4 Trigonometric Functions. Chapter Contents. Trigonometric Equations. Angles and Their Measurement. The Sine and Cosine Functions

Transcription

1 _CH04_st.qd 0/5/4 5:33 PM Page 06 Jones & Bartlett Learning.. gualtiero boffi/shutterstock, Inc. 4 Trigonometric Functions Chapter Contents 4. Angles and Their Measurement 4.8 Trigonometric Equations 4. The Sine and Cosine Functions 4.9 Simple Harmonic Motion 4.0 Jones & Bartlett Learning, Right Triangle Trigonometr 4.3 Jones Graphs & of Bartlett Sine andlearning, Cosine Functions 4.4 Other Trigonometric Functions 4. Applications of Right Triangles 4.5 Sum and Difference Formulas 4. Law of Sines 4.6 Product-to-Sum and Sum-to-Product 4.3 Law of Cosines 4.7 Inverse Trigonometric Functions Jones & Bartlett Learning, Formulas 4.4 Limit Concept Revisited FOR SALE OR The Chapter 4 Review Eercises

2 Jones & Bartlett Learning.. 4. Angles and Their Measurement Verte Terminal side Terminal side Initial side (b) Standard position FIGURE 4.. Initial and terminal sides of an angle INTRODUCTION We begin our stud of trigonometr b discussing angles and two methods of measuring them: degrees and radians. As we will see it is the radian measure of an angle that enables us to define trigonometric functions on sets of real numbers. Jones & Bartlett Verte Angles Learning, An angle is formed b two half-ras, or half-lines, which have a common FOR SALE endpoint, OR called the verte. We designate one ra the initial FOR side SALE of the OR angle DISTRIBUT and the Initial side other the terminal side. It is useful to consider the angle as having been formed b a rotation from the initial side to the terminal side as shown in FIGURE 4..(a). An angle (a) Two half-ras is said to be in standard position if its verte is placed at the origin of a rectangular Jones & Bartlett Learning, coordinate sstem with its initial side Jones coinciding & Bartlett with the Learning, positive -ais, as shown in Figure 4..(b). Degree Measure The degree measure of an angle is based on the assignment of 360 degrees (written 360 ) to the angle formed b one complete counterclockwise rotation, as shown in FIGURE 4... Other angles are then measured in terms of a 360 angle, with a angle Jones being & Bartlett formed b Learning, of a complete 360 rotation. If the rotation is counterclockwise, the FOR measure SALE will be OR positive; if clockwise, the measure is negative. For eample, the angle in FIGURE 4..3(a) obtained b one-fourth of a complete counterclockwise rotation will be 4(360 ) FOR SALE Shown OR in Figure 4..3(b) is the angle formed b three-fourths FOR of SALE a complete OR DISTRIBUT clockwise rotation. This angle has measure 3 4(360 ) Jones & Bartlett Learning, FOR SALE OR (a) 90 angle (b) 70 angle FIGURE 4.. Angle of FIGURE 4..3 Positive measure in (a); 360 degrees negative measure in (b) θ θ Coterminal Angles Comparison of Figure 4..3(a) with Figure 4..3(b) shows Jones & Bartlett that the Learning, terminal side of a 90 angle coincides with the Jones terminal & Bartlett side of a 70 Learning, angle. LL When two angles in standard position have the same terminal sides we sa the are θ 360 coterminal. For eample, the angles u, u360, and u360 shown in FIGURE 4..4 are coterminal. In fact, the addition of an integer multiple of 360 to a given angle FIGURE 4..4 Three results in a coterminal angle. Conversel, an two coterminal angles have degree measures that differ b an integer multiple of 360. coterminal angles EXAMPLE Angles and Coterminal Angles For a 960 angle: (a) Locate the terminal side and sketch the angle. (b) Find a coterminal Jones angle & between Bartlett 0 and Learning, 360. (c) Find a coterminal FOR angle between SALE OR 360 and Angles and Their Measurement 07

3 Jones & Bartlett Learning.. Solution (a) We first determine how man full rotations are made in forming this angle. Dividing 960 b 360 we obtain a quotient of and a remainder of 40. Equivalentl we can write (360) 40. Thus, this angle is formed b making two counterclockwise rotations before completing of another rotation. As illustrated in FIGURE 4..5(a), the terminal side of Jones lies & in Bartlett the third Learning, quadrant. (b) Figure 4..5(b) shows that the angle 40 is coterminal with a 960 angle. (c) Figure 4..5(c) shows that the angle 0 is coterminal with a 960 angle FOR SALE OR 0 (a) FOR SALE (b) OR (c) FIGURE 4..5 Angles in (b) and (c) are coterminal with the angle in (a) Minutes and Seconds With calculators it is convenient to represent fractions of degrees b decimals, such as 4.3. Traditionall, however, fractions of degrees were Jones epressed & Bartlett in minutes Learning, and seconds, where 5 60 minutes (written 60r)* () and r 5 60 seconds (written 60s). () For eample, an angle of 7 degrees, 30 minutes, and 5 seconds is epressed as 7 30r5s. Jones & Bartlett Learning, Some calculators have a special DMS ke Jones for converting & Bartlett an angle Learning, given decimal FOR SALE OR degrees to Degrees, Minutes, and Seconds (DMS FOR notation), SALE and OR vice versa. The following eample shows how to perform these conversions b hand. EXAMPLE 08 CHAPTER 4 TRIGONOMETRIC FUNCTIONS Using () and () Convert: (a) 86.3 to degrees, minutes, FOR and seconds, SALE OR (b) 7 47r3s to decimal notation. Solution In each case we will use () and (). 3 (a) Since 0.3 represents 00 of and 5 60r, we have (0.3)(60r) r. 8 Now 3.8r5 3r 0.8r, so we must convert 0.8r to seconds. Since 0.8r represents 0 of r and r5 60s, we have 86 3r 0.8r5 86 FOR 3r SALE (0.8)(60s) OR r 48s. Hence, r48s. *The use of the number 60 as a base dates back to the Bablonians. Another eample of the use of this base in our culture Jones in the & Bartlett measurement Learning, of time ( hour 5 60 minutes and minute 5 60 seconds).

4 Jones & Bartlett Learning.. (b) Since 5 60r, Jones it follows & Bartlett that r5 ( Learning, 60). Similarl, s 5 ( 60)r5 ( 3600). Thus we have 7 47r3s r 3s ( 60) 3( 3600) < Jones & Bartlett Thus we Learning, see that 7 47r 3s < Radian Measure Another measure for angles is radian measure, which is generall used in almost all applications of trigonometr that involve calculus. The radian measure of an angle u is based on the length of an arc on a circle. If we place the verte of the angle u at the center of a circle of radius r, then u is called a central angle. As we know, an angle u in standard position can be viewed as having been formed b the initial side rotating from the positive -ais FOR to the SALE terminal OR side. The region formed b the initial and terminal sides with a central angle u is called a circular sector. As shown in FIGURE 4..6(a), if the initial side of u traverses a distance s along the circumference of the circle, then the radian measure of u is defined b Jones & Bartlett Learning, u5 s (3) r. In the case when the terminal side of u traverses an arc of length s along the circumference of the circle equal to the radius r of the circle, then we see from (3) that the measure of the angle u is radian. See Figure 4..6(b). Terminal side Terminal side r r r r θ s Initial side θ Initial side Jones & Bartlett Learning, r (a) (b) FIGURE 4..6 Central angle in (a); angle of radian in (b) The definition given in (3) does not depend on the size of the circle. To see this, all s' we need do is to draw another circle centered at the verte of u of radius rrand subtended arc length See FIGURE r' s θ sr Because the two circular sectors are similar the r ratios s/r and sr/rr are equal. Therefore, regardless of which circle we use, we obtain the Jones & Bartlett same radian Learning, measure for u. FOR SALE OR In equation (3) an convenient unit of length ma be FOR used for SALE s and OR r, but DISTRIBUT the same unit must be used for both s and r. Thus, FIGURE 4..7 Concentric s (units of length) circles u(in radians) 5 r (units of length) appears to be a dimensionless quantit. For eample, if s 5 4 in. and r 5 in., then the radian measure of the angle is u5 4 in. in. 5, where is simpl a real Jones number. & Bartlett This is the reason Learning, wh sometimes the word radians is omitted when an angle is measured FOR SALE in radians. OR We will come back to this idea in Section Angles and Their Measurement 09

5 Jones & Bartlett Learning.. One complete rotation of the initial side of u will traverse an arc equal in length to the circumference of the circle pr. It follows from (3) that one rotation 5 s r 5 pr 5 p radians. r We have the same convention as before: An angle formed b a counterclockwise rotation & is Bartlett considered Learning, positive, whereas an angle formed b a clockwise Jones & rotation Bartlett is nega- Learning, LL Jones FOR tive. SALE In FIGURE OR 4..8 we illustrate angles in standard position of p/, FOR p/, SALE p, OR and 3p DISTRIBUT radians, respectivel. Note that the angle of p/ radians shown in 4..8(a) is obtained b one-fourth of a complete counterclockwise rotation; that is 4 (p radians) 5 p Jones radians. & Bartlett Learning, The angle shown in Figure 4..8(b), obtained b one-fourth of a complete clockwise rotation, is p/ radians. The angle shown in Figure 4..8(c) is coterminal with the angle shown in Figure 4..8(d). In general, the addition of an integer multiple of p radians to an angle measured in radians results in a coterminal angle. Conversel, an two coter- angles measured in radians Jones will & Bartlett differ b an Learning, integer multiple of p. minal 3 FIGURE 4..8 Angles measured in radians EXAMPLE 3 A Coterminal Angle Find an angle between 0 and p radians that is coterminal with u5p/4 radians. Sketch the angle. Jones & Bartlett Learning, 3 Solution Since p,p/4, 3p, we subtract the equivalent of one rotation, or p FOR SALE 4OR radians, to obtain 4 FIGURE 4..9 Coterminal angles in Eample 3 Jones Alternativel, & Bartlett we Learning, can proceed as in part (a) of Eample Jones and divide: & Bartlett p/4learning, 5 LL FOR p 3p/4. SALE OR Thus, an angle of 3p/4 radians is coterminal with FOR u, as SALE illustrated OR DISTRIBUT in FIGURE Conversion Formulas Although man scientific calculators have kes that convert between degree and radian measure, there Jones is an & eas Bartlett wa to Learning, remember the relationship between the two measures. Since the circumference of a unit circle is p, one complete rotation has measure p radians as well as 360. It follows that p radians or 80 5p radians. (4) If we interpret (4) as 80( ) Jones 5p( & radian), Bartlett then Learning, we obtain the following two formulas for converting between degree FOR and radian SALE measure. OR p 4 0 CHAPTER 4 TRIGONOMETRIC FUNCTIONS (a) (b) (c) (d) p 5p 4 8p 4 5 3p 4.

