the seasons, periods of high activity etc. They fall into the category of"circadian rhythms " or, a much abused term, biorhythms.

Examples of circadian rhythms are: heart rate, core body temperature, bloodpressure, sugar content of the blood, neural activity, amount of certain hormones inthe bloodstream, etc.Just to show you an example of how these rhythms are used, we present you with

an abstract of research being carried out on Alzheimer’s disease (we will explain thespecial terminology used in the abstract in Section 4.3):

1

Circadian changes related to sundowns and sleep-wake disturbances inAlzheimer’s diseaseLadislav Volicer, David Harper, and Andrew SatlinAbstractCircadian rhythms of core body temperature and motor activity were measured in

25 patients with diagnosis of probable Alzheimer’s disease (AD) and 9 healthycontrols. AD subjects had lower diurnal motor activity, higher percentage of nocturnalactivity, lower interdaily stability of motor activity and delayed acrophase whencompared with controls. AD subjects had also higher mesor temperature, increasedamplitude of the temperature rhythm and delayed acrophase when compared withcontrols.AD subjects were divided according to the occurrence of sundowning determined

by staff reports. Severity of sundowning was associated with later acrophases of bothmotor activity and core body temperature, decreased correlation of circadian rhythm ofbody temperature with 24 hr. cycle, and decreased amplitude of temperature variation.These data indicate that sundowning is related to a disturbance of circadian rhythmscaused by AD.

The functions they ended with are in the graphed below:

2

In general, if we know that we have a periodic function and have data to try toconstruct a mathematical model, it still remains a difficult task to complete, and onewhere the use of the trigonometric functions, to be introduced in the next section, playan essential role.In this course we will only study periodic functions that arise from trigonometric

functions (which are essential for many other studies, whenever angles andmeasurements are needed, as for example, in orthopedics, to determine when a jointis within normal or expected ranges or not, etc.).

4.1 Radian measure of angles and the sine and cosinefunctions.This subject is one with a long history, with the first recorded works dating back to

the ancient Egypt, works that were influenced by knowledge acquired by theBabylonians (transmitted, supposedly, by oral traditions). The subject matter is thestudy of angles and measurements that van be done with them.Definition of angle. An angle is generated by the rotation of a closed half-line

(ray) about its endpoint. The angle is said to be positive if the rotation iscounterclockwise, negative if the rotation is clockwise (ad zero if there is no rotation atall). The endpoint is called the vertex of the angle.

3

To measure angles we will usually do it with the radian measure of an angle, whichwe describe below:Radian Measure of an Angle. Consider an angle and assume that the closed half

line that generates it is the nonnegative x-axis in the coordinatized plane (equal unitsin both axes). Take the point 1,0 and follow its trajectory as the angle is generated;the path it traces lies in the circle centered at the origin with radius one, from now oncalled the unit circle. The distance traveled by this point is the radian measure of theangle if the angle is positive, zero if the angle is zero, and its additive inverse if theangle is negative.The point x0,y0 where 0,1 ends up in the unit circle is called the terminal point of

the angle.

Two angles are said to be coterminal if they have the same terminal point.Examples.1. A positive angle of a full revolution has radian measure 2. Since we have

already been told that an angle of a full revolution measured in degree has measure360, the it follows that one degree measured in radians has measure 2

360 180.

Suppose that is an angle with terminal point x0,y0.

4

Definition. cos x0, sin y0.Immediate Identities:1. Since the point x0,y0 is in the unit circle, we must have that x02 y02 1, thus, if

is any angle:cos2 sin2 1.

2. Since the angle has as terminal point x0,y0,cos cossin sin.

3. Since adding or subtracting any number of full revolutions about the origin yieldthe same terminal point, we have, for any R:

cos 2k cos,k Z, in radians.sin 2k sin,k Z, in radians.

4. Define the functions cos : R R, sin : R R in the obvious way (that iscos x0, from the terminal point and the same for sin), then both of these functionsare periodic of period 2.5. The sign of cos, sin depends on the quadrant where the terminal point lies.

If the terminal point lies in the first quadrant, then cos 0, sin 0, if it lies in thesecond quadrant, cos 0, sin 0, if it lies in the third quadrant,cos 0, sin 0 and if it lies in the fourth quadrant, cos 0, sin 0.6.

cosk 1, if k Z is evencosk 1if k Z is oddsink 0

7. cos 2 k 0,k Z,

sin 2 2k 1,k Z

sin 32 2k 1,k Z

5

Identity 1, cos2 sin2 1, is a fundamental fact which is used in manysituations. One example is that of its use in proving the so-called trigonometricidentities.

Examples of simple trigonometric identities:

1. Show that sin2t3sint4

1cos2tsint sint4

sint , whenever 1 cos2t sint 0.

To do this we see that using Identity 1 in the denominator, and factoring thenumerator (as we would factor a quadratic polynomial), we get:

sin2t 3sint 41 cos2t sint

sint 4sint 11 1 sin2t sint

sint 4sint 1sin2t sint

sint 4sint 1sintsint 1

sint 4sint .

