Analytical Trigonometry With Applications 1

LESLIE ELIZABETH BUCK

ANALYTICAL TRIGONOMETRY WITH APPLICATIONS

PREPARATION FOR CALCULUS PART I

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ii

Analytical Trigonometry with Applications: Preparation for Calculus Part I 1st edition © 2018 Leslie Elizabeth Buck & bookboon.com ISBN 978-87-403-2275-0 Peer review by Theodore Koukounas, Maria Alzugaray and Jean Nicolas Pestieau


iii

ANALYTICAL TRIGONOMETRY

CONTENTS

WITH APPLICATIONS

CONTENTS 1

Basic Definitions and Theorems

1

1.1

Angle Measurement and Arc Length

1

1.2

The Pythagorean Theorem

10

1.3

Right Triangle Trigonometry

14

1.4

Fundamental Trigonometric Identities

49

2

Graphs of Trigonometric Functions

48

2.1

Domain and Range of a Trigonometric Function

48

2.2

Graphs of Sinusoidal Functions

52

2.3

Graphs of Tangent, Cotangent, Secant and Cosecant

57

2.4

Transformations of the Graphs of Trigonometric Functions

64

2.5

Inverse Trigonometric Functions

89

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ANALYTICAL TRIGONOMETRY WITH APPLICATIONS

CONTENTS

3

Trigonometric Identities and Equations

107

3.1

Review of Fundamental Trigonometric Identities

107

3.2

Sum and Difference of Angles Identities

115

3.3

Double- and Half-Angle Identities

126

3.4

Product-to-Sum and Sum-to-Product Identities

135

3.5

Solving Trigonometric Equations

141

4

Applications of Trigonometry

153

4.1

Right Triangle Trigonometry

153

4.2

The Law of Sines

158

4.3

The Law of Cosines

167

4.4

The Area of a Triangle

171

Endnotes

177

Index

182


v


LINKS TO SELECT FIGURES, TABLES AND DEFINITIONS

WITH APPLICATIONS

LINKS TO SELECT FIGURES, TABLES AND DEFINITIONS Definition 1.3.3 Trigonometric functions of a point, (x, y), on the circumference of a circle of radius r

17

Figure 1.3.13 Reference triangles

31

Table 1.4.1 Fundamental trigonometric identities

44

Table 2.1.1 Domain and range of trigonometric functions

51

Table 2.5.3 Domain and range of inverse trigonometric functions

99

Table 3.2.1 Summary of sum/difference of angles identities

122

Table 3.3.2 Summary of double- and half-angle identities

130

Table 3.4.1 Product-to-sum and sum-to-product identities

139

Table 4.2.1 Summary for Law of Sines SSA case

164


vi

A special thanks to

Teodore Koukounas, Maria Alzugaray and Jean Nicolas Pestieau for their wonderful insights and helpful criticism.


vii


BASIC DEFINITIONS AND THEOREMS

WITH APPLICATIONS

Chapter 1 Basic Definitions and Theorems Learning Objectives:    

Review basic terminology Review the Pythagorean theorem Review definitions of trigonometric functions from a right t riangle and a circle Review the fundamental trigonometric identities

Section 1.1 Angle Measurement and Arc Length A ray is a half line. It has an initial point, but no terminal point. It i s infinite in one direction. Two rays that initiate from the same point form an angle in two dimensions. This point is called the vertex of the angle. An angle is a circular measure delimited by its rays.

Figure 1.1.1 Angle

An angle that initiates from the positive x-axis is in standard position. A positive angle opens in the counter-clockwise direction. A negative angle opens in the clock-wise direction.

Figure 1.1.2 Angles in standard position

Perpendicular rays form a right angle. A straight angle is composed of rays that digress from their common vertex in opposite directions. Coincident rays form a zero angle. An acute angle measures


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WITH APPLICATIONS

between a zero angle and a right angle. An obtuse angle measures between a right angle and a straight angle. A reflex angle measures larger than a straight angle (positive or negative).

Figure 1.1.3 Some particular angles in standard position

Angle Measure The most common units of angle measure are degrees and radians. A degree measures 1/360 of one complete revolution. That is, one revolution measures 360 (degrees). A straight angle measures 180 and a right angle measures 90 . A zero angle measures 0 . 







Definition 1.1.1 Degree

A degree is 1/360 of one single and complete revolution about a point.

Degree measure becomes more precise if finer units of measurement called minutes and seconds are used. One degree equals 60 minutes and one minute equals 60 seconds just like on a clock.

1 revolution = 360





1 = 60'

1' = 60"

This measurement is called DMS, that is, Degrees Minutes Seconds. Precisely measured angles may also be expressed in decimal degrees.

Example 1.1.1 Express the angle measure 33 28' 15" in decimal degrees. Note this DMS measurement has 6 significant figures or digits. 

First convert seconds to minutes: 15" =

15" 1

⋅



1 60"

=

1 4

= 0.25′

Combine minutes and convert to degrees: 

28.25 =

28.25 1

⋅

1° 60

≈ 0.4708°

33 28' 15" ≈ 33.4708 (decimal degrees retaining 6 significant digits) 




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WITH APPLICATIONS

Example 1.1.2 Express the angle measure 124.389 in DMS. 

Subtract the whole degrees (124 ) from the fractional degrees. Convert the fractional degrees to minutes. 

 60 0.389° = (0.389°) 1°  = 23.34′

Subtract the whole minutes. Convert the fractional minutes to seconds.

0.34 = 0.314′ ⋅ 610" = 20.4" 124.389



≈

124 23' 20" 

Radians

Definition 1.1.2 Radian

Radian measure is defined to be the circular arc length an angle subtends divided by the radius of that arc.

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WITH APPLICATIONS

The arc length and the radius should be measured in consistent units. Radian measure is, therefore, dimensionless. Radians are real numbers. To measure angles using arc length is natural and logical. Thus, it is the angle measure of the System International (SI). An arc length of a circle (see Figure 1.1.4) is proportional to its radius. There are infinitely many arcs of different length that may be drawn for a single angle. Hence, arc length alone cannot measure an angle. However, if the measure of any arc drawn through a particular angle is divided by the length of its radius, the result is constant. This constant is the angle measure. That is,  =



where  denotes the angle measure (in radians), s denotes the length of the arc and r denotes the length of the radius.

Figure 1.1.4 Arc length

In a circle of constant radius, the measure of its central angle is proportional to the length of the arc subtended by that angle. The constant of proportionality is the length of the radius. If the radius changes, the arc length changes proportionally for the same angle. Angle measure is proportional to arc length.

If the length of the arc is equal to the length of its radius, then the angle subtended by this arc is equal to 1 radian.   =



 =



=1

The following link provides a dynamic illustration of radian measure: Visualize radian measure by Lucas V. Barbosa.

The circumference of a circle is given by

C = 2 r where C is the circumference and r is the radius. The circumference is the arc length of a complete circle, that is, the arc length of a circle is  = 2. So by the definition of radian measure,   =



2 =



= 2


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WITH APPLICATIONS

where  is the central angle. This shows that a complete revolution of 360 = 2. It follows that a straight angle measures  radians, that is, 180 = . This yields a conversion factor for changing from degree measure to radian measure and vice versa. 

°

°

=





=

Similarly, it follows that a right angle measures /2 radians or 90 = /2.

Example 1.1.3 Convert 72 to radians. 72° 72° =

1

 ⋅

2

180°

=

5

This illustrates a useful concept known as unit analysis. The units indicate which conversion factor should be used. Use the factor that divides the units being converted. In the above example, degrees  were being converted to radians. The factor, , divides 72 degrees by 180 degrees. The result, °

 

, is thus unitless, that is, it is in radians. Radians are a unitless measure.

Example 1.1.4 Convert 135 to radians. 135° 135° =

1

 ⋅

3

180°

=

4

When converting DMS to radian measure, it is necessary to convert DMS to decimal degrees and then follow the above steps.

Example 1.1.5 Convert 23 17' 58" to radians. 23.299°

′

23° 17 58" ≈ 23. 299° =

 ⋅

1

180°

≈ 0.12944 ≈ 0.40664

Example 1.1.6 Convert /3 to degrees.  3

=

 180° ∙ = 60° 3 

Note in this example we used the reciprocal of the conversion factor in order to retain degree units.

Example 1.1.7 Convert 7/6 to degrees. 7 6

=

7 180° ∙ = 210° 6 

Example 1.1.8 Convert 2.4 to degrees.


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WITH APPLICATIONS

2.4 =

2.4 180° ∙ ≈ 1 40° 1 

No units are written when the angle measure is radians. When an angle measure is written without units, it is known to be in radians. Two significant digits were retained in the above solution.

Arc Length The definition of the radian provides a useful relationship between an arc length, its radius and the subtended angle. This relationship may be used as a formula for computing arc lengths of circles.  =



Reformulating:  = 

where s denotes arc length,  denotes the angle (in radians by definition) and r denotes the radius as previously stated. The units of s and r must be consistent in this formula.

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WITH APPLICATIONS

Example 1.1.9 Find the arc length subtended by a 30 angle in a circle of radius 2 meters.

 = 30°= 30° ⋅ 1 180° 6  = 62  = 3  Example 1.1.10

Find the arc length subtended by an angle of

 in a circle of radius 12 centimeters. 

 = (53)12  = 20  Example 1.1.10 Two gears rotate together as shown in Figure 1.1.5. The larger gear has a radius of 9 inches while the smaller gear has a radius of 4 inches.

Figure 1.1.5 Gears

a) After the smaller gear makes a complete revolution of 2  radians, through what angle does the larger gear turn? The total arc length of the smaller gear is

=24 = 8 . As the smaller gear completes a rotation, the larger gear rotates a distance (arc length) of 8π inches. The angle through which the larger gear turns is

 =  = 89. b) How many turns (revolutions) will the larger gear complete if the smaller gear rotates 29 turns? The angle through which the larger gear turns is

 = 2989 = 232 9 . Download free eBooks at bookboon.com

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WITH APPLICATIONS

The number of turns (each of arc length 2π) is 232 9

1 ⋅

116

2

=

≈ 12. 9

9

The larger gear makes nearly 13 complete revolutions. Area of a Sector

The area of a circle is given by  = 



where A is the area and r is the radius. The area of a semicircle is   =







or half that of a circle. Table 1.1.1 shows the pattern continued. Notice that in each successive row above, the angle is halved and likewise the area is halved. With inductive reasoning, the area of a sector may be shown to be   =







where A is the area of the sector,  is the central angle in radians and r is the radius. central angle circle

2

semi-circle



/4

1/8r 2

. . .

. . .



½ r 2

eighth circle





½r 2 ¼r 2



general sector of angle

r 2



/2

quarter circle

. . .

area

Table 1.1.1 Area of a sector

Figure 1.1.6 Area of a sector


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WITH APPLICATIONS

Example 1.1.12 Find the area of a sector of radius 1.2 feet subtended by an angle of 2/3 radians.

 = 12 (23)1.2 ft = 0.48 ft Example 1.1.13 Find the area of a sector of radius 6 kilometers subtended by an angle of 210.

=210°= 76  = 12 (76)6 km = 21 km Example 1.1.14 A town code states that no landscape features may exceed 4 feet in height within a 20-foot radius of the apex of a street corner.

Figure 1.1.7 Apex of a corner

Approximate the area of Mr. Greenblat’s property subject to this town code.

 ≈ 12 220 ft = 100 ft


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WITH APPLICATIONS

Section 1.2 The Pythagorean Theorem The Pythagorean theorem relates the lengths of the sides of a right triangle in Euclidean geometry. More specifically, it states that the sum of the squares of the legs of any right triangle equal the square of the hypotenuse. If a and b represent the lengths of the legs of a right triangle and c represents the length of the hypotenuse, then

a2 + b2 = c2 . This theorem is named after the Greek mathematician Pythagoras (ca. 570 BC — ca. 495 BC) although there is evidence to suggest that knowledge of this relation predates the Pythagoreans. It may be the most well known theorem in mathematics and probably has more distinct and varied proofs than any other theorem. A simple geometric proof utilizes Figure 1.2.1 below.

Figure 1.2.1 A geometric illustration of the Pythagorean theorem

"Pythagore". Licensed under Creative Commons Attribution-Share Alike 3.0 via Wikimedia Commons Two dynamic visualizations of the geometry of this theorem may be found at:

Geometric proof of the Pythagorean theorem

"Pythag anim". Licensed under Creative Commons Attribution-Share Alike 3.0 via Wikimedia Commons

Proposition 47 of Euclid’s Elements provides an elaborate proof based upon the congruence of triangles and the areas of rectangles. It is speculated that Pythagoras used a proof based on similar triangles and proportionality (Maor 2007, p39).


10



WITH APPLICATIONS

A simple algebraic proof utilizes the following geometric figure of a square of side length a + b with another square of side length c inscribed inside (Gardner 1984, p. 154).

Figure 1.2.2 Another geometric illustration of the P ythagorean theorem

The area of each right triangle formed around the inner square is ½ab. The area of the large square is equal to (a + b)2. It is also equal to the area of the inner square plus the areas of the four triangles, that is, c2 + 4(½ab). So, (a + b)2 = c2 + 4(½ab) a2 + 2ab + b2 = c2 + 2ab a2 + b2 = c2 .

Example 1.2.1 The legs of a right triangle measure 3 feet and 4 feet. Find the length of the hypotenuse.

c2 = 32 + 42 = 25 Solving, c = ± 5. However, c = 5 since the length of the hypotenuse cannot be negative.

Example 1.2.2 The hypotenuse of a right triangle is 13 centimeters. One leg measures 7 centimeters. Find the length of the other leg.

(13)2 = 72 + b2 b2 = (13)2 – 72 = 169 – 49 = 120

 = √ 120 = √ 4√ 3 0 = 2√ 30 . Example 1.2.1 above illustrates a Pythagorean triple (3, 4, 5). Definition 1.2.1 Pythagorean triple

A Pythagorean triple is a set of three integers satisfying the Pythagorean theorem.

There is a specific relationship between the integers of a Pythagorean triple. This was formalized in Euclid’s Elements.


11



WITH APPLICATIONS

If a = m2 – n2 then b = 2mn and c = m2 + n2

where m and n are integers with m > n. It is left as an exercise for the reader to show that substituting the aforementioned expressions for a, b and c into the Pythagorean theorem yields equality.

Example 1.2.3 Let m = 5 and n = 4. Then a = (5)2 – (4)2 = 25 – 16 = 9, b = 2(5)(4) = 40 and c = (5)2 + (4)2 = 41.

There have been many other formulas derived for the generation of Pythagorean triples since the time of Euclid (300 BC). There are sixteen Pythagorean triples for c < 100. They are: ( 3, 4, 5) (12, 35, 37) (33, 56, 65)

( 5, 12, 13) ( 9, 40, 41) (48, 55, 73)

( 8, 15, 17) (28, 45, 53) (13, 84, 85)

( 7, 24, 25) (11, 60, 61) (36, 77, 85)

(20, 21, 29) (16, 63, 65) (39, 80, 89)

(65, 72, 97)

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The Pythagorean theorem has great application. This will be illustrated extensively throughout this text. The Distance Formula

The distance formula is an application of the Pythagorean theorem.

Figure 1.2.3 Distance between two points

c2 =

a2

+

b2

D2 = ( x1 – x2)2 + ( y1 – y2)2 or

 =  (1 −  ) + (1 −  ) Example 1.2.4 Find the distance between the points (-2, 5) and (3, 7) in t he xy-plane.

 =  (1 −  ) + (1 −  )  =  (−2 − 3) + (5 − 7)  = √25+ 4  = √ 29


13



WITH APPLICATIONS

Section 1.3 Right Triangle Trigonometry The trigonometric functions are defined to be ratios of the lengths of t he sides of a right triangle. With three sides of a triangle, there are six possible combinations and, as such, six t rigonometric functions. Definition 1.3.1 Trigonometric functions

Let  be an acute interior angle of any right triangle. Let a represent the length of the leg of the triangle adjacent to angle  . Let o represent the length of the leg of the triangle opposite  . Let h represent the length of the hypotenuse (the longest side, opposite the right angle). The six possible ratios of any two sides of this triangle are:

sin() =

cos() = tan() =



csc() =

ℎ 

sec() =

ℎ 

cot () =



ℎ  ℎ   

where sin(  ) is the sine function, cos(  ) is the cosine function, tan(  ) is the tangent function, csc(  ) is the cosecant function, sec(  ) is the secant function and cot(  ) is the cotangent function.

Figure 1.3.1 Relationship between the sides of a right triangle

The argument of each trigonometric function is  or an independent variable representing an angle. When an angle (or the equivalent) is input to a trigonometric function, the output is a ratio of two sides of a right triangle. A note concerning notation: when writing function notation always include the argument of the function. Sin is a topic of theology, not trigonometry. Sin is a trigonometric function.

Example 1.3.1

a) Find the values of the six trigonometric functions of  for the right triangle illustrated in Figure 1.3.2.


14



WITH APPLICATIONS

Figure 1.3.2

The length of the hypotenuse is found using the Pythagorean theorem. It may also be found by recalling one of the Pythagorean triples, (3, 4, 5).

sin() = 35

csc() = 53 sec() = 54 cot() = 43

cos() = 45 tan() = 34

b) Find the values of the six trigonometric functions of  for the right triangle illustrated in Figure 1.3.2.

sin() = 45

csc() = 54

cos() = 35 tan() = 43

sec() = 53 cot() = 34

The values of the trigonometric functions vary with the angle. It is of note in the above example that  +  = /2.  and  are complementary angles. Sine and cosine are sin(  ) = cos(  ) and cofunctions. In other words, the sine of an angle is equal to the cosine of that angle s complement. ’

() = −  If an angle,  is evaluated within any of the trigonometric functions, the same result is obtained if the complement of said angle,

2 −, is evaluated within the cofunction. There are three pairs of

trigonometric cofunctions: the sine and the cosine (stated above); the secant and the cosecant; and the tangent and the cotangent.

() =  −

() =   − 

Definition 1.3.2 Cofunction

A cofunction is the function of the complement of an angle.


15



WITH APPLICATIONS

If a right triangle is placed within a circle centered at (0, 0), keeping the angle of interest adjacent to the axis of the domain (typically called the x-axis), the trigonometric functions may be redefined in terms of the coordinates of the plane and the radius of the circle.

Figure 1.3.3 Right triangle within a circle of radius r centered at (0,0)

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Definition 1.3.3 Trigonometric functions of a point, (x, y), on the circumference of a circle of radius r

sin() =

 

csc() =

 

cos() =

 

sec() =

 

tan() =

 

cot () =

 

where x is the length of the leg of a right triangle that is adjacent to the angle of interest, y is the length of the leg opposite said angle, r is the length of the hypotenuse of the triangle and  is a central angle in standard position.

Please note that ( x, y) is a point on the circumference of the circle defined by the equation

x 2 + y2 = r 2 . This definition of the trigonometric functions is essentially a renaming of components or a change of perspective. Refer back to the triangle in Figure 1.3.3 to see that x2 + y2 = r 2 is also the Pythagorean theorem. Definitions 1.3.2 and 1.3.3 may be considered classic definitions of the trigonometric functions. These functions may also be defined using a tangent line to a circle. The naming of the tangent function is not a coincidence, but rather an intentional label suggestive of what it represents. Likewise, the word sine comes from a corruption of the Sanskrit word for chord. The sine is still sometimes referred to as the half-chord. The following video will help clarify how the tangent function relates to the tangent line. It provides a deeper understanding of the trigonometric functions and a more eloquent explanation of each function’s period , range and domain. These properties are essential in understanding the graphs of these transcendental functions in the next chapter. Tangent line definitions

Example 1.3.1 Find the values of the six trigonometric functions corresponding to the point (1, √ 3 ) in the xy plane.

Refer to Figure 1.3.4 below. Find the radius of the circle (hypotenuse of the triangle) passing through the point using the Pythagorean theorem. x2 + y2 = r 2 (1)2 + (√ 3 )2 = r 2 4 = r 2 2=r


17



WITH APPLICATIONS

Figure 1.3.4 for example 1.3.1

 sin() = 23

csc() =  2 =  2

 3 = ∙ 3  3

2 3 3

cos() = 12

sec() = 21 = 2

 tan() = 13 = √ 3

cot () =  1 =  1

 3 = ∙ 3  3

 3

3

3

3

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.

.

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18




WITH APPLICATIONS

Rationalize any irrational denominators using the multiplicative identity.

Example 1.3.2 Find the values of the six trigonometric functions corresponding to the point ( -1, √ 3 ) in the xy plane.

Refer to Figure 1.3.5. Find the radius of the circle passing through the point using the Pythagorean theorem. x2 + y2 = r 2 (1)2 + (√ 3 )2 = r 2 4 = r 2 2 = r Note that the length of each leg of the right triangle is positive, despite the signs on the coordinates of the point. Thus |-√ 3| = √ 3 was used in the Pythagorean theorem above.


 sin() = 23

csc() =  2 =  2

cos() = − 12

sec() = − 21 = −2

tan() = −13 = −√ 3

1 cot () = −1 = −  

3



3

 3 = ∙ 3  3

√  

 3 = ∙ 3  3

 − 33

Note that the absolute values of all six functions remain the same as in the previous example. The values of the trigonometric functions in Example 1.3.1 and the values of the functions in Example 1.3.2 differ only in sign, if at all. This is due to the symmetry of the circle and the signs of the coordinates in the xy-plane. If the same triangle is moved into quadrants 3 or 4 of the xy-plane, the pattern continues.


19



WITH APPLICATIONS

Example 1.3.3 Find the values of the six trigonometric functions corresponding to the point ( -1, -3) in the xy-plane. Refer to Figure 1.3.6. The radius of the circle is 2 as it was in the two last examples.

Figure 1.3.6 for Examples 1.3.3 and 1.3.4

sin() = − 23

csc() = 2 = −  2

cos() = − 12

sec() = − 21 = −2

tan() = −−13 = √ 3

1 cot () = −1 =  



 3 = ∙ 3  3

− 3



− 3

− 23 3 

 3  3 = ∙ 3  3 3

Example 1.3.4 Find the values of the six trigonometric functions corresponding to the point (1, -3) in the xy-plane.