6 Jones & Bartlett Learning.. Using a calculator to carr out the divisions in (5) and (6), we find that < radian Jones and & Bartlett radian < Learning, Although we will continue to use the FOR terms SALE radian OR and radians it is common to use the abbreviation rad for both words. EXAMPLE 4 Conversion Between Degrees and Radians Convert: (a) 0 to radians (b) 7p/6radians to degrees (c) radians to degrees. Solution (a) To convert from degrees to radians we use (5): 0 5 0( ) 5 0 # a p Jones 80 radianb 5 p & Bartlett Learning, LL FOR 9 radian. SALE OR DISTRIBUT (b) To convert from radians to degrees we use (6): 7p 6 radians 5 7p # Jones ( radian) & Bartlett 5 7p 6 6 a80 Learning, p b 5 0. (c) We again use (6): TABLE 4.. Conversion Jones Between & Bartlett Degrees Learning, and Radians FOR SALE OR 5 p 80 radian radian 5 a 80 (6) p b approimate answer rounded to two decimal places radians Jones 5 # ( & radian) Bartlett 5 Learning, # a 80. p b 5 a 360 p b < 4.59 Table 4.. provides the radian and degree measure of the most commonl used Degrees angles. p p p p Radians p FOR SALE OR Terminolog You ma recall from geometr that FOR a 90 SALE angle is OR called DISTRIBUT a right angle and a 80 angle is called a straight angle. In radian measure, p/ is a right angle and p is a straight angle. An acute angle has measure between 0 and 90 (or between 0 and p/ radians); and an obtuse angle has measure between 90 and 80 Jones & Bartlett Learning, (or between p/ and p radians). Two Jones acute & angles Bartlett are said Learning, to be complementar if their sum is 90 (or p/ radians). Two positive angles are supplementar if their sum is 80 (or p radians). The angle 80 (or p radians) is a straight angle. An angle whose terminal side coincides with a coordinate ais is called a quadrantal angle. For eample, 90 (or p/ radians) is a quadrantal angle. A triangle that contains a right angle is called a right triangle. The lengths a, b, and c of the sides of a right triangle satisf the Pthagorean Jones theorem & Bartlett a b Learning, 5 c, where c is the length of the side opposite the right angle (the FOR hpotenuse). SALE OR 4. Angles and Their Measurement (5)

7 Jones & Bartlett Learning.. EXAMPLE 5 Jones Complementar & Bartlett and Learning, Supplementar Angles (a) Find the angle that is complementar FOR SALE to u OR (b) Find the angle that is supplementar to f5p/3radians. FIGURE 4..0 Length of arc s determined b a central angle u Solution (a) Since two acute angles are complementar if their sum is 90, we find the angle that is complementar to u574.3 is 90 u (b) Since two positive angles are supplementar if their sum is p radians, we find the angle that is supplementar to f5p/3radians is Arc Length In man applications it is necessar to find the length s of the arc subtended b a central angle u in a circle of radius r. See FIGURE From the definition s r θ of radian measure given in (3), Jones & Bartlett r Learning, Jones & Bartlett Learning, u (in radians) 5 s r. B multipling both sides of the last equation b r we obtain the arc length formula s 5 ru. We summarize the result. THEOREM 4.. Arc Length Formula For a circle of radius r, a central angle of u radians subtends an arc of length EXAMPLE 6 CHAPTER 4 TRIGONOMETRIC FUNCTIONS pf5p p 3 5 3p 3 p 3 5 p 3 radians. Finding Arc Length Find the arc length subtended b a central angle of (a) radians in a circle of radius 6 inches, (b) 30 in a circle of radius feet. Solution (a) From the arc length formula (7) with radians and inches, we have s 5 ru 5 # u5 r So the arc length is inches. Students often appl the arc length formula incorrectl b using degree measure. Remember s 5 ru is valid onl if u is measured in radians. s 5 ru (7) (b) We must first epress 30 in radians. Recall that 30 5p/6radians. Then from the arc length formula (7) we have s 5 ru 5()(p/6) 5 p. So the arc length is Jones p & < Bartlett 6.8 feet. Learning, Area of a Circular Sector The area of, sa, a quarter circle is the fractional amount one-fourth of the total area pr, that is, the area is 4 pr. This reasoning carries over in finding the area of an circular sector. If u in radians is the central angle of the Jones & Bartlett Learning, circular sector shown in Figure 4..0, then Jones the area & ABartlett of the sector Learning, is simpl the fractional amount u/p of the total area of the circle. FOR Thus SALE OR A 5 u p (pr ) 5 r u. Note for the quarter circle eample, Jones & u5p/ Bartlett and Learning, the fractional amount of the total area is (p/)/p 5 4. We summarize FOR this SALE result in OR the net theorem.

8 Jones & Bartlett Learning.. THEOREM 4.. Jones Area & Bartlett of a Circular Learning, Sector For a circle of radius FOR r, the area SALE A of OR a circular sector with central angle u measured in radians is given b EXAMPLE 7 Eercises 4. Answers FOR SALE to selected OR odd-numbered problems begin on page ANS 4. A 5 (8) r u Area of a Circular Sector A 4-inch pizza is cut into 8 slices. Let us assume that the pizza is a perfect circle and Jones & Bartlett Learning, that the slices are eactl the same Jones size. Find & the Bartlett area of one Learning, slice. Solution One slice of pizza is a circular FOR sector SALE with radius OR r 5 7 inches. The central angle of the sector is 360 / We convert this central angle from degree measure to radian measure: # p 80 < p Jones & Bartlett Learning, 4 radian. FOR SALE A 4-inch OR pizza is usuall cut Then b (8) the area of FOR the sector, SALE or slice, OR is into eight slices (or pieces) Tobik/ShutterStock, Inc. A 5 # 7 # p # p p 8 in < 9.4 in. In Problems 6, draw the given angle in standard position. Bear in mind that the lack of a degree smbol ( ) in an angular measurement indicates that the angle is measured in radians FOR 3. SALE 35 OR p 5p 7p p 3 3. p Jones 4. 3p & Bartlett Learning, In Problems 7 0, epress the given angle in decimal notation r7s rs r r Jones & Bartlett In Problems Learning, 4, epress the given angle in terms Jones of degrees, & Bartlett minutes, and Learning, seconds. LL FOR SALE. OR FOR SALE OR DISTRIBUT In Problems 5 3, convert the given angle from degrees to radians Jones & Bartlett Learning, Jones 3. & Bartlett 30 Learning, In Problems 33 40, convert the given angle from radians to degrees. p p p 7p p 37. Jones 38. 7p & Bartlett Learning, Angles and Their Measurement 3

9 Jones & Bartlett Learning.. In Problems 4 44, find the angle between 0 and 360 that is coterminal with the given angle Find the angle between 360 and 0 that is coterminal with the angle in Problem 4. Jones 46. & Find Bartlett the angle Learning, between 360 and 0 that is coterminal Jones with the & angle Bartlett in Learning, LL FOR SALE Problem OR 43. In Problems 47 5, find the angle between 0 and p that is coterminal with the given angle. Jones & Bartlett Learning, 7p 47. 9p 48. FOR SALE 49. OR 5.3p p Find the angle between p and 0 radians that is coterminal with the angle in Problem Find the angle between p FOR and SALE 0 radians OR that is coterminal with the angle in Problem 49. In Problems 55 6, find an angle that is (a) complementar and (b) supplementar to Jones the & given Bartlett angle, Learning, or state wh no such angle can be found p p p 5p FOR SALE OR In Problems 63 and 64, find both the degree and the FOR radian SALE measures OR of the angle formed b the given rotation. Refer to Figures 4.. and three-fifths of a counterclockwise rotation 64. five and one-eighth clockwise rotations In Problems 65 68, find the measure FOR SALE of a central OR angle u in a circle of radius r that subtends an arc length s. Give u in (a) radians and (b) degrees. 65. r 5 5 ft, s ft 66. r 5 0 in, s 5 36 in 67. r 5 9 m, s 5 5 m 68. r 5 0 cm, s 5 90 cm In Problems 69 7, find the arc length s subtended b a central angle u in a circle of radius r. 69. u53 radians, r 5 5 in 70. u5.5 radians, r 5 4 cm Jones & Bartlett Learning, 7. u530, r 5 m 7. Jones u55, & Bartlett r 5 6 ft Learning, In Problems 73 76, find the area of the circular sector having the given radius r and central angle u. 73. r 5 3 ft, u57. radians 74. r 5 8 in, u5 p 3 radians 75. r 5 6 m, u530 FOR SALE 76. OR r 5 cm, u575 4 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

10 Jones & Bartlett Learning.. In Problems 77 and 78, the light red region in the given figure is portion of a circle. Find the area A of the region if u is measured in (a) radians, and (b) degrees R r r θ θ FIGURE 4.. Region in Problem 77 FIGURE 4.. Region in Jones & Bartlett Problem 78Learning, Applications 79. Analog Clock Consider the analog clock shown in FIGURE What are the Jones & Bartlett Learning, degree and the Jones radian measures & Bartlett of the Learning, angle between two adjacent hour tick marks on the clock FOR face? SALE OR What are the degree and the radian measures of the angle between two adjacent 9 3 minute tick marks on the analog clock face in Figure 4..3? What are the degree and the radian measures of the smallest positive angle 7 5 formed b the hands of the analog clock in Figure 4..3 at (a) 8:00, (b) :00, 6 FOR SALE OR and (c) 7:30? FIGURE 4..3 Analog clock in 8. What are degree and the radian measures of the angle through which the 3 Problems 79 8 minute hand on the analog clock in Figure 4..3 rotates in (a) 4 hour, and (b) 3.5 hours? 83. Planet Earth The Earth rotates on its ais once ever 4 hours. How long does it take the Earth to rotate through Jones an angle & Bartlett of (a) 40 Learning, and (b) p/6 radians? 84. Planet Mercur The planet Mercur FOR completes SALE OR one rotation on its ais ever 59 das. Through what angle (measured in degrees) does it rotate in (a) da, (b) hour, and (c) minute? 85. Angular and Linear Speed If we divide (7) b time t we get the relationship v 5 rv, where v 5 s/t is called the linear speed of a point on the circumference of a circle and v5u/tis called the angular speed of the point. A communications satellite is placed in a circular geosnchronous orbit 35,786 km above the surface of the Earth. The time it takes the satellite to make one full revolution Planet Mercur in Poblem 84 around the Earth is 3 hours, 56 minutes, 4 seconds and the radius of the Earth is 6378 km. See FIGURE Jones & Bartlett (a) Learning, What is the angular speed of the satellite in Jones rad/s? & Bartlett Learning, LL FOR SALE OR (b) What is the linear speed of the satellite in km/s? FOR SALE OR DISTRIBUT Satellite 86. Pendulum Clock A clock pendulum is.3 m long and swings back and forth along a 5-cm arc. Find (a) the central angle and (b) the area of the sector through which the pendulum sweeps in one swing. Jones & Bartlett Learning, 87. Sailing at Sea A nautical mile Jones is defined & Bartlett as the arc Learning, length subtended on the surface of the Earth b an angle of measure minute. If the diameter of the Earth is 797 miles, find how man statute (land) miles there are in a nautical FIGURE 4..4 Satellite in Problem 85 mile. 88. Circumference of the Earth Around 30 B.C.E. Eratosthenes calculated the circumference of the Earth from the following observations. At noon on the longest da of Jones the ear, & the Bartlett Sun was Learning, directl overhead in Sene, while it was inclined 7. from FOR the vertical SALE in OR Aleandria. He believed the two cities to be on Courtes of Mariner 0, Astrogeolog Team, and USGS 4. Angles and Their Measurement 5

11 Jones & Bartlett Learning.. Aleandria the same longitudinal line and assumed that the ras of the Sun are parallel. Thus 7. Ras he concluded that the arc from Sene to Aleandria was subtended b a central stades from angle of at the center FOR of the SALE Earth. See OR 7. FIGURE At that time the distance Sun from Sene to Aleandria was measured as 5000 stades. If one stade feet, Sene find the circumference of the Earth in (a) stades and (b) miles. Show that Earth Eratosthenes data gives a result that is within 7% of the correct value if the polar FIGURE 4..5 Earth in Jones & diameter Bartlett of Learning, the Earth is 7900 miles (to the nearest mile). Problem 88 FOR 89. SALE Circular OR Motion of a Yo-Yo A o-o is whirled around in FOR a circle SALE at the end OR of DISTRIBUT its 00-cm string. (a) If it makes si revolutions in 4 seconds, find its rate of turning, or angular speed, in radians per second. (b) Find the speed at which the o-o travels in centimeters per second; that is, its linear speed. 90. More Yo-Yos If there is a knot in the o-o string described in Problem 79 at a point 40 cm from the o-o, find (a) the angular speed of the knot and (b) the linear speed. 9. Locating a Cell Phone As shown in the accompaning photo, man cell phone antennas have a Jones triangular & Bartlett shape. The Learning, reception sectors a, b, and g corresponding to the three sides FOR of SALE the antenna OR are shown in FIGURE A cell Yo-Yo in FOR SALE Problems OR 89 and 90 tower with a triangular antenna is located at the common center of the circles in the figure; the circles have a radius mile, miles, 3 miles, and so on. Suppose a cell phone signal is detected in the b sector approimatel 5.3 miles from the antenna. Determine the area of the circular band where the cell phone is located bounded between radius 5 miles and radius 6 miles. (A more precise location of a cell phone can be obtained using either triangulation between three cell towers or GPS.) Monke Business Images/ShutterStock, Inc. b sector Triangular cell phone FIGURE 4..6 Reception sectors in Jones & Bartlett antenna Learning, Problem 9 9. Circular Motion of a Car Tire As shown in FIGURE 4..7 the diameter of a car tire is 6 inches. Suppose the car is driven.5 miles. What is the corresponding radian measure of the angle through which the tire turns? 93. Circular Motion of a Car Tire An automobile with 6-in. diameter tires is traveling at a rate of 55 mi/h. (a) Find the number of revolutions per minute that its tires are making. (b) Find the angular speed of its tires in radians per minute. 6 inches 94. Diameter of the Moon The average distance from the Earth to the Moon as FIGURE 4..7 Car tire in given b NASA is 38,855 miles. If the angle subtended b the Moon at the ee Jones Problems & Bartlett 9 and Learning, 93 of an observer on Earth Jones is 0.5, & then Bartlett what is Learning, the approimate diameter of the Moon? FIGURE 4..8 is not FOR to scale. SALE OR Aleksangel/ShutterStock, Inc. 6 CHAPTER 4 TRIGONOMETRIC FUNCTIONS Darrl Brooks/ShutterStock, Inc. g sector a sector