2. Show that sin4t 2sin2tcos2t sin2t3sin2t 2.Here we use the fundamental identity to express cos2t as 1 sin2t.

sin4t 2sin2tcos2t sin4t 2sin2t1 sin2t sin4t 2sin2t 2sin4t sin2t3sin2t 2.

Special angles.We have seen already that one degree measured in radians has measure

180 .By the same token, one radian has degree measure equal to 180

.We will always use radian measure whenever dealing with trigonometric functions.

However, we will make use of the degree measure of an angle to calculate specialvalues of the trigonometric functions, so that we can use facts learned before abouttriangles and their properties (for example that the angles in a triangle add up to 180.1.

4 , this radian measure corresponds to an angle of 45 and when considering

sin 4 , cos4 we first observe that the terminal point lies in the first quadrant, so both

are positive, and that, considering the triangle with vertices 0,0, cos 4 , sin4 and

cos 4 , 0, we have an isosceles right triangle with hypothenuse equal to 1. Thus

6

cos 4 sin4 0, and :

cos2 4 sin2 4 1

2cos2 4 1

cos2 4 12 ,cos

4 0

cos 4 12.

Thus, cos 4 sin4

12. This is the way we will state solutions to our

problems, so we will not require "rationalizing the denominator ".2.

3 (the degree measure is 60). In this case we extend our triangle (by adding a

side at the point cos 3 , sin3 to an equilateral triangle with the base in the x axis

(draw a picture). Then the height of the triangle is sin 3 and the midpoint of the baseis cos 3 . It follows that cos

3

12 and so:

cos2 3 sin2 3 1

14 sin2 3 1, sin

3 0

sin2 3 34

sin 3 32 .

3. 6 (this corresponds to a measure of 30

). In this case take the right triangle withvertices at 0,0, cos 6 , 0 and cos

6 , sin

6 and "flipped it about the x axis. You

get an equilateral triangle . Since the point cos 6 , 0 is the midpoint of the sideopposite the origin, we must have that sin 6

12 , and by the same type of reasoning

we had in 2, cos 6 32 .

4. 23 (this corresponds to 120). Here we observe that 23 2

6 , and we use

this information to observe that if we consider the angle 2 , which has terminal point in

the positive y axis, then we have to add the angle 6 to it, and if we view this using

this axis as reference point, then the value of sin 3 cos6

32 , and that

|cos 23 | cos 23 sin 6 12 .

Use of the terminal point information.

The position of the terminal point determines if the values of sin, cos of thatparticular angle are positive or negative. If one value is given, then the value of theother is readily obtainable.

7

Examples.1. An angle is such that its terminal point lies in the first quadrant and sin 1

4 .Find cos.Solution.Since the terminal point lies in the first quadrant, both of the trigonometric functions

of this angle have positive value and cos2a sin2 1. We conclude that:

cos2 116 1,cos 0

cos 1 116 15

4 .

2. An angle has terminal point in the third quadrant and cos 34 . Find sin.

Solution.The same type of reasoning tells us that sin 0, so that:

cos2 sin2 1, sin 0

sin 1 916

716 7

4 .

3. Let be an angle with terminal side in the third quadrant and such thatcos 35 .Find sin.Since the terminal point is in the Third quadrant, sin 0.

cos2 sin2 1, sin 0925 sin2 1

sin2 1 925 16

25

sin 1625 45 .

Exercises 4.1

1. Find sin 56 .2. Find cos 54 .

8

3. Find the terminal point for the angle with measurement of 136 .

4. Find the terminal point of 83 .

5. An angle has terminal point in the third quadrant and cos 44 . Find sin.6. An angle has terminal point in the fourth quadrant and sin 34 . Find

cos.7. Suppose that an angle is such that cos 0. Which are the quadrants

where the terminal point may lie?8. Let be an angle such that cos 0, sincos 0.Which are the quadrants where

the terminal point may lie?9. Let 0,2, cos sin.Which are the quadrants where the terminal point

may lie?10. If cos 6

7 , which are the possible values for sin?11. Prove that cost

1sint sintcost 1

cost , whenever cost 0.12. Show that cos4t sin4t 1 2cos2t sin2t.13. Prove that cos4t 5cost 4 sin2tsin2t 3.14. Find all solutions of the equation cos4t 5cost 4 0.15. Find all solutions of the equation sin2t 1

2 sint 12 0.

16. Prove that sintcost 2 1

cos2t 1, t

2 k,k Z.

4.2 Addition and subtraction formulas.

In this section we concentrate in finding the values of the functions sine andcosine for the sum of two angles, say ,. What we are after is expressing the valueof the desired trigonometric function in terms of the values of these functionsevaluated at and .Theorem 4.2.1. Let a,b be angles. Then

cosa b cosacosb sina sinb.Proof.We consider first the case when a b.