Refer to Figure 1.3.6. The radius is still 2.  sin() = − 23

csc() = 2 = −  2

 3 = ∙ 3  3

cos() = 12

sec() = 21 = 2

− 3

tan() = −1 3 = −√ 3

cot () = 1 = −  1



− 3

 3 = ∙ 3  3

− 233 

 − 33

Observe how the signs of the ratios vary with each trigonometric function and with each quadrant of the plane. In the first quadrant, the range values of all six trigonometric functions are positive. It is no coincidence that x > 0 and y > 0 in quadrant I. In the second quadrant, only the sine function and its reciprocal function, the cosecant, are positive. The definitions of the sine and the cosecant involve y and r . r is the radius of a circle and, as such, is positive. In quadrant II, y > 0 and x < 0. Hence a ratio of y and r is positive. The definitions of the other four trigonometric functions involve


20



WITH APPLICATIONS

ratios of x and r or x and y. Therefore, the remaining four functions yield a negative output for angles terminating in quadrant II. Continuing t his line of reasoning, the only trigonometric functions with positive output for angles terminating in quadrant III are the tangent and cotangent functions since x < 0 and y < 0 in this quadrant. In quadrant IV, only the cosine function and its reciprocal function, the secant, are positive. Figure 1.3.7 summarizes these facts employing a convenient mnemonic device: ASTC. A is for ‘All’ positive in quadrant I, S is for ‘Sine’ including its reciprocal function, cosecant, positive in quadrant II, et cetera. The acronym may stand for ‘All Should Try Calculus.’ Use your imagination.

Figure 1.3.7 ASTC

> Apply now REDEFINE YOUR FUTURE

AXA GLOBAL GRADUATE PROGRAM 2015 p o t s n o n o t o h P © -


21




WITH APPLICATIONS

The previous four examples utilize a reference triangle. Refer back to Figures 1.3.5 and 1.3.6. The notation  ref indicates the reference angle. It is the same in all four examples. It is the interior angle of the right triangle that is adjacent to the domain axis.  , however, the argument of each trigonometric function, is the central angle of the circle and is not necessarily equal to  ref . In Example 1.3.1,  ref is equal to  . In Example 1.3.2,  ref is supplementary to  . In Example 1.3.3,  ref is part of the angle denoted  . The angle,  , is in standard position and varies from example to example.  is the angle being evaluated in each trigonometric function. If  ref is evaluated in any of the trigonometric functions the correct absolute value is obtained. In other words, if  and  ref are defined as in Figures 1.3.5 and 1.3.6, with  in standard position and  ref adjacent to the domain axis, then sin( ref ) = |sin |. The same is true for the other five trigonometric functions. Verify this in the previous examples.

 <  < ° If an equilateral triangle of side length 2 is bisected, then the reference triangle used in the previous four examples may be derived (see Figure 1.3.8). An equilateral triangle is equiangular and the measure of each interior angle is /3 or 60. The bisected angle is then /6 or 30. An angle bisector in an equilateral triangle will also bisect the base at a right or 90 angle. This creates two equal right triangles, each one being the reference triangle used in Examples 1.3.1 through 1.3.4.

Figure 1.3.8 Reference triangle

The angles π/6 and π/3 are both reference angles. They can be used to find the values of trigonometric functions at other angles. The length of leg y above may be found using the Pythagorean theorem.

  +   =  (1) +   = (2)  = 3  = √ 3 Examples 1.3.1 through 1.3.4 could be restated as follows:

Example 1.3.1A

Find the exact values of the six trigonometric functions of

 . 

Refer back to Figure 1.3.4 as necessary. Refer to the reference triangle in Figure 1.3.8 and apply t he definitions of the six trigonometric functions. Rationalize denominators as necessary. Download free eBooks at bookboon.com

22



WITH APPLICATIONS

sin(3) =  23

csc(3) =  23 =  23 ∙   33 = 2 33

cos(3) = 12

sec(3) = 21 = 2

tan(3) =  13 = √ 3

cot(3) =  13 =  13 ∙   33 =  33

Example 1.3.2A

Find the values of the six trigonometric functions of Refer to back to Figure 1.3.5.

 .

 = 23.  = 3.

sin(23) =  23

csc(23) =  23 =  23 ∙   33 = 2 33

cos(2π3) = − 12

sec(23) = − 21 = −2

 3 tan(2π3) = −1 = −√ 3

1 ∙ √ 3 = − √ 3 cot (23) = −1 = − 3 √ 3 √ 3 √ 3

Note that

cos(3) = |cos(23)| = |− 12| = 12.

If the reference triangle is moved into quadrants 3 or 4 of the xy-plane, angles evaluated in the trigonometric functions.

Example 1.3.3A

Find the values of the six trigonometric functions of

   and  may be

 4 and  = . . In Figure 1.3.6,  =  3 3 

sin(4π3) = −  23

csc(4π3) = −2 3 = −  23 ∙   33 = − 2 33

cos(4π3) = − 12

sec(4π3) = − 21 = −2

tan(4π3) = −−1 3 = √ 3

cot(4π3) = −−1 3 =  13 ∙   33 =  33

Example 1.3.4A

Find the values of the six trigonometric functions of

 5 and  = . . In Figure 1.3.6,  =  3 3 


23



WITH APPLICATIONS

sin(5π3) = −  23

csc(5π3) = −2 3 = −  23 ∙   33 = − 2 33

cos(5π3) = 12

sec(5π3) = 21 = 2

tan(5π3) = −1 3 = −√ 3

cot(5π3) = −1 3 = −  13 ∙   33 = −  33

 

 6

The compliment of reference angle is reference angle . The trigonometric functions may be evaluated at

  7   , , , and by moving this reference triangle around the circle with adjacent 6 6 6 6 6

to the domain axis.

Example 1.3.5 Find the values of the six trigonometric functions corresponding to the points ( , 1), ((- , -1) and ( , -1) in the xy-plane. Refer to Figure 1.3.9. Find the radius of the circle.

√ 3

√ 3

√ 3


24

√ 3, 1),




WITH APPLICATIONS

x2 + y2 = r 2

√ 3 ) + (1) = r

(

2

2

2

4 = r 2 2=r


a) For coordinates (

√ 3, 1) or  = π:

sin() = 12

csc() = 21 = 2

cos() =  23

sec() =  23 =  23 ∙   33 = 2 33

tan() =  13 =  13 ∙   33 =  33

cot() =  13 = √ 3

Note that any of these values could have also been obtained using the cofunction identities.

sin6 = 12 =cos2 − 6=cos3 csc6 = 2 = sec 2 − 6=sec3 tan6 = 1 =cot2 − 6=cot3 √ 3 Download free eBooks at bookboon.com

25



WITH APPLICATIONS

b) For coordinates (-√ 3, 1) or  =

5π : 6

sin() = 12

csc() = 21 = 2

cos() = − 23

sec() = 2 = −  2



 3 2 3 = − ∙ 3 3  3

− 3

tan() = 1 = −  1

 3 = ∙ 3  3

− 3

c) For coordinates (-√ 3, -1) or  =

 − 33

 cot () = −1 3 = −√ 3

7π : 6

sin() = − 12

2 csc() = −1 = −2

cos() = − 23

sec() = 2 = −  2



− 3

1 tan() = −1 =   − 3

 3 2 3 = − ∙ 3 3  3

 3  3 = ∙ 3  3 3

d) For coordinates (√ 3, -1) or  =

cot () = −−13 = √ 3 

π : 6

sin() = − 12

2 csc() = −1 = −2

 cos() = 23

sec() = √ 23 =  2

 3 2 3 = 3 ∙ 3  3

1 tan() = −1 = −   3

 3  3 = − ∙ 3 3  3

 cot () = −13 = −√ 3

A short video illustrating the last five examples.

Another reference triangle may be derived through the bisection of a unit square along its diagonal.


26



WITH APPLICATIONS

Figure 1.3.10 Reference triangle

The hypotenuse is obtained by applying the Pythagorean theorem. The reference angle is π/4. Positioning this reference triangle within a circle centered at the origin around the four quadrants enables the evaluation of angles

 3 5 4

,

4

,

4

and

7 4

in the trigonometric functions.


27




WITH APPLICATIONS

Example 1.3.6 Find the values of the six trigonometric functions corresponding to the points (1, 1), (-1, 1), (-1, -1) and (1, -1) in the xy-plane. Refer to Figures 1.3.10 and 1.3.11.


 

a) For coordinates (1, 1) or  = :

sin() = 12 = 12 ∙ √ 22 = √ 22 √ √ √ cos() = 1 = 1 ∙ √ 2 = √ 22 √ 2 √ 2 √ 2 tan() =  = 1

csc() = √ 12 = √ 2 sec() = √ 12 = √ 2 cot() = 1

Note that any of these values could have also been obtained using the cofunction identities.

sin4 = √ 22 =cos2 − 4=cos4 csc4 = √ 2 =sec2 − 4=sec4 tan4 = 1 = cot2 − 4=cot4


28



WITH APPLICATIONS

b) For coordinates (-1, 1) or  =

sin() =

3π : 4

1 1 √ 2 √ 2 = ∙ = 2 √ 2 √ 2 √ 2

cos() = −

1 1 √ 2 √ 2 =− ∙ = − 2 √ 2 √ 2 √ 2

−1

sin() =

1 1 √ 2 √ 2 =− ∙ = − 2 √ 2 √ 2 √ 2

1

sin() =

csc() = −

√ 2 = −√ 2 1

sec() = −

√ 2 = −√ 2 1

7 : 4

−1 1 √ 2 √ 2 =− ∙ = − 2 √ 2 √ 2 √ 2 1

√ 2 = −√ 2 1

cot () = 1

() = 1 = 1

d) For coordinates (1, -1) or  =

sec() = −

5 : 4

−1 1 √ 2 √ 2 =− ∙ = − 2 √ 2 √ 2 √ 2

cos() = −

√ 2 = √ 2 1

cot () = −1

tan() = 1 = −1

c) For coordinates (-1, -1) or  =

csc() =

1

√ 2

√ 2

cos() = √ 2 = √ 2 ∙ √ 2 = 2 −1

csc() = −

√ 2 = −√ 2 1

√ 2 sec() = 1 = √ 2

cot () = −1

tan() = 1 = −1

All of the information derived thus far is summarized in Table 1.3.1 and in Figures 1.3.12 and 1.3.13.


29



WITH APPLICATIONS

Function

 =

 =

/6



/4



 =

/3



Sine

1 2

√ 2 2

√ 3 2

Cosine

√ 3 2

√ 2 2

1 2

Tangent

√ 3 3

1

√ 3

Cosecant

2

√ 2

2√ 3 3

Secant

2√ 3 3

√ 2

2

Cotangent

√ 3

1

√ 3 3

Table 1.3.1 Values of the six trigonometric functions at the reference angles

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30




WITH APPLICATIONS

Figure 1.3.12 Coordinates and associated angles on the circumference of a circle of radius r

Figure 1.3.13 Reference triangles

The trigonometric functions are defined as ratios of the lengths of the sides of any right triangle. The previous discussion illustrates how the coordinates on the circumference of a circle centered at the origin may also be used to define these functions. Figure 1.3.12 has 4 additional points: (r, 0), (0, r), (-r, 0) and (0, -r). These points are associated with 

the angles, 0, ,  and 2

3 2

, respectively. These angles are known as the quadrantal angles. They

define the four quadrants of the xy-plane. With coordinates at each of the four quadrantal angles, it is possible to find exact values of the trigonometric functions at each of these angles.


31



WITH APPLICATIONS

Definition 1.3.4

Quadrantal angle

A quadrantal angle is an angle in standard position with a terminating ray coincident with an axis.

Example 1.3.7

Find the exact values of the six trigonometric functions at

sin0 = 0 =0 cos0 =  =1 tan0 = 0 = 0

0, 2 , and  radians. csc0 does not exist dne sec0 =  = 1 cot0 dne

sin2 =  =1 cos2 = 0 =0 tan2 

csc2 = 1 sec2 cot2 = 0 = 0

sinπ = 0 =0 cos = − =−1 tan = 0 =0

cscπ dne

sin(32) = − =−1 cos(32) = 0 =0 tan(3π2) dne

csc(32)=−1 sec(3π2)dne cot(32) = 0 = 0

sec = −1 cotπ dne

The coordinates on the circumference of a circle define the trigonometric functions. Any point on the circumference of a circle may be used to find the values (if they exist) of the six trigonometric functions. The position of the point determines the value of the trigonometric function. If two angles in standard position terminate at the same point and are evaluated inside a trigonometric function, their values will be the same.


32



WITH APPLICATIONS

There are infinite angles in standard position with terminal rays that coincide with the same point on

sin(π6) = 12, then so does sin(6 +2)= 12 and sin(6 +4)= 12 and sin(6 +6)= 12 and sin(6 −2)= 12, et cetera. In other words, the trigonometric functions are a circle. Therefore, if

periodic. After a certain period, the values in the range of each function repeat. This is reasonable based upon the symmetry of a circle.

For angle values greater than or equal to 0 but less than 2, a trigonometric function cycles through all its range values at least once. For angles greater than 2  or angles less than 0, there will be repetition of these range values since the positions on the circumference of the circle are revisited. Definition 1.3.5 Period of a trigonometric function

The period of a trigonometric function is the shortest interval of its domain that will correspond to one complete cycle through its range.

A complete cycle through the range of a trigonometric function is represented by a continuous set of points along the circumference of a circle. The length of this cycle will depend upon the specific function. For example, the set of range values for the sine function is governed by the y-coordinates of these points along the circumference. Each value in the range of the sine function is the ratio of a y-coordinate to the radius of the circle.

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33




WITH APPLICATIONS

The values of the y-coordinates on a circle of radius r vary from – r to r (Figure 1.3.12). Example 1.3.7 shows the

sin− 2 = − = −1 and sin2 =  = 1. This means that the range of the sine

function is [-1, 1]. In order to be guaranteed to cover the entire range of the sine function starting at any point in the sine’s domain, a complete revolution around the circle or 2 radians is necessary. Therefore, the period of the sine function is 2.

( + ) = () , where k is an integer. In other words, if an angle is increased or decreased by any multiple of 2, its value in the sine function remains constant. The range value of a trigonometric function is determined by the position of coordinates on a circle. Please note that , unless k is zero.

 + 2 ≠ 

A similar argument demonstrates that the range of the cosine is [-1, 1] and its period is also 2.

( + ) = () , where k is an integer. The range of the tangent function is not as easily determined. The range of the tangent function is defined to be the ratio of the y-coordinate to the x-coordinate on the circumference of a circle. Please refer back to Figure 1.3.12. With a variable in the denominator, there are angles outside of the domain for a tangent function. For instance, at

 = 2 the x-coordinate is zero, and as such,  = 2 is

not in the domain of the tangent function since division by zero is not defined. The tangent function is undefined at . Likewise, the tangent function is undefined at . The x-coordinate is

 = 2

 = − 2

non-zero for all other values of  between on the open interval

− 2 and . That is to say, the tangent function is defined

− 2 , 2. This is an interval of the tangent function’s domain. Refer to Chapter

2, Section 2.1 for more on the domain of a trigonometric function.

− 2 , 2, the range of the tangent   function is infinite. For example, at  = 0 , the ratio of the coordinates, =   = 0. Zero is in the If ratios of the x and y-coordinates are considered over the interval

 

range of the tangent function. As  increases from 0 toward , the values of the y-coordinates increase while the values of the x-coordinates simultaneously decrease. Under these conditions, the entire ratio or quotient increases without bound. A much simpler example may help illustrate this phenomenon.

Example 1.3.8 Determine the behavior of the quotients when 10 is divided by successively smaller values.

10 = 10 1 10 = 100 0.1 10 = 1000 0.01 10 = 10,000 0.001

10 = 100,000 0.0001 10 = 1,000,000 0.00001 10 = 10,000,000 0.000001 10 = 100,000,000 0.0000001 Download free eBooks at bookboon.com

34



WITH APPLICATIONS

This pattern may, of course, be continued infinitely. The quotient values tend toward infinity. This illustration is much simpler than tangent function range values since the numerator of these ratios is held constant. If the numerator were to increase simultaneously, the ratios would increase more rapidly.  2

If this same analysis is applied to the angle values ranging from 0 toward − , the ratios increase in magnitude but in the negative direction or toward -∞. Therefore, the range of the tangent function is  2

 2

(-∞, ∞). It is clear that a distance in the domain of  radians, that is − to radians, corresponds to the entire set of reals or (-∞, ∞). The period of the tangent function is  radians.

( +  ) = () , where k is an integer. Similar arguments show the period of the cotangent function to be  radians, and the periods of the secant and the cosecant functions to be 2  radians.

( +  ) = () , where k is an integer; ( +   ) = () , where k is an integer; ( +   ) = () , where k is an integer. Table 1.3.2 summarizes the periods of the trigonometric functions.

Function

Period

Sine

2

Cosine

2

Tangent



Cosecant

2

Secant

2

Cotangent



Table 1.3.2 Periods of the six trigonometric functions

The periodic identities may be used to find exact values of the trigonometric functions at particular reflex angles.


35



WITH APPLICATIONS

Example 1.3.9 Determine the exact values of the following trigonometric functions:

a) sin(405 ) 

b) cos(-240 ) 

76

c) tan

11 3

d) csc

− 34

e) cot

f) sec(840 ) . 

Solutions:

 sin405° =sin45°+360° =sin45° = √ 22 =1

 cos−240° =cos120°−360° =cos120° =−cos60° = − 12 =−1  tan(76)=tan6 +=tan6 = √ 33 =1 5 +2)=csc(5) = −csc = − 2 = − 2√ 3 =1  csc(11 )=csc( 3 3 3 3 √ 3 3  cot(− 34)=cot4 −=cot4 = 11 =1 =−1  sec840° =sec120°+720° =sec120° = −sec60° = − 21 = −2  = 2 �e Graduate Programme for Engineers and Geoscientists

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36




WITH APPLICATIONS

In the previous examples, the periodic identity was used. Also, consideration was given to the quadrant where the angle terminated. For example in part b), the angle of -240 terminates in quadrant II where the range values of the cosine function are negative. After applying the periodic identity, the angle of 120 terminates in quadrant II as well. The cosine of 120 is equal to the negative cosine of 60 due to the quadrant of angle termination. Therefore, the reference angle of 60 was used to determine the absolute value and a negative sign was appended due to the quadrant of termination. Please note that -240 ≠ 120 ≠ 60 . However, cos(−240°) = cos(120°) = −cos(60°). 















Example 1.3.10 Determine the exact values of the six trigonometric functions of an angle in standard position terminating at the point (-2, 5).

The angle is not one of the three reference angles, nor is it a quadrantal angle. It is not necessary to determine the angle. Find the radius of the circle passing through the point using the Pythagorean theorem. x2 + y2 = r 2 (2)2 + (5)2 = r 2 29 = r 2

√ 29 = 

sin() =

5 √ 29 5√ 29 ∙ = 29 √ 29 √ 29

csc() =

√ 29 5

cos() =

−2 √ 29 −2√ 29 ∙ = 29 √ 29 √ 29

sec() = −

tan() =

5 5 =− −2 2

cot () =

√ 29 2

−2 2 = − 5 5

Example 1.3.11 Determine the exact values of the six trigonometric functions of an angle in standard position terminating at the point (-2, -√ 2 ).

It is not necessary to determine the angle in order to evaluate the trigonometric functions. Find the radius of the circle passing through (-2, -√ 2). x2 + y2 = r 2 (2)2 + (√ 2 )2 = r 2 6 = r 2

√ 6 = 


37



WITH APPLICATIONS

sin() =

−√ 2 √ 6 2√ 3 √ 12 √ 3 ∙ =− =− =− 6 6 3 √ 6 √ 6 csc() =

2√ 3 √ 6 √ 2 √ 12 ∙ = − = − = −√ 3 2 2 −√ 2 √ 2

cos() =

−2 √ 6 −2√ 6 √ 6 ∙ = =− 6 3 √ 6 √ 6

sec() = −

tan() =

−√ 2 √ 2 = −2 2

cot () =

√ 6 2

−2 2 √ 2 2√ 2 = ∙ = = √ 2 2 −√ 2 √ 2 √ 2

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38

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WITH APPLICATIONS

Section 1.4 Fundamental Trigonometric Identities A mathematical identity is always true or equal for all values in its domain. Definition 1.4.1 Trigonometric Identity

A trigonometric identity is an equation involving one or more trigonometric functions that is always true for all values in the domain of said function(s).

The fundamental trigonometric identities are the building blocks for the establishment of other useful identities. These fundamental identities follow easily from the definitions of the trigonometric functions and the Pythagorean theorem.

Quotient Identities csc() = Likewise,



1 1 =  =  sin()  sin() =

( ≠ 0) (1)

1

(2)

csc()

Similar identity relationships hold for the cosine and secant functions and for the tangent and cotangent functions. 1 cos() = (3) sec()

sec() =

tan() =

cot () =

1

(4)

cos() 1

(5)

cot() 1

(6)

tan()

 1    = sin() tan() = = ∙ =    1 cos()   

( ≠ 0) (7)

Similarly,

cot () =

cos() sin()


39

(8)



WITH APPLICATIONS

Pythagorean Identities

sin =  → sin =  cos =  → cos =  2 + 2 = 2 Substituting into the Pythagorean theorem,

cos2 + sin2 = 2 2cos2+ 2sin2=2 2cos2 + 2sin2 = 2 2 2 2 cos2 +sin2 =1

9

To obtain another relationship, divide by the square of the cosine,

cos2 + sin2 = 1 cos2 cos2 cos2 2 2 sin 1 1 + (cos) = (cos) 1 + tan2 = sec2 10 Dividing by the square of the sine yields another Pythagorean identity involving the cotangent and cosecant functions,

cos2 +sin2 = 1 cos2 + sin2 = 1 sin2 sin2 sin2 2 2 cos 1 (sin ) + 1 = (sin) cot2 + 1 = csc2 11 Please note the notation,

sin2 means sin2. This applies to all six trigonometric functions.


40



WITH APPLICATIONS

Cofunction Identities

The cofunction identities were established in Section 1.3. If an angle,  is evaluated within any of 

the trigonometric functions, the same result is obtained if the complement of said angle,

2 −, is

evaluated within the cofunction. This is a direct result of the trigonometric definitions. There are three pairs of trigonometric cofunctions: the sine and the cosine; the secant and the cosecant; and the tangent and the cotangent.

sin() =cos2 − sec() = csc2 − tan() = cot2 −

cos() =sin2 − csc() = sec2 − cot() = tan2 −

(12) (13) (14)

Let y be the length of the side of a right triangle opposite angle  . Let x be the length of the side of a right triangle adjacent to angle  . Then side x is opposite angle

π−  and side y is adjacent to angle

−  (see Figure 1.3.3). Angle −  is the complement of angle  .  

sin() = ℎ =  = ℎ =cos2 − 

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WITH APPLICATIONS

Periodic Identities The periodic identities were established in Section 1.3. k is an integer value.

sin( + 2) = sin()

(15)

cos( + 2) = cos()

(16)

tan( +  ) = tan()

(17)

csc( + 2) = csc()

(18)

sec( + 2) = sec()

(19)

cot ( +  ) = cot( )

(20)

Odd/even Identities Definition 1.4.1 Odd/even functions

A function is even if it satisfies the condition: (−) = (), for all  in the domain of  . A function is odd if it satisfies the condition: (−) = −(), for all  in the domain of .