12 Jones & Bartlett Learning.. Jones & Bartlett Learning, 38,855 mi 0.5º Observer on Earth Moon FIGURE 4..8 The curved red arc represents FOR the SALE OR DISTRIBUT approimate diameter of the Moon For Discussion Jones & Bartlett Learning, 95. Intersection of Circles Each Jones of the circles & Bartlett in FIGURE Learning, 4..9 has its center on an ais, passes through the origin, and FOR has radius SALE r. OR (a) Construct a right triangle in the figure and then use that triangle to epress the area A of the intersection of the circles, the ellow region in the figure, as a function of r. (b) Find the area Jones of the & intersection Bartlett Learning, of the circles FOR ( SALE 5) 5 5 OR and ( 5) 5 5. FIGURE 4..9 Intersecting circles in Problem The Sine FOR and SALE Cosine OR Functions t cos t P(t) = (cos t, sin t) sin t Jones & Bartlett + = Learning, FIGURE 4.. Coordinates of P(t) are (cost, sint) INTRODUCTION Originall, the trigonometric functions were defined using angles in right triangles. A more modern approach, and one that is used in calculus, is to define the trigonometric functions on sets Jones of real numbers. & Bartlett As we Learning, will see, the radian measure for angles is ke in making these definitions. Trigonometric Functions For each real number t there corresponds an angle of t radians in standard position. As shown in FIGURE 4.. we denote the point of intersection of the terminal side of the angle t with the unit circle b P(t). The and coordinates of this point Jones give us & the Bartlett values of Learning, the si basic trigonometric functions. The -coordinate of P(t) is FOR called SALE the sine OR of t, while the -coordinate of P(t) is called the cosine of t. Jones & Bartlett DEFINITION Learning, 4.. Sine and Cosine Functions Jones & Bartlett Learning, LL FOR SALE Let OR t be an real number and P(t) 5 (, ) be the point FOR of intersection SALE OR of DISTRIBUT the unit circle with the terminal side of the angle of t radians in standard position. Then, the sine of t, denoted sin t, and the cosine of t, denoted cos t, are sin t 5 () and cos t 5 () Since to each real number t there corresponds a unique point P(t) 5 (cos t, sin t), we have just defined two functions the sine function and the cosine function each with domain the set Jones R of real & Bartlett numbers. Learning, Four additional trigonometric functions are defined in terms of the FOR coordinates SALE of OR P(t) 5 (, ). 4. The Sine and Cosine Functions 7

13 Jones & Bartlett Learning.. DEFINITION 4.. Jones Tangent, & Bartlett Cotangent, Learning, Secant, and Cosecant FOR SALE Functions OR The tangent, cotangent, secant, and cosecant functions of the real number t are tan t 5, 0 cot t 5, 0 (3) (4) sec t 5 (5), 0 and csc t 5 Jones & Bartlett Learning, (6), FOR 0 SALE OR Using sin t 5 and cos t 5 in (3) (6) of Definition 4.. we obtain the impor- identities: tant FOR tan t 5 sin SALE t OR cot t 5 cos t (7) cos t sin t sec t 5 csc t 5 (8) cos t sin t. Jones & Bartlett Learning, LL FOR Because SALE of OR the role plaed b the unit circle in Definitions 4.. FOR and SALE 4.., OR the si DISTRIBUT trigonometric functions are referred to as the circular functions. For the remainder of this section and the net we are going to eamine the sine and cosine functions in detail. We will come back to the tangent, cotangent, secant, and Jones & Bartlett Learning, cosecant functions in Section 4.4. A number of properties of the sine and cosine functions follow from the fact that P(t) 5 (cos t, sin t) lies on the unit circle. For instance, the coordinates of P(t) must satisf the equation of the circle: 5. Substituting 5 cos t and Jones 5 sin t & gives Bartlett an important Learning, relationship between the sine and the cosine called the Pthagorean FOR identit: SALE OR (cos t) (sin t) 5. From now on we will follow two standard practices in writing this identit: (cos t) and (sin t) will be written as cos t and sin t, respectivel, and the sin t term will be written first. (9) THEOREM 4.. For all real numbers t, Pthagorean Identit sin Jones t cos & Bartlett Learning, t 5 (0) Again, if P(, ) denotes a point on the unit circle (9), it follows that the coordinates of P must satisf the inequalities # # and # #. Because 5 cos t and 5 sin t we Jones have the & following Bartlett bounds Learning, on the values of the sine and cosine functions. 8 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

14 Jones & Bartlett Learning.. THEOREM 4.. Jones Bounds & Bartlett on Learning, the Values of Sine and Cosine For all real numbers t, FOR SALE OR # sin t # and # cos t # Jones & Bartlett The foregoing Learning, inequalities can also be epressed as Jones 0 sin t 0 #& Bartlett and 0 cos tlearning, 0 #. Thus, LL (0, ) FOR SALE for eample, OR there is no real number t such that sin t 5 FOR 3. SALE OR DISTRIBUT II I sin t > 0 sin t > 0 Domain and Range From the preceding observations we have the sine and cosine cos t < 0 cos t > 0 functions f (t) 5 sin t and g(t) 5 cos t each with domain the set R of real numbers and range the interval [, ]. (, 0) Jones (, 0) sin t < 0 & sin Bartlett t < 0 Learning, cos FOR t < 0 cos SALE t > 0 OR Signs of the Circular Functions FOR The signs SALE of the OR function values sin t and cos t III IV are determined b the quadrant in which the point P(t) lies, and conversel. For eample, if sin t and cos t are both negative, then the point P(t) and terminal side of the corresponding angle of t radians must lie in quadrant III. FIGURE 4.. displas the signs (0, ) FIGURE 4.. Algebraic signs of sin t of the cosine and sine functions in each of the four quadrants. Jones & and Bartlett cos t in the Learning, four quadrants EXAMPLE FOR Using SALE the OR Pthagorean Identit Given that cos t 5 3 and that P(t) is a point in the fourth quadrant, find sin t. Solution Substitution of cos t 5 into the Pthagorean identit (0) gives sin t A 3B 5 or sin t (cos 3, sin 3) 3 radians Since sin t is the -coordinate Jones & Bartlett of P(t), a Learning, point the LL FOR SALE fourth OR quadrant, we must take the negative square root for FOR sin tsale : OR DISTRIBUT (, 0) 8 sin t 5 Å 9 5! 3. Jones & Bartlett Learning, EXAMPLE Sine and Jones Cosine & Bartlett of a Real Learning, Number FIGURE 4..3FOR The point SALE P(3) OR Use a calculator to approimate sin 3 and cos 3 and give a geometric interpretation of these values. P( ) = (cos, sin ) Solution From a calculator set in radian mode, we obtain cos 3 < and P( = (0, ) ) = P(0) = sin 3 < These values represent the - and -coordinates, respectivel, of the Jones &(cos Bartlett, sin ) Learning, (cos 0, sin 0) point of intersection of the terminal side of the angle of 3 radians in standard position FOR SALE = (, 0) OR = (, 0) with the unit circle. As FOR shown SALE in OR FIGURE 4..3, this point lies in the second quadrant P( ) = (cos, sin ) because p/, 3,p. This would also be epected in view of Figure 4.. since cos 3, 3 P( ) = the -coordinate, is negative and sin 3, the -coordinate, is positive. (cos 3, sin 3 ) = (, 0) = (0, ) Values Corresponding to Unit Circle Intercepts As shown in FIGURE Jones & Bartlett Learning, 4..4, the LL FIGURE 4..4 Sine and cosine values FOR SALE - and OR -intercepts of the unit circle give us the values of FOR the sine SALE and cosine OR DISTRIBUT functions for quadrantal angles for the real numbers corresponding to quadrantal angles listed net. Values of the Sine and Cosine For t 5 0: sin and cos 0 5 For t 5 p sin p and cos p : For t 5p: sin p50 and cos p5 3p For t 5 3p sin 3p Jones and cos : & Bartlett 5 Learning, The Sine and Cosine Functions 9

15 Jones & Bartlett Learning.. Periodicit In Section Jones 4. we & saw Bartlett that for Learning, an real number t, the angles of t radians and t 6 p radians are coterminal. Thus the determine the same point (, ) on the unit circle. Therefore sin t 5 sin(t 6 p) In other words, the sine and cosine functions repeat their values ever p units. It also Jones follows & Bartlett that for Learning, an integer n: sin(t np) 5 sin t and cos(t np) 5 cos t. () DEFINITION 4..3 for ever real number t. FOR SALE OR Even Odd Properties The smmetr of the unit circle endows the circular functions with several additional properties. For an real number t, the points P(t) and P(t) = (cos t, sin t) P(t) on the unit circle are located on the terminal side of an angle of t and t radians, respectivel. These two points will alwas be smmetric with respect to the -ais. t Jones FIGURE & Bartlett 4..5 illustrates Learning, the situation for a point P(t) ling Jones in the& first Bartlett quadrant: Learning, LL t FOR TheSALE -coordinates OR of the two points are identical; however, the FOR -coordinates SALE OR have DISTRIBUT equal magnitudes but opposite signs. The same smmetries will hold regardless of which quadrant contains P(t). Thus, for an real number t, cos(t) 5 cos t and P( t) = (cos ( t), sin ( t)) sin(t) 5sin t. Appling the definitions of even and odd functions from Section. we have the following result. FIGURE 4..5 Coordinates Jones of & P(t) Bartlett Learning, and P(t) THEOREM 4..4 Even and Odd Functions The cosine function is even and the sine function is odd. That is, for ever real number t, cos(t) Jones 5 cos & t Bartlett and Learning, sin(t) 5 sin t (5) 0 CHAPTER 4 TRIGONOMETRIC FUNCTIONS Periodic Functions and cos t 5 cos(t 6 p). Jones & Bartlett Learning, A nonconstant function f is said to be periodic Jones if there & Bartlett is a positive Learning, number p such that f (t) 5 f (t p) (3) for ever t in the domain of f. If p is the smallest positive number for which (3) is true, then p is called the period of the function f. The equations in () impl that the sine and the cosine functions are periodic with period p # p. To see that the period of sin t is actuall p, we observe that there is onl one point on the unit circle with -coordinate, namel, P(p/) 5 (cos(p/), sin(p/)) 5 (0, ). Therefore, sin t 5 onl for t 5 p, p Jones 6 p, p & Bartlett Learning, LL 6 FOR 4p, SALE OR DISTRIBUT and so on. Thus the smallest possible positive value of p is p. Jones & Bartlett Learning, THEOREM 4..3 Period of the Jones Sine and & Bartlett Cosine Learning, The sine and cosine functions are periodic with period p. Therefore, sin(t p) 5 sin t and cos(t p) 5 cos t () (4)