9

Consider the triangle with vertices at 0,0, cosb, sinb and cosa, sina and theone with vertices at 0,0, 1,0 and cosa b, sina b. The last triangle is a rotationof the first triangle so that the side joining 0,0 and cosa, sina lies on the x-axis.It follows that the sides opposite to the origin have equal length, concluding that:

cosa cosb2 sina sinb2 cosa b 12 sina b2

cosa cosb2 sina sinb2 cosa b 12 sina b2

cos2a 2cosacosb cos2b sin2a 2sina sinb sin2b cos2a b 2cosa b 1 si2 2cosacosb sina sinb 2 2cosa b

cosa b cosacosb sina sinb.This proves the case a b. If b a, then:

cosa b cosa bcosa b cosb acosa b cosbcosa sinb sina

and we get the same result.

Consequences:1. Let a,b be angles. Then

cosa b cosa b cosacosb sina sinb cosacosb sina sinb.

2. Let be an angle. Then;

10

cos 2 coscos

2 sin sin

2 sin.

cos 2 coscos

2 sin sin

2 sin

sin 2 cos

2

2 cos

sin 2 cos

2

2 cos coscos sin sin

cos.3. Let a,b be angles. Then

sina b sinacosb cosa sinb, sina b sinacosb cosa sinb.Proof.

sina b cosa b 2

cosa 2 cosb sina

2 sinb

sinacosb cosa sinb.

sina b sina b sinacosb cosa sinb sinacosb cosa sinb.

Examples.1. Find cos 23 cos

3 coscos

3 sin sin

3

12 .

With these tools we can get enough points to graph both sine and cosine.

1050-5-10

2

1

0

-1

-2

x

y

x

y

graph of cosineFor the graph of sine we have:

11

1050-5-10

2

1

0

-1

-2

x

y

x

y

graph of sineWe have seen that if , are angles, then:

cos coscos sin sincos coscos sin sinsin sincos cos sinsin sincos cos sin.

These formulas are useful in many contexts.The first way we can use them is to find the exact value of trigonometric functions

for angles that can be expressed as sums or differences of those which we alreadyknow their values.

Example 1. Find cos 512 .

Observe that the degree measure of this angle is 512

180 75 so that we can

write 512

4 6 . Thus:

cos 512 cos4

6

cos 4 cos6 sin

4 sin

6

12

32 1

212

3 12 2

.

12

Double Angle Formulas.

If is an angle thencos2 cos2 sin2sin2 2sincos.

Example 2. Find sin 23 .

sin 23 sin23

2sin 3 cos3

2 3212

32 .

Half Angle Formulas.

Let be an angle and observe that 2 2 . Hence:

cos cos2 2 cos2 2 sin

2 2 .

This, together with the identity cos2 sin2 1, shows that, for cos2 2 :

cos cos2 2 1 cos2 2

2cos2 2 1

2cos2 2 cos 1

cos2 2 cos 1

2 .

To find which of the two roots of the right hand side we must use, we have toconsider where that terminal point of the angle lies.

Example 3. Find cos 8 . the terminal side of this angle is in the first quadrant, socos 8 0. Also

8 1

24 . Hence:

13

cos2 8 cos 4 1

2

12 1

2

1 22 2

.

Therefore, since cos 8 0,

cos 8 1 22 2

.

How about sin2 2 ?

cos cos2 2 sin2 2

1 sin2 2 sin2 2

1 2sin2 2

2sin2 2 1 cos

sin2 2 1 cos

2Again, the sign of sin 2 will depend on the terminal point of the angle

2 .

For example sin 8 0, sin2 8

1cos 4 2 . We conclude that:

sin 8 1 1

2

2 2 12 2

.

Exercises 4.2

1. Find sin 56 , cos56 .

2. Let be an angle. Find cos 6 , sin

6 in terms of cos and sin.

3. Let be an angle. Find cos3, sin3 in terms of sin and cos.4. If 0 , find sin 2 and cos

2 in terms of cos.

5. If 2, find sin 2 and cos2 in terms of cos.

6. Find sin 8 , cos8 .

7. Find sin 512 , cos512 .

8. Find sin 16 , cos

16 .

9. Find sin 8 cos8 .

14

10. Find sin 512 cos512 .

11. Find all angles (measured in radians) such that sin 32 .

12. Find all angles (measured in radians) such that cos 32 .

13. Find all angles (measured in radians) such that sin 32 .

14. Find all angles (measured in radians) such that cos 32 .

15. Prove that sin4t 4sintcost1 2sin2t.16. Prove that cos3t 1cos2t

2 cost.17. Prove that cos3t cost4cos2t 3.

18. Show that if 0 then sin2 4 1 cos1

2

2 .19. Show that if cost sint then 1 1

2 sin2t sin3tcos3tsintcost .

20. Show that 2cos2t 1 cos2t sin2t.21. Prove that 1 cos5t sin3t 2sin5tcos5t 4sin2t.22. Show that cos2xcosx 2cosx 1

cosx .