Even functions are symmetric with respect to the y-axis or the vertical, dependent axis. Odd functions are symmetric with respect to the origin. Two of the trigonometric functions are even: cosine and its reciprocal, the secant function. The other four are odd.

cos(−) =

 

= cos()

Cosine is even.

(21)

Refer to Figure 1.4.1. The x-coordinate remains constant under a reflection of  over the x-axis. Therefore, the cosine is an even function. A reflection of  over the x-axis, however, negates the value of the y-coordinate on the point defining the terminal ray of - . The sine function is odd.

sin(−) =

−  = − = −sin()  

Sine is odd.

Figure 1.4.1 Odd/even identities


42

(22)



WITH APPLICATIONS

tan(−) =

sin(−) cos(−)

=

−sin() cos()

= −tan()

(23)

Based upon the previous results, the tangent function is odd. The corresponding reciprocal functions of sine, cosine and tangent bear the same odd/even properties, respectively.

csc(−) = −csc() sec(−) = sec() cot (−) = −cot ()

odd

(24)

even

(25)

odd

(26)

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43




WITH APPLICATIONS

Summary Fundamental Trigonometric Identities

1 sin() = csc() 1 cos() = sec() 1 tan() = cot() sin() tan() = cos()

Quotient Identities

Odd/even Identities

1 cot() = tan() cot() = cos() sin()

sin() =cos2 − cos() =sin2 −  sec() = csc2 − csc() = sec2 −  tan() = cot2 − cot() = tan2 −  sin(+2) = sin() = csc() cos(+2) = cos() = sec() tan( +  ) = tan() = cot()

Periodic Identities

1 sec() = cos()

cos() +sin() = 1 1+tan() = sec() cot() + 1 =csc()



1 csc() = sin()

cos(−) =() sin(−) =−sin() tan(−) = −tan()

Table 1.4.1 Fundamental Trigonometric Identities


44

csc(+2) sec(+2) cot(+) sec(−) =sec() csc(−) =−csc() cot(−) =−cot()



WITH APPLICATIONS

Example 1.4.1 Establish the identity:

sec() csc()

1

=

cot()

Solution: Work one side of the identity and arrive at the other side. Make sequential substitutions using equal functions or expressions derived from the list of fundamental identities until the other side of the identity is achieved.

sec() csc()

=

sec() 1

∙

1 csc()

=

sec() 1

∙ sin() =

1 cos()

∙ sin() =

sin() cos()

= tan() =

1 cot()

It may be helpful to keep in mind the goal, that is, the other side of the identity while making progressive substitutions.


sec 2 () − tan2 () = 1 Solution: Consider using a Pythagorean identity when proving identities involving trigonometric functions raised to powers, especially if the power is two.

sec 2 () − tan2 () = tan2 () + 1 − tan2 () = 1


csc()sin() = cos() sec(4 − ) Solution:

1 csc()sin() sin() ∙ sin() 1 = = = cos(4 − ) = cos( − 4) = cos() sec(4 − ) sec(4 − ) sec(4 − ) This example utilized the even identity of the cosine function and the periodic identity of the cosine function with k = -2.


csc() − cot () =

1 csc() + cot()

Solution:

1 csc() + cot()

=

1

∙

csc() − cot()

csc() + cot() csc() − cot()

=

csc() − cot() csc 2 () − cot 2 ()


45

= csc() − cot ()



WITH APPLICATIONS

It is perfectly acceptable to work the right side of the identity and arrive at the left side. The conjugate binomial of the denominator, csc() − cot(), is used here to simplify the fraction. Note that using csc() − cot() = 1 csc() − cot() is an application of the multiplicative identity. In addition, csc 2 () − cot 2 () = 1 from the Pythagorean identity.


46




GRAPHS OF TRIGONOMETRIC FUNCTIONS

WITH APPLICATIONS

Chapter 2 Graphs of Trigonometric Functions Learning Objectives:       

Understand domain and range of trigonometric functions Review the graphs of standard trigonometric functions Review concepts of amplitude, period and frequency with respect to trigonometric graphs Review transformations in the plane Understand transformations of trigonometric function graphs, in particular, phase shift Review inverse functions in general Introduce inverse trigonometric functions

Section 2.1 The Domain and Range of a Trigonometric Function The domain of the sine function is the entire set of reals. This results from its definition as the ratio of a real number, y, to a non-zero constant, r . Likewise, the domain of the cosine function is the set of real numbers, that is, (-∞, ∞). The range of the sine and the cosine functions is the set of real numbers from -1 to 1 inclusive or the closed interval [-1, 1]. A discussion of the rationale that was presented in section 1.3 is reiterated here: the set of range values for the sine function is governed by the y-coordinates of the points along the circumference of a circle

sin = . Each value in the

range of the sine function is the ratio of a y-coordinate to the radius of the circle. The values of the y-coordinates on a circle of radius r vary from – r to r (Figure 2.1.1, formerly Figure 1.3.12). So

sin− π2 = − = −1 is the minimum value in the range and sin2 =  = 1 is the maximum value. This logic applies to the range of the cosine function as well.

The other four trigonometric functions have discontinuities in their domains by definition. The range of the tangent function is defined to be the ratio of t he y-coordinate to the x-coordinate on the

tan = . With a variable in the denominator, there are values outside  of the domain for the tangent function. For instance, at  = , the x-coordinate is zero, and as such, 2   = 2 is not in the domain of the tangent function. The tangent function is undefined at  = 2.  Likewise, the tangent function is undefined at  = − . The x-coordinate is non-zero for all other 2   values of between − and . That is to say, the tangent function is defined on the open interval 2    − 2 , 2. This is one interval of the tangent function’s domain. The tangent is undefined at all odd  or  = 2 +1 , where k is an integer. If these values are removed from the set of multiples of 2  circumference of a circle



reals, the domain of the tangent function is found to be

⋯∪(− 52 ,− 32 ) ∪ (− 32 ,− 2)∪− 2 , 2 ∪ (2 , 32 ) ∪ (32 , 52 ) ∪ (52 , 72 ) ∪ ⋯

− 2 , 2, the range is infinite. For   = 0. Zero is in the range example, at  = 0 (the point (r , 0) on a circle of radius r ), the ratio =   If ratios of the y to x-coordinates are considered over the interval

 

of the tangent function. As  increases from 0 toward , the values of the y-coordinates increase while the values of the x-coordinates simultaneously decrease. Under these conditions, the entire


48



WITH APPLICATIONS

ratio or quotient increases without bound. The quotient values tend toward infinity. Refer back to Example 1.3.8 for a more simplified example of this phenomenon. On the following page, a link is provided to a video illustrating the range of the tangent function.

Figure 2.1.1 (formerly Figure 1.3.12) Coordinates and associated angles on the circumference of a circle

If this same analysis is applied to the angle values ranging from 0 toward

− 2, the ratios increase in

magnitude but in the negative direction, that is, toward -∞. Therefore, the range of the tangent function is (-∞, ∞). Similar arguments show the domain of the cotangent function to be all real numbers except integer multiples of π or , where k is an integer. The domain of the cotangent function is

 =  ⋯∪ −3,−2∪ −2,− ∪ −,0 ∪ 0, ∪ ,2 ∪ 2,3 ∪ 3,4 ∪ ⋯

And like its reciprocal cofunction, the tangent function, the range of the cotangent is the set of reals or (-∞, ∞).

cos = 0 sec = 1⁄cos  = 2 +

The secant function is undefined at any values of  that solve the equation, . That is, the secant is undefined wherever the cosine function is equal to zero. Zeros of the cosine function are not in the domain of the secant function. Recall the quotient identity, . The zeros of the cosine function are odd multiples of π/2, so the secant function is undefined at , where k is an integer. If these values are removed from the set of reals, the domain of the

1 2

secant function is

⋯∪(− 52 , − 32 ) ∪ (− 32 , − 2)∪− 2 , 2 ∪ (2 , 32) ∪ (32 , 52)∪ (52 , 72) ∪ ⋯ Download free eBooks at bookboon.com

49



WITH APPLICATIONS

Please note that the secant and the tangent functions have the same domain. If the range values of the secant function are analyzed over the domain interval r / x

− 2 , 2, the ratio

sec =  varies from infinitely large positive values as  approaches – π/2 ( x-coordinate

approaches 0) to the value of 1 when  = 0 ( x = r ). The maximum value of x is r when  = 0. This corresponds to a minimum value of 1 in the range of the secant function. As  increases toward π/2, x decreases toward zero again and the ratio r / x increases toward ∞. Therefore, an interval of the secant’s range is [1, ∞). This is not the entire range since period. Recall the period of the secant is 2π.

− 2 , 2 is only half of this function’s

If the range values of the secant function are analyzed over the domain interval

2 , 32 , the ratio r / x

varies from infinitely large negative values as  decreases toward π/2 ( x-coordinates are negative as they approach 0) to the value of -1 when  = π ( x = -r ). The minimum value of x is -r when  = π. This corresponds to a maximum value of -1 in the range of the secant function. As  increases toward 3π/2, x decreases toward zero again. The x-coordinates are negative and become progressively smaller. Therefore, the ratios of r to x increase toward -∞. The other interval of the secant’s range is (-∞, 1]. The full range of the secant function is (-∞, -1][1, ∞).

Similar to the secant function, the cosecant function is undefined for angles that terminate at points on a circle with zero y-coordinates. In other words, the cosecant function is undefined at multiples of π or , where k is an integer. The domain of the cosecant function is

 = 

⋯∪ (−3,−2) ∪ (−2,−) ∪ (−,0) ∪ (0, ) ∪ (,2) ∪ (2,3) ∪ (3,4) ∪ ⋯ Please note that the cosecant has the same domain as the cotangent. The range of the cosecant function is the same as the range of the secant function, that is, (-∞, -1][1, ∞). The following video should be helpful in understanding range and domain of trigonometric functions. Understanding minimizes, even eliminates, memorization. It is recommended that the video, ‘Tangent Line Definitions’ from Chapter 1, Section 1.3 on defining trigonometric functions by the tangent line be reviewed prior to the following video.

A video illustrating the range and domain of the trigonometric functions.

Any point on a circle of radius r defines the values of the six trigonometric functions. Any point, ( x, y), in the plane is a point on the circumference of circle centered at the origin of some radius. Therefore, any point in the plane may define the trigonometric functions.


50



WITH APPLICATIONS

Trigonometric function

Domain

Range

Sine

(-∞, ∞) (-∞,

[-1, 1]

Cosine

(-∞, ∞) (-∞,

[-1, 1]

Tangent

⋯∪(− 32 , − 2)∪− 2 , 2 ∪ (2 , 32 ) ∪ (32 , 52 )∪ ) ∪ ⋯

(-∞, (-∞, ∞)

Cosecant

Secant

Cotangent

⋯ ∪ −2,− −2,− ∪ −,0 −,0 ∪ 0, 0,  ∪ ,2 ,2 ∪ 2,3 2,3 ∪ ⋯ ⋯∪(− 32 , − 2)∪− 2 , 2 ∪ (2 , 32 ) ∪ (32 , 52 )∪ ) ∪ ⋯ ⋯ ∪ −2,− −2,− ∪ −,0 −,0 ∪ 0, 0,  ∪ ,2 ,2 ∪ 2,3 2,3 ∪ ⋯

(-∞, (-∞, -1][1, ∞)

(-∞, (-∞, -1][1, ∞)

(-∞, (-∞, ∞)

Table 2.1.1 Summary of domain and range

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Section 2.2 Graphs of Sinusoidal Functions Definition 2.2.1 Sinusoidal function

A sinusoidal function is a smooth, wave function that oscillates symmetrically symmetrically between a crest and a trough in a repetitive fashion. It is named after the sine function.

Consider graphing the trigonometric functions in the xy the xy-plane. -plane. The independent variable will be x be x and the dependent variable, y variable, y.. In the xy the xy-plane, -plane, x x and and y y are are real variables. If we define the sine to be a function of x of x,, then it mandates that the angle x angle x be be measured in radians. Recall, radians are real numbers. Imagine the infinite angles of a circle linearized linearized onto a horizontal axis (the abscissa or domain axis). It is reasonable, but not necessary, necessary, to scale this axis in terms of . The ordinate axis (range axis) represents the ratios that define the trigonometric functions. Defined in this way, the trigonometric functions become real-valued real-valued functions and may be represented graphically in the t he real plane. The sine and the cosine cosine are examples examples of sinusoidal sinusoidal functions. They They are smooth, smooth, continuous functions over the set of real numbers. The domain domain of the sine and the cosine function is ( -∞, ∞).

Example 2.2.1 ( ) = sin( sin( ) over 2 periods of its domain. Graph (

One quarter of the period or π/2 is a useful scale for the x the x-axis. -axis. Extreme values and zeros of the sine curve are easily identified and plotted using this scale. These values correspond to quadrantal angles (0, π/2, π, 3π/2, etc.). The etc.). The sine function takes on the values of -1, -1, 0, or 1 at these multiples of π/2.

y

x

Figure 2.2.1 Sine curve

The graph in Figure 2.2.1 uses a scale of π/4. This scale still allows easy location of the quadrantal angles and, as such, simplifies the location and plotting of both extrema and zeros ( x-intercepts). x-intercepts). In


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this graph, maxima are located at (π/2, 1) and ((-3π/2, 1). Minima are Minima are located at (-π/2, (-π/2, -1) and (3π/2, 1). Zeros are located at multiples of π. The graph has one intercept, the origin. It should be noted that every point on the curve corresponds to a point on t he circumference circumference of a circle centered at the origin. This causes the graphs of the trigonometric functions to have symmetry and periodicity. One period of the t he sine function is i ndicated on the interval (0, 2π). The period of a standard sine curve is any distance of 2π in the domain. domain. For example, the interval (-π/2, (-π/2, 3π/2) is also one period of the sine. Likewise, (-π, (-π, π) represents one period of the sine function. The curve is symmetric with respect to the origin. Recall that t he sine is an odd function. The range is clearly [-1, 1].

Example 2.2.2 () = cos( cos( ) over 2 periods of it’s domain. Graph ( domain.

The cosine is the cofunction of the sine, sine, that is, cos(π/2 – x) x) = sin( x) x) and sin(π/2 – x) x) = cos( x). x). The graph of the cosine function is a horizontal translation or shift of the sine curve π/2 radians in the negative direction.

y

x

Figure 2.2.2 Cosine curve

In this graph, maxima are located at (-2π, (-2π, 1), (0, 1) and (2π, 1). Minima are Minima are located at (-π, (-π, -1) and (π, -1). At each of these points, the sine curve has zeros. Refer back to Figure 2.2.1. Zeros of the cosine function are located at odd multiples of π/2. At these points, the sine function has extrema (maxima extrema (maxima or minima). The y The y-intercept -intercept of the cosine graph is located at (0, 1). An example of one period of the cosine is indicated from 0 to 2π 2π in the graph. graph. This corresponds to one revolution of 2π radians in a circle. The cosine curve has y has y-axis -axis symmetry and is an even function. The range is [-1, 1]. The graphs of sine and cosine should be closely compared. Figure 2.2.3 enables this comparison. The horizontal shift of π/2 radians is clear in this figure.


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Figure 2.2.3 Sine curve with cosine comparison

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Every point on either curve above corresponds to a point on the circumference of a circle centered at the origin. The graphs of the trigonometric functions have symmetry and periodicity. One period of both functions is indicated on the interval (0, 2π). The period of a standard sinusoidal trigonometric function is any distance of 2π in the domain. Sinusoidal functions are wave functions. They have crests or peaks at maximum values and minimum values at the bottom of each trough or valley. In this area, mathematical terminology coincides with that of physics and telecommunications.

Definition 2.2.2 Amplitude

Amplitude is the vertical distance between any peak or valley of a wave and its point of equilibrium. It is the height or depth of the wave.

For the sine and cosine curves graphed in Figures 2.2.1 and 2.2.2, the amplitude is 1 (see Figure 2.2.4).

Figure 2.2.4 Amplitude of a sine curve

Definition 2.2.3 Frequency

The frequency of a wave curve is the number of times the curve repeats during its standard period.

The frequency for the standard sine or cosine curve is 1. One complete crest and trough occurs over one period of a sinusoidal curve when the frequency is equal to 1. In other words, one complete revolution of the circle occurs over one period of a sinusoidal curve. This concept, along with


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amplitude, will be further clarified in Section 2.4, Transformations of the Graphs of Trigonometric Functions. The following summarizes the general form of a sinusoidal function. Definition 2.2.4 General Sinusoidal Function in Standard Position

() =  sin() and () =  cos() where |A| is the amplitude and || is the frequency.

For each graph in this section, |A| = 1 and  = 1.


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Section 2.3 Graphs of Tangent, Cotangent, Secant and Cosecant The graphs of the remaining four trigonometric functions have vertical asymptotes corresponding to the undefined values in their domains.

Example 2.3.1 Graph

 = tan over 4 periods of it’s domain.

Recall the domain of the tangent function:

⋯∪ (− 52 , − 32 ) ∪ (− 32 , − 2) ∪ − 2 , 2 ∪ (2 , 32) ∪ (32 , 52) ∪ (52 , 72) ∪ ⋯ The tangent curve has vertical asymptotes at points not included i n the domain (odd multiples of π/2). The period of the tangent function is a distance of π radians in the domain as indicated in Figure 2.3.1. The tangent function cycles through its entire range from -∞ to ∞, over one period. The curve has no extrema. Zeros occur at multi ples of π. The graph has origin symmetry. Recall the tangent is an odd function. The scale is π/4. Observe the periodicity of the graph. For example, tan(3π/4) = tan(π/4) = tan(5π/4) = 1. The distance between -3π/4 and π/4 is π radians or 1 period of the tangent function.

Figure 2.3.1 Tangent curve

The graph of the tangent’s cofunction, the cotangent, is a horizontal translation of π/2 radians in the plane. See Figure 2.3.2. Since the tangent and cotangent are also reciprocal functions, a reflection in the graph is observed in addition to the translation (see Figure 2.3.3).


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Example 2.3.2 Graph () = cot () over 4 periods of it’s domain .

Recall the domain of the cotangent function:

⋯ ∪ (−2, −) ∪ (−,0) ∪ (0, ) ∪ (,2) ∪ (2, 3) ∪ ⋯ The discontinuities in the domain determine the locations of vertical asymptotes in the graph. Like the tangent, the cotangent curve has no extrema, is symmetric to the origin and is an odd function. Likewise, its period is π radians and its range is (-∞, ∞). In contrast to the tangent function, the cotangent function has zeros at odd multiples of π/2 (locations of the asymptotes of the tangent curve). Note that the cotangent curve has a vertical asymptote at the y-axis or the line x = 0.

Figure 2.3.2 Cotangent curve

It is helpful to view the graphs of trigonometric cofunctions together. In Figure 2.3.3, the tangent is graphed with the cotangent. The similarities between the curves may be easily observed along with the defining horizontal shift between cofunctions.


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Figure 2.3.3 Tangent with cotangent comparison


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Example 2.3.3 Graph () = csc() over 2 periods of it’s domain .

Recall the domain of the cosecant function:

⋯ ∪ (−2, −) ∪ (−,0) ∪ (0, ) ∪ (,2) ∪ (2, 3) ∪ ⋯ It determines the location of the vertical asymptotes on the graph. Please note that the definition of the cosecant function also reveals the location of asymptotes. Since csc( x) = r / y, the y-coordinate is zero at the quadrantal angles of 0, π, 2π, -π, etc., that is, at any multiple of π. These are the locations of the vertical asymptotes on the cosecant curve. The cosecant has minima (-3π/2, 1) and (π/2, 1) in Figure 2.3.4. Please note that the horizontal distance between these points is 2π, the period of the cosecant. It has relative maxima at the points (π/2, -1) and (3π/2, -1). Other maxima may be found by adding multiples of 2π to the x-coordinates of these points. The cosecant has no zeros. In fact, it has no i ntercepts at all. It is an odd function with origin symmetry.

Figure 2.3.4 Cosecant curve

It is useful to compare the cosecant with its cofunction, the secant. The reader may wish to graph these functions together on the same set of axes to visualize the horizontal shift of π/2 radians. Another useful comparison is to graph a t rigonometric function with its reciprocal function. The reciprocal function of the cosecant is the sine. Recall the identity:

csc() =

1 sin()

or sin() =

1 csc()

.

It is revealing to graph the sine curve coincidentally with the cosecant curve. See Figure 2.3.5. The sine function shares extreme values with the cosecant. Maxima of the sine are minima of t he cosecant. Maxima of the cosecant are minima of the sine. Zeros of the sine curve coincide with


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vertical asymptotes of the cosecant curve. The range of the cosecant function, along with i ts periodic nature is evident in its graph. Recall the range of the cosecant to be: (-∞, -1][1, ∞) Understanding the relationship between the graphs of the sine and cosecant functions, simplifies the task of graphing a cosecant curve.

Figure 2.3.5 Cosecant with sine comparison

Example 2.3.4 Graph

 = sec over 2 periods of it’s domain.

Recall the domain of the secant function:

⋯∪ (− 32 , − 2) ∪ − 2 , 2 ∪ (2 , 32) ∪ (32 , 52) ∪ ⋯ Like the tangent function, the graph of the secant function has vertical asymptotes at odd multiples of π/2 (Figure 2.3.6). The secant function is not defined at these values. If these values are approached in the domain, the range values of the secant increase without bound toward ∞ or -∞. Asymptotic behavior is seen in the secant’s graph near odd multiples of π/2. Please note the locations of extrema. These are offset or shifted horizontally from the positions on the secant’s cofunction, the cosecant. The secant has no zeros or intercepts. It is an even function with y-axis symmetry.


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Figure 2.3.6 Secant curve

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As with the cosecant, it is useful to compare the secant with its with its reciprocal function, the cosine. Refer to Figure 2.3.7. Recall the identity:

sec() =

1 cos()

 cos() =

1 sec()

.

The secant curve shares points with the cosine in the same way that the cosecant curve shares points with the sine. Naturally, the same relationship exists between the zeros of the cosine and asymptotes on the graph of the secant, as it did between the sine and cosecant curves. Graphing a cosine curve may facilitate the graphing of a secant curve.

Figure 2.3.7 Secant with cosine comparison


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Section 2.4 Transformations of the Graphs of Trigonometric Functions Definition 2.4.1 Transformation of the plane

A transformation of the plane is a function that maps one set of points in the plane (perhaps a geometric shape or the graph of a relation) to another set of points in the plane.