16 Jones & Bartlett Learning.. The following additional properties of the sine and cosine functions can be verified b considering the smmetries of appropriatel chosen points on the unit circle. cos(p t) 5cos t and sin(p t) 5 sin t (8) P t = cos ( For eample, to justif the properties in (6) of Theorem 4..5 for 0, t,p/, ( ( sin t, ( t consider FIGURE Since the points P(t) and P(p/ t) are smmetric with respect to the line 5, we can obtain the coordinates of P(p/ t), b interchanging the coordinates of P(t). Thus, = Jones & Bartlett cos t 5 sina p Learning, P(t) = (cos t, sin t) FOR SALE tb OR and sin t 5 cosap tb. (, 0) The special properties of the sine and cosine functions in Theorem 4..5 will become quite useful as soon as we determine additional values for sin t and cos t in the interval Jones & Bartlett [0, p). Learning, Using results from plane geometr we will Jones now find & the Bartlett values of Learning, the sine and LL FIGURE 4..6 Geometric justification FOR SALE cosine OR functions for t 5p/6, t 5p/4, and t 5p/3. of (6) in Theorem 4..5 Finding sin(p/4) and cos(p/4) We draw an angle of p/4 radians (45 ) in standard position and locate and label P(p/4) 5 (cos(p/4), sin (p/4)) on the unit circle. As shown in FIGURE Jones & Bartlett 4..7, weformaright triangle b dropping a perpendicular from Learning, P( 4) = (cos 4, sin 4) P(p/4) to the -ais. Since the sum of the angles in an triangle is p radians (80 ), the third angle of this triangle is also p/4 radians, hence the triangle is isosceles. Therefore the coordinates of P(p/4) are equal; that is, cos(p/4) 5 sin(p/4). It follows from the 4 Pthagorean identit (0) ( ( THEOREM 4..5 Additional Properties Jones & Bartlett Learning, cosap Jones tb 5 sin t and sina p & Bartlett Learning, LL tb 5 cos t (6) FOR SALE OR DISTRIBUT ( ( cos(t p) 5cos t and sin(t p) 5sin t (7) Jones sin p 4 & cos Bartlett p 4 5 Learning, that p cos 4 5. Dividing b and taking the square root, we obtain cos(p/4) 56!/. Since P(p/4) FIGURE 4..7 The point P(p/4) lies in the first quadrant, both coordinates must be positive. So we have found the (equal) coordinates of P(p/4): p cos 4 5! Jones p and sin 4 5! & Bartlett Learning, LL FOR SALE. OR DISTRIBUT Finding sin(p/6) and cos(p/6) We construct two angles of p/6 radians (30 ) in the first and fourth quadrants, as shown in FIGURE 4..8, and label the points of intersection P Jones & Bartlett Learning, with the unit circle P(p/6) and Q, respectivel. Jones & Bartlett B drawing Learning, perpendicular line segments FOR SALE 6 OR from P and Q to the -ais, we obtain two FOR congruent SALE right OR triangles because each triangle has a hpotenuse of length and angles of 30, 60, and 90. Since the 90 angles form O 6 Q a straight angle, these two right triangles form an equilateral triangle ^POQ with sides of length. Since sin(p/6) is equal to half of the vertical side of ^POQ, we have Jones & Bartlett plearning, sin 6 5 FOR SALE. FIGURE OR 4..8 The point P(p/6) 4. The Sine and Cosine Functions

17 Jones & Bartlett Learning.. From this result and the Pthagorean identit (0) we find the value of cos(p/6): a FOR b cos p SALE OR 6 5 implies p cos p or cos 6 5!3. lies in the first quadrant. Finding sin(p/3) and cos(p/3) We draw angles of p/6 and p/3 in standard P( 3 position and locate and label the points P(p/6) and P(p/3), as shown in FIGURE P We then construct two congruent triangles b dropping perpendiculars to 3 ( 6 the - and -aes, respectivel. It follows from the congruence of these triangles that Jones & Bartlett Learning, 6 FOR SALE OR p cos 3 5 sin p 6 5 p and FOR sin SALE 3 5 cos por 6 5!3. We take the positive square root here because P(p/6) ( ( The foregoing results also follow from (6) of Theorem 4..5 with t 5p/6. We summarize the values of the sine and cosine functions corresponding to the basic fractional multiples of Jones p that we & Bartlett have determined Learning, so far. FIGURE 4..9 The point P(p/3) Values of the Sine and Cosine (Continued) p p For t 5 p sin and cos 6 5!3 6 5 Jones & Bartlett Learning, 6 : Jones & Bartlett Learning, LL p p For t 5 p sin and cos 4 5! 4 5! 4 : p p For t 5 p sin and cos !3 3 : Reference Angle As we noted at the beginning of this section, for each real number t there is a unique angle of t radians in standard position that determines the point P(t), with coordinates (cos t, sin t), on the unit circle. As shown in FIGURE 4..0, the terminal side of an angle of t radians (with P(t) not on an ais) will form an acute angle with the -ais. We can Jones then locate & Bartlett an anglelearning, of tr radians in the first quadrant that is congruent to this acute angle. FOR TheSALE angle ofor tr radians is called the reference angle for t. Because of the smmetr of the unit circle, the coordinates of P(tr) will be equal in absolute value to the respective coordinates of P(t). Hence sin t 56 sin t r and cos t 56 cos t r. As the following eamples will show, reference angles can be used to find the trigonometric function values of an integer multiple of p/6, p/4, and p/3. P(t) P(t ) t P(t ) t t t t t P(t ) P(t) FIGURE 4..0 Reference angle FOR tr is an SALE acute angle OR P(t) CHAPTER 4 TRIGONOMETRIC FUNCTIONS

18 Jones & Bartlett Learning.. Jones & Bartlett Learning, P 3 3 ( = (, ( 5 3 P =, 3 FIGURE 4.. Reference angle in part (a) of Eample 3 ( Jones & Bartlett given value Learning, of t. ( ( FOR SALE OR ( ( ( P =, 4 ( ( ( 5p sin 3 5sin p 3 5!3 5p and cos 3 5 cos p 3 5. Jones & Bartlett Learning, (b) The point P(3p/4) lies in the third quadrant and has reference angle p/4, as shown in FIGURE 4... Therefore, P( 3 = (, 4 FIGURE 4.. Reference angle in 3p sina part (b) of Eample 3 4 b 5sin p 4 5! 3p and cosa 4 b 5cos p 4 5! Jones & Bartlett Learning,. LL ( ( EXAMPLE 3Jones & Using Bartlett Reference Learning, Angles Find eact values of sin FOR t and SALE cos t for OR the given real number t. (a) t 5 5p/3 (b) t 53p/4 Solution In each part we begin b finding the reference angle corresponding to the (a) From FIGURE 4.. we find that an angle of t 5 5p/3 radians determines a point P(5p/3) in the fourth quadrant and has the reference angle tr5 p/3 radians. After adjusting the signs of the coordinates of P(p/3) 5 (/,!3/) to obtain the fourth quadrant point P(5p/3) 5 (/,!3/), we find that FOR reference SALE angle OR T T Sometimes, in order to find the trigonometric values of multiples of our basic fractions of p we must use periodicit or the even-odd function properties in addition Jones & Bartlett Learning, to reference numbers. EXAMPLE 4 Using Periodicit and a Reference Angle Find eact values of Jones sin t and & cos Bartlett t for t 5Learning, 9p/6. Solution Since 9p/6 is greater than p, we rewrite 9p/6 as an integer multiple of p plus a number less than p: 9p 6 5 4p 5p 6 5 (p) Jones 5p 6. & Bartlett Learning, LL 3 From the periodicit equations () with n 5 and t 5 5p/6 we know that 5 3 P( = (, P( = (, 6 sin(9p/6) 5 sin(5p/6) and cos(9p/6) 5 cos(5p/6). Net we see from FIGURE that the reference angle for 5p/6 is p/6. Since P(5p/6) is a second quadrant Jones & Bartlett Learning, 6 point, we have ( ( ( ( 9p sin 6 5 sin 5p 6 5 sin p 6 5 Jones & FIGURE Bartlett 4..3 Learning, Reference angle in FOR Eample SALE 4OR 9p andcos Jones & Bartlett 6 5 cos 5pLearning, 6 5cos p 6 5!3. 4. The Sine and Cosine Functions 3

19 Jones & Bartlett Learning.. EXAMPLE 5 Jones Using & the Bartlett Even Odd Learning, Properties Find eact values of sin t and cos FOR t for SALE t 5p/6. OR Solution Since sine is an odd function and cosine is an even function, See (5) in Theorem p sina 6 b 5sin p 6 5 p and cosa 6 b 5 cos p 6 5!3 Jones & Bartlett. Learning, LL This problem could also have been solved b using a reference angle. Trigonometric Functions of Angles In this section we have defined sine and cosine functions of the real number t b using the coordinates of a point P(t) on the unit circle. It is now possible to define the trigonometric functions of an angle u. For FOR SALE OR an angle u, we simpl let where the real number t is the radian measure of u. As mentioned in Section 4., it is common to omit the word radians Jones when & Bartlett measuring Learning, an angle. So we write sin(p/6) for both the sine of the real number FOR p/6 SALE and for OR the sine of the angle of p/6 radians. Furthermore, since the values of the trigonometric functions are determined b the coordinates of the point P(t) on the unit circle, it reall does not matter whether u is measured in radians or in degrees. For eample, regardless of whether we are given u5p/6 radians or u530, the point on the unit circle corresponding to this angle in standard position is A!3/, /B. Thus, p sin 6 5 sin CHAPTER 4 TRIGONOMETRIC FUNCTIONS sin u 5sin t and cos u 5cos t, p!3 and cos 5 cos Eercises 4. Answers to selected odd-numbered problems begin on page ANS 4.. Given that cos t 5 and that is a point in the second quadrant, find. Given that sin t 5 5 P(t) sin t. 4 and that P(t) is a point in the second quadrant, find cos t. 3. Given that sin t 5 and Jones that P(t) & Bartlett is a point Learning, the third quadrant, find 4. Given that cos t cos t. and that is a point in the fourth quadrant, find 5. If sin t 5 find all possible values of 6. If cos t 5 3 7, 4 FOR P(t) SALE OR sin t. cos t. 0, find all possible values of sin t. 7. If cos t 50., find all possible values of sin t. Jones 8. & If Bartlett sin t 5 0.4, Learning, find all possible values of cos t. 9. If sin t cos t 5 0, find all possible values of sin t and cos t. 0. If 3 sin t cos t 5 0, find all possible values of sin t and cos t. In Problems 4, find the eact value of (a) sin t and (b) cos t for the given value of t. Do not use a calculator. Jones & Bartlett Learning,. t 5p/. Jones t 5 3p & Bartlett Learning, 3. t 5 8p 4. FOR t 53p/ SALE OR In Problems 5 6, for the given value of t determine the reference angle tr and the eact values of sin t and cos t. Do not use a calculator. 5. t 5 p/3 Jones 6. t 5& 4p/3 Bartlett Learning, 7. t 5 5p/4 8. t 5 3p/4 9. FOR t 5 SALE p/6 OR 0. t 5 7p/6

20 Jones & Bartlett Learning... t 5p/4 Jones &. Bartlett t 57p/4 Learning, 3. t 55p/6 4. t 5p/6 5. t 55p/3 6. t 5p/3 In Problems 7 3, find the given trigonometric function value. Do not use a calculator. 7. sin(p/3) 8. cos(7p/6) 9. cos(7p/4) 30. sin(9p/) 3. cos 5p 3. sin(3p/3) In Problems 33 38, justif the given statement with one of the properties of the trigonometric functions. 33. sin p5sin 3p 34. cos(p/4) 5 sin(p/4) Jones & Bartlett Learning, 35. sin(3 p) 5sin(3 p) Jones 36. & Bartlett cos 6.8pLearning, 5cos 4.8p 37. cos cos(0.43) FOR 38. SALE sin(p/3) OR 5 sin(p/3) In Problems 39 46, find the given trigonometric function value. Do not use a calculator. 39. sin 35 Jones & Bartlett Learning, 40. cos cos 0 4. sin cos sin(80 ) 45. sin(60 ) 46. cos(300 ) Jones & Bartlett In Problems Learning, 47 50, find all angles t, where 0 # t, p, Jones that satisf & Bartlett the given Learning, condition. LL FOR SALE 47. OR sin t cos t 5 FOR SALE OR DISTRIBUT 49. cos t 5!/ 50. sin t 5 In Problems 5 54, find all angles u, where 0 #u,360, that satisf the given condition. 5. cos u5!3/ FOR 5. SALE sin u5 OR 53. sin u5!/ 54. cos u5 Applications 55. Free Throw Under certain conditions the maimum height attained b a basketball released from a height h at an angle u measured from the horizontal with an initial velocit v is given b 5 h (v 0 0 sin u)/g, where g is the acceleration due to gravit. Compute the maimum height reached b a free throw if h 5.5 m, v u564.47, and g m/s m/s,. Jones & Bartlett 56. Putting Learning, the Shot The range of a shot put released Jones from & a height Bartlett h above Learning, the LL FOR SALE OR ground with an initial velocit v 0 at an angle u to the FOR horizontal SALE can OR be approimated DISTRIBUT b Pétur Ásgeirsson/ShutterStock, Inc. R 5 v 0 cos u (v g 0 sin u"v 0 sin ugh), Free throw where g is the acceleration due to gravit. If v m/s, u540, and g m/s, compare the ranges achieved FOR SALE for the OR release heights (a) h 5.0 m and (b) h 5.4 m. (c) Eplain wh an increase in h ields an increase in R if the other parameters are held fied. (d) What does this impl about the advantage that height gives a shot putter? 57. Acceleration Due Jones to Gravit & Bartlett Because Learning, of its rotation the Earth bulges at the equator and is flattened FOR at the poles. SALE As OR a consequence, the acceleration due to gravit is 4. The Sine and Cosine Functions 5