4.3 Simple harmonic motion and cosinor analysis

We have discussed the addition and subtraction identities for the sine and cosinefunctions. As we have seen, one use of them is to compute the values of thesefunctions for angles that can be written as a sum or difference of angles for which wealready know the values of sine and cosine.One very used fact is that:

cos coscos sin sin.This usefulness follows from the following reasoning:Many functions appear in the following form: if B,C are real numbers which are not

both zero (in mathematical symbology: if B2 C2 0) if 0, and if D is a realnumber, consider the function f : R R,

ft Bcost C sint D.An important question is: how does it graph look like?We re-write the function in the following form:

ft B2 C2 BB2 C2

cost CB2 C2

sint D.

Now we see that the point in the plane BB2C2

, CB2C2

is a point in the unit circle,

so there exists an angle 0 0,2 such that cos0 BB2C2

, sin0 CB2C2

.

Then we have:

15

ft B2 C2 cos0cost sin0 sint C

B2 C2 cos0 t C

B2 C2 cost 0 C

B2 C2 cost 0 C.

If we set A B2 C2 and t0 0 , we get:

ft Acost t0 C.This proves that the graph of any such function is the graph of a cosine function

which is multiplied by a positive constant and translated both horizontally andvertically.The coefficient A 0 is called the amplitude can be regarded as the "amount of

variation " between the highest and lowest points in the graph, the period p 2

indicates the length of the intervals over which repetition occurs.The period p is determined in the following way, recalling that 2 is the smallest

positive period of the cosine function:ft p ft, t R

Acost p t0 C Acost t0 C, t cost t0 p cost t0, t R.

From the fact that 2 is the smallest positive period of cosine, it follows that:p 2

p 2 .

The number 1p

2 is called the frequency and indicates the rapidity of"oscillation ", the number t0 is called the phase shift and shows how far away fromzero is the horizontal shift and C indicates how far up or down the function has beenshifted.We first will treat some examples to clarify the specific terms.

Example 1. Consider the function f : R R, ft 2 3 cos2t 2sin2t 6.Here we have that A 12 4 16 4, so our first step is to write it as:

ft 4 32 cos2t 12 sin2t 6

4cos 6 cos2t sin6 sin2t 6

4cos2t 6 6

4cos2t 12 6

16

Example 2. Let f : R R, ft 4cos3t 4sin3t 2.A\ 16 16 4 2

ft 4 2 12cos3t 1

2sin3t 2

Since the cosine is negative and the sine is positive the angle has terminal point inthe second quadrant, so the angle we choose is 3

4 .

ft 4 2 cos 34 cos3t sin 34 sin3t 2

4 2 cos3t 34 2

4 2 cos3t 14 2

52.50-2.5-5

7.5

5

2.5

0

-2.5

x

y

x

y

Graph of last example

In biology, more particularly, in the study of circadian rhythms, when the dataindicates that the function being studied seems to be of sinusoidal type, that is, itseems to reflect the graph of a shifted cosine function, the term "cosinor analysis " isused. Of course, the first task is to determine if the periodic function being studiedseems to be sinusoidal or if it clearly is not. The following graphs illustrate this point:

17

52.50-2.5-5

10

7.5

5

2.5

0

-2.5x

y

x

y

Non-sinusoidal graph

52.50-2.5-5

6.25

5

3.75

2.5

1.25

x

y

x

y

sinusoidal plotWhen the phenomenon being studied appears to be sinusoidal, then the biologists

usually try to fit a function of the type

ft Acos 2t t0P M.

This is what is called "cosinor analysis ", a special case of the "time series "technique.In this setting, M is called the "mesor ", an acronym for "midline estimating statistic

of rhythm ", A is the amplitude, P 2 2P

is the period and t0 is called the "acrophase "

which measures the timing of the maximum for the cosine function (or, how far away

18

from zero does the first maximum occur). The plots of periodic functions in theAlzheimer’s study presented in the introduction to periodic functions.Of course, once the determination is made that cosinor analysis is the appropriate

tool to use, then data (obtained from observations or experiments) is used todetermine the mesor, amplitude, period and acrophase. This is not an easy task andresearchers work hard to determine them by using statistical and curve fittingtechniques that are beyond the scope of this course; the objective in presenting thishere is to show how some periodic functions are studied when they arise inapplications to biology.We will content ourselves with dealing with functions that are presented to us and

interpreting the terms that appear using this terminology.For example, in the article Age-dependent Changes in 24 hour rhythms...by P.

Cano, D.P. Cardinali et al, which can be accessed through BioMed Central(http://bmc.ub.uni-ptsdam.de/1472-6793-1-14) one of their sets of data yields thefollowing information:Mesor: 149, amplitude 69, Acrophase 2.5 hours, period 24 hours. Thus the

function they arrived at is:

ft 69cos t 2.512 149.

40200-20-40

300

250

200

150

100

50

0

x

y

x

y

Plot of Cano et al

We have already discussed the setting in which biologists do cosinor analysis, andwe have discussed that it is functions of the type f : R R given by:

ft Acos 2t t0P M.

19

We will always choose t0 0, that is the acrophase is always nonnegative, andalways A 0,P 0.

Notice that in this setting, the maximum value Q attained isQ A M

while the minimum value q isq A M

Notice thatQ q 2A

A Q q2 .