Some examples of transformations include translations, rotations, dilations or scales, reflections, shears, projections, etc. Transformations have applications in kinematics and robotics, computer graphics, optics, signal processing, engineering and, of course, mathematics and physics. The focus of this discussion is on the 3 transformations summarized in Tables 2.4.1 and 2.4.2 below:

Type

Operator

Colloquial

1. Translation

Addition

Shift; move point for point

2. Dilation (scale)

Multiplication

Magnify/compress

3. Reflection

Negation

Flip or rotate about an axis or line

Table 2.4.1 Types of transformations and associated operators

Transformation occurs to the:

Direction

Affect is:

Independent variable (usually x)

Horizontal

Counter-intuitive (not expected)

Dependent variable (usually y)

Vertical

Intuitive

Table 2.4.2 Affect of transformations

A translation or a shift moves the entire graph, point for point, a fixed distance. A horizontal translation moves the curve left or right and a vertical translation moves it up or down. A shift of the curve in any other direction may be achieved by a combination of horizontal and vertical translations.

Example 2.4.1 Find the new function that results after the graph of () = 2  − 5 is shifted

a) 3 units up

b) 2 units down

c) 4 units left

Solution: a) Translating a function vertically affects only the y-coordinates. In other words, translating vertically affects the dependent variable of a function. Since  = (), shifting any function up 3 units means  +  = () + . In this example, () +  = 2  − 5 + . The operation associated with translation is addition. Graph () and () + 3 to visualize the translation. Observe the shifting of key points such as the


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extrema and the intercepts. The y-intercept of () is (0, 0). After the transformation, the y-intercept of () + 3 is (0, 3). b) () −  = 2  − 5 − . Graph the original function along with the transformed function and compare. c) Translating a function horizontally affects only the x-coordinates. Translating horizontally affects the independent variable of a function. Horizontal transformations are counter-intuitive, that is, to move the curve left, add; to move the curve right, subtract. To shift () left 4 units, add 4 to the independent variable.

( + ) = 2( + )  − 5( + ). Graph and compare. (0) = 0 or (0, 0) is a point on () . This point becomes (-4, 0) after translating 4 units left. That is, (−4) = 0 . Evaluating the transformed function at -4: (−4 + ) = 2(− 4 + )  − 5(− 4 + ) = 2(0)  − 5(0) = 0 . The result is zero. (-4, 0) is a point on the graph of ( + ). Recall that only the x-coordinate is affected.


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(1) = −3 or (1, -3) is a point on () . After translating 4 units to the left, (1, -3) becomes (-3, -3). Evaluating the transformed function at x = -3:

(−3+) = 2(−3+)  − 5(−3+) = 2(1)  − 5(1) = 2 − 5 = −3. (-3, -3) is a point on the graph of ( + ). A dilation or scaling stretches or compresses the entire graph by a certain factor.

Example 2.4.2 Find the new function that results after the graph of

() =  −  is stretched three times

a) vertically

b) horizontally.

Solutions:

 = (), () = ( − ) =

a) Dilating a function vertically affects the y-coordinates or the dependent variable. Since dilating any function 3 times vertically means . In this example,  . The operation associated with dilation is multiplication.

 = ()

 −  Graph () and 3() to visualize the dilation. Observe the stretching of key points. An x-intercept on the graph of () is (0, 0). After the transformation, this intercept is still (0, 0). This does not illustrate the dilation. But it should be noted that x-intercepts of a function remain unaffected by a vertical dilation. (2, 2) is another point on the graph of . After dilating 3 times vertically, this point becomes (2, 6) (only the y-coordinate is affected).

()

b)



() = 13  − 13  = 19   − 13 .

The horizontal dilation is counter-intuitive (similar to the previous horizontal translation). Tripling the independent variable results in a horizontal compression of the curve by a factor of 3. This example requires the stretching of the function 3 times in the horizontal direction. In order to achieve this, 3 must divide the value of the independent variable in the function. The point (1, 0) is transformed into the point (3, 0). The point (3, 6) is transformed into the point (9, 6). Verify this.

A reflection or a flip rotates the entire graph about a line. A reflection about the x-axis is a vertical reflection and, as such, affects the dependent variable. A reflection about the y-axis is a horizontal reflection and affects the independent variable. The operation associated with reflection is negation.

Example 2.4.3 Find the new function that results after the graph of

() =   is reflected

a) over the x-axis

b) over the y-axis.

Solutions: a)

– () = − 

This transformation will affect all dependent coordinates ( y values) of the graph. It is a vertical


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reflection. The point (1, e) is transformed into the point (1, -e). The point (0, 1) is transformed into the point (0, -1). b) () =  − This transformation will affect all independent coordinates ( x values) of the graph (a horizontal affect). The y-intercept is not affected by this reflection. The point (0, 1) remains (0, 1). The point (2, e2) is transformed into the point (-2, e2). It is often of interest and very useful, in the study of transformations, to understand which properties of the set (graph) are invariant under the specific transformation. For example, distances between points are preserved or invariant under a translation or a reflection. The distances do not change. On the contrary, the distance between points is not preserved under a dilation of the set. The distance between points is not invariant under dilation. Recall the general form of the sinusoidal function:

() =  sin() and () =  cos(). A change in amplitude (| A|) causes a vertical dilation of the curve while a change in frequency (||) results in a horizontal dilation.

Example 2.4.4 Graph () = 3sin().

The transformation occurs outside or to the dependent variable. Therefore the affect is vertical and intuitive. The operation is multiplication, so t his is a dilation. The standard sine curve is vertically stretched by a factor of 3. This affects the amplitude and extrema. The amplitude is now 3. The zeros remain unchanged. The period is not affected by this transformation.

Figure 2.4.1 () = ()


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Example 2.4.5 Graph () = cos(3).

The independent variable is being multiplied, so this is a horizontal dilation. The affect of a horizontal transformation is counter-intuitive. Multiplication by 3 compresses the standard cosine curve 3 times. This affects zeros, extrema and the period of the function. The period of the standard cosine is 2π. The graph of the cos(3x) in Figure 2.4.2 shows 3 periods over this same distance. This illustrates the idea of frequency. The new cosine curve is 3 times as frequent as the original cosine. This means that the period of the standard cosine cur ve, 2π, is divided by 3. The period of the new transformed cosine is 2π/3. This illustrates a general formula for the period of a sinusoidal function:

=

2 ||

where T is the period and  is the frequency. Note that when || = 1, as it is in the standard cosine curve, the period is 2π. It should also be noted that the dilation in this example did not affect the amplitude (| A| = 1) nor did it affect the y-intercept.

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Figure 2.4.2

() =()

Example 2.4.6

Graph

() =cos13 .

The dilation in this example is still horizontal, yet the factor of 1/3 has the affect of stretching the standard cosine 3 times. This has no affect upon the amplitude or the y-intercept. It does stretch the period 3 times making it 6π. If the formula from Example 2.4.5 is used

the same result is obtained.

2 = 2 = 3(2) = 6  = || 1 3

In addition to the period, each x-coordinate of the original cosine’s zeros and extrema has been tripled by this transformation. Refer back to Figure 2.2.2.


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Figure 2.4.3

() = 

Example 2.4.7 Graph

() =sin().

This is another example of a horizontal dilation compressing the standard sine curve by a factor of π. The frequency of the new curve is π. Applying the period formula,

2 = 2 = 2  = ||  the transformed period is found to be 2. It is not necessary that the period have a factor of π or be irrational. As mentioned previously, a good x-axis scale is one quarter the period or smaller. In figure 2.4.4, an x-axis scale of ¼ is used. A scale of ½ would have sufficed. It is desirable to be able to plot the characteristic or defining points of a curve. For trigonometric functions, these points are the intercepts, extrema and asymptotes (if any). These occur at period quarters (formerly quadrantal angles). Please note in Figure 2.4.4 that the extrema and zeros have been divided by a factor of π when compared to those of the standard sine curve in Figure 2.2.1.


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Figure 2.4.4 () = ()

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Example 2.4.8

() = −2 cos12  .

Graph

Multiple transformations are applied to the same function in this example. The leading negative sign reflects the cosine in the vertical direction (the negative is outside), that is, over the x-axis. The amplitude is |-2| or 2. The factor of -2 causes two different transformations. The negative reflects and the 2 dilates. There is also a horizontal stretch 2 times. When considering a function with multiple transformations, it is advisable to separate them. The amplitude is doubled along with the period:

2  = |2 | = 1 = 4. The x-coordinates of the 2

extrema and zeros have been doubled. The y-intercept is affected in this example, both by the doubling of amplitude and by the reflection over the x-axis. See Figure 2.4.5.

Figure 2.4.5

() = −  


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Definition 2.4.2 Phase Shift

The displacement of a sinusoidal curve from its standard position is called a phase shift or a phase offset. It is typically measured in time units or radians.

When two waveforms are in phase they coincide in time or space. By definition, the standard sine and cosine curves are out of phase a horizontal distance of π/2 radians. In Figure 2.4.6, the horizontal distance between pairs of corresponding points on these curves is π/2. For example, the maximum point (0, 1) on the cosine is translated π/2 radians from the maximum point (π/2, 1) on the sine. The cofunction identity reveals this fact.

sin() =cos2 −

cos() =sin2 − 

If brought into phase, these two functions would coincide. Using this identity, it is possible to translate any cosine function into a sine function and vice versa.

Figure 2.4.6 (formerly Figure 2.2.4) Phase shift between a sine and a cosine curve

Example 2.4.9 Rewrite

() =2cos(4) as a sine function. () =2cos(4) =2sin2 −4.

Another physical example of phase shifts are the distribution of time zones across the globe. A manmade construct, they do a good job of modeling the natural variation of daylight hours as our planet rotates its course about the sun. The current time in New York City, USA is out phase with that of Copenhagen, Denmark by 6 hours.


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Example 2.4.10

 =sin− 12  as a cosine function.

Rewrite

Solution:

 =sin− 12 =cos2 + 12 .

For purpose of this text, a horizontal translation of a sinusoidal curve is a phase shift. Sinusoidal curves are simple waveforms. In comparison, music is a good example of a complex waveform.

 will be used to designate phase shift. The general sinusoidal formulas become  =sin( + ) + and  =cos( + )+ where |A| is the amplitude, || is the frequency, − is the phase shift and D is the vertical translation.  is negative due to the counter-intuitive nature of the horizontal transformation. When − > 0 ( < 0), the shift is in the positive direction. When − < 0, it shifts in the negative direction. Please note that when A =  = 1 and  = D = 0, the above become the equations of the standard sinusoidal curves. The symbol

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Example 2.4.11 Graph

() = cos(2 −).

It is advisable to separate the two transformations, that is, factor the horizontal dilation of 2 from the phase shift (horizontal translation) of +π/2. The equation becomes

() = cos(2 −) = cos2 − 2. The amplitude is 1, the frequency is 2 and the period is π. A good scale is π/4. The phase shift should also be considered when choosing scale. The standard cosine is shifted π/2 radians in the positive direction, that is, the phase shift is +π/2. - = - π/2, therefore, = π/2. Since this is a multiple of π/4, π/4 is a good choice. The extrema and zeros will coincide with a scale of this size.



Figure 2.4.7



() = ( − )

Looking at the graph, it is readily apparent that this graph may also be expressed as a compressed and shifted sine function of the form:

() = sin2 − 4 = sin2 − 2. Example 2.4.12

() = 13 sin(3 + 2)

Graph

.


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The amplitude is 1/3, the frequency is 3, the period is 2/3π and the phase shift is -2π/3 (factor). The negative sign indicates the direction of the shift. Recall that horizontal translations are counterintuitive.

Figure 2.4.8 () =

 ( + ) 

Upon inspection, it is clear that it would have been simpler t o express this function as () = 1 sin(3). The extrema and range were changed by these transformations. The range has become [3

1/3, 1/3].

Example 2.4.13 Graph () = sin(− + 2) + 1 = sin(2 − ) + 1.

The amplitude is 1, the frequency is 1 and the period is 2π. Still, this is a tricky one to scale. One quarter the period is π/2, however, the phase shift is +2 (factor the minus) so this should be added to π/2 to locate significant quadrantal points, that i s, extrema and zeros. The graph is also vertically shifted up 1 unit. This affects all characteristic points and the range of the sine function. The range is now [0, 2]. The quadrantal points or extrema are labeled in the graph. In order to obtain exact values for these points, start with 2 + π/2 ≈ 3.571. Locate this minimum. Since the period is 2π, the x-coordinate of the previous maximum is π less than this value or 2 + π/2 – π = 2 – π/2 ≈ 0.429. The y-intercept may be obtained by solving the equation () = 0. Solving trigonometric equations is a topic of Chapter 3.


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Figure 2.4.9

f ( x ) = sin(- x + 2) + 1

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Transformations may be applied to the tangent, cotangent, secant and cosecant functions. These are not sinusoidal functions, so terminology such as amplitude, frequency and phase shift do not apply to these curves. The coefficient of the independent variable still divides the period of the standard function.

Example 2.4.14

Graph

 = tan12  + 3.

Since the period of the standard tangent function is π, the formula for period of the tangent function (and the cotangent) is

 = ||. The period of  = tan12  + 3 is  = || = 1 = 2 . The 2

horizontal shift is -2/3π (counter -intuitive). A good scale is then |π/2 – 2π/3| = |-π/6| = π/6. The scale in Figure 2.4.10 is actually π/12. Zeros and asymptotes will fall on multiples of π/6. The tangent curve has also been stretched 2 times in the horizontal direction. The shift combined with the stretch affects vertical asymptotes and domain. The vertical asymptotes fall on certain odd multiples of π/3 and are separated by a distance of 2π (the period). Two are shown of Figure 2.4.10. The domain of this tangent curve is:

5 )∪ (− 5 , ) ∪ ( , 7 ) ∪ (7 , 13) ∪ ⋯ ⋯∪(− 11 ,− 3 3 3 3 3 3 3 3

Figure 2.4.10

 =   + 


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Example 2.4.15 Graph

 =csc3− − 2.

The period is 2π/3. The formula for the period of the cosecant (and secant) function is the same as the formula for the sine and cosine. See Table 2.4.3 at the end of this section. The horizontal shift is π/3. π/6 should make a good scale. Note that all asymptotes are shifted along with the curve itself. The cosecant is shifted down 2 units affecting range, zeros and extrema. Examples of extrema are labeled in Figure 2.4.11. The range is easily identified as (-∞, -3] [-1, ∞). Please note the affect on domain (position of asymptotes) due to the horizontal compression. Compare the domain in Figure 2.4.11 to the domain of the standard cosecant curve.

∪

Figure 2.4.11

 =−

Example 2.4.16

 = − 12 cot+ 4.

Graph

The period is π (unchanged). The horizontal shift is – π/4. π/4 should make a good scale. The cotangent is also vertically reflected and compressed. This reflection makes it appear more like a tangent curve. It may readily be expressed as

 = 12 tan− 4.

It is a good idea to verify coordinates of the graph by evaluating them in the function. For example, the point (3π/4, ½) on the graph in Figure 2.4.12 can be verified as follows:

(34) = − 12 (34 − 4) = − 12 2 = − 12 −1 = 12


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Figure 2.4.12

() = −  + 

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General Sinusoidal Equations

Variables

f ( x) = Asin(( x + )) + D

|A| = amplitude || = frequency -  = phase shift D = vertical translation

=

f ( x) = Acos(( x + )) + D

|A| = amplitude || = frequency -  = phase shift D = vertical translation

=

Non-Sinusoidal Equations

Variables

f ( x) = Atan(( x + )) + D

|A| = vertical dilation || = horizontal dilation  = horizontal translation D = vertical translation

=

f ( x) = Acot(( x + )) + D


=

f ( x) = Asec(( x + )) + D


=

f ( x) = Acsc(( x + )) + D


=

Table 2.4.3 Summary of formulas

Example 2.4.17 Find a sinusoidal function for the curve in Figure 2.4.13.


81

Period

2 ||

2 ||

Period

 ||

 ||

2 ||

2 ||



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Figure 2.4.13 Unidentified sinusoidal curve

Solution: A cosine function is chosen based upon the look of the graph (the y-intercept is not at zero). Either sine or cosine could be used to define a sinusoidal curve with an equation. The general cosine function from Table 2.4.3 is

 =  cos( + ) + . The amplitude is clearly 3/2. A = -3/2, however, as the graph could have been reflected about the xaxis. A phase shift could also be used to achieve the desired affect, however, phase shifts can be more difficult to express. If A is allowed to be negative, a phase shift is unnecessary, that is,  = 0 . D is also zero (no vertical shift). In order to determine , the period must be found. Peak to peak, the period is observed to be 2π/3. The period may also be determined by examining zeros on the graph.

=

2 ||

2 2 = → || = 3 3 || Since no apparent horizontal reflection exists,  = 3 as well. The equation becomes

3 3  = − cos(3 + 0) + 0 →  = − cos3 . 2 2


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Solution: A sine function is chosen based upon the look of the graph. It has clearly been shifted up 1 unit, so D = 1. Without this shift, the y-intercept would have been at the origin. The general sine function from Table 2.4.3 is

 =sin(+)+.  = ±1

Clearly the amplitude is 1. This means that in the equation. A = -1 means the standard sine curve was reflected about the x-axis. Since sine is an odd function, a reflection about the x-axis is indistinguishable from a reflection about the y-axis. So either A or may be negative In order to determine , the period must be found. Peak to peak, the period is observed to be 4π.



One solution is



.

 = ||2 4= ||2 → || = 24 = 12  =−sin +1.  =sin−  +1. ≠1  =sin12 +2+1=sin12 ++1. Another possible solution is

Instead of a reflection, a phase shift could be used to achieve the desired affect. Phase shifts are generally more complicated, especially if . Another possible solution is,


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Solution: A sine function is chosen based upon the look of the graph. It appears to be a sine curve shifted π/8  . The general sine function is radians to the right.

=− 8

 =sin( + ) + . The amplitude is observed to be 3. There are no apparent reflections and no vertical shift, D = 0. Peak to peak, the period is observed to be π.

2  = ||

2 → || = 2 = 2  = ||  A solution is

 =3sin2− 8=3sin2− 4. There are other natural solutions including

many involving the cosine function.

Sinusoidal motion is also referred to as simple harmonic motion.


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Definition 2.4.3 Simple Harmonic Motion

Simple harmonic or sinusoidal motion is a type of smooth, periodic motion where the restorative force is proportional to the displacement in accordance with Hooke’s Law given by F = -kx

where F is the restoring force, k is a constant and x is the distance displaced.

A good physical example of simple harmonic motion is a mass on a spring that is displaced from its equilibrium state or pulled and released. k is called the spring constant, a measure of its stiffness. The force F , working to bring the mass back to its equilibrium state, is proportional to the distance it was pushed or pulled. If this system does not lose energy, as would happen in reality, the oscillation of the mass would trace out a sinusoidal curve as illustrated in the following video: Simple Harmonic Motion

The circle in the video is called a phase portrait . The vertical axis represents velocity and the horizontal axis represents position. Position and velocity are out of phase, shifted π/2 radians like the sinusoidal curves (Figure 2.4.16). The curve tracing out i n the video represents the position of the mass. At its highest point (peak) or at its lowest point, the velocity of the mass is zero (ycoordinate is zero), for an instant, as it changes direction. The x-intercepts of the position curve represent the maximum velocity of the object before the spring begins to decelerate it back again to a zero velocity followed by a change in direction.

Figure 2.4.16 Time series showing phase shift between velocity and position of an oscillating spring

Example 2.4.20


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Develop a model for the motion of a mass oscillating on a spring. The time of oscillation is 5 seconds after the mass is pulled down 3 inches from its resting position. Solution: The time of oscillation represents the period of the curve. So the frequency is

2 → || = 2 . 5 = || 5 Since time ‘begins’ after the mass is pulled downward, its position is negative 3 inches from equilibrium. So it is a good idea to use a cosine function.

 = cos +  +   = cos(25  + ) +  The independent variable used is t, suggestive of what it represents, time. P (t ) is position. There is no phase shift and no vertical shift.

 = cos(25 ) The displacement of the mass represents the amplitude or |A| = 3. Since time begins when the spring is pulled downward, a reflection about the x-axis is required.

 = −3cos(25 ) 0 = −3cos0 = −3

5 = −3cos2 = −3

At t = 0, and after 5 seconds, , the position of the object is at its initial position once again (its time of oscillation is 5 seconds). Figure 2.4.17 shows the model.


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Section 2.5 Inverse Trigonometric Functions Definition 2.5.1 Inverse Function

A function g is called the inverse of a function f , if y = f ( x) implies x = g ( y) and x = g ( y) implies y = f ( x).

The inverse of a function reverses that function’s process. If f brings the value x from its domain to the value y in its range, then its inverse or anti-function, g , must bring the value y back from f ’s range to the value x in f ’s domain. The functions f and g are said to be inverses of each other.

Figure 2.5.1 Conceptual diagram of a function and its inverse

The inverse of a function f is denoted f -1. Using this notation, Definition 2.5.1 could be restated: Definition 2.5.2 Inverse Function (restated)

If f -1 is the inverse of a function f , then (  −1) =

 −1( ) = 

for all  in the domain of  and its inverse, −1. Replace g with f -1 in definition 2.5.1. x = g ( y) becomes x = f -1( y). Substitute y = f ( x) into x = f -1( y):

 =  =  −1 =  −1( ). This is the case where composition is commutative,

 °  −1 = (  −1) =  −1( ) =  −1°  = . Please note in Figure 2.5.1 that the range of f -1 is the domain of f . Likewise, the domain of f -1 is the range of f . The domain and range of a function and its inverse are essentially swapped . This is


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necessary for the inverse to perform its intended function. This definition leads to a symmetric relationship between the graph of a function and its inverse. The swapping of range and domain causes a reflection about the line y = x between the graph of a function and its inverse.

Example 2.5.1 Graph f ( x) = e x and f -1( x) = ln( x) on the same set of axes.

Figure 2.5.2 Symmetry between f ( x ) = e x and f -1( x ) = ln( x )

f ( x) = e x and f -1( x) = ln( x) are two well-known inverse functions. Notice the symmetry about the li ne y = x. The definition of inverse holds, that is,

 −1 =   =  = l n   =  −1( ). This symmetry enables the inverse to be obtained graphically. The inverse of a function need not exist. Or, it might exist under certain conditions such as a restriction on the domain.

Example 2.5.2 Find the inverse of f ( x) = x2 graphically.

First graph f . Then reflect it over the line y = x.