21 Jones & Bartlett Learning.. North pole not a constant 980 cm/s, but varies with latitude. As shown in FIGURE 4..4, the latitude of a point on the Earth is an angle f measured (usuall in degrees, minutes, Jones & Bartlett 90 N Learning, 75 N seconds) north (N) or south (S) from the equatorial plane. Based on satellite studies, a mathematical model for the acceleration due to gravit is given 60 N b Latitude 45 N 30 S Prime meridian 30 N 5 E 0 W Longitude 5 N 05 W90 W 30 W5 W 75 W 60 W 45 W Equator 5 S 0 longitude 0 latitude g lat sin f sin f. Jones &(a) Bartlett Find g lat Learning, at Meico Cit, Meico (9.4 N), Los Angeles, Jones CA &(34.05 Bartlett N), Learning, LL FOR SALE New York OR Cit, NY (40.70 N), and Fairbanks, AK (64.83 FOR N). SALE OR DISTRIBUT (b) At what latitude is a minimum? A maimum? South pole For Discussion 90 S Jones & Bartlett Learning, 58. Discuss how it is possible to determine Jones without & a calculator Bartlett that Learning, the point FIGURE 4..4 Latitude in Problem 57 P(6) 5 (cos 6, sin 6) lies in the fourth quadrant. FOR SALE OR 59. Discuss how it is possible to determine without the aid of a calculator that both sin 4 and cos 4 are negative. 60. Is there a real number t satisfing 3 sin t 5 5? Eplain wh or wh not. 6. Is there an angle u satisfing Jones cos & u Bartlett 5? Eplain Learning, wh or wh not. 6. Suppose f is a periodic function with period p. Show that F() 5 f (a), a. 0, is periodic with period p/a. g lat g lat 4.3 Graphs of Sine and Cosine Functions INTRODUCTION One wa to further our understanding of the trigonometric functions is to eamine their graphs. In this section we consider the graphs of the sine and cosine functions. Graphs of Sine and Cosine In Section FOR 4. SALE we sawor that the domain of the sine function f (t) 5 sin t is the set of real numbers (`, `) and the interval [, ] is its range. Since the sine function has period p, we begin b sketching its graph on the interval [0, p]. We obtain a roughsketchofthegraph given in FIGURE 4.3.(b) b considering various positions of the point P(t) on the unit circle, as shown in Figure 4.3.(a). As t varies Jones from 0 to & p/, Bartlett the value Learning, sin t increases from 0 to its maimum value. But as t variesfor from p/ SALE to 3p/, OR the value sin t decreases from to its minimum value. We note that sin t changes from positive to negative at t 5p. For t between 3p/ and p, we see that the corresponding values of sin t increase from to 0. The graph of an periodic function over an interval of length equal ( 5 P 4 6 CHAPTER 4 TRIGONOMETRIC FUNCTIONS ( P ( ( P 4 ( ( ( 3 P ( 4 f(t) = sin t, 0 t (a) Unit circle Jones &(b) Bartlett One ccle of Learning, sine graph FIGURE 4.3. Points P(t) on a circle corresponding to points on the graph Jones & Bartlett Learning, t

22 Jones & Bartlett Learning.. to its period is said to be one ccle of its graph. In the case of the sine function, the graph over the interval [0, p] in Figure 4.3.(b) is one ccle of the graph of f (t) 5 sin t. Note: Change of smbols From this point on we will revert to the traditional smbols and when graphing trigonometric functions. Thus, f (t) 5 sin t will either be written f () 5 sin or simpl 5 sin. Jones & Bartlett TheLearning, graph of a periodic function is easil obtained Jones b repeatedl & Bartlett drawing Learning, one ccle LL FOR SALE of its OR graph. In other words, the graph of 5 sin on, sa, FOR thesale intervals OR [p, DISTRIBUT 0] and [p, 4p] is the same as that given in Figure 4.3.(b). Recall from Theorem 4..4 of Section 4. that the sine function is an odd function since f () 5 sin() 5sin 5f (). In other words, if (, ) is a point on the graph of f, then so is (, ). Thus, from Theorem.. it follows that the graph of 5 sin shown in FIGURE 4.3. is smmetric with respect to the origin. = sin FIGURE 4.3. Graph of 5 sin One ccle B working again with the unit circle we can obtain one ccle of the graph of the cosine function g() 5 cos on the interval [0, p]. In contrast to the graph of f () 5 sin where for the cosine function we have FIGURE Jones & Bartlett Learning, f (0) 5 f (p) 5 0, Jones & Bartlett g(0) 5 Learning, g(p) 5. shows one ccle (in red) of 5 cos on [0, p] FOR along SALE with the OR etension of that ccle (in blue) to the adjacent intervals [p, 0] and [p, 4p]. We see from this figure that the graph of the cosine function is smmetric with respect to the -ais. This is a consequence of g being an even function: g() 5 cos() 5 cos 5 g(). = cos One ccle FIGURE Graph of 5 cos Intercepts In this and subsequent courses in mathematics it is important that ou know the -coordinates of the -intercepts of the sine and cosine graphs in other words, the zeros of f () 5 sin and g() 5 cos. From the sine graph in Figure 4.3. we see that the zeros of the sine function, or the numbers for which sin 5 0, are FOR SALE 5 0, 6p, OR 6p, 63p, c. 4.3 Graphs of Sine and Cosine Functions 7

23 Jones & Bartlett Learning.. These numbers are integer multiples of p. From the cosine graph in Figure we see that cos 5 0 when 56p/, 63p/, 65p/, c. These numbers are odd-integer multiples of p/. We summarize the foregoing discussion. Properties of the Sine and Cosine Functions The domain of f () 5 sin and of g() 5 cos is the set of all real numbers (`, `). The range of f () 5 sin and of g() 5 cos is the interval [, ] on the -ais. The period of f () 5 sin and of g() FOR 5 cos SALE is the OR number p. The sine function f is an odd function, and so its graph is smmetric with respect to the origin. The cosine function g is an even function, and so its graph is smmetric with respect to the -ais. The zeros of the sine FOR function SALE f are OR the numbers 8 CHAPTER 4 TRIGONOMETRIC FUNCTIONS Note in () that if n is an integer, then n is an odd integer. Using the distributive law, the zeros of the cosine function are often written as 5p/ np. Jones & Bartlett Learning, Variation of the Graphs As we did in Jones Chapters & Bartlett and 3 we Learning, can obtain variations FOR SALE OR of the basic sine and cosine graphs through rigid FOR and nonrigid SALE OR transformations. For the remainder of the discussion in this section we will consider graphs of functions of the form 5 A sin(b C) D or 5 A cos(b C) D, (3) where A, B, C, and D are real constants. Graphs of 5 A sin and 5 A cos We begin b considering the special FOR SALE = sin OR cases of (3): 5 A sin and 5 A cos. 5 np, n 5 0, 6, 6, c. The zeros of the cosine function g are the numbers Jones & Bartlett Learning, 5 (n )p/, n 5 0, 6, 6, c. Jones & Bartlett () Learning, LL The multiple A can be either positive or negative, but does not affect the period of the 3 function; in other words, the period of both 5 A sin and 5 A cos is p. For 0A0. graphs of these functions can be interpreted as a vertical stretch of the graphs = sin FOR SALE OR of 5 sin or 5 cos ; when 0, 0A0, the graphs are a vertical compression of the basic sine or cosine graphs. The number 0A0 is called the amplitude of the functions or of their graphs. For eample, as FIGURE shows we obtain the graph of FIGURE Vertical stretch of the graph of 5 sin 5 sin b stretching the graph of 5 sin verticall b a factor of. Note that the Jones & Bartlett Learning, maimum and minimum values of 5 sin Jones occur & at Bartlett the same -values Learning, as the maimum and minimum values of 5 sin. In general, FOR the maimum SALE OR distance from an point FOR SALE OR on the graph of 5 A sin or 5 A cos to the -ais is 0A0. The amplitude of the basic functions 5 sin and 5 cos is 0 A 0 5. In general, if a periodic function f is continuous, then over a closed interval of length equal to its period, f has both a maimum value M and a minimum value Jones m. The & Bartlett amplitude Learning, is defined b amplitude FOR SALE 5 OR [M m]. (4) ()

24 Jones & Bartlett Learning.. Jones & Bartlett Learning, = cos = cos 3 FIGURE Graph of function in Eample As the net eample shows, when A, 0 the graphs are also reflected in the -ais. EXAMPLE Verticall Compressed/Reflected Graph Graph 5 cos. Solution With the identification A 5 we see that the amplitude of the function is 0A0 5 0 Because and is negative, the graph of A0, A cos is the FOR SALE graph OR of. 5 cos compressed verticall b a factor of FOR and then SALE reflected OR in DISTRIBUT the -ais. The graph of 5 cos on the interval [0, p] is shown in red in FIGURE Jones & Bartlett Learning, Vertical Stretch/Compression The graphs of the functions 5 A sin and 5 A cos have amplitude 0A0 and period p. The graphs of 5 A sin and 5 A cos are the graphs Jones of & 5 Bartlett sin and Learning, 5 cos stretched verticall if 0A0., and compressed verticall if 0, 0A0,. FIGURE Graph of function in Eample Graphs of 5 A sin D and 5 A cos D The graphs of 5 A sin D and 5 A FOR cos SALE D OR DISTRIBUT are the graphs of 5 A sin and 5 A cos shifted verticall, up for D. 0 and down for D, 0. The amplitude of the graph of either 5 A sin D or 5 A cos D is still 0 A 0. EXAMPLE Verticall Shifted Sine Graph Graph 5 sin. Solution The graph of 5 sin is the graph of 5 sin given in Figure shifted up unit. From the identification A 5, the amplitude is 0A0 5. Because the maimum and minimum FOR SALE of OR 5 sin occur, respectivel, at 5p/ and = + sin 5 3p/, a rigid upward translation of the graph does not change the latter numbers but increases the maimum and minimum b unit. We see in FIGURE that the maimum of 5 sin is 3 at 5p/and the minimum of 5 sin is at Jones 5 3p/. Using the function values M 5 3 and m 5, we can verif the amplitude 3 & Bartlett Learning, FOR SALE of OR 5 sin using (4): [M m] 5 [3 ()] 5. Note that the range [, 3] of the function 5 sin is the range [, ] of Jones & Bartlett Learning, 5 sin shifted up unit on the Jones -ais. & Bartlett Learning, Graphs of 5 A sin B and 5 A cos B We now consider the graph of 5 A sin B and 5 A cos B. Throughout the discussion we ma assume that B. 0. Because p is the period of both 5 A sin and 5 A cos a ccle of the graphs of 5 A sin B and Jones 5 A cos & BBartlett begins at Learning, 5 0 and will start to repeat its values when B 5 p. Dividing the FOR last equalit SALE b OR B shows that the period of each of the functions 4.3 Graphs of Sine and Cosine Functions 9

25 Jones & Bartlett Learning.. 5 sin B and 5 cos B is p/b and so the graph of each function over the interval [0, p/b] is one ccle of its graph. If 0, B,, then the period p/b is greater than p, and we can we can characterize the ccle of either 5 sin B or 5 cos B on [0, p/b] as a horizontal stretch of the graphs of 5 sin and 5 cos on the interval [0, p]. On the other hand, if B., then the period p/b is less than p, and so the graphs on [0, p/b] can be interpreted as a horizontal compression of Jones the & graphs Bartlett of the Learning, functions 5 sin and 5 cos on [0, p]. Jones Finall, & Bartlett we can easil find SALE the OR -coordinates of the -intercepts of the graphs FOR of SALE 5 A sin OR B and DISTRIBUT Learning, LL FOR 5 A cos B b replacing the smbol b B in () and () and solving for. The net eample illustrates these concepts. Jones & Bartlett Learning, EXAMPLE 3 Horizontall Jones Compressed & Bartlett Sine Graph Learning, Careful here: sin sin With the identification B 5 the period of the sine function 5 sin is p/b 5 p/ 5p. Therefore one ccle of the graph is completed on the interval [0, p]. FIGURE shows that two ccles of the graph of 5 sin (in red) are completed on the interval [0, p] whereas the graph of 5 sin (in blue) has completed onl one ccle. Interpreted in terms Jones of transformations, & Bartlett the Learning, graph of 5 sin on [0, p] is a horizontal compression of the FOR graph of SALE 5 sin OR on [0, p]. To find the -intercepts of the graph of 5 sin we solve the equation sin 5 0. From () with replaced b, we get 5 np or 5 np. FOR B SALE letting nor 5 0, 6, 6, 63, , the zeros of 5 sin FOR are SALE 5 0, OR 6 p, DISTRIBUT 6p, 6 3 p, 6p, and so on. As we see in Figure 4.3.7, the -intercepts on the nonnegative -ais are the points (0, 0), (p/, 0), (p, 0), (3p/, 0), (p, 0),.... = sin = sin FOR 4SALE OR 4 4 One ccle of = sin One ccle of = sin FIGURE Graph of function in Eample 3 Horizontal Stretch/Compression The graphs of the functions 5 A sin B and 5 A cos B 30 CHAPTER 4 TRIGONOMETRIC FUNCTIONS for B. 0 have amplitude 0A0 and period p/b. The graphs of 5 A sin B and 5 A cos B are the graphs of 5 A sin and 5 A cos stretched horizontall if 0, Jones B,, & and Bartlett compressed Learning, horizontall if B..