Observation: the mesor is the number that is the midpoint of the intervaldetermined by the minimum value and the maximum value.

Example 1. Consider the function ft 3cos 2t1/27 5 (time measured in days).Determine the amplitude, acrophase, period and mesor.Solution.This can be read directly from the function:amplitude 3, acrophase 1/2, period 7, mesor 5.

Example 2. Data determine that a certain phenomenon is a sinusoidal function oftime and cosinor analysis is to be used. The period is one day, the maximum valueattained is 30, the minimum value is 10 and the value at 0 is 25. Determine thefunction that is searched for.Solution.From the data, A 10,M 20,P 1, so the function is of the form:

ft 10cos2t t0 20.We only have to find t0.Notice that f0 10cos2t0 20 25.Thus cos2t0 cos2t0 1

2 . since we are searching for the smallest positivenumber for which cos2t0 1

2 , we conclude that

2t0 3

t0 16 .

20

The function we are searching for is:

ft 10cos2t 16 20.

The acrophase is 16 of a day.

Problems.1. Given the function ft 10cos3t 10sin3t 5, write it as

ft Acos3t t0 Dand state the amplitude and period.2. Same for the function ft 3 cost 3sint 5.3. Same for the function ft 3 cost 3sint 5 .4. Analysis of data has determined that the cosinor method leads to the function

ft 5cos t 25 6, t in days.

Determine the amplitude, period, acrophase and mesor.5. Same for the function ft 7cos t1/25 10.6. Data shows that a certain phenomenon can be modeled using the cosinor

method and that the maximum value is 110, the minimum value is 10, the period istwo hours and the value at zero is 25 3 60. Find the function, determine theacrophase.7. Same for: maximum value 40, minimum value is 20, the period is three hours

and the value at zero is 30 102.

8. A circadian rhythm is being modeled with cosinor analysis. The period is fourhours, the amplitude is 80, the minimum value is 90 and the value at zero is 170. Findthe function determined by this method.9. If there is a function determined by the cosinor method and you know the

amplitude and minimum value, how can you find the mesor?10. If there is a function determined by the cosinor method and you know the

amplitude and maximum value, how can you find the mesor?

4.4 The other trigonometric functions

We will enter a brief discussion of the functions that you have called tangent,cotangent, secant, cosecant. All of these functions are obtained using quotients whereeither sine or cosine in the denominator.Since division by zero is mathematical anathema, we need to find the sets which

will be the appropriate domains for these functions.

21

Notice that cos 0 when 2 k,k Z. We define:

D1 x R : there is no k Z such that x 2 k.

Also sin 0 when k,k Z. we define:D2 x R : there is no integer k such that x k.

We define:1. tangent, tan : D1 R, tanx sinx

cosx .Its graph looks like:

52.50-2.5-5

50

25

0

-25

-50

x

y

x

y

Graph of a portion of tangentFor example:

tan 4 sin 4 cos 4

1

tan 6 sin 6 cos 6

13

tan 3 sin 3 cos 3

3 .

2. cotangent, cot : D2 R, cotx cosxsinx .

3. secant, sec : D1 R, secx 1cosx .

22

52.50-2.5-5

50

25

0

-25

x

y

x

y

Part of the graph of secant4. cosecant, csc : D2 R, cscx 1

sinx .

Some identities:Recall that

sin2x cos2x 1sin2xcos2x

1 1cos2x

tan2x 1 sec2x.Also recall that sin sincos cos sin,cos( coscos sin sin.Let , be angles such that tangent is defined on both of them as well as on .

Then:

tan sin cos

sincos cos sincoscos sin sin .

tan tan1 tan tan .

Another important observation is that tanx tanx.

Exercises 4.4

23

1. Find tan 512 .2. Find sec 512 .3. Show that if tan and tan2 are defined, then tan2 sin2

2cos21.

3 Find tan 43 .4. Find sec 43 .5. Show that if

4 k2 ,k an integer, then sec2

12cos21

.

6. Find all angles (measured in radians) such that tan 1. 07. Find all angles (measured in radians) such that tan 3 .8. Find all angles (measured in radians) such that tan 1

3.

9. Find all angles (measured in radians) such that tan 0.10. Find all angles (measured in radians) such that tan 0.11. An angle has terminal side in the second quadrant and tan 3. Then

sin ?12. Prove that tan2t sect sec2t sect 1. Which values of t must be

excluded?13. Prove that sec2t sec2t

2sec2t. Which values of t must be excluded?

14. Prove that tan2t sec2t 2sec2t 1.15. Prove that tantcott

sectcsct 1costsint .

16. Prove that tan2t sin2t sin4tcos2t

.

17. Prove that cott csct sint1cost .

18. Prove that cos3tsin3t

costsint 2sin2t2 .

19. Find sec 8 .20. Find tan

12 .