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Figure 2.5.3 Finding the inverse of f ( x ) = x 2

The dark green reflection is not a function. It fails the vertical line test, which means it does not satisfy the definition of a function.

Definition 2.5.3 Function

A function is a relation that uniquely associates each value from its domain with exactly one value in its range.

The dark green relation does not uniquely associate 1 with a value in its range. 1 is mapped to 1 and to -1 (both (1, 1) and (1, -1) lie on the reflected dark green curve). That is not a unique mapping. If the dark green curve were a function and it had function notation, such as g ( x), then g (0) = 0, but g (1) = ? Does it equal 1 or -1? A function must have a unique range output for each domain input. In order for a function to be inverted , it must be one-to-one (1:1). A 1:1 function passes a horizontal line test: no horizontal line drawn through its graph intersects the curve more than once. The quality of being 1:1 enables the reflection about y = x to result in a curve that represents a function. A horizontal line reflected about y = x becomes vertical. If a curve passes a horizontal line test, i t will pass a vertical line test upon reflection over the line y = x. The function x2 is not 1:1 (see Figure 2.5.3). So it has no inverse. But it is known that √  is the inverse of x2. The square root function is used to solve quadratics and the square function is used to solve equations involving square roots. This seems to be a contradiction. It is not. The fact is that the inverse of x2 exists when its domain is restricted to non-negative values (see Figure 2.5.4). In this way, its reflection is a function, the inverse function  −1 () = √  .


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Figure 2.5.4

() =   and its inverse, −() = √ 

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None of the trigonometric functions are 1:1. Their domains must be restricted in order to define their inverses. Recall the graph of y = sin( x) (Figure 2.2.1). There are infinite ways to slice this curve or to restrict its domain in order to obtain a 1:1 graph. When restricting domain, it is important that the entire set of range values be preserved. And it is desirable to retain the domain values corresponding to the first quadrant, that is, to retain the domain interval (0, π/2). On this interval, all six trigonometric functions are positive. This is more of a preference, but it simplifies the function. When these criteria are adhered to, there is only one way to slice each trigonometric function. Reiterating this criteria: 1) The entire range is preserved 2) The domain interval (0, π/2) is retained (give or take a few points) and 3) The curve slice is 1:1 In this way, each trigonometric inverse is unique.

Example 2.5.3 Find the inverse of f ( x) = sin( x).

Figure 2.5.5 shows the 1:1 slice of the sine curve. See that the previous check list is satisfied.

Figure 2.5.5 1:1 Slice of the sine curve

Figure 2.5.6 shows the reflection of t his slice over the line y = x. The reflected curve defines the inverse sine curve or f ( x) = sin-1( x). The -1 in this notation is not an exponent. It is notation to denote inverse.


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Figure 2.5.6 1:1 Slice of the sine curve with its inverse

Please note sin-1( x) ≠ csc( x).

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The restricted domain of the sine function, [[-π/2, π/2, π/2] (endpoints included), becomes the range of range of the inverse sine. The domain of the inverse i nverse sine is [-1, 1]. Please note how the domain and range swap range swap between the sine sine and its inverse. inverse. Function

Domain

Range

f ( x x)) = sin( x x))

[-π/2, [π/2, π/2] (restricted)

[-1, 1]

f -1( x) x) = sin-1( x) x)

[-1, 1]

[-π/2, [π/2, π/2]

Table 2.5.1

Domain and range of sine sine and and inverse inverse sine

Example 2.5.4 Find the inverse of f of f ( x) x) = cos( x x). ).

Figure 2.5.7 shows the 1:1 slice of the cosine curve following the domain restriction criteria. The domain has been restricted to [0, π]. The cosine is then reflected about the li ne ne y y = = x x to to obtain the graph of the inverse cosine.

Figure 2.5.7 1:1 Slice of the cosine curve with its inverse

Note the swapping the swapping of of range and domain between the function and its inverse. Also note the swapping of x of x and and y y coordinate coordinatess between corresponding reflected points. For examp example, le, the point (π, 1) on the cosine is reflected to the position, (-1, (-1, π) on the graph of graph of the t he arccosine (inverse cosine). cosine). -1 The y The y-intercept -intercept of f of f ( x) x) = cos ( x) x) must be the point (0, π/2). Why?

Function

Domain

Range

f ( x) x) = cos( x x))

[0, π] (restricted)

[-1, 1]

f -1( x) x) = cos-1( x x))

[-1, 1]

[0, π]

Table 2.5.2 Domain and range of cosine and inverse cosine


95



WITH APPLICATIONS

The remaining four inverses will have horizontal asymptotes since all the non-sinusoidal trigonometric function graphs have vertical asymptotes.

Example 2.5.5 Find the inverse of f of f ( x x)) = tan( x x). ).

It makes sense to take a slice of the curve between asymptotes (if possible). Following the established criteria, the restricted domain becomes (-π/2, (-π/2, π/2). Figure 2.5.8 shows the tangent tangent slice slice with its inverse.

Figure 2.5.8 1:1 Slice of the tangent curve with its inverse

A vertical line reflected over y over y = = x x becomes becomes horizontal. For example, the vertical asymptote x asymptote x = π/2 on the tangent curve becomes the horizontal asymptote y asymptote y = π/2 on its inverse. inverse.

Example 2.5.6 Find the inverse of f of f ( x) x) = cot( x x). ).

As with the tangent, it is possible to t ake a 1:1 slice of the cotangent curve between vertical vertical asymptotes. The restricted domain is (0, π). This interval, (0, π), is the range of range of the inverse cotangent. Note that the x the x-axis -axis is a horizontal asymptote of the inverse and the y the y-axis -axis is a vertical asymptote of the cotangent. Notice the asymptote reflections and the coordinate reflections of selected points.


96



WITH APPLICATIONS

Figure 2.5.9 1:1 Slice of the cotangent curve with its inverse

Figure 2.5.10 and 2.5.11 show the t he secant and cosecant slices cosecant slices with with their respect inverses. Due to t he nature of the secant and cosecant curves, it is necessary to slice to slice each each of these curves to include a vertical asymptote. No curve slice between asymptotes asymptotes is 1:1 for these two functions, if it retains the entire range. Take note of asymptote and coordinate reflections in these figures. Similar to the cotangent, in Figure 2.5.11, it is somewhat difficult to see the asymptotes of the cosecant and its inverse as they coincide with t he he x x and and y y-axes. -axes. Table 2.5.3 summarizes the domain and range of each trigonometric function and its inverse.


97



WITH APPLICATIONS

Figure 2.5.10 1:1 Slice of the secant curve with its inverse

Figure 2.5.10 1:1 Slice of the cosecant curve with its inverse


98



WITH APPLICATIONS

Function

Domain

Range

f ( x) = sin( x)

[-π/2, π/2] (restricted)

[-1, 1]

f -1( x) = sin-1( x)

[-1, 1]

[-π/2, π/2]

f ( x) = cos( x)

[0, π] (restricted)

[-1, 1]

f -1( x) = cos-1( x)

[-1, 1]

[0, π]

f ( x) = tan( x)

(-π/2, π/2) (restricted)

(-∞, ∞)

f -1( x) = tan-1( x)

(-∞, ∞)

(-π/2, π/2)

f ( x) = cot( x)

(0, π) (restricted)

(-∞, ∞)

f -1( x) = cot-1( x)

(-∞, ∞)

(0, π)

f ( x) = sec( x)

[0, π/2) U (π/2, π] (restricted)

(-∞ -1] U [1, ∞)

f -1( x) = sec-1( x)

(-∞ -1] U [1, ∞)

[0, π/2) U (π/2, π]

f ( x) = csc( x)

[-π/2, 0) U (0, π/2] (restricted)

(-∞ -1] U [1, ∞)

f -1( x) = csc-1( x)

(-∞ -1] U [1, ∞)

[-π/2, 0) U (0, π/2]

Table 2.5.3

Domain and range of inverse trigonometric functions

Understanding how and why the trigonometric functions are sliced to create inverse functions eliminates the need to memorize. By inspecting the standard trigonometric curve, it is possible to determine the 1:1 slice. This determines the restricted domain that becomes the range of the inverse. It is necessary to understand the range and domain of functions when evaluating them.


99



WITH APPLICATIONS

Example 2.5.7

Find

sin−1 (√ 22).

Solution: The range of the arcsine (inverse sine) is [ -π/2, π/2]. The answer must lie on this interval. Since the

sin4 = √ 22 ,sin−1 (√ 22) = 4 . Please note that sin34 = √ 22 , as well. However, 3π/4 cannot be the solution since it does not lie on t he interval [-π/2, π/2] (the range of the arcsine). Recall the whole point of slicing the trigonometric curves was to obtain functions, with each input having exactly one output.


100




WITH APPLICATIONS

Example 2.5.8 Find sin-1(-1).

Solution: The range of the arcsine (inverse sine) is [-π/2, π/2]. The answer must lie on this interval. Since the sin(-π/2) = -1, sin-1(-1) = -π/2. It is also true that sin(3π/2) = -1, however, 3π/2 cannot be the solution since it does not lie on the interval [-π/2, π/2] (the range of the arcsine).

Example 2.5.9 Find tan-1(-1).

Solution: The range of the arctangent is (-π/2, π/2). Since the tan(-π/4) = -1, tan-1(-1) = -π/4. This is the only solution since the arctangent is a function.

Example 2.5.10 Find csc-1(1/2).

Solution: The domain of the inverse cosecant is (-∞ -1] U [1, ∞). Since ½ does not lie on this interval, that is, ½ is not in the domain of the inverse cosecant, there is no solution to this example.

Example 2.5.11 Find sec-1(2).

Solution: The range of the inverse secant is [0, π/2) U (π/2, π]. Since the sec(π/3) = 2, sec-1(2) = π/3.

Example 2.5.12

Find sec(sec-1(2)). Solution: From the previous example, sec-1(2) = π/3. Then sec(sec -1(2)) = sec(π/3) = 2.

Example 2.5.13 Find cos(cos-1(-1)).

Solution: The range of the arccosine is [0, π]. The cos(π) = -1, so cos-1(-1) = π. It then follows that cos(cos-1(-1)) = cos(π) = -1.


101



WITH APPLICATIONS

(  ) =  ( ) = 

−1 −1 Do not assume that will always hold when dealing with functions that are not one-to-one. The domain and range should always be considered. Example 2.5.13 Find cos-1(cos(3π/2)).

Solution: cos(3π/2) = 0. But 3π/2 is not in the restricted domain of the cosine. In other words, 3π/2 is not in the range of the arccosine. So, cos-1(cos(3π/2)) = cos -1(0) = π/2 not 3π/2. In this example,

 −1( ) ≠ .

Example 2.5.14 Find cot-1(cot(-π/6)).

Solution: -π/6 is not in the restricted domain of the cotangent.

cot−1 cot 6 = cot−1(√ 3) = 56 ≠

 6 . The output of an inverse trigonometric function represents an angle in the restricted domain.


102




WITH APPLICATIONS

Example 2.5.15 Find cos-1(sin(5π/6)).

Solution:

cos− sin56=cos− 12= 3 .

Example 2.5.16

sincos−   23.

Find

Solution:

Hint: Let

sin cos−  √ 23=sin56= 12 . cos−   23=→cos=   23 .

While there are many solutions to this equation, only 5π/6 lies on [0, π].

Example 2.5.17 Find

tan(cot−(√ 3)).

Solution:

tan(cot−(√ 3))= tan6= √ 13 = √ 33 .

Example 2.5.18

tan− cot43.

Find

Solution:

tan− cot43= tan− √ 13= 6 .

Example 2.5.19

sec− sin .

Find

Solution:

  22

sec− sin 34= sec−  √ 22 .

is not in the domain of the inverse secant. This problem has no solution.

Evaluating inverse sine, inverse cosine and inverse tangent on a typical scientific calculator or graphing utility is straightforward as these functions are built into most calculators. The other inverse trigonometric functions are typically not built in.


103



WITH APPLICATIONS

Example 2.5.20 Find

sec−(√ 2).

Solution:



Define a variable for the solution. is a good choice as it represents the range of an inverse trigonometric function.

sec−(√ 2) = 

Rewrite this as a secant function, that is, invert the equation.

sec = √ 2 sec Rewrite the secant as a cosine since this function is built into the calculator.

Invert equation one more time,

1 = √ 2 cos 1 cos cos =  √ 2 cos cos−  √ 12 =  .

 is the variable representing representing the answer. sec−(√ 2) = cos−  √ 12 ≈ 2.36. This provides insight into evaluating inverse trigonometric functions that are not built into the t he calculator. Simply take the reciprocal of the argument.

Example 2.5.21 Find

cot−5.

Solution:

cot−5 = tan− 15 ≈ 0.1974 .

Example 2.5.22 Find an algebraic expression for

sincos−, -1 < x sin < x < < 1. Assign a variable to the inverse function such as cos − =  . Invert this equation, cos =  . Draw cos = , to illustrate this relationship: a triangle using the definition of the cosine function cos 1 y



x


104



WITH APPLICATIONS

Use the Pythagorean Theorem Theorem to find the length of the side opposite ,

 = √1   2 .

sin((cos−) = sin( sin sin() =  1   2 =  and applying the definition of sine to t he triangle, that is, sin sin(() =  . The algebraic expression, √1   2 , is the solution. Cleary, -1 < x < x < < 1 is the t he domain of this expression. substituting cos −

Example 2.5.23 cos((tan− ). Find an algebraic expression for cos

Let tan− 

=  . Invert this equation, tan = . Draw the triangle:

r x

 Note tan = in the triangle. 1

 1

The Pythagorean theorem gives the length of the hypotenuse, 

cos(tan− ) = cos( cos( cos() =

= √  2 + 1

1 . √  2 + 1

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105




TRIGONOMETRIC IDENTITIES AND EQUATIONS

WITH APPLICATIONS

Chapter 3 Trigonometric Identities and Equations Learning Objectives:    

Review fundamental trigonometric identities Learn techniques for proving identities Derive new and useful identities from the fundamental identities Solve trigonometric equations

Section 3.1 Review of Fundamental Trigonometric Identities A mathematical identity is a statement that is always true or equal for all values in the domain of its variable(s). Definition 3.1.1 (formerly 1.4.1) Trigonometric Identity

A trigonometric identity is an equation involving one or more trigonometric functions that is always true for all values in the domain of said function(s).

The above definition is repeated from Chapter 1, Section 1.4, Fundamental Trigonometric Identities. The summary of fundamental identities from Chapter 1 is also repeated in Table 3.1.1. Use this as a reference. These identities are used to establish new identities. Some strategies for proving identities are as follows: 1. 2. 3. 4. 5. 6. 7.

Apply the multiplicative identity – multiply by a ratio equal to 1 Apply the additive identity – add zero, that is add and subtract equal amounts on one side Substitute with an established identity Factor Multiply by the conjugate binomial in conjunction with the multiplicative identity Distribute Combine rational expressions

These strategies will be illustrated by example. It is important to begin at one side of the identity and arrive at the other side. Note: In proving identities, the equality is in question. The proof is attempting to show the two sides are equal. Therefore, use of any properties of equality is invalid. In order to add equal amounts to both sides of an equation or multiply both sides by equal amounts, equality must first be established. It is illogical to use properties of equality when trying to prove equality. It is circular reasoning. Equality cannot be used to show equality.


107



WITH APPLICATIONS

Fundamental Trigonometric Identities

1 sin() = csc() 1 cos() = sec() 1 tan() = cot() sin() tan() = cos()

Quotient Identities

1 sec() = cos() 1 cot() = tan() cot() = cos() sin()

cos() +sin() = 1 1+tan () =sec() cot () +1=csc ()


sin() =cos2 − sec() = csc2 − tan() = cot2 −


1 csc() = sin()

cos() =sin2 −  csc() = sec2 −  cot() = tan2 − 

sin(+2) = sin() = csc() cos(+2) = cos() = sec() tan(+) = tan() = cot()

Periodic Identities

cos(−) =cos() sin(−) =−sin() tan(−) = −tan()

Odd/even Identities

csc(+2) sec(+2) cot(+)

sec(−) =sec() csc(−) =−csc() cot(−) =−cot()

Table 3.1.1 (formerly Table 1.4.1) Fundamental Trigonometric Identities

Example 3.1.1 Prove the identity:

secsin = tan . Download free eBooks at bookboon.com

108



WITH APPLICATIONS

Solution:

secsin

=

=

1 cos

sin

Substitute the identity sec() =

1 cos()

sin Multiply

cos

= tan

Substitute the identity tan() =

sin() cos()

Example 3.1.2 Prove the identity: (1 − sin)(1 + sin) = cos 2  .

Solution:

(1 − sin)(1 + sin)

= 1 − sin2  = cos 2 ()

Distribute Substitute cos2 () + sin2 () = 1

→ cos 2 () = 1 − sin 2 ()

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109




WITH APPLICATIONS

Sometimes it is necessary to rearrange an identity before making a substitution. Example 3.1.3 Prove the identity: cot 4   cot 2 csc2  = cot 2  .

Solution:

cot 4   cot 2 csc 2 

= cot 2 (cot 2   csc 2 )

Factor

= cot 2 (cot 2   (cot 2 ()  1))

Substitute the identity cot 2 ()  1 = csc 2 ()

= cot 2 (cot 2   cot 2 ()  1)

Distribute the minus

= cot 2 (1)

Subtract

= cot 2 

Multiply

Example 3.1.4

Prove the identity:

 −  + 

=

− +

.

Solution:

csc  1 csc  1

1  1 sin() = 1  1 sin()

Substitute csc() =

1  sin() sin() = 1  sin() sin() =

=

1  sin() sin()

1

sin()

Combine rational expressions

∙

sin()

Invert and multiply

1  sin()

1  sin()

Reduce the ratio

1  sin()

It is important to keep in mind the expression trying to be obtained. This can give insight into a strategy. In Example 3.1.4, the identity contained only sine and cosecant functions. Thus the quotient identity, csc() =

1 , is a good candidate for substitution. sin()

Note: ‘cross-multiplication’ is only applicable to a proportion. Two equal ratios constitute a proportion. Equality has not yet been established, so ‘cross-multiplication’ is not valid. The exercise is to prove equality. This is why it is important to work one side, and only one side, of the identity. It does not matter which side is selected to begin the proof. Work left to right or right to left. It is often simpler to begin with the more complicated expression, regardless of which side it is on.


110



WITH APPLICATIONS

Example 3.1.5

Prove the identity: sec = tan +

cos . 1 + sin

Solution:

tan +

cos 1 + sin

= tan +

= tan +

= tan + = tan +

=

=

sin cos

+

cos 1 + sin

∙

1 − sin

Apply the multiplicative identity in conjunction with the conjugate of the denominator

1 − sin

cos(1 − sin)

Multiply only the denominators

1 − sin2  cos(1 − sin)

Substitute cos2 () + sin2 () = 1

cos 2  1 − sin

Reduce the ratio

cos 1 − sin

Substitute tan() =

cos

sin() cos()

1 Combine rational expressions

cos

Substitute sec() =

= sec

1 cos()

In the above example, only the denominators were multiplied in the second step. It can be a good strategy to make one change at a time and take a look at what happened before making another change. Multiplying both the numerator and denominator could cloud the next steps.

tan +

cos(1 − sin) 1 − sin2 

= tan +

cos − cossin cos 2 

In this example, it might not throw off the direction of the proof, but it certainly adds an additional unnecessary step. This example illustrates another important idea related to keeping the goal in mind or keeping in mind the expression trying to be obtained. The sec, a single-term expression, was the goal. The expression begun with was two-termed. When this is the case, it should be understood that these two terms must be combined in order to arrive at a single term, the sec. An alternative approach to the last example is:

tan +

cos 1 + sin

=

=

=

sin cos

+

cos

Substitute tan() =

1 + sin

sin(1 + sin) cos(1 + sin)

+

cos 2  cos(1 + sin)

sin(1 + sin) + cos 2  cos(1 + sin)


111

sin() cos()

Obtain a common denominator Combine the rational expressions



WITH APPLICATIONS

=

=

=

sin + sin2  + cos 2 

Simplify the numerator

cos(1 + sin) sin + 1

Substitute cos2 () + sin2 () = 1

cos(1 + sin) 1

Reduce the ratio

cos

Substitute sec() =

= sec

1 cos()

There is more than one way to prove an identity. The better way is the one that is understood. Proving identities can be challenging. Knowledge and practice will help. Patience and tenacity are essential. Creativity is a plus.

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112




WITH APPLICATIONS

Example 3.1.6

Prove the identity: 1 = 2cos  

tancot . tan  cot

Solution:

2cos  

tan  cot tan  cot

sin cos  cos sin  = 2cos   sin cos  cos sin

Substitute tan() =

and cot () =

sin   cos  sincos = 2cos   sin   cos  sincos



= 2cos  

sin() cos() cos() . sin()

Combine rational expressions

sin   cos  Invert and multiply

sin   cos 

= 2cos   sin   cos 

Substitute cos  ()  sin () = 1

= cos    sin 

Combine like terms

=1


It is inadvisable to automatically convert all terms in an identity to sine and cosine functions. While this is always possible, it can overly complicate expressions and cause unnecessary confusion. There are also times when it is a good strategy to retreat . If the expressions do not appear to be heading in the right direction, perhaps becoming burdensome or unwieldy, it may helpful to start over, perhaps even set the proof aside to try again at another time. As a novice, it is possible to get ‘lost’ in the proof . Take breaks, but do not quit altogether.

Example 3.1.7

Prove the identity:

 −





2

+

= 2 .

Solution:

1 1  sin



1 1  sin

=

= =

=

1  sin (1  sin)(1  sin)



1  sin (1  sin)(1  sin)

2

Get common denominators Combine rational expressions

(1  sin)(1  sin) 2

1  sin2 

Multiply

2 cos2 


= 2sec

Substitute sec =


113

1 cos



WITH APPLICATIONS

Example 3.1.8

− = 1  22 . Prove the identity:    −  Solution:

2cos2  1 cos4   sin4 

=

(2cos2  1)

2

cos2 sin2cos2+sin2 

Factor

2cos  12 =  cos   sin 

Substitute cos   +sin  = 1

2cos   12 =  cos   1 + cos 

Substitute sin  = 1 

2cos  12 = 2cos   1

Combine like terms

= 2cos   1

Reduce

= 21sin    1

Substitute cos   =

= 12sin 

Distribute and combine like terms

cos  

1  sin

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Section 3.2 Sum and Difference Identities The fundamental trigonometric identities may be used to derive an infinite number of trigonometric identities, many having practical applications.

cos12  were needed, reference angles would be insufficient. cos12  ≠  cos6. π/12 is a difference of reference angles, that is, π/3 – π/4 and cos12  = cos3 − 4. But cos3 − 4≠cos3−4. A new identity is required. cos −  =coscos+sinsin If the exact value of

The following video illustrates a geometric derivation of t his identity. In the video, the angles are positive and in standard position although the identity holds for all angles. Proof of the cosine difference of angles identity

Once the difference of angles identity for cosine is established, the sum of angles identity easily follows by negating the angle beta and applying the odd/even identities for sine and cosine.

cos −  =coscos+sinsin cos−− =coscos−+sinsin− cos +  =coscos−sinsin Example 3.2.1

Find the exact value of Solution:

cos12 =cos3 − 4 = cos3cos4+sin3sin4 = 12 ∙ √ 22 + √ 23 ∙ √ 22 = √ 42 + √ 46 = √ 2 +4 √ 6 .