26 Jones & Bartlett Learning.. FOR SALE OR = cos 4 4 EXAMPLE 4Jones & Horizontall Bartlett Learning, Compressed Cosine Graph Find the period of 5FOR cos 4SALE and graph OR the function. Solution Since B 5 4, we see that the period of 5 cos 4 is p/4 5p/. We conclude that the graph of 5 cos 4 is the graph of 5 cos compressed horizontall. To graph the function, we draw one ccle of the cosine graph with amplitude on the interval [0, Jones & Bartlett Learning, p/] and then use periodicit to etend the graph. Jones & Bartlett FIGURE shows Learning, four complete OR ccles of 5 cos 4 (in red) and one ccle of 5 cos FOR (in SALE blue) on OR [0, p]. DISTRIBUT Notice LL = cos FOR SALE that 5 cos 4 attains its minimum at 5p/4since cos 4(p/4) 5 cos p5and its maimum at 5p/since cos 4(p/) 5 cos p 5. FIGURE Graph of function in Eample 4 In the case when B, 0 in either Jones 5& A sin Bartlett B or Learning, 5 A cos B, we can use the even/odd properties, (5) of Section 4., FOR to rewrite SALE the OR function with positive B. This is illustrated in the net eample. EXAMPLE 5 Horizontall Stretched Sine Graph Find the amplitude Jones and period & Bartlett of 5 sin( Learning, ). Graph the function. Solution Since we require FOR B SALE. 0, we OR use sin() 5sin to rewrite the function as 5 sin( ) 5sin. = sin With the identification A 5, the amplitude is seen to be 0A Now with Jones & Bartlett B 5 we Learning, find that the period is p/ Hence we can interpret the ccle of 5sin 5 4p. 3 FOR 4 SALE OR on [0, 4p] as a horizontal stretch and a reflection FOR SALE in the OR -ais DISTRIBUT (because = sin A, 0) of the ccle of 5 sin on [0, p]. FIGURE shows that on the interval [0, 4p] the graph of 5sin FIGURE Graph of function in (in red) completes one ccle whereas the graph of Eample 5 5 sin (in blue) completes two ccles. Graphs of A sin(b C) and FOR SALE A cos(b OR C) We have seen that the basic graphs of 5 sin and 5 cos can be stretched or compressed verticall: 5 A sin and 5 A cos, shifted verticall: FOR 5 A sin SALE D OR and 5 A cos D, and stretched or compressed horizontall: 5 A sin B D and 5 A cos B D. Jones & Bartlett The graphs Learning, of 5 A sin(b C) and 5 A cos(b FOR SALE C) OR DISTRIBUT are the graphs of 5 A sin B and 5 A cos B shifted horizontall. And finall, the graphs of 5 A sin(b C) Jones D and & Bartlett 5 A cos(b Learning, C) D are the graphs of 5 A sin(b C) and FOR 5 SALE A cos(b OR C) shifted verticall. Since the last case is straightforward, we are going to focus on the graphs of 5 A sin(b C) and 5 A cos(b C) in the remaining discussion. For eample, we know from Section. that the graph of 5 cos( p/) is the basic cosine graph shifted p/ Jones units to the & Bartlett right. In FIGURE Learning, the graph of 5 cos( p/) (in red) on the interval FOR [0, p] SALE is one OR ccle of 5 cos on the interval [p/, 3p/] 4.3 Graphs of Sine and Cosine Functions 3

27 Jones & Bartlett Learning.. (in blue) shifted horizontall p/ units to the right. Similarl, the graphs of 5 sin( p/) and 5 sin( p/) are the basic sine graphs shifted p/ units to the left and to the right, respectivel. See FIGURES 4.3. and = cos ( 3 = cos ( FIGURE Horizontall shifted FIGURE 4.3. Horizontall shifted FIGURE 4.3. Horizontall shifted cosine Jones graph& Bartlett Learning, sine graph Jones & sine Bartlett graph Learning, B comparing the red graphs in Figures with the graphs in Figures 4.3. and we see that: 3 CHAPTER 4 TRIGONOMETRIC FUNCTIONS = sin = sin + 3 ( ( In other words, we have graphicall verified the identities = sin ( 3 = sin the cosine graph shifted p/ units to the right is the sine graph, the sine graph shifted Jones p/ units & Bartlett to the left Learning, is the cosine graph, and the sine graph shifted p/ FOR units SALE to the right OR is the cosine graph reflected in the -ais. Jones & Bartlett plearning, cosa (5) b 5 sin, sina p b 5 cos, and sina Jones p& Bartlett Learning, LL b 5cos. FOR SALE OR DISTRIBUT For convenience we now consider the graph of 5 A sin(b C), for B. 0. Since the values of sin(b C) range from to, it follows that A sin(b C) varies between A and A. That is, the amplitude of 5 A sin(b C) is 0A0. Also, Jones & Bartlett Learning, as B C varies from 0 to p, the graph Jones will complete & Bartlett one Learning, ccle. B solving FOR SALE OR B C 5 0 and B C 5 p, we find that one FOR ccle SALE is completed OR as varies from C/B to (p C)/B. Therefore, the function 5 A sin(b C) has the period Everthing said in this paragraph also holds for 5 A cos(b C). p C a C B B b 5 p B. Moreover, if f () 5 A sin B, then f a C B b 5 A sin Ba C b 5 A sin(b C). B The result in (6) shows that the graph of 5 A sin(b C) can be obtained b shifting & the Bartlett graph of Learning, f () 5 A sin B horizontall a distance 0C0 /B. Jones The number & Bartlett C/B is called Learning, LL Jones FOR the phase SALE shift OR of the graph of 5 A sin(b C). If C/B, 0 FOR the shift SALE is to the OR right DISTRIBUT whereas if C/B. 0, the shift is to the left. EXAMPLE 6 Equation of Shifted Cosine Graph Jones & Bartlett Learning, The graph of 5 0 cos 4 is shifted p/ Jones units to the & Bartlett right. Find Learning, its equation. Solution We first identif f () 5 0 cos 4 and B 5 4. Because we want to shift the graph of f to the right the phase shift is C/B 5p/. Then the analogue of (6) for the cosine function is: f a p b 5 0 Jones cos 4a & Bartlett p b or Learning, 5 0 cosa4 p 3 b. ( (6)

28 Jones & Bartlett Learning.. As a practical matter, if we are given the equation 5 A sin(b C) (or 5 A cos(b C) ), then the phase shift C/B of the graph of can be obtained b factoring the number B from B C: 5 A sin(b C) 5 A sin Ba Jones & Bartlett The foregoing Learning, information is summarized net. Horizontall Shifted Sine and Cosine Graphs The graphs of the functions 5 A sin(b C) and Jones 5 A & cos(b Bartlett C), Learning, B. 0, (7) are the graphs of 5 A sin B FOR and SALE 5 A cos OR B shifted horizontall b 0C0 /B units. The number C/B is called the phase shift of the graph. The horizontal shift is to the right if C/B, 0 and to left if C/B. 0. The amplitude of each function in (7) is 0A0 and the period of each function is p/b. Jones & Bartlett Learning, The range of FOR each function SALE OR in (7) is the interval [0A0, 0A0 ] on the -ais. 3 3 The numbers 0A0 and p/b are also referred to as the amplitude and period of the graphs of the functions in (7). FOR SALE EXAMPLE OR 7 Horizontall Shifted Sine FOR Graph SALE OR DISTRIBUT Graph 5 3 sin( p/3). Solution For purposes of comparison we will first graph 5 3 sin. The amplitude = 3 sin ( 3 of is 0A0 Jones & Bartlett Learning, 5 3 sin 5 3 and its Jones period is & p/b Bartlett 5 p/ Learning, 5p. Thus one ccle of 5 3 sin is completed on the interval FOR [0, p]. SALE We etend OR this graph to the adjacent interval [p, p] as shown in blue in FIGURE Net, we rewrite 5 3 sin( p/3) b factoring from p/3: ( = 3 sin 5 3 sina p b 5 3 sin a 3 p Jones & Bartlett Learning, 6 b. FIGURE Graph of function in FOR Eample SALE 7OR From the last form we see that the phase shift is p/6, 0. This means that the graph of the given function, shown in red in Figure 4.3.3, is obtained b shifting the graph of 5 3 sin to the right 0C0 /B 5 0 (p/3)0/ 5p/6units. Remember, this means that if (, ) is a point on the blue graph, then ( p/6, ) is the corresponding point on the red graph. For eample, 5 0 and 5pare the -coordinates of two -intercepts of the blue graph. Thus 5 0 p/6 5p/6and 5pp/6 5 7p/6 are -coordinates of the -intercepts of the red or shifted graph. These numbers are indicated b the arrows in Figure Note that one ccle of the red graph is completed on the interval [p/6, 7p/6]. EXAMPLE 8 Horizontall FOR Shifted SALE Graphs OR C B b. Determine the amplitude, the period, the phase shift, and the direction of horizontal shift for each of the following functions. (a) 5 5 cosa5 Jones 3p p (b) 58 sina b & Bartlett Learning, 4 b 4.3 Graphs of Sine and Cosine Functions 33

29 Jones & Bartlett Learning.. Solution (a) We first make the identifications A 5 5, B 5 5, and C 53p/. Thus the amplitude is 0A0 5 5 and the period is p/b 5 p/5. The phase shift can be computed either b C/B 5 (3p/)/5 53p/0 or b rewriting the function as 3p 5 5 cos 5a 0 b. Jones The & last Bartlett form Learning, indicates that the graph of 5 5 cos(5 Jones 3p/) & is Bartlett the graph Learning, of LL FOR 5SALE 5 cos 5 OR shifted 3p/0 units to the right. (b) Since A 58 the amplitude is 0A With B 5 the period is p/b 5 p/ 5p. B factoring from p/4, we see from p 58 sina 4 b Jones 58 sin & a Bartlett p 8 b Learning, 34 CHAPTER 4 TRIGONOMETRIC FUNCTIONS that the phase shift is p/8. 0 and so the graph of 58 sin( p/4) is the graph of 58 sin shifted p/8 units to the left. EXAMPLE 9 Jones Horizontall & Bartlett Shifted Learning, Cosine Graph Graph 5 cos(p p). Solution The amplitude of 5 cos p is 0A0 5 and the period is p/p 5. Thus = cos( + ) one ccle of 5 cos p is completed on the interval [0, ]. In FIGURE two ccles Jones of & the Bartlett graph of Learning, 5 cos p (in blue) are shown. The -intercepts Jones of & this Bartlett graph correspond SALE to the OR values of for which cos p 5 0. B (), this implies FOR p 5 SALE (n OR )p/ DISTRIBUT Learning, LL FOR 3 4 or 5 (n )/, n an integer. In other words, for n 5 0,,,,,3,... we get 56, 6 3, 6 5, and so on. Now b rewriting the given function as = cos 5 cos p( ) FIGURE Graph Jones of function & Bartlett in Learning, Eample 9 FOR SALE OR we see the phase shift is. 0. Thus the graph of FOR 5 SALE cos(por p) (in red) in Figure 4.3.4, is obtained b shifting the graph of 5 cos p to the left unit. This means that the -intercepts are the same for both graphs. EXAMPLE 0 Alternating Current IFOR The current I (in amperes) in a wire of an alternating-current circuit is given b I(t) = SALE 30 sin 0 OR t 30 I(t) 5 30 sin 0pt, where t is time measured in seconds. Sketch one ccle of the graph. What is the maimum value of the current? t Solution The graph has amplitude 30 and period p/0p 5 Therefore, we sketch one ccle of the basic sine curve on the interval [0, Jones & Bartlett Learning, 60], as shown Jones in FIGURE & Bartlett From Learning, LL 30 FOR the figure SALE it is OR evident that the maimum value of the current is I 5 FOR 30 amperes SALE and OR occurs DISTRIBUT at t 5 40 since FIGURE Graph of current in Eample 0 I a 40 b 5 30 sina0p # 40 b 5 30 sin p Eercises 4.3 Answers to selected odd-numbered problems begin on page ANS 5. In Problems 6, use the techniques of shifting, stretching, compressing, and reflecting to sketch at least one ccle of the graph of the given function.. 5 cos Jones. 5 & Bartlett cos Learning, 3. 5 sin sin 5. FOR 5 SALE 4 OR cos 6. 5 sin