4.5 Solving triangles

In our incursion into periodic and trigonometric functions, our emphasis has been,so far, in determining functions based on the cosine in order to study periodicfunctions which often appear in biology.One traditional use of trigonometry is in "solving triangles ", by which we mean

given data from a triangle, determine the rest of the characteristics of the triangle. Ofcourse the scope is too broad as stated and we will concentrate on some techniquesthat will provide answers when the right information is available.These problems also find many uses in biology (and medicine) but the angles that

24

appear require, almost always, the use of calculators (or tables), a feature we areconsciously avoiding since our interest is in the mathematical thought behind thetechniques, not in button pushing on a calculator nor the use of specializedsoftware,something that almost anyone can do without knowing the principlesinvolved.When dealing with triangles, the information we need or seek takes the form of the

length of the sides and the measurement of the angles made by two sides thatintersect. These triangles are usually drawn without the use of a coordinatized planebut, in order to prove some of our assertions we will make use of the Cartesiancoordinates and will position the triangle with one of the vertices at the origin in orderto facilitate the proofs.The facts that will be taken for granted in this discussion are: the angles in a

triangle are always assumed to be positive, the sum of the angles in any triangle is radians (180), similar triangles have the same corresponding angles and all thequotients of corresponding sides are equal. We usually refer to the "side opposite thevertex " or the "side opposite the angle " ; we distinguish between right triangles (oneangle measure

2 radians), acute triangles (all angles measure strictly between zeroand

2 and obtuse triangles (one angles is greater than2 . if we do not want to

specify the type of a triangle that is not a right triangle we will call it an "oblique "triangle. We assume that you know how to compute areas of triangles.When considering right triangles, the sides that meet in a right angle are called the

"legs "of the triangle and the side opposite the right angle is the hypothenuse (andPythagoras ’Theorem applies).Given a triangle, we assume it is well known what the height of a vertex is, what

the medians are and that all these set of three lines intersect at one point.Suppose that we have a right triangle with vertices A,B,C, C the vertex where the

right angle is c is the hypothenuse and a,b are the legs. Let be the angle at vertex A

If we set this triangle in the Cartesian plane, with vertex A at the origin, vertex C in

25

the x axis and the hypothenuse (or its extension) intersecting at point Q in the unitcircle ( so it is the terminal point of the angle ), we see by similarity of triangles, that:

cos ac

sin bc

tan ab .

Example 1. We have a right triangle with one angle measuring 6 radians and with

opposite side measuring 5 inches. Find the length of the hypothenuse.Solution.Let b be the length of the other leg (in the same units). Then:

tan 6 13

5b 1

3

b 5 3 .Now, we know that the hypothenuse is equal to the square root of the sum of the

squares of the legs (Pythagoras’ Theorem), so if we denote by h the length of thehypothenuse is:

h 52 5 3 2

25 75

100 10 inches.

Note. Once you choose the units, the results will always be the same numericalvalue with that unit system.There are many problems involving right triangles and all of them are dealt with the

same ease (be it the height of a tree when its shadow has a given length and theangle of incidence of the sun light is given, if the length of a ladder is given togetherwith the information that it rests on a vertical wall and the distance from the base ofthe ladder to the wall is also given, etc. etc.)Now we will concentrate on triangles that are not right triangles.

Law of Sines. Consider a triangle with vertices A,B,C, corresponding angles (corresponding to vertex A), ,, and sides a (opposite to A ),b,c. Then:

sina

sinb

sinc .

Proof. We will illustrate a rigorous proof using a particular triangle, as follows:

26

We see that, if h is the length of the perpendicular line from vertex B to the side AB,then:

h c sinh a sin.

This implies thatc sin a sinsinc sin

a .

Now consider the perpendicular (height) from C to the side (or its prolongation) tothe side AB, call it h :From this we get that:

From this we get:h b sinh c sin.

27

It follows that:sinb

sinc .

Putting all our equalities together, we get:sina

sinb

sinc .

There are several situations where this result allows us to "solve " for the trianglegiven. We illustrate them with examples.

Example 1. (One side and two angles given) A triangle ABC is such that a 15,

4 , 3 .Find b,c.

Solution.Since 7

12 and 712 5

12 , we need to compute

sin sin 4 6 sin

4 cos

6 cos

4 sin

6

12

32 1

212 3 1

2 2.

Thus:sina

sinb

3 130 2

12 b

b 303 1

.

sina

sinc

3 130 2

32c

2c 30 2 33 1

c 15 63 1

.

When two sides and an opposite angle are given, then we may have that the"opposite angle " is too big for the remaining side and then no solution exists, asillustrated below:

28

On the other hand there could be two triangles that meet our data, as illustratedbelow:

Or, it could happen that there is only one solution.

Example 2. For the triangle ABC we are given: 6 ,a 3,c 7. Find, if possible,

b.Solution.From the law of sines we get:

sin 6 3

sin7

sin 76 1.

No triangle will meet ours specifications. There is no solution.

29

Example 3. Solve the triangle ABC with the following specifications:

6 ,a 10,c 2 10.Solution.

sin 6 10

sin10 2

120

sin10 2

sin 10 220 1

2.