Example 3.2.2


cos12 .

cos712.

 +  = coscos−sinsin cos(7 )=cos 12 3 4 3 4 3 4 = 12 ∙ √ 22 − √ 23 ∙ √ 22 = √ 42 − √ 46 = √ 2 −4 √ 6 . Download free eBooks at bookboon.com

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WITH APPLICATIONS

Example 3.2.3

Prove the identity

sin = cos2 − .

Solution:

cos2 −  = cos2cos + sin2 = 0 ∙ cos + 1 ∙ sin = sin.

There are sum and difference of angles identities for each trigonometric function.

sin = cos2 −  sin −  = cos(2 −  − )  = cos −  +  2  = cos( −  + ) 2  = cos2 − cos − sin2 − sin = sincos − cossin

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The sine difference of angles identity is

sin −  =sincos−cossin . The sine sum of angles identity follows directly by negating the angle beta and applying the odd/even identities for sine and cosine.

sin +  =sincos+cossin. A visual proof lies in the diagram of Figure 3.2.1. Although the angles are acute in the diagram, the identities hold for all angles alpha and beta.

Figure 3.2.1 Geometric proof Sum of Angles Identities for sine and cosine - Trigonography.com

Example 3.2.4

sin7 12.  +  = sincos+cossin sin(7 )=sin 12 3 4 3 4 3 4 = √ 23 ∙ √ 22 + 12 ∙ √ 22 = √ 46 + √ 42 = √ 6 4+ √ 2 .


Example 3.2.5

Find the exact value of

sin5 12. Download free eBooks at bookboon.com

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WITH APPLICATIONS

Solution:

7  2)=sin(7)coscos(7)sin sin(5 )=sin( 12 12 12 12 6 12 6 =  6 4  2 ∙  23   2 4  6 ∙ 12 18  √ 6  √ 2  √ 6 = √ 18 8 8 6 2 = 3 2  28 6

= 2 2 8 2 6 =  2 4  6 . An alternative solution:

3  2)=sincoscossin sin(5 )=sin( 12 12 12 4 6 4 6 =  22 ∙  23   22 ∙ 12 =  46   42 =  6 4  2 . This solution is obviously simpler. Note that

7  sin5 12=sin12=cos12. Why?

Hint: Sketch these angles in a circle. What is the

cos11 12  equal to?

The tangent function has sum and difference identities.

sin sin   = sincoscossin tan tan   = cos cos   coscossinsin +  ∙  =  +  =   −    −  1 = tan ∙  cos cossesec c = 1 an andd sins sinsecec = sin 1 cos tan   costan ∙ sec = tan  tan = sin cos  sintan sec 1  tantan  tan . tan tan   = 1tan  tantan the Tangent Sum of Angles identity.


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WITH APPLICATIONS

The Tangent Difference of Angles identity is derived using the fact that the tangent is an odd function, that is, using the odd identity for tangent:

tan( tan( − ) =

tan tan − tan tan

. 1 + tan tanta tan n

The Sum and Difference of Angles identities for all six trigonometric functions are summarized summarized in Table 3.2.1. The derivations of the last three identities are left as an exercise for the reader. It should be noted that the Sum Sum and Difference Difference of Angles Identities for the cotangent, cotangent, secant secant and cosecant cosecant have narrow application but are included for interest and derivation practice. The Sum and Difference of Angles identities for sine and cosine are also referred to as Ptolemy’s Identities Identities as as they result from his theorem. Ptolemy’s Theorem may Theorem may be used to prove the Sum and Difference of Angles Identities for the sine and cosine functions. Theorem 3.2.1 Ptolemy’s Theorem

If a quadrilateral is inscribed in a circle, the product of its diagonal lengths equals the sum of the products of the opposite opposite sides. If the vertices of the quadrilateral are labeled ABCD labeled ABCD,, then then Ptolemy’s theorem becomes the equation AC equation AC BD = BD = AB AB CD + CD + AD AD BC .

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119




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Figure 3.2.1 Cyclic quadrilateral illustrating Ptolemy’s theorem

Example 3.2.6


tan11 12 . 9 + 2)=tan(3 + ) tan(11 )=tan( 12 12 12 4 6 tan 34  + tan 6 = 1 − tan 34  tan 6 −1 + √ 33 −1 + √ 33 = = √ 3 3 √ 1 − −1 ∙ 3 1 + 3 −1 + √ 33 3 −3 + √ 3 = √ 3 ∙ 3 = 3 + √ 3 1+ 3 = −33 ++√ √ 33 ∙ 33 −− √ √ 33 √ 3 − 3 = −12+6√ 3 = −2+ √ 3 = −9+6 9−3 6

The periodic property of the trigonometric functions makes evaluating

tan   much simpler.

 =tan(3 − 4)=tan −  tan(11 )=tan− 12 12 12 12 4 3   − tan tan = 4  3 1 + tan4tan3 Download free eBooks at bookboon.com

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WITH APPLICATIONS

− √ 3 = 1 − √ 3 = 1 +1 1( √ 3) 1 + √ 3 = 11 −+ √ √ 33 ∙ 11 −− √ √ 33 = 1 −12−√ 33+ 3 = 4 −−22√ 3 =−2+ √ 3 Example 3.2.7


tan23 12 .

Solution:

11 + 12 = tan11 +  tan23 =tan 12 12 12 12 11 + tan tan 12 = 1 − tan11 12 tan = −2+1 −√ 30+ 0 = −2+ √ 3 . The

11. (Why?) tan23 = tan 12 12

Another application of identities is the discovery or creation of new identities.

Example 3.2.8 Show

sin +  +sin −  =2sincos .

Solution:

sin +  +sin −  =sincos+cossin+sincos−cossin=2sincos.


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WITH APPLICATIONS

Function

Sum of Angles Identity

Sine

sin(  ) = sincos  cossin

Cosine

cos(  ) = coscos  sinsin

Tangent

Difference of Angles Identity

tan  tan

tan(  ) =

cot (  ) =

Secant

sec(  ) =

Cosecant

csc(  ) =

cos(  ) = cos  sinsin

tan(  ) =

1  tantan

Cotangent

sin(  ) = sincos  cossin

cotcot  1

tan  tan 1  tantan

cot (  ) =

cot  cot

csccsc

sec(  ) =

cotcot  1

csccsc

cotcot  1 cot  cot

sec sec 1  tantan

csc(  ) =

cot  cot

secsec tan  tan

Table 3.2.1 Summary of Sum/Difference of Angles Identities

Of course, identities apply only to values ( or  here) in each function’s respective domain .

Example 3.2.9

Prove

− +

=

 −  + 

.

Solution:

cot cot  1 cot  

= cot(  ) =

=

cos(  ) sin(  ) coscos  sinsin sincos  cossin

Example 3.2.10

Prove

 +   − 

=

 +  −

.


122

.



WITH APPLICATIONS

Solution:  +   − 

=

1 1 + cot cot 1 1 − cot ∙ cot

=

cot  cot cotcot cotcot  1 cotcot

=

=

cot  cot cotcot

∙

Substitute tan() =

1 . cot()

Add the rational expressions.

cotcot

Invert & multiply.

cotcot  1

cot  cot

Reduce.

cotcot  1


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WITH APPLICATIONS

Example 3.2.11

Show

sec . tan  csc   = 1sec tantan

Solution:

sin   ∙ 1 tan  csc   = cos   sin   sec . = cos1  =sec   = 1 sectantan

Example 3.2.12

Show

− =  . + +

Solution:

− = − +    

Substitute

cot   = cotcot1 cot  cot .

cotcot = cotcot1 ∙ 1 cotcot1 = cot  cot csccsc . = csc   Example 3.2.13

Simplify the difference quotient

Invert & multiply.

Reduce. Substitute after rearrangement of

csccsc csc   = cot  cot

+ − . 

Solution:

 +  −  = sincosℎcossinℎsin  ℎ

= sincosℎsincossinℎ ℎ = sincosℎ1cossinℎ ℎ cossinℎ = sincosℎ1  ℎ ℎ sinℎ) =sin(cosℎ1 )cos( ℎ ℎ

Substitute

sin   =sincoscossin .

Commute.

Factor.

Split ratios (decompose the fractions). Factor.


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WITH APPLICATIONS

Understanding this example, and Example 3.2.14 that follows, will be helpful in an introductory Calculus course. Example 3.2.14

Simplify the difference quotient

+−. 

Solution:

 + − = coscosℎsinsinℎcos ℎ  = coscosℎcossinsinℎ ℎ = ℎ1ℎ ℎ sinsinℎ = coscosℎ1  ℎ ℎ sinℎ). =cos(cosℎ1 )sin( ℎ ℎ

Br ain power

Substitute

cos   =coscossinsin.

Commute.

Factor.

Split ratios. Factor.

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Section 3.3 Double- and Half-Angle Identities Recall the Sum of Angles identity for sine:

sin +  = sincos + cossin If

 = , this simplifies to:

sin +  = sincos + cossin sin2 = 2sin sin2 = 2sincos.

or

the double-angle identity for sine. Similarly,

cos +  = coscos−sinsin cos +  = cos −sinsin cos2 =cos−sin.

the double-angle identity for cosine. Due to the ease of Pythagorean identity substitution, the double-angle cosine identity is typically presented in 3 ways:

cos2 =cos−sin =1−2sin =2cos − 1.

cos =1−sin Substitute sin  =1−cos   Substitute

The first identity contains both sinusoidal functions, while the second is expressed in just sine and the last in just cosine. It can be helpful to have these options.

Example 3.3.1

If

8 sin = 15 17 and cos= 17, find the exact values of a)

sin2

b)

cos2

Solution: a)

8  = 240 . sin2 = 2sincos = 215   17 17 289

 8  b) cos2 =2cos  − 1 = 2   − 1 =

17

sin(2) = c) tan2 = cos(2)

240 289 − 161 289

128 − 289 = − 161 . 289 289 289

= − 240 161 .

The signs of the answers indicate that

 < 2 <  . 


126

c)

tan2.



WITH APPLICATIONS

The tangent function has its own double-angle identity,

+ tan = 2tan . tan(2) = 1tan − tantan 1 − tan c) alternatively:

2 15 2tan 8 = tan(2) = 1 − tan = 1 − 15 8

15 4 = − 240 . 161 − 161 64

Table 3.3.1 summarizes the double-angle identities for circular/trigonometric functions.

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127




WITH APPLICATIONS

Double-Angle Identities

sin2=2sincos cos2 =cos−sin  =1−2sin  =2cos −1 tan2= 1 −2tantan  cot cot2= 2cot−1  csc sec2= cot−1  csc 1 csc2= 2cot = sin2

Sine

Cosine

Tangent

Cotangent

Secant

Cosecant

Table 3.3.1 Double-Angle Identities

Every double-angle identity provides a half-angle identity.

cos2=1−2sin cos=1−2sin (2) sin (2)= 1− cos 2 sin(2)= ±√ 1 − cos 2 . ± 

The double signs

  sin 2

Replace with .

Solve for

.

Apply inverse square root function.

indicate that there are two Half-Angle Sine identities: a positive and a



negative. The value of the angle combined with the range of the sine function determines the sign used in the identity.


128



WITH APPLICATIONS

Example 3.3.2






sin12 .

 sin2  3 3 √ √ cos  1 − 1 − 1 −   3 2 2 − √   6  2 2  sin12 = 2 = 2 = 2 ∙ 2 = 4 = 2 −2 √ 3 . cos12 .

is half of . The angle

terminates in the first quadrant so the

is positive.

Example 3.3.3


Solution: This may be found using the half-angle cosine identity or by using of the cosine difference of angles identity (Example 3.2.1). Find t he half-angle identity for cosine:

cos2=2cos−1 cos=2cos (2)−1 cos (2)= 1 c2os cos(2)=± 1  c2os  3 3 √ √ 1  c o s  1  1     cos12=  2 6 = 2 2 = 2 2 ∙ 22 =  2 4√ 3 =  2 2 √ 3 . √  + √  cos12 

In example 3.2.1, the

was found to be

. Verify that these two answers represent the

same irrational number.

The half-angle tangent identity may be derived applying the half-angle identities for sine and cosine along with the quotient identity for tangent.

   1− cos   s i n    tan (2)= cos 22 = 1 c22os = 1−1 ccooss tan(2)= ± 11− ccooss Download free eBooks at bookboon.com

129



WITH APPLICATIONS

tan1112.  23 1   23 2  2   3 11 ( ) 1  cos 1  11 6 tan 12 =  1  cos(116) = 1   23 = 1   23 ∙ 2 = 2   3 .  1112 t a n −+√ 

Example 3.3.4


Solution:

Note the leading negative before the radical sign. The angle

terminates in the second quadrant

so the value of the tangent function should be negative. Also, in Example 3.2.6, the

found to be

+√ 

was

. Show that both expressions represent the same irrational number. Table 3.3.2

summarizes the identities of this section.

Double-Angle Identities

sin2=2sincos cos2 =12si =cossinn =2cos1 tan2= 1 2tatnan  c o t cot2= 2c1 ot sec2= coctsc1  c s c csc2= 2cot = sin12

Half-Angle Identities

sin (2)= 1 c2os cos (2)= 1 cos2  tan (2)= 11 coscos    cot (2)= 1cos 1cos sec (2)= 1cos2  csc (2)= 1cos2 

sin(2)=± 1  cos2  cos(2)=± 1  c2os tan(2)=± 11 ccooss cos cot(2)=± 11cos sec(2)=± 1 c2os csc(2)=± 1cos2 

Table 3.3.2 Summary of double- and half-angle identities


130



WITH APPLICATIONS

Example 3.3.5

Show

(2) tan2 = sin cos(2).

Solution:

sin2 = 2sincos cos2 cos − sin  2sincos sec = cos − sin ∙ sec 1  2sincos = 1 − sin cos1 = 1 −2tan tan = tan2

Substitute respective double-angle identities.

Multiplicative identity.

Multiply and reduce.

Substitute

tan = sin cos.

Substitute tangent double-angle identity.


131




WITH APPLICATIONS

Example 3.3.6

Prove

tan22 . cos2 = 11+− tan 

Solution:

1 − tan = 1 − tan 1 + tan sec   cos  1 − tan = 1 ∙ 1 = cos − sin = cos2

Substitute

1 + tan = sec.

Invert and multiply.

Multiply and reduce. Substitute

cos2 = cos2 − sin2.

Example 3.3.7

Show

tan2 = 1+sincos(()) .

Solution:

sin2 = 2sincos 1 + cos2 1 + 2cos − 1 = sincos cos = sin cos = tan

Substitute double-angle identities. Simplify and reduce.

Reduce.

Substitute

sin = tan. 1 + cos 2

tan = sin cos.

Substitute half-angle identity for tangent.

Example 3.3.8 Show

sin3 = −4sin + 3sin .

Solution:

sin3 = sin2 +  = sin2cos + cos2sin = 2sincoscos + 1 − 2sinsin = 2sincos + sin − 2 Download free eBooks at bookboon.com

132

Sine sum of angles identity Substitute respective double-angle identities. Multiply.



WITH APPLICATIONS

= 2sin(1−sin)+ sin −2sin cos = 1 −sin. = −4sin +3sin. Substitute

Multiply and combine.

Example 3.3.9 A projectile is launched with an initial velocity of 400 m/s at an angle of elevation of 3π/8. Assuming gravity is 9.81 m/s2, how far would the unimpeded projectile travel?

Solution: The range of a projectile is given by the equation

sin2   =  where R is the horizontal range of the projectile, v0 is the initial velocity and g is the gravitational constant.

 √ 2 m 3 sin 4  160000 s   2   400 m/s  = 9.81 m/s = 9.81 m/s ≈ 11,533 meters.

Example 3.3.10 The fuselage of an airplane undergoes various stresses, that is, forces acting upon it. Normal stresses act upon the surface in a perpendicular direction. Shear stress acts upon the surface in a parallel direction. These stresses need to be understood by designers for proper function and safety. By examining an infinitesimally small square of the fuselage, these stresses may be determined. Figure 3.3.1 illustrates this section of the fuselage with its associated normal and shear stresses. represents shear stress acting across the surface and represents normal stresses that may stretch or compress the surface. Along with these principal stresses, additional stresses may act at other angles. In order to calculate these stresses, stress transformation equations are required.



Figure 3.3.1 Stress diagram of infinitesimally small s ection of the fuselage


133





WITH APPLICATIONS

Stress Transformation Equations

′ = ( +2 ) +  −2 cos2 + sin2 ′ = ( +2  )−  −2 cos2 − sin2 ′ = − −2 sin2 +cos2 If the normal stress in the horizontal direction is 65 megaPascals (mPA) and 85 mPa in the vertical direction with 30 mPA of shear stress, what are the stresses at an angle of 40 ?

°

′ = (65 +2 85) + (65 −2 85)cos80° +30sin80° =75−100.174 + 300.985 ≈ 102 mPa. ′ = (65 +2 85) − (65 −2 85)cos80° −30sin80° =75+100.174 − 300.985 ≈ 47 mPa. ′ = −(65 −2 85)sin80° +30cos80° = 100.985 + 300.174 ≈15 mPa.


134




WITH APPLICATIONS

Section 3.4 Product-to-Sum and Sum-to-Product Identities Like the other identities established in this chapter, the Product-to-Sum and Sum-to-Product identities are for convenience and simplification. It may be desirable to switch from a sum to a product, or vice versa, in order to simply a formula or solve an equation.

sin( + ) = sincos + cossin + sin( − ) = sincos − cossin sin( + ) +sin( − ) = 2sincos sin( + ) = sincos + cossin − ( − ) = sincos − cossin sin( + ) −sin( − ) = 2cossin These manipulations of the Sine Sum/Difference of Angles identities give two Product-to-Sum identities for combinations of sine and cosine. One is sufficient for converting products of both sine and cosine, so the first one is typically used. Likewise,

cos( + ) = coscos − sinsin + cos( − ) = coscos + sinsin cos( + ) +cos( − ) = 2coscos cos( − ) = coscos + sinsin − cos( + ) = coscos − sinsin cos( − ) −cos( + ) = 2sinsin yield a Product-to-Sum identity for cosine and one for sine. A ratio of these two identities reveals a Product-to-Sum identity for the tangent function:

2sinsin = cos( − ) −cos( + ) . tantan = 2coscos cos( + ) +cos( − ) Example 3.4.1


sin8cos8.

Solution:

sin8cos8 = 12 sin8 + 8 + 12 sin8 − 8 = 12 sin4 + 12 sin(0) = 12 ∙ √ 22 = √ 42 .


135



WITH APPLICATIONS

Example 3.4.3


 tan5 12tan12.

Solution:

5 −  −cos5 +   cos−cos 1 − 0 cos 5  12 12 12 12 = 3 2 =2 =1. tan12tan12 = 5    cos2+cos3 0 + 1 cos12 + 12 +cos5 − 2 12 12

Example 3.4.4

Show

( + ) + cos( − ) . cot = cos sin( + ) − sin( − )

Solution:

cos +  +cos −  = 2coscos sin +  −sin −  2cossin = cos sin =cot

Substitute respective product-to-sum identities for sine and cosine

Reduce Substitute

cot = cos sin

It can be helpful at times to convert sums into products. One application of this is in solving trigonometric equations (see section 3.5).

 −  =  and  +  =  in the Product-to-Sum identity for sine: Product-to-Sum identity: cos −  −cos +  =2sinsin cos − cos =2sin +2 sin −2  Substituting:  =  +    =  −  after solving the system of equations  −  =  and  +  = : Let

2

2

−= + += 2  =  +  →  =  +2 .

 Substituting  =  and  =  for consistency in the identity form, cos −cos =2sin +2 sin −2  cos−cos = 2sin  +  sin  −  The derivation of is similar.

2

2

the Sum-to-Product identity for cosine. Or


137



WITH APPLICATIONS

cos −cos = −2sin ( +2 ) sin ( −2 ) after applying the Odd identity for sine. The Product-to-Sum and Sum-to-Product identities for the sine, cosine and tangent functions are summarized in Table 3.4.1.

Example 3.4.5

Show

+  . tan+tan= sin coscos

Solution:

sin + sin tan+tan= cos cos = sincos+cossin coscos = sin+ coscos .

Substitute.

Add the rational expressions.

Sine sum of angles identity.


138




WITH APPLICATIONS

Example 3.4.6 Prove

sincos+cos3 = sin2cos2. sincos+cos3 = sin2cos(42 ) cos(−22) = 2sincoscos2 = sin2cos2

Solution:

Product-to-Sum Identities

Sum-to-product identity. Simplify, commute and apply the cosine even identity. Sine double-angle identity.

Sum-to-Product Identities

sin+sin = 2sin( + 2 )cos( − ) 2 2sinsin = cos − −cos + sin−sin = 2sin( − 2 )cos( + 2) cos+cos = 2cos( + 2 )cos( − ) 2 2coscos = cos ++cos − cos−cos = −2sin( +2  )sin( − 2) 2sincos = sin + +sin − + tan +tan = sin coscos

os − −cos + tantan = ccos + +cos −

Table 3.4.1 Product-to-Sum and Sum-to-Product Identities

Example 3.4.7 In acoustics, the interference pattern between two sounds can be modeled and understood using a Sum-to-Product identity.

cos+cos = 2cos( + 2 )cos( − 2) cos21+cos2 = 2cos(2 1 +2   )cos(2 1 −2  ) Download free eBooks at bookboon.com

139



WITH APPLICATIONS

where f 1 is the frequency of one sound in Hertz (Hz), f 2 is the frequency of the other sound and t is time in seconds.

 +    −  

, the average of the two frequencies, is called the modulation frequency. If

the frequencies are close,

is small and is perceived by the human ear as an amplitude

modulation or a change in amplitude. The

cos(2 1 2 2 )

cos(2 1 2 2 )    cos(2 1 2 2 )=1 cos(2 1 2 2 )=0

term acts as an envelop for the

term and controls the interference. When

waves are in phase and interference is constructive. When

, the sound

, the sound waves

are in antiphase (off by π radians) and the interference is destructive.