30 Jones & Bartlett Learning.. In Problems 7 0, the given figure shows one ccle of a sine or cosine graph. From the figure, determine A and D and write an equation of the form 5 A sin D or 5 A cos D for the FOR graph. SALE OR FIGURE Graph for Problem 7 4 FOR SALE OR DISTRIBUT 4 FIGURE Graph for Problem Jones & Bartlett Learning, FIGURE Graph for Problem 0 FIGURE Graph for Problem 9 In Problems 6, use () and () of this section to find the -intercepts for the graph of the given function. Do not graph.. 5 sin p. 5cos cos sin(5) 5. 5 sina p 6. 5 cos( p) 4 b In Problems 7 and 8, find the -intercepts of the graph of the given function on the interval [0, p]. Then find all intercepts using periodicit sin 8. 5 cos In Problems 9 4, the given figure shows one ccle of a sine or cosine graph. From FOR SALE the figure, OR determine A and B and write an equation of the FOR form SALE 5 A OR sin BDISTRIBUT or 5 A cos B for the graph Jones & Bartlett Learning, 3 FIGURE 4.3. Graph for FIGURE Graph Jones for & Bartlett Learning, Problem 0 Problem Graphs of Sine and Cosine Functions 35

31 Jones & Bartlett Learning.. 36 CHAPTER 4 TRIGONOMETRIC FUNCTIONS.. FIGURE 4.3. Graph for FOR SALE Problem OR 3 FIGURE Graph for Problem Jones & Bartlett Learning, Jones & 3Bartlett Learning, FIGURE Graph for FIGURE Graph for FOR SALE OR Problem 4 Problem 3 In Problems 5 30, find the amplitude and period of the given function. Sketch at least one ccle of the graph. Jones & Bartlett Learning, LL FOR 5. SALE 5 4 sin OR p sin FOR SALE OR DISTRIBUT cos p sin() cosa p Jones & Bartlett b Learning, In Problems 3 36, (a) sketch one ccle of the graph of the given function. (b) Find the amplitude 0A0 b inspection of the function. (c) Find the maimum value M and the minimum value m of the function on the interval in part (a). (d) Then use (4) to verif the amplitude 0A0 of the function. (e) Give the range of each function cos Jones & Bartlett 3. Learning, 53 3sin sin p cos p sin p sin 3 In Problems 37 and 38, from the given figure determine A, B, and D and write an FOR equation SALE of the OR form 5 A sin B D or 5 A cos B D for FOR the graph. SALE OR DISTRIBUT 5 5 cos FIGURE Graph for FIGURE Problem 37 Graph for Jones & Bartlett Problem Learning, 38

32 Jones & Bartlett Learning.. In Problems 39 48, find the amplitude, period, and phase shift of the given function. Sketch at least one ccle of the graph sina p sina3 p 6 b 4 b 4. 5 cosa p 4. 5 cosa p Jones & Bartlett Learning, 4 b Jones & Bartlett 6 b Learning, LL FOR SALE 43. OR 5 4 cosa 3p sina p b FOR SALE 4 b OR DISTRIBUT sina 46. 5cosa p 3 b pb Jones & Bartlett Learning, sinap cosap 4p 3 p 3 b 3 b In Problems 49 and 50, write an equation of the function whose graph is described in words. 49. The graph of Jones 5 cos & is Bartlett verticall stretched Learning, up b a factor of 3 and shifted down b 5 units. One ccle FOR of SALE 5 cos OR on [0, p] is compressed to [0, p/3] and then the compressed ccle is shifted horizontall p/4 units to the left. 50. One ccle of 5 sin on [0, p] is stretched to [0, 8p] and then the stretched ccle is shifted horizontall p/ units to the right. The graph is also compressed 3 verticall b a factor of 4 and then reflected in the -ais. In Problems 5 54, find horizontall shifted sine and cosine functions so that each function satisfies the given conditions. Graph the functions. 5. Amplitude 3, period p/3, shifted b p/3 units to the right Jones & Bartlett Learning, 5. Amplitude 3, period p, shifted Jones b p/4 units & Bartlett to the left Learning, 53. Amplitude 0.7, period 0.5, shifted b 4 units to the right Amplitude period 4, shifted b /p units to the left 4, In Problems 55 and 56, graphicall verif the given identit. 55. cos( p) 5cos 56. sin( p) 5sin Applications 57. Pendulum The angular displacement u of a pendulum from the vertical at time t seconds is given b u(t) 5u 0 cos vt, where u 0 is the initial displacement at t 5 0 Jones & Bartlett seconds. Learning, See FIGURE For v5 rad/s and Jones u 0 5p/0, & Bartlett sketch two Learning, ccles of LL the resulting function. θ Fahrenheit Temperature Suppose that θ p T(t) sin (t 8), 0 # t # 4, is a mathematical model of the Fahrenheit temperature at t hours FIGURE Pendulum in after midnight on a certain da of the week. Problem 57 (a) What is the temperature at 8 A.M.? (b) At what time(s) does T(t) 5 60? (c) Sketch the graph of T. (d) Find the maimum Jones & and Bartlett minimum Learning, temperatures and the times at which the occur. 4.3 Graphs of Sine and Cosine Functions 37

33 Jones & Bartlett Learning Depth of Water The depth d of water at the entrance to a small harbor at time t is modeled b a function of the form where A is one-half the difference between the high- and low-tide depths; p/b, Jones & B Bartlett. 0, is the Learning, tidal period; and C is the average depth. Assume Jones that & the Bartlett tidal Learning, LL FOR SALE period is OR hours, the depth at high tide is 8 feet, and the depth FOR at SALE low tide OR is DISTRIBUT 6 feet. Sketch two ccles of the graph of d. 60. Hours of Dalight The number H of dalight hours per da in various locations in the world can be modeled b a function of the form H(t) 5 A sin B(t Jones C) & Bartlett D, Learning, where the variable t represents the number of das in a ear corresponding to a specific calendar date (for eample, Februar corresponds to t 5 3 das), and A, B, C, and D are positive constants. In this problem we construct a model for Los Angeles, CA for the ear 07 (not a leap ear) using data obtained from the U.S. Naval Observator, Jones Washington, & Bartlett D.C. Learning, (a) Find the amplitude A if FOR 4.43SALE is the maimum OR number of dalight hours at the summer solstice and if 9.88 is the minimum number of dalight hours at the winter solstice. (b) Find B if the function H(t) is to have the period 365 das. Jones &(c) Bartlett For Los Learning, Angeles the ear 07, we choose C Jones Eplain & Bartlett the signifi-learningcance OR of this number. [Hint: C has the same units as t.] FOR SALE OR DISTRIBUT LL Sunset in Los Angeles FOR SALE (d) Find D if the number of dalight hours at the vernal equino for 07 is.4 and occurs on March 0. (e) What does the model H(t) predict to be the number of dalight hours on Januar? On June? On August? On December? Jones & Bartlett Learning, (f) Using a graphing utilit to obtain the Jones graph & of Bartlett H(t) on the Learning, interval [0, 365]. egd/shutterstock, Inc. Calculator/Computer Problems In Problems 6 64, use a graphing utilit to investigate whether the given function is periodic. 6. f () 5 sina 6. f () 5 b sin 63. f () 5 (cos ) 64. f () 5 sin Jones For & Discussion Bartlett Learning, FOR In Problems SALE OR 65 and 66, describe in words how ou would obtain FOR the graph SALE of the OR DISTRIBUT given function b starting with the graph of 5 sin (Problem 65) and the graph of 5 cos (Problem 66) sin( p) cos( p) FOR SALE OR In Problems 67 and 68, find the period of the given FOR function. SALE OR 67. f () 5 sin sin 68. In Problems 69 and 70, discuss and then sketch the graph of the given function. 69. f () 5 0 sin 0 FOR SALE 70. OR f () 5 0 cos 0 38 CHAPTER 4 TRIGONOMETRIC FUNCTIONS d(t) 5 A sin Bat p b C, 3 f () 5 sin 5 cos

34 Jones & Bartlett Learning Other Trigonometric Functions INTRODUCTION Recall that the remaining four trigonometric functions are the tangent, cotangent, secant, and cosecant functions and are denoted, in turn, as tan, cot, sec, and csc. We saw in Section 4. that b using () and () of Definition 4.. in (3) (6) of Definition 4.. we can epress these four new functions in terms of sin and cos : tan 5 sin cot 5 cos cos sin sec 5 csc 5 Jones cos & Bartlett sin Learning,. () () Domain and Range Because the functions in () and () are quotients, we know from Definition.6. that the domain of each function consists of the set of real numbers ecept those Jones numbers & Bartlett for which Learning, the denominator is zero. We have seen in () of Section 4.3 that cos FOR 5 0SALE for 5 OR (n )p/, n 5 0, 6, 6, c, and so the domain of tan and of sec is 5 0 (n )p/, n 5 0, 6, 6, c6. Similarl, from () of Section 4.3, sin 5 0 for 5 np, n 5 0, 6, 6, c, and so it follows that the domain of cot and of csc is 5 0 np, n 5FOR 0, 6, SALE 6, c6. OR DISTRIBUT We know that the values of the sine and cosine are bounded, that is, 0 sin 0 # and 0 cos 0 #. From these last inequalities we have 0 sec 0 5 Jones ` (3) cos ` & 5 Bartlett $ Learning, FOR SALE 0 cos OR 0 and 0 csc 0 5 ` (4) sin ` 5 $. 0 sin 0 tan < 0 tan > 0 cot < 0 cot > 0 Jones & Bartlett II sec < Learning, 0 sec > 0 I Recall that an inequalit Jones such & Bartlett as (3) means Learning, that sec $ or sec #. Hence the csc > 0 csc > 0 range of the secant function is (`, ] < [, `). The inequalit in (4) implies that the cosecant function has the same range (`, ] < [, `). When we consider tan > 0 tan < 0 the graphs of the tangent and cotangent functions we will see that the have the same range: cot > 0 cot < 0 (`, `). III sec < 0 sec > 0 IV If we interpret as an angle, then FIGURE 4.4. illustrates the algebraic signs of the csc < 0 csc < 0 Jones & Bartlett tangent, Learning, cotangent, secant, and cosecant functions in Jones each of& the Bartlett four quadrants. Learning, This LL FIGURE 4.4. Signs of tan, cot FOR, SALE is easil OR verified using the signs of the sine and cosine FOR functions SALE OR displaed DISTRIBUT in sec, and csc, in the four quadrants Figure 4... Jones & Bartlett Learning, EXAMPLE Eample Jones 5 of & Section Bartlett 4. Learning, Revisited Find tan, cot, sec, and csc for 5p/6. Solution In Eample 5 of Section 4. we saw that p sina 6 b Jones 5sin p& 6 5 Bartlett Learning, p and cosa 6 b 5 cos p 6 5! Other Trigonometric Functions 39

35 Jones & Bartlett Learning.. Therefore, b the definitions in () and (): p tana 6 b 5 FOR SALE OR /!3/ 5!3, cota p 6 b 5!3 / 5!3, / TABLE tan cot sec csc p 6 p 4 p 3 p seca 6 b 5!3/ 5!3, csca p 6 b 5 p Jones / 5. & Bartlett Learning, 0!3!3 Table 4.4. summarizes some important values of the tangent, cotangent, secant, and cosecant and was constructed using values of the sine and cosine from Section 4.. A!3 0!3 dash in the table indicates that the trigonometric function is not defined at that particular value of. Jones & Bartlett Learning,!!3 FOR SALE OR Identities The tangent is related to the secant b a useful identit. If we divide the!!3 Pthagorean identit b cos, we see that sin cos 5 (5) sin (6) cos cos cos 5 cos. FOR Similarl, SALE dividing OR (5) b sin ields an identit relating the cotangent FOR with SALE the cosecant: OR DISTRIBUT sin sin cos sin 5 sin. Jones & Bartlett Learning, Using the laws of eponents, sin cos 5 a sin FOR SALE cos b 5 tan, cos 5 a OR cos b 5 sec, cos sin 5 acos sin b 5 cot, sin 5 a sin b 5 csc, we see that (6) and (7) can be written FOR SALE in a simpler OR manner: tan 5 sec cot 5 csc. Jones Since & Bartlett the last two Learning, identities are direct consequences of sin Jones cos & 5Bartlett the too Learning, are LL FOR called SALE Pthagorean OR identities. Finall, note that the tangent and cotangent function are related b the reciprocal identit cot 5 cos sin 5 5 Jones sin & tan Bartlett. Learning, cos FOR SALE OR (7) Summar For future reference, especiall for the work in the net section, we pause here to summarize a small collection of identities that are so basic to the stud of trigonometr that the are known Jones collectivel & Bartlett as the Learning, fundamental identities. You should firml commit these identities FOR to memor. SALE OR 40 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