There are two angles between 0 and with this value of sine, namely 4 and

34 . Consider first

4 .

Thus 6 4

512 ,

sinb

sin 6 10

3 12 2

b 120

b 20 3 12 2

b 10 3 12

.

If we use 34 , then

6

34 11

12 . sin1112 sin

12 sin

12 .

sin 12 sin

4

6 sin

4 cos

6 cos

4 sin

6

12 32

12

3 12 2 .

sin 6 10

sinb

120

3 12 2 .

b

b 10 3 12 .

Example 4. Solve the triangle ABC given that a 5 32 ,

3 ,c 5.Solution.

30

5 3232

5sin

sin 5 3

25 32

sin 1

2 .

In this case we have only one solution, with 2 , and so

6 .Thus

bsin

5 3232

b 5 2 .

Law of Cosines. Consider a triangle with vertices A,B,C, corresponding angles,, and opposite sides a,b,c. Then:

a2 b2 c2 2bccos.Proof.For this proof we will put the triangle in the Cartesian Plane, with vertex A at the

origin (coordinates 0,0,C on the x axis (coordinates b, 0 and vertex B in theupper half plane ( coordinates x0,y0). Here is a picture of a particular case:

We see that:

31

x0 ccosy0 c sin.

Thus:

a ccos b2 c sin2

a2 c2 cos2 2cbcos b2 c2 sin2 c2 b2 2cbcos.

This proves the result. Now we only have to familiarize with some of itsapplications; observe that the result is very dependent on our labeling of vertices,angles and sides.

Example1. Given the triangle ABC where b 25 cm., c 30 cm. and 6 , find

a.Solution.From the law of cosines we get that:

a2 252 303 2.cos 6 25 30

a2 625 900 750a2 875,a 0

a 875 5 35 cm.

Exercises 4.5

1. Suppose that a triangle is such that a 10,c 20, 60. Then b ?2. From the roof of a building, an observer can see the top of a skyscraper with an

angle of elevation of 4 radians and the base of the skyscraper with an angle of

depression of 6. if the two buildings are 300 feet apart. Find the height of the

skyscraper.3. Let ABC be a triangle such that b 15,c 20,

3 . Find a, sin and sin.4. Let ABC be a right triangle (

2 ) such that 6 and a 10. find ,b,c.

5. Let ABC be a triangle such that a 10,b 20, 3 . Find c.

6. Let ABC be a triangle such that a 10, 4 ,

3 . Find b,c,.

7. Solve, if possible, the triangle ABC given that 3 ,a 14,c 30 (if there are

no solutions, prove it, if there are two solutions, find both).8. Given a triangle ABC such that

6 ,b 10,c 20, find a.

4.6 Inverse trigonometric functions

32

We will only consider the functions sin, cos, tan. We know that sin, cos are periodic ofperiod 2 and that the domain of tan is D1, and that

2 ,2 D1. The conclusion

from these preliminary observations is that none of these functions have an inversefunction in the domain in which they were defined.It is, however, very useful to find angles with given values of one of their

trigonometric functions. In order to do this we will restrict the domain of each of thesein order to make this objective possible.Let us first consider the function sin. If we restrict it to the new domain

2 ,2 ,

then we see that all values of the function (that is, its range) are covered by values ofthe function on this interval and it is very intuitive, from the unit circle representation ofthe points cos. sin that the function is one-to-one on this interval. Thus weconsider the function sin restricted to the domain

2 ,2 . that is we now consider

sin : 2

2 1,1 with (approximate) graph:

10.50-0.5-1

1

0.5

0

-0.5

-1

x

y

x

y

Graph of sinThus, we accept that there is, for this particular case, an inverse function

sin1 : 1,1 2 ,

2 . Thus sin

1 12 6 , sin

1 32

3 , sin

1 32

3 , and so

on ( no addition of multiples of 2).

33

10.50-0.5-1

1.5

1

0.5

0

-0.5

-1

-1.5

x

y

x

y

Graph if sin1Since the restriction we are considering of the function sin is the sine function as

defined before (with a different domain), we always have that sinsin1x x forx 1,1. However, we must be careful with the reverse composition, as the followingexamples show:Example 2. Find sin1sin 23 .Solution.Since 2

3 2 , we must find an angle whose radian measure is between

2 and

2 that has the same sine. Since

23

3 , we conclude that :

sin1sin 23 3 .

Example 3. Find sin1sin 136 .Solution.Since 13

6 does not belong to 2 ,

2 , we must search for one with radian

measure in this interval and with the same value of sine. Observe that 136 2

6we immediately see that:

sin1sin 136 6 .

Example 4. Find sin1sin 53 .Solution.We must proceed as above. Now we observe that 53 2

3 , concluding that:

sin1sin 53 3 .

Now we concentrate on the function cosine. For our purposes we restrict it to the

34

interval 0, and consider it as a function from 0, onto 1,1, and we have thefollowing approximate graph:

32.521.510.50

2

1

0

-1

-2

x

y

x

y

Graph of the restriction of cosineWe see that it is reasonable to infer that the function is one-to-one, and we will

make this assumption (which is valid, as a matter of fact). Thus we get the inverse ofthis restriction of cosine and write it as cos1 : 1,1 0,.