The distinctive warbling sound of the Lakota Flute exemplifies this phenomenon. To the unaided, human ear the warbling sounds like a changing pitch. It is actually different harmonics coming in and out of phase. “The natural overtone series of harmonics in music is the basis for our scales and foundations of music theory.” according to Robert Barta, Professor of Computer Science and music aficionado at Suffolk County Community Coll ege. If f 1 = 20 Hz at t = 1 second in a sustained note blown from a Lakota Flute, find f 2 such that the harmonics are in phase. Solution:

cos(2 1  2 )=1

for the waves to be in phase.

This means that

 1=1 cos2 20 2  −  cos0=1   −  = 1 cos2=1 

should be noted that

may be to equal zero since as well, so

. So f 2 should also be 20 Hz. It

or f 2 = 18 Hz is also a solution.

This is typically the case when solving tri gonometric equations. Due to the periodic nature of these functions, equations involving them have infinite solutions. That is, of course, if the domain is not intentionally restricted or naturally restricted by the constraints of t he problem. In the previous example, the domain for f 2 is naturally restricted to [0, ∞). The upper limit of this domain is not technically infinite. Theoretically, it should be limited to the Debye constant (Hill, T, 1986). So the actual solution set to Exercise 3.4.7 should be finite. The following section explores trigonometric equations in greater detail.


140



WITH APPLICATIONS

Section 3.5 Solving Trigonometric Equations Equations involving trigonometric functions may have infinite solutions due to the periodic nature of these functions. If the domain is restricted intentionally or naturally by the constraints of the problem, then the solution set may be finite. Regardless, all solutions must exist within the domain of the function(s).

Example 3.5.1 Solve cos = 1 .

Solution: Refer back to Example 3.4.7 of Section 3.4, Product-to-Sum and Sum-to-Product identities. This actual equation was solved within the context of the Lakota Flute problem. It is much simpler here and has no restriction on the domain imposed by some context. The equation has infinite solutions. Any value of  for which the cosine is equal to 1 i s a solution.

cos(0) = 1 cos(2) = 1 cos(4) = 1 cos(6) = 1 cos(−2) = 1 cos(−4) = 1 et cetera, ad infinitum. The periodic identity of the cosine is relevant here.

cos( + 2) = cos() Any integer multiple of 2π when added to a solution yields yet another solution.

cos(0 + 2  ) = cos(0) = 1 cos(0 + 2(0)) = cos(0) = 1 cos(0 + 2(1)) = cos(2) = 1 cos(0 + 2(2)) = cos(4) = 1 cos(0 + 2(3)) = cos(6) = 1 cos(0 + 2(−1)) = cos(−2) = 1 cos(0 + 2(−2)) = cos(−4) = 1

A concise way to represent all solutions i s  = 0 + 2, that is,  = 2, where k is an integer. This is the solution set. Figure 3.5.1 illustrates this concept.


141



WITH APPLICATIONS

Figure 3.5.1 cos(x) = 1

The cosine graph intersects the line y = 1 at infinite points. These points are  is an integer.

= 0 + 2π, where

k

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Example 3.5.2  Solve

cos  = 1 .

Solution: This quadratic form may be separated into two di stinct equations using the square root property:

cos = 1 and cos = −1. cos=1 has already been solved in the previous example. cos=−1 has solutions at odd multiples of π (refer back to Figure 3.5.1). =−1 3=−1 5 = −1

et cetera. Therefore, the general solution for both equations (the solution to the original equation) is then any integer multiple of π or

{  =  |  is an integer}.

Example 3.5.3  Solve

4cos  = 3 .

Solution: Solve for the cosine function,

cos = 34 .

Apply the Square Root property in order to find two linear trigonometric equations:

cos=± √ 23 . Find all solutions on the domain interval [0, 2π). Figure 3.5.2 assists this process.

= √ 3

Figure 3.5.2 (formerly Figure 1.3.8) Reference triangle

cos(6) =  23 from the reference triangle. This is a first quadrant solution. The cosine is also

positive in quadrant IV. If the reference angle is placed in quadrant IV, the resulting angle is 11π/6, that is,

 3 cos(11 ) = 6 2 . Rotating the reference triangle into quadrants II and III the solutions to the


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 23 cos56= −  23 cos76= −  23 . cos=− 6 , 56 , 76 , 116 .

linear equation on [0, 2π) are

are found:

and

Thus the solutions

Figure 3.5.3 (formerly Figure 1.3.9) Reference angle placed in each quadrant

 and   and . 6 + = { = 56 +}where  is an integer .

Now generalize the solution. The angles

are π radians apart (antiphase). So

integer multiple of π is also a solution. The same is true for



plus any

The general solution is:

cos−sin=0. (cos –sin)(cos+sin)=0, so cos2cos –si−1 n=0foandr coscos−si +snin=0.. cos −si n=0 2cos −1=0 cos= 12 cos=± √ 12 =± √ 22 .

Example 3.5.4 Solve

Solution:

Factoring yields it can be solved this way, it may be simpler to substitute


144

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WITH APPLICATIONS

As with the first example, a graph can be useful in verifying solutions. For instance, graphing f ( x) = sin(2 x) and the horizontal line y = 0 reveals a solution at each intersection of the t wo graphs. Since y = 0 is actually the x-axis, it is not necessary to graph this line since discerning the xintercepts of the curve f ( x) = sin(2 x) would suffice. Setting a proper scale based upon the solution(s) found on [0, 2π) is helpful. In the above example, a scale of π/2 or a smaller multiple of π/2 makes the solutions evident.

Example 3.5.6 Solve

sin(2) +2cos−sin=1.

Solution: It is useful to separate the different trigonometric functions into factors. Then the zero factor property may be applied and 2 separate linear equations may be solved. First, substitute the DoubleAngle identity for sine:

2sincos+2cos−sin=1

Subtract 1 from both sides and factor by grouping:

2sincos+2cos−sin−1=0 2cos(sin+1)−1(sin+1)=0 (2cos−1)(sin+1)=0 Separate into linear equations:

2cos−1=0 and sin+1=0

Solve individually for each trigonometric function:

cos= 12 and sin=−1. Solutions for

cos= 12 on [0, 2π) are 3 , 53 . The reference angle is π/3 and the cosine function’s

range is positive for angles corresponding to quadrants I and IV. Only the quadrantal angle, 3π/2, is a s olution for on [0, 2π) is

sin(2) +2−sin=1

sin=−1 on [0, 2π). The solution set for

{3 , 53 , 32 } . The general solution is each of these base solutions plus any multiple of 2π or

{3 +2, 53 +2, 32 +2 } . It is possible to express this more concisely by combining the first two expressions, but some clarity is lost. The solutions to this example are shown visually in the following two figures: Figure 3.5.4 and 3.5.5.


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WITH APPLICATIONS

Figure 3.5.4 Terminating angle positions for solutions to Example 3.5.6

Figure 3.5.5 Intersecting curves illustrating solutions to Example 3.5.6

In Figure 3.5.5, it is difficult to discern the 2 two solutions at 3π/2 and 5π/3. Zooming in on this region reveals that the only two intersections here are indeed at 3π/2 and 5π/3. The graph partially confirms solutions and aids in determining if any were missed. True confirmation, however, comes from inputting any found solutions into the original equation.


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WITH APPLICATIONS

tan2=−  33 .

Example 3.5.7

Solve

Solution:

tan2= − 11+− coscos(()) =−  33 . tan2 −  33. tan2 3 √  1−1+ cos = cos 3 1−1+ cos 3 1 = = cos 9 3 3(1− cos)=1+cos 3− 3cos=1+cos 2=4cos 12 =cos. ={3 , 53}.

Substitute the Half-Angle identity for tangent: is used because it is given that angle



is negative. It equals

Since

The negative sign

is negative, the

must terminate in quadrant II or IV.

On [0, 2π), solutions to the last equation are

One of these is an extraneous solution.

When the square function is applied (inverse of the square root function), the solution set was expanded. The radical equation has fewer solutions than the squared equation due to restrictions on the domain (see Section 2.5, Inverse Trigonometric Functions). It should be evident that π/3 is the extraneous solution. Recall that

tan2

is negative so theta cannot be an angle terminating in the

first quadrant. If this is not readily evident, check each solution in the original equation. And graph f ( x) = tan( x/2) and the line y =

−  33

to examine the intersections.

The general solution set is

{= 53 +2}

where k is an integer.

csc=cot+1 .

Example 3.5.8 Solve

Solution: With two different trigonometric functions in one equation, several strategies are appropriate. 1. Make a substitution(s) to eliminate one more functions creating a single function equation. 2. Factor the equation, hopefully separating the trigonometric functions into different factors. 3. Some combination of 1 & 2 (this approach was applied in Example 3.5.7). The first strategy is appropriate for this example.


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csc=cot+1 cot+1=cot+1 cot−cot=0 This is a quadratic form in the cotangent function. It is analogous to x2 – x = 0.

The solutions on [0, 2π) are

cotcot−1=0 cot=0 and cot−1=0   = + 2 , 32 , 4 , 54. 2 { = 4 +}. =csc = The general solution is

A sketch of the

angles in a circle may help clarify the general solution. A graph of illustrates these solutions.

cot+1

and

sin+sin7=0.Find all solutions on [0,2.

Example 3.5.9 Solve

Solution:

sin+sin=2sin+2cos−2 −7 sin+sin7=2sin+7 cos 2 2 =0 2sin4cos−3=0.

Use the Sum-to-Product identity:

Using the zero factor property and the even identity for cosine, the equation becomes 2 simpler equations:

sin4=0 and cos3=0. sin0=sin=sin2=sin3=⋯=sin=0 4=0, 4=, 4=2, 4=3, 4=4, 4=5, 4=6, 4=7, 4=8,⋯ =0, = 4 ,  = 2 ,  = 34 , =, = 54 ,  = 32 ,  = 74 ,  =2, ⋯ 0, 4 , 2 , 34 ,, 54 , 32 , 74  cos2=cos32=cos52=cos72=cos92=⋯=cos(2+12 )=0 3 = 2 , 3= 32 , 3= 52 , 3= 72 , 3= 92 , 3= 112 ,⋯  = 6 ,  = 2 ,  = 56 ,  = 76 ,  = 32 ,  = 116 ,⋯ .

It is known that

where k is an integer so

But 2π is not on the interval [0, 2π). So the solutions so far are

.

Also,

making


150



APPLICATIONS OF TRIGONOMETRY

WITH APPLICATIONS

Chapter 4 Applications of Trigonometry Learning Objectives:   

Solve right and oblique triangles Apply the Law of Sines and the Law of Cosines to practical problems Find areas of triangles using alternative formulas

Section 4.1 Right Triangle Trigonometry Trigonometry may be used to solve triangles, that i s, to find lengths of sides or measures of angles given certain information. For instance, if the measure of angle A and the length of side b are known in the right triangle in Figure 4.1.1, then the measure of the other angle and the length of the remaining two sides may be determined.

Example 4.1.1

B c a A= 40° b = 100 Figure 4.1.1

Angles A and B are complimentary so B = 90° – A = 50°. The length of side a may be found using the tangent function and side c using the cosine function.

tan(40°) =

 →  = 100tan(40°) ≈ 84 100

cos(40°) =

100 100 →= ≈ 131  cos(40°)

Of course, the sine function could be used to find c after a has been determined.

sin(40°) =

84 84 →= ≈ 131  sin(40°)

Or, the Pythagorean theorem could have been used to determine c from a and b.

842 + 1002 = 17056 =  2 →  = √17056 ≈ 131 When possible, it is good practice to choose methods that only use given values. Any error in a calculated value will be propagated in further calculations. The tangent and the cosine functions were chosen because they did not require the use of any calculated values.


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Example 4.1.2 Solve right the triangle in Figure 4.1.3 if B = 62° and c = 45 meters.

B c a A b Figure 4.1.3

sin(62°) = cos(62°) =



A = 90° – 62° = 28°.

→  = 45 sin(62°) ≈ 40 meters.

45 

→  = 45 cos(62°) ≈ 21 meters.

45

It is a good crosscheck to verify that the longest side is opposite the largest angle and the shortest side is opposite the smallest angle. If the Pythagorean theorem is used to check the lengths, expect rounding errors to influence the calculation, causing it to be somewhat off.

Example 4.1.3 Solve right the triangle in Figure 4.1.4, if a = 12 and b = 5.

B c a A b Figure 4.1.4

c = 13 since 5, 12, 13 is a Pythagorean triple. The Pythagorean theorem could also be used to find the length of side c.

tan( ) =

12 5

= 2.4 →  = − (2.4) ≈ 67°

B = 90° – 67° = 23°.

Example 4.1.4 Jack and Jill went up the stairwell with a flat screen TV and got stuck. Jack did not listen to Jill when she suggested that they use a little measurement and trigonometry before starting up the stairs. The television was too long to clear the turns in the stairwell. See Figure 4.1.5.


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WITH APPLICATIONS

45°

Figure 4.1.5 ‘Flat screen television’ stuck in a stairwell

Assuming the walls on either side of each step of the stairway are parallel, the width of the hall is 3/4 meter and the angle between the television and the wall at the top of the stairs is 45°, how long is the television? Solution: The large, right triangle in Figure 4.1.5, whose hypotenuse represents the length of the television set, can be divided into a square and two smaller right triangles as i n Figure 4.1.6.

45°

Figure 4.1.6

The triangle in the top right involves the given angle of 45° and the known hallway width. The hypotenuse of this triangle makes up half the length of the television and may be determined using the sine function: 0.75 meters sin 45° =

0.5

0.75 →  =

0.5sin45°

≈ 2.1 meters

where l represents the length of television. The hypotenuse is half l because the triangle in the top right is congruent to the triangle in the bottom left of Figure 4.1.6. Why? Assumptions of congruence or equality should never be made based upon appearance. Mathematical justification should always be found and verified. The real question is, why are Jack and Jill trying to move such an expensive television?

Example 4.1.5 A land surveyor places a total station theodolite at point A in Figure 4.1.7. A rodman holds a mirror at point B and the distance AB is found to be 128 meters. The angle of declination, , is also


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Section 4.2 The Law of Sines Some triangles are just not right. They are oblique. The Pythagorean theorem does not apply to oblique triangles. The definitions of the trigonometric functions are based upon right triangles, not oblique triangles. Definition 4.2.1 Oblique Triangle

An oblique triangle is not right. It has no right angle.

Oblique triangles may have all acute, non-right angles or two acute, non-right angles and one obtuse angle. B c a

A

C b Figure 4.2.1 Random oblique triangle

Throughout this chapter, angles will be represented using capital letters and side lengths will be represented using lower case. Angles and their opposing sides will be designated with the same variable, different case as in Figure 4.2.1.

Example 4.2.1 A mariner is trying to determine her distance from land by taking two different sightings of a nearby lighthouse (Figure 4.2.2). At the first sighting, the angle of elevation, , is found to be 27° 48'. After traveling for a certain time, the navigator knows her travelled distance is 50 km. At this point she takes another sighting and the angle of elevation,  , is measured as 34° 55'. How far is she from the lighthouse over water?

Figure 4.2.2 Lighthouse sightings

Without knowledge of the height of the lighthouse or perhaps another component, this problem cannot readily be solved using right triangles. Fortunately, another strategy exists, using the Law of Sines. One side and two angle measures are known for the oblique triangle in Figure 4.2.2. This triangle is isolated for clarity in Figure 4.2.3.


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WITH APPLICATIONS

 ° − 

Lighthouse



 50 km

x km Figure 4.2.3

Theorem 4.2.1 The Law of Sines Identity

In a triangle, the ratio of the sine of an interior angle to the length of its opposite side is proportional to the ratio of the sine of any other interior angle to the length of its opposite side and is equal to the diameter of the circumscribed circle (2r ) or its reciprocal. That is to say,

or

   = = = 2r sin sin sin 1 sin sin sin = = =    2

where a, b and c are the sides of the triangle and A, B and C are the respective angles opposite those sides.

Solution:

 = 27° 48 = 27.8° and  = 34° 55′ ≈ 34.9° 180° −  = 145.1°  = 180° − 145.1° − 27.8° = 7.1° since the interior angles of a triangle sum to 180°. The Law of Sines gives:

50 km ℎ = sin(7.1°) sin(27.8°) where h is the hypotenuse of the right triangle (Figure 4.2.3) involving  and the height of the lighthouse (assuming it is perpendicular to the horizon).

ℎ=

50(sin(27.8°)) ≈ 189 km sin(7.1°)

Now using right triangle trigonometry:


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WITH APPLICATIONS

 = ℎsin = (189 km) sin(34.9°) ≈ 108 km from land. Before the advent of hand-held computers (calculators and/or cell phones) and Global Positioning Systems, navigators carried five-figure tables of trigonometric values much like those found in the indices of mid-to-late 20th century mathematics and engineering texts.

Proof of the Law of Sines Identity:

Figure 4.2.4 shows two random oblique triangles and their respective heights. B

B

c

a h

c

h a

A

C b

A

C b

Figure 4.2.4 Oblique triangles

Figure 4.2.4 illustrates:

sin =

ℎ 

→ ℎ = sin and sin =

ℎ 

→ ℎ = sin.

The above relationships are true for either triangle in Figure 4.2.4 since sin = sin( − ). The sine of angle C and its supplement may be shown to be equal using the Difference of Angles identity for sine: sin( − ) = sincos − cossin = sin. It follows that

ℎ = sin = sin → 

sin

=

 sin

.



, draw an altitude from angle C and proceed in a   similar fashion. All three proportions are equal by t he transitive property of equality. To show the other proportion,

=



Figure 4.2.5 shows the circumscribed circle of a random oblique triangle ABC . Triangle ADC is inscribed in the same circle such that it shares side b with triangle ABC and side AD is a diameter of the circle. Thales’ theorem states that under these conditions, angle C of triangle ADC must be right. It follows from the definition of sine

sin =

 

=

 2

→ 2 =

 sin

.

Angle B and angle D must be congruent since they subtend t he same chord, b, and the same arc AC (Figure 4.2.5). So sin D = sin B and   2 = = . sin sin Finally,


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WITH APPLICATIONS

Example 4.2.

84°

Solve the triangle:

4

a

29°

C b

The measure of angle C = 180° − 84° − 29° = 67°. The Law of Sines gives:

4 sin(67°)

4 sin(67°)

=

=



→  =

4sin(29°) ≈ 2.1 sin(67°)

→  =

4sin(84°) ≈ 4.3. sin(67°)

sin(29°)

 sin(84°)

The Law of Sines is used in a technique called triangulation. Triangulation is the process forming triangles to locate something. NASA uses a form of t riangulation in conjunction with radio waves to locate spacecraft. Geologists use it to locate the epicenter of an earthquake. Surveyors use it for property boundary surveys.

Example 4.2.3 A land surveyor needs to accurately locate a position some distance off from his current position. See Figure 4.2.6. In order to find the distance from point A to point C , the surveyor first measures a small distance to point B. The distance to point B is made small for accuracy. With a Theodolite (survey instrument), accurate angle measures from both points, A and B, may be made.

C B

A Figure 4.2.6 Triangulation

Length AB is measured to be 12 kilometers. Angle A is measured to be 84.3° and angle B is found to be 72.6°. Angle  = 180° − 84.3° − 72.6° = 23.1°. Now the Law of Sines may be used to find the distance AC .

12 km sin(23.1°)

=

 sin(72.6°)

→  =

12sin(72.6°) sin(23.1°)

≈ 29.2 km.


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WITH APPLICATIONS

Example 4.2.4

Solve the triangle:

B

c = 5

a = 3 h

°

A = 41

C b

The Law of Sines gives:

sin41° = sin → sin = 5sin41° →  = sin− (5sin41°) ≈ sin−1.09. 3 5 3 3 If an attempt to evaluate the arcsine of 1.09 is continued, a calculator should give an error message, perhaps, ‘Domain Error’. Recall the domain of the inverse sine function is [-1, 1] (range of the sine − has no solution since 1.09 is not in its domain. Furthermore, this problem has function). no solution and illustrates an important aspect of applying the Law of Sines. The following video explains in detail. The Law of Sines Ambiguous Case

sin 1.09


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WITH APPLICATIONS

Example 4.2.4 was an example of the ‘no triangle possible’ case. Based upon the given information, the height of the triangle is ℎ = 5sin(41°) ≈ 3.3 and, as such, side a is not long enough to reach the base and form a triangle. Table 4.2.1 summarizes the possible cases when two sides of an oblique triangle are known and an angle not interior to those sides is also known.

Law of Sines Case: Angle Side Side

a

b h

B Case:

Determination:

No triangle possible

b < h = asin( B)

Right triangle

b = h = asin( B)

Ambiguous: 2 oblique triangles possible

h = asin( B) < b < a

1 oblique triangle

b > a

Table 4.2.1 Summary for Law of Sines SSA case

Example 4.2.5 Solve the triangle:

B

c = 5

a = 2.5 h

A = 30°

C b

Since this is the SSA case and the length of side a is less than side c, the height should be determined.

ℎ = 5sin(30°) = 2.5 Since a = h, triangle ABC must be a right triangle. Solve by the methods of section 4.1, that is, use right triangle trigonometry. Law of Sines will work, but it will take longer. Angle B is the complement of angle A, so B = 60°. b =  5 − (2.5) = √ 18.75 ≈ 4.3 by the Pythagorean theorem.


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WITH APPLICATIONS


B

c = 6

a = 4 h

A = 30°

C b

Solution: Since this is the SSA case and the length of side a is less than side c, determine the height.

ℎ = 6sin(30°) = 3 Since h = 3 < a = 4 < c = 6, this is the ambiguous case. Apply the Law of Sines. Triangle #1:

Using the given ratio, that is,

sin(30°) 4

=

 (°)

sin( )

,

→ sin( ) =

6

6sin(30°) 4

=

3

3 →  = sin− ≈ 48.6°. 4 4

Angle  = 180°  48.6°  30° = 101.4°.

4 sin(30°)

=

 sin(101.4°)

→  =

4sin(101.4°) sin(30°)

≈ 7.8.

B1 = 101.4°

c = 6

a = 4 h = 3

A = 30°

C 1 = 48.6° b1 = 7.8

Triangle #2:

Angle C 2 is the supplement of C 1, that is, C 2 = 180° – C 1 = 180° – 48.6° = 131.4°. Angle  = 180°  131.4°  30° = 18.6°.  Using the given ratio, that is, , (°)

4 sin(30°)

=

 sin(18.6°)

→   =

4sin(18.6°) sin(30°)

B2 = 18.6°


165

≈ 2.6.