10.50-0.5-1

3

2.5

2

1.5

1

0.5

0

x

y

x

y

Graph of cos1

We observe that cos1 12

4 , cos1 1

2 3

4 , cos1 12

23 , cos

1 32

6 ,

and so on.

35

Again we always have that:coscos1x x.

However we must be careful with cos1cos, and remember that the angle weget belongs to the interval 0,.

Example 5. Find cos1cos 3 .

Solution.Since cos

3 cos3 , we have:

cos1cos 3 cos1cos 3 3 .

Example 6. Find cos1cos 53 .Solution.53 2

3 so we conclude that:

cos1cos 53 cos1cos2 3

cos1cos 3

3 .

Example 7. Find sincos1 25 .Solution.Since cos1 25 0,, sincos

1 25 0.

sin2cos1 25 cos2cos1 25 1

sin2cos1 25 425 1

sin2cos1 25 2125

sincos1 25 215 .

As it regards the function tangent, we restrict it to the interval 2 ,

2 and recall

that its range is R. Thus tan1 : R 2 ,

2 .

36

1050-5-10

2.5

1.25

0

-1.25

-2.5

x

y

x

y

Graph of tan1tan11

4 , tan1 3

3 and so on.

Example 8. Find tan1tan 23 .Solution.The terminal point of 2

3 is in the second quadrant, so the value of tangent isnegative. We conclude that:

tan1tan 23 tan1tan 3

tan1tan 3

3 .

Example 9. Find sintan16.Solution.We know that tan16 0, 2 (why?), so both cosine and sine of this angle are

positive.

37

tantan16 6sintan16costan16

6

sintan161 sin2tan16

6

sin2tan16 361 sin2tan1637sin2tan16 36

sin2tan16 3637

sintan16 3637 6

37.

There are many other examples we could show you, and some will be included inthe exercise set that follows. We now concentrate on some of the uses of thesefunctions when dealing with cosinor analysis and combinations of sine and cosine.

Example 10. A circadian rhythm of period 24 hours has been determined to fit thecosinor analysis method. The maximum value is 175, the minimum value is 105 andthe value at zero is 112. Find the amplitude, mesor and acrophase and show thefunction obtained.Solution.

ft Acos 2t t024 M

Acos t t012 M

A 1751052 35,M 140,

38

ft 35cos t t012 140

f0 112

35cos t012 140 112

35cos t012 28

cos t012 2835

t012 cos1 2835

t0 12 cos1 2835

ft 35cost 12

cos1 2835 12 140.

Example 11. Consider the function ft 3cos2t 5sin2t 5. write it in the formft Acos3t t0 5.Solution.

ft 34 334

cos2t 534

sin2t 5

The point 334, 5

34 is in the fourth quadrant so we choose the angle

2 cos1 334 0, cos0 3

34, sin0 5

34, concluding that

ft 34 cos2t 2 cos1 334

5

ft 34 cos2t cos1 334

5.

Exercises.1. Find sin1sin 106 .2. Find cos1cos 83 .3. Find sincos1 23 .4. Find costan14.5. Find sincos1 34 .6. Find cossin1 13 .7. A given circadian rhythm can be represented using the cosinor method, it has

period of 24 hours a maximum value of 190, a minimum value of 120 and the value attime zero is 132. Find the amplitude, mesor and acrophase.8. A given circadian rhythm can be represented using the cosinor method, it has a

39

period of 12 hours, maximum value of 30, minimum value 10 and at zero its value is11. Find the amplitude, mesor and acrophase.9. Write the function ft 4cost 4sint 6 in the form

ft Acost t0 6.10. Write the function ft 2cos 2 t 3sin

2 t 5 in the form

ft Acost t0 5.11. Find tansin1 1

5.

12. Find sinsin1 13 cos1 15 .

13. Prove that if t 1,1 then sin1t cos1t 2 .

14. Prove that if t R then tan1t sin1 t1t2

.

15. Solve if possible, the triangle ABC given that 3 ,a 20,b 10 (if there are

no solutions, prove it, if there are two solutions, find both).16. Solve, if possible, the triangle ABC given that

6 ,a 15,c 20 (if there areno solutions, prove it, if there are two solutions, find both).16. Solve, if possible, the triangle ABC given that

6 ,b 5,c 10 (if there areno solutions, prove it, if there are two solutions, find both).17. Consider the function f : R R, ft 3cos5t 6sin5t 15 in the form

ft Acoswt t0 C, to 0,2.18. Consider the function f : R R, ft 4cos5t 6sin5t 10 in the form

ft Acoswt t0 C, to 0,2.19. Consider the function f : R R, ft 3cost 2sint 10 in the form

ft Acoswt t0 C, to 0,2.20. Consider the function f : R R, ft 5cos3t 2sin3t 5 in the form

ft Acoswt t0 C, to 0,2

40