WITH APPLICATIONS

Section 4.3 The Law of Cosines B

c

a h

A

C ccos A

acosC b

Figure 4.3.1 Law of Cosines

By definition, an altitude or height of a triangle forms a right angle with it s base. In Figure 4.3.1, the height of the triangle apportions the base, b, into two segments equal to ccos A and acosC . This is according to the definition of cosine. The sum of these two lengths is the length of side b, that is,  = cos + cos.

Multiplying by b   = cos + cos.

Dropping an altitude from angle A gives  = cos + cos.

Multiplying by a  = cos + cos.

Repeating for angle C   = cos + cos.

Adding the first two equations and subtracting the last gives:  +   −   = cos + cos + cos + cos − cos − cos  +   −   = cos + cos   =  +   − 2cos

and the Law of Cosines is obtained. Please note that the Law of Cosines is the Pythagorean theorem when angle C = π/2. In addition, it is important to note that the side opposite the angle in the cosine function is the isolated square.   =   +   − 2cos

This side

is opposite

this angle in the Law of Cosines.


167



WITH APPLICATIONS

Theorem 4.3.1 The Law of Cosines Identity

  =  +   2cosC relates the sides of a triangle denoted a, b and c where angle C is opposite side c.

Example 4.3.1 (formerly 4.2.7) Solve the triangle:

B

c = 6

a

A = 30°

C b = 7.8

Applying the Law of Cosines:

 = 6 + (7.8)  2(6)(7.8)cos(30°) ≈ 15.78 and following order of operations. So  ≈ √ 15.78 ≈ 4.0. At this point, it is possible to use the Law of Sines. But it also possible to have an ambiguous case result using this strategy. Continuing with the Law of Cosines:

6 = 4 + (7.8)  2(4)(7.8)cos 36 = 76.84  62.4cos 40.84 = 62.4cos cos = Angle 

40.84 ≈ 0.6545 →  = cos − 0.6545 ≈ 49.9° 62.4

= 180°  30°  49.9° = 100.1°.


B

c = 6

a = 7

A

C b=8

Applying the Law of Cosines:

7 = 6 + 8  2(6)(8)cos


168



WITH APPLICATIONS

Example 4.3.3 The Law of Cosines is applicable to mechanics. Please view and/or interact with the following GeoGebra manipulative:

Piston simulation by Ken Schwartz

A spark in the piston of a cylinder in a car’s engine fires and moves a 9 cm connecting rod that rotates a crankshaft with 4 cm pins in the counter-clockwise direction. The crankshaft, in turn, moves the piston back down into the cylinder, creating another combustion event and continuing the cycle. A spark is produced only when the piston is fully extended into the cylinder, that is, at position 0 + 2πk radians during the crankshaft’s rotation, where k is an integer (see section 3.5, Solving Trigonometric Equations). What is the length of the piston stroke, d , when  , the rotation angle of the crankshaft, equals π/3? Solution:



Let d = the distance from the center of the crankshaft to the top of piston (piston stroke). Let = the angle of the crankshaft’s rotation. According to the Law of Cosines:

 =  +  −2cos 9 = 4 +  −24cos3 81=16+  −8(12) 65= −4 a quadratic equation. Completing the square:

69= −4+4 69 =  − 2   = 2 ± √ 69  ≈ 10.3 cm. Obviously, only the positive solution is applicable due to the natural restriction imposed on t he domain by the context of the problem. The length of the piston stroke, d , is dependent upon the length of the connecting rod and pins.


170



WITH APPLICATIONS

Section 4.4 The Area of a Triangle The commonly used formula for the area of a triangle is half the length of the base times the height or A = 1/2bh. If the height is not available or difficult to measure, it can be useful to have formulas involving angles.

c

c h



h



b

b

Figure 4.4.1

In Figure 4.4.1,

ℎ = sin. The area formula becomes  =  ℎ =  sin. In other words, the area

of a triangle may be found if two sides and the angle between them are known.

Example 4.4.1 Find the area of the triangle:

7 mm

π/6

9 mm

 = 12 9 mm7 mmsin6 = 634 mm = 15.75 mm. This modified area formula motivates an alternative proof of the Law of Sines: B

c

A

a

b

C

Figure 4.4.2

Refer to Figure 4.4.2. The area of this triangle may be expressed in several different ways depending on which altitude is used.

 = 12 sin = 12 sin = 12 sin. Multiplying by 1:

2 ( = 1 sin = 1 sin = 1 sin)  2 2 2 Download free eBooks at bookboon.com

171



WITH APPLICATIONS

2 si n  si n  si n  = = =     .

Another formula for the area of triangle involves only the lengths of the sides of the triangle. It is called Heron’s Formula.

Theorem 4.4.1

Heron’s Formula

The area of a triangle, A, is related to its semiperimeter, s, by

 =√ (−)(−)(−) =  + 2 + 

where a, b and c are the lengths of the sides of the triangle and

.

Proof of Heron’s Formula:

The Law of Cosines states:

Rearranging:

Applying the square property:

 = + −2cos.     + − cos= 2 .  ( + −)     + −  cos = 2  = 4 .

A Pythagorean Identity states:

sin= √ 1 −cos     ( ) + −  sin= 1 − 4 sin=  4 −(4+  −). Factoring and simplifying:

Rearranging:

Factoring again:

      ( 2−( + − ))(2+ + − ) √ sin= . 2       (  −( −2+ ))( +2+ − ) √ sin= . 2 Download free eBooks at bookboon.com

172



WITH APPLICATIONS

And factoring once more:

sin=   −−2+ −. sin=   −−+−+−++ . 2  = 12 sin  = 12    −−+−+−++ 2  = 14   −−+−+−++ +− ++ +−  = (−− )( )  2 2 2 2 .

Recall the area of the triangle may be given by:

Substituting

=  + 2 + ,

 = −−−= −−−.

Example 4.4.2 Find the area of the triangle:

4.5 km

5.6 km

8.9 km

=  + 2 +  = 192 =9.5  = −−−= 9.550.63.9=√111.15≈10.5 . Example 4.4.3 Why can’t a triangle be constructed using lengths 1, 2 and 3 for its sides?

Solution:


173



WITH APPLICATIONS

The area of the second figure, the degenerate triangle, is zero. A non-degenerate triangle cannot be constructed using any random set of side lengths. The lengths of the sides of a non-degenerate triangle must satisfy the strict case of the triangle inequality.

Theorem 4.4.2 The Triangle Inequality

In any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side. If a, b and c represent the lengths of the sides of a triangle, then a + b > c, a + c > b and b + c > a.

For the degenerate triangle in example 4.4.3, 1 + 3 > 2 and 2 + 3 > 1, but 1 + 2 = 3. This makes it degenerate. The Triangle Inequality is important in Analysis and Topology. It is a consequence of the Law of Cosines. Proof of the Triangle Inequality: The Law of Cosines states

  =  +   − 2 cos . The cos  ≥ −1 due to its range. Therefore, − cos  ≤ 1 and

  =  +   − 2 cos  ≤  +   + 2 = ( + ) . If

  ≤ ( + ) then

 ≤  + . The sense of the inequality is preserved since c and a + b are positive (Sarkar).


175



ENDNOTES

WITH APPLICATIONS

Endnotes 1. George Johnston Allman (1889). Greek Geometry from Thales to Euclid (Reprinted by Kessinger Publishing LLC 2005 ed.). Hodges, Figgis, & Co. p. 26. ISBN 1-4326-0662-X. "The discovery of the law of three squares, commonly called the "theorem of Pythagoras" is attributed to him by – amongst others – Vitruvius, Diogenes Laertius, Proclus, and Plutarch ..." 2. Heath, Sir Thomas (1921). "The 'Theorem of Pythagoras'". A History of Greek Mathematics (2 Vols.) (Dover Publications, Inc. (1981) ed.). Clarendon Press, Oxford. p. 144 ff . ISBN 0-486-240738. 3. Euclid's Elements: web page version using Java applets from Euclid's Elements by Prof. David E. Joyce, Clark University. 4. Maor, Eli. The Pythagorean Theorem: A 4,000-year History, p. 61 (Princeton University Press, 2007). 5. Gardner, M. "The Pythagorean Theorem." Ch. 16 in The Sixth Book of Mathematical Games from Scientific American. Chicago, IL: University of Chicago Press, pp. 152-162, 1984. 6. "cofunctions." Dictionary.com Unabridged. Random House, Inc. 11 Oct. 2014. . 7. The Trigonographer (28 September 2015). "Angle Sum and Difference for Sine and Cosine". Trigonography.com. Retrieved 28 May 2017. 8. Krishnavedala. Illustration of how a phase portrait would be constructed for the motion of a simple pendulum. November 2014. 9. Beer.F.P. , Johnston.E.R. (1992). Mechanics of Materials , 2nd edition. McGraw-Hill 10. Winckel, Fritz (1967). Music, Sound and Sensation: A Modern Exposition, p.134. Courier. 11. Clint Goss; Barry Higgins (2013). "The Warble". Flutopedia. Retrieved 2013-03-06. 12. J. W. Coltman (October 2006). "Jet Offset, Harmonic Content, and Warble in the Flute". Journal of the Acoustic Society of America. 120 (4): 2312 – 2319. PMID 17069326. doi:10.1121/1.2266562. 13. Hill, T. L. An Introduction to Statistical Thermodynamics. New York: Dover, 1986. 14. Weisstein, Eric W. "Triangle Inequality." From MathWorld --A Wolfram Web Resource. http://mathworld.wolfram.com/TriangleInequality.html 15. Coxeter, H. S. M. and Greitzer, S. L. Geometry Revisited . Washington, DC: Math. Assoc. Amer., pp. 1 – 3, 1967 16. Livio, Mario (2002). The Golden Ratio: The Story of Phi, The World's Most Astonishing Number . New York: Broadway Books. ISBN 0-7679-0815-5. 17. Yates, R. C.: A Handbook on Curves and Their Properties, J. W. Edwards (1952), "Evolutes." p. 206 18. https://en.wikipedia.org/wiki/Logarithmic_spiral


177


ENDNOTES

WITH APPLICATIONS

19. Weisstein, Eric W. "Logarithmic Spiral." From MathWorld --A Wolfram Web Resource. http://mathworld.wolfram.com/LogarithmicSpiral.html 20. Conway, J. H. and Guy, R. K. "Euler's Wonderful Relation." The Book of Numbers. New York: Springer-Verlag, pp. 254-256, 1996. 21. Conway, John Horton, and Guy, Richard (1996). The Book of Numbers (Springer, 1996). ISBN 978-0-387-97993-9. 22. Sandifer, C. Edward. Euler's Greatest Hits (Mathematical Association of America, 2007). ISBN 978-0-88385-563-8 23. Weisstein, Eric W. "Phasor." From MathWorld --A Wolfram Web Resource. http://mathworld.wolfram.com/Phasor.html 24. Woo, Eddie. Parametrics Part I 25. B. L. van der Waerden, A History of Algebra, Springer Verlag, NY 1985. 26. Barret, G., Bartkovich, K., Compton, H., Davis, S., Doyle, D., Goebel, J. and Gould, L. Contemporary Precalculus Through Applications: Functions, Data Analysis, and Matrices, Janson Publications 1991. 27. Famous Curves Index. http://www-history.mcs.st-andrews.ac.uk/history/Curves/Curves.html 28. M. E. Baron. The Origins of the Infinitesimal Calculus, Pergamon Press, 1969. 29. W. Dunham. Journey Through Genius, Wiley , 1990. 30. V. J. Katz. A History of Mathematics: An Introduction, 2nd Edition, Addison Wesley, 1998. 31. J. J. O’Connor and E.F. Robertson, Charles de Bouvelles; http://www.history.mcs.st-andrews.ac.uk/Mathematicians/Bouvelles.html http://www.history.mcs.st-andrews.ac.uk/Mathematicians/Torricelli.html http://www.history.mcs.st-andrews.ac.uk/HistoryTopics/Brachistochrone.html 32. Guo, James C.Y. Theoretical Fluid Mechanics: Laminar Flow Velocity Profile. University of Colorado, Denver

33. Daugherty, Robert L., Franzini, Joseph B. and Finnemore, E. John. Fluid Mechanics with Engineering Applications, 8th Edition, McGraw Hill, 1985. 34. Kalman, Dan. Ellipse, Mathematical Association of America (MAA), Journal of Online Mathematics and its Applications (JOMA), Volume 8 (2008). 35. hyperphysics.phy-astr.gsu.edu/hbase/kepler.html (©C.R. Nave, 2017). 36. Lunin, Stephen V. Globoid Gear Technology, www.zakgear.com/Wormoid.html (2007). 37. https://en.wikipedia.org/wiki/Introductio_in_analysin_infinitorum 38. https://en.wikipedia.org/wiki/Complex_plane 39. https://en.wikipedia.org/wiki/Argument_(complex_analysis)


178


ENDNOTES

WITH APPLICATIONS

40. Needham, Tristan. Visual Complex Analysis. Oxford University Press, 1997. 41. Azad, Kalid. https://betterexplained.com/articles/intuitive-understanding-of-eulers-formula 42. https://en.wikipedia.org/wiki/Hyperbolic_navigation 43. math.sci.ccny.cuny.edu/document/show/2685 44. Libii, Josué Njock. The Use of Conic Sections in Basic Mechanics Courses: Some Examples. Purdue University Fort Wayne. 45. Althoen, S. and Mclaughlin, R. Gauss-Jordan Reduction: A Brief History. February 1987. http://macs.citadel.edu/chenm/240.dir/12fal.dir/history4.pdf 46. Whitford, D. E. and Klamkin, M. S. “On an Elementary Derivation of Cramer's Rule”, American Mathematical Monthly, vol. 60, pp.186 – 7, 1953. 47. Strang, Gilbert. Linear Algebra and its Applications. 3rd Edition, Harcourt Brace Jovanovich, 1988. 48. Campbell, David K. (25 November 2004). "Nonlinear physics: Fresh breather". Nature. 432 (7016): 455 – 456. Bibcode:2004Natur.432..455C. doi:10.1038/432455a. ISSN 0028-0836. 49. Sullivan, Michael and Sullivan, Michael, III. Precalculus Enhanced with Graphing Utilities. 6th Edition. Pearson, 2013. 50. Kennedy, John. Some Polynomial Theorems. 51. Hahn, Harry K. "The Ordered Distribution of Natural Numbers on t he Square Root Spiral". arXiv:0712.2184 52. Nahin, Paul J. (1998), An Imaginary Tale: The Story of [the Square Root of Minus One], Princeton University Press, p. 33, ISBN 0-691-02795-1 53. Adamchik, Victor. Mathematical Induction. https://www.cs.cmu.edu/~adamchik/21127/lectures/induction_1_print.pdf 54. Wilder, Raymond L. Introduction to the Foundations of Mathematics. 2nd Edition. Courier Corporation, p. 120.

55. Wolfe, John H., Mueller, William F. and Mullikin, Seibert, D. Industrial Algebra and Trigonometry. McGraw Hill, 1945.


179



INDEX

WITH APPLICATIONS

Index A abscissa, 52 acoustics, 139 algebraic expression, 104, 105 alpha, 117 ambiguous case, 163, 165, 168 amplitude, 48, 55, 56, 67, 68, 69, 72, 74, 75, 76, 78, 81, 82, 84, 8 5, 87, 140 angle, I, 2, 4, 41, 126, 127, 128, 129, 130, 131, 132, 136, 139, 145, 146, 149, 160, 162, 164, 165, 168, 169, 177 angle bisector, 22 angle of declination, 156 angle of elevation, 133, 158 arc length, 3, 4, 6, 7, 8 arccosine, 95, 101, 102 arctangent, 101 area of a sector, 8 argument, 14, 22, 104 asymptote, 58, 96, 97

B beta, 115, 117 binomial, 46, 107

C chord, 17, 160 circle, ii, 1, 4, 7, 8, 16, 17, 19, 20, 22, 24, 27, 31, 32, 33, 34, 37, 48, 49, 50, 5 2, 53, 55, 86, 118, 119, 150, 159, 160, 161 circular reasoning, 107 circumference, ii, 4, 17, 31, 32, 33, 34, 48, 49, 50, 53, 55 circumscribed, 159, 160 cofunction , 15, 25, 28, 41, 44, 49, 53, 57, 60, 61 , 73, 108 common denominators, 113 commutative, 89 complement, 15, 41, 164 completing the square, 170 complimentary, 153 composition, 89 congruent, 156, 160 conjugate, 46, 107, 111 continuous, 33, 52 contradiction, 91 coordinate, 33, 34, 42, 48, 50, 60, 65, 66, 69, 76, 86, 96, 97 cosecant, 14, 15, 20, 35, 40, 41, 50, 60, 61, 63, 78, 79, 97, 98, 101, 110, 119 cosine, 14, 15, 21, 34, 37, 39, 40, 41, 42, 43, 45, 48, 49, 52, 53, 54, 55, 63, 68, 69, 72, 73, 74, 75, 79, 82, 85, 87, 95, 102, 103, 104, 113, 115, 117, 119, 126, 129, 135, 136, 137, 138, 141, 142, 143, 145, 146, 150, 153, 167 cotangent, 14, 15, 21, 35, 39, 40, 41, 49, 50, 57, 58, 59, 78, 79, 96, 97, 102, 119, 150 counter-intuitive, 65, 66, 68, 74, 76, 78

D Debye constant, 140 decimal degrees, 2, 5 degenerate, 174, 175 degree, 2 Degrees Minutes Seconds, 2 denominator , 34, 46, 48, 111, 151 dependent variable, 52 diagonal, 26, 119 diameter , 159, 160 difference quotient, 124, 125 dilation, 64, 66, 67, 68, 69, 70, 75, 81


182


INDEX

WITH APPLICATIONS

discontinuities, 48, 58 distance formula, 13 domain, 16, 17, 22, 24, 33, 34, 35, 39, 42, 48, 49, 50, 51, 52, 53, 55, 57, 58, 60, 61, 78, 79, 89, 90, 91, 93, 95, 96, 97, 99, 101, 102, 103, 105, 107, 122, 140, 141, 143, 149, 163, 170 Domain Error, 163

E epicenter of an earthquake, 162 equilateral triangle, 22 equilibrium state, 86 even function, 42, 53, 61 extraneous solution, 149 extrema, 52, 53, 57, 58, 61, 65, 67, 68, 69, 70, 72, 75, 76, 79

F frequency, 48, 55, 56, 67, 68, 70, 74, 75, 76, 78, 81, 87, 140 fundamental trigonometric identities, 1, 39, 107, 115 fuselage, 133

G GeoGebra, 170 Global Positioning Systems, 160 graph, 52, 53, 56, 57, 58, 60, 61, 63, 64, 65, 66, 67, 68, 75, 76, 79, 82, 84, 85, 90, 91, 93, 95, 142, 146, 147, 149, 150, 151 graphing utility, 103

H harmonics, 140 Heron’s Formula, 172

Hertz, 140 horizontal line test, 91 hypotenuse, 10, 11, 14, 15, 17, 27, 105, 156, 159

I identity, 19, 37, 39, 40, 45, 46, 49, 60, 63, 73, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 119, 126, 128, 129, 131, 136, 137, 138, 139, 141, 150 independent variable, 65, 66 infinite, 1, 33, 34, 48, 52, 93, 115, 140, 141, 142 infinity, 35, 49 inscribed, 11, 119, 160 integer multiples, 49 intercepts, 52, 60, 61, 65, 66, 70, 86, 146 interval, 33, 34, 48, 50, 53, 55, 93, 96, 100, 101, 143, 145, 150 invariant, 67 inverse, 48, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 9 9, 100, 101, 102, 103, 104, 128, 149, 163 irrational, 19, 70, 129, 130

L Lakota Flute, 140, 141 land surveyor, 156, 162 Law of Cosines, I, 153, 166, 167, 168, 169, 170, 172, 175 Law of Sines, I, ii, 153, 158, 159, 160, 162, 163, 164, 165, 166, 168, 171

M maxima, 53, 60 maximum, 48, 50, 55, 73, 76, 86 mechanics, 170 megaPascals, 134


183


INDEX

WITH APPLICATIONS

minima, 53 minimum, 48, 50, 55, 76 modulation, 140 music theory, 140

N navigator, 158 normal stress, 134 numerator , 35, 111, 112

O oblique, 153, 158, 160, 161, 164 odd function, 53, 57, 58, 60, 84, 119 odd/even properties, 43 one-to-one, 91, 102 ordinate axis, 52 oscillation, 86, 87 overtone, 140

P parallel, 133, 156 period, 17, 33, 34, 35, 48, 50, 52, 53, 55, 57, 58, 60, 67, 68, 69, 70, 72, 75, 76, 78, 79, 82, 84, 85, 87 period of a trigonometric function, 33 periodicity, 53, 55, 57 perpendicular, 133, 159 phase portrait, 86, 177 phase shift, 48, 73, 74, 75, 76, 78, 81, 82, 84, 86, 87 point of equilibrium, 55 projectile, 133 properties of equality, 107 proportion, 110, 160 Ptolemy’s t heorem, 119 Pythagorean theorem, 1, 10, 11, 12, 13, 15, 17, 19, 22, 27, 37, 39, 105, 153, 155, 158, 164, 167 Pythagorean triple, 11, 155

Q quadrant, 20, 37, 93, 129, 130, 143, 144, 149 quadrantal angle, 32, 37, 146 quadrantal angles, 31, 52, 60, 70 quadratic form, 143, 150 quadrilateral, 119, 120

R radians, 3, 4, 5 radius, ii, 3, 4, 6, 7, 8, 9, 16, 17, 19, 20, 24, 31, 33, 37, 48, 50 range, 17, 20, 33, 34, 35, 37, 48, 49, 50 , 51, 52, 53, 57, 58, 61, 76, 79, 89, 91, 93, 95, 96, 97, 99, 100, 101, 102, 104, 128, 133, 146, 163, 175 rational expressions, 107, 110, 111, 113, 123, 138 reciprocal, 5, 20, 42, 43, 49, 57, 60, 63, 104, 159 reference angle, 22, 24, 27, 37, 143, 145, 146 reference triangle, 22, 23, 24, 26, 27, 143 reflection, 42, 57, 66, 67, 72, 79, 82, 84, 87, 90, 91, 93 restricted domain, 102 right triangle, 1, 10, 11, 14, 15, 16, 17, 19, 22, 31, 41, 153, 156, 159, 164 rounding errors, 155

S scale, 52, 57, 64, 70, 75, 76, 78, 7 9, 146, 151 secant, 14, 15, 21, 35, 39, 41, 42, 4 9, 50, 60, 61, 63, 78, 79, 97, 98, 101, 103, 104, 119


184

Analytical Trigonometry With Applications 1

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