36 Jones & Bartlett Learning.. Fundamental Jones Trigonometric & Bartlett Learning, Identities Pthagorean identities: sin cos 5 (8) tan 5 sec (9) cot 5 csc (0) Quotient identities: tan 5 sin cot 5 cos () cos sin Reciprocal identities: sec 5 cos csc 5 sin cot 5 tan EXAMPLE FOR Using SALE a OR Pthagorean Identit Given that csc 55and 3p/,, p, determine the values of tan and cot. Solution We first compute cot. It follows from (0) that cot 5 csc. For 3p/,, p, we see from Figure 4.4. that cot must be negative and so we take the negative square root: cot 5 "csc 5 "(5) 5!4 5!6. Using cot 5 /tan, we have rationalizing the denominator T tan 5 cot 5!6 5!6. In Eample, given the information csc 55and 3p/,, p, we could easil find the values of the remaining five trigonometric functions. One wa of proceeding would be to use csc 5 /sin to find sin 5 /csc 5 Then we use sin cos 5. 5 to find cos. After we have found cos, the remaining three trigonometric functions Learning, can be obtained from () and (). Jones & Bartlett Periodicit Because the sine and cosine functions are p periodic, each of the functions in () and () have a period p. But from (7) of Theorem 4..5 we have Also see Problems 55 and 56 sin( p) tan( p) 5 Jones & Bartlett (3) cos( p) 5 sin Learning, 5 tan. Eercises 4.3. cos Thus (3) implies that tan and cot are periodic with a period p #p. In the case of the tangent function, tan 5 0 onl if sin 5 0, that is, onl if 5 0, 6p, 6p, and so on. Therefore, the smallest positive number p for which tan( p) 5 tan is p 5p. The cotangent function Jones has & the Bartlett same period Learning, since it is the reciprocal of the tangent function. 4.4 Other Trigonometric Functions 4 ()

37 Jones & Bartlett Learning.. THEOREM 4.4. Period Jones & of Bartlett the Tangent Learning, and Cotangent The tangent and cotangent functions FOR SALE are periodic OR with period p. Therefore, tan( p) 5 tan and cot( p) 5 cot (4) for ever real number for which the functions are defined. THEOREM 4.4. Period of the Secant and Cosecant The secant and cosecant functions are periodic with period p. Therefore, sec( p) 5 sec and Jones csc( & Bartlett p) 5Learning, csc (5) for ever real number for which the functions are defined. Even-Odd Properties Jones Because & the Bartlett cosine Learning, function is even and the sine function is odd, each of the remaining four trigonometric functions is either even or odd. for ever real number for which the functions are defined, PROOF: We prove the first entries in (6) and (7). Because cos() 5 cos and sin() 5sin, This is a good time to review (7) of Section 3.5. THEOREM CHAPTER 4 TRIGONOMETRIC FUNCTIONS Even and Odd Functions The tangent, cotangent, and cosecant functions are odd functions, whereas the secant Jones function & Bartlett is an even Learning, function. That is, tan() 5tan and cot() 5cot (6) sec() 5 sec and csc() 5csc (7) tan() 5 sin() cos() 5 sin cos 5sin Jones & Bartlett Learning, 5tan cos, FOR SALE sec() 5 cos() 5 OR cos 5 sec proves, in turn, that tan is an odd function and sec is an even function. Graphs of 5 tan and 5 cot The numbers that make the denominators of tan, cot, sec, and csc equal to zero correspond to vertical asmptotes of their graphs. For eample, we encourage ou to verif, using a calculator, that p tan S ` as S and tan S ` as S p. Jones & Bartlett Learning, Jones & Bartlett Learning, FOR SALE OR In other words, 5p/and 5p/are vertical asmptotes. The graph of 5 tan on the interval (p/, p/) given in FIGURE 4.4. is one ccle of the graph of 5 tan. Using periodicit we etend the ccle in Figure 4.4. to adjacent intervals of length p, as shown in FIGURE The -intercepts of the graph of the tangent function are (0, 0), Jones FIGURE 4.4. & Bartlett One ccle Learning, of the graph (6p, 0), (6p, 0), c, and Jones the vertical & Bartlett asmptotes Learning, of the graph are 56p/, offor 5 tan SALE OR 63p/, 65p/, c.

38 Jones & Bartlett Learning.. = tan = cot 3 3 FIGURE Graph of 5 tan FIGURE Graph of 5 cot The graph of 5 cot is similar Jones to the & graph Bartlett of the tangent Learning, function and is given in FIGURE In this case, the graph of 5 cot on the interval (0, p) is one ccle of the graph of 5 cot. The -intercepts of the graph of the cotangent function are (6p/, 0), (63p/, 0), (65p/, 0), c, and the vertical asmptotes of the graph are 5 0, 6p, 6p, 63p, c. Because 5 tan and 5 cot are odd functions, their graphs are smmetric with respect to the origin. Graphs of 5 sec FOR and SALE 5OR csc For both 5 sec and 5 csc we know that 0 0 $, and so no portion of their graphs can appear in the horizontal strip,, of the Cartesian plane. Hence the graphs of 5 sec and 5 csc have no -intercepts. Both 5 sec and 5 csc have period p. The vertical asmptotes Jones & Bartlett for the Learning, graph of 5 sec are the same as 5 tan Jones, namel, & Bartlett 56p/, Learning, 63p/, LL FOR SALE 65p/, OR c. Because 5 sec is an even function, its graph FOR is SALE smmetric OR with DISTRIBUT respect to the -ais. On the other hand, the vertical asmptotes for the graph of 5 csc are the same as 5 cot, namel, 5 0, 6p, 6p, 63p, c. Because 5 csc is an odd function, its graph is smmetric with respect to the origin. One ccle of the graph of 5 sec on [0, p] is etended to the interval [p, 0] b periodicit (or -ais smmetr) in FIGURE Similarl, in Jones Jones & Bartlett Learning, FIGURE & Bartlett we etend Learning, one ccle of 5 csc on (0, p) to the interval (p, 0) b periodicit FOR SALE (or origin OR smmetr). = sec = csc FIGURE Graph of sec FIGURE Graph of csc Transformations and Graphs Similar to the sine and cosine graphs, rigid and Jones & Bartlett Learning, nonrigid transformations can be applied Jones to the & graphs Bartlett of Learning, 5 tan, 5 cot, 5 sec, and 5 csc. For eample, a function such FOR as SALE 5 A tan(b OR C) D can be analzed in the following manner: vertical stretch/compression/reflection T 5 A tan(b C) D c horizontal stretch/compression b changing period vertical shift T c horizontal shift 4.4 Other Trigonometric Functions 43 (8)

39 Jones & Bartlett Learning.. If B. 0, then the period of 5 A tan(b C) FOR and SALE OR 5 A cot(b C) is p/b, (9) whereas the period of 5 A sec(b C) and 5 A csc(b C) is p/b. (0) As we see in (8), the number A in each case can be interpreted as either a vertical stretch or a compression of a graph. However, ou should be aware of the fact that the functions Of the si trigonometric functions, onl the FOR sine and cosine functions have an amplitude. in (9) SALE and (0) OR have no amplitude, because none of the functions FOR has a SALE maimum OR and DISTRIBUT a minimum value. EXAMPLE 3 Comparison of Graphs Find the period, -intercepts, and vertical asmptotes for the graph of 5 tan. Graph FOR SALE OR the function on [0, p]. Solution With the identification B 5, we see from (9) that the period is p/. Since tan 5 sin /cos, the -intercepts of the graph occur at the zeros of sin. From () of Section 4.3, sin 5 0 for 5 np so FOR that SALE 5 OR np, n 5 0, 6, 6, c. That is, 5 0, 6p/, 6p/ 5p, 63p/, 64p/ 5 p, and so on. The -intercepts are (0, 0), (6p/, 0), (6p, 0), (63p/, 0), c. The vertical asmptotes of the graph occur at zeros of cos. From () of Section 4.3, the numbers for which cos 5 0 are Jones found & Bartlett in the following Learning, manner: p p 5 (n ) so that 5 (n ), n 5 0, 6, 6, c. 4 That is, the vertical asmptotes are 56p/4, 63p/4, 65p/4, c. On the interval [0, p], the graph of 5 tan has three intercepts (0, 0), (p/, 0), and (p, 0) and two Jones & Bartlett Learning, vertical asmptotes 5p/4and 5 3p/4. Jones In FIGURE & Bartlett 4.4.7, we Learning, have compared the FOR SALE OR graphs of 5 tan and 5 tan on the interval. FOR The graph SALE of OR 5 tan is a horizontal compression of the graph of 5 tan (a) = tan on [0, ] (b) = tan on [0, ] FIGURE Graph of function in Eample 3 EXAMPLE 4 Comparison of Graphs Compare one ccle of the graphs of 5 tan and 5 tan( p/4). Solution The graph of 5 tan( p/4) is the graph of 5 tan shifted horizon- p/4 units to the right. Jones The intercept & Bartlett (0, 0) for Learning, the graph of 5 tan is shifted to tall (p/4, 0) on the graph of 5 tan( FOR p/4). SALE OR The vertical asmptotes 5p/and 44 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

40 Jones & Bartlett Learning.. 5p/for the graph Jones of 5 & tan Bartlett are shifted Learning, to 5p/4 and 5 3p/4 for the graph of 5 tan( p/4). In FIGURES 4.4.8(a) and 4.4.8(b) we see, respectivel, that a ccle of the graph of 5 tan on the interval (p/, p/) is shifted to the right to ield a ccle of the graph of 5 tan( p/4) on the interval (p/4, 3p/4) Jones & Bartlett Learning, 4 (a) Ccle of = tan (b) Ccle of = tan ( /4) on ( /, /) on ( /4, 3 /4) FIGURE Graphs of functions in Eample 4 As we did in the analsis of the graphs of 5 A sin(b C) and 5 Jones & Bartlett A cos(b Learning, C), we can determine the amount of horizontal Jones & shift Bartlett for graphs Learning, of functions OR such as 5 A tan(b C) and 5 A sec(b FOR C) b SALE factoring OR the DISTRIBUT number LL FOR SALE B. 0 from B C. EXAMPLE 5 Two Shifts and Two Compressions Jones & Bartlett Learning, Graph 5 sec(3 p /). Solution Let s break down the analsis of the graph into four parts, namel, b transformations. (i) One ccle of the graph of 5 sec occurs on [0, p]. Since the period of 5 sec 3 is p/3, one ccle of its graph occurs on the interval [0, p/3]. In other words, the graph of 5 sec 3 is a horizontal compression of the graph of 5 sec. Since sec 3 5 /cos 3, the FOR vertical SALE asmptotes OR occur at the zeros of cos 3. Using () of Section 4.3, we find p p 3 5 (n ) or 5 (n ), n 5 0, 6, 6, c. Jones 6 & Bartlett Learning, LL FOR SALE FIGURE OR 4.4.9(a) shows two ccles of the graph 5 sec 3; FOR one ccle SALE on [ OR p/3, DISTRIBUT 0] and another on [0, p/3]. Within those intervals the vertical asmptotes are 5p/, 5p/6, 5p/6, and 5p/. (ii) The graph of 5 sec 3 is the graph of 5 sec 3 compressed verticall b a Jones & Bartlett Learning, factor of and then reflected in the Jones -ais. See & Bartlett Figure 4.4.9(b). Learning, (iii) B factoring 3 from 3 p/, we FOR see from SALE OR 5 seca3 p b 5 sec 3a p 6 b that the graph of 5 Jones sec(3 & Bartlett p /) Learning, is the graph of 5 sec 3 shifted p/6 units to the right. B shifting FOR the two SALE intervals OR [ p/3, 0] and [0, p/3] in Figure 4.4.9(b) 4.4 Other Trigonometric Functions 45

41 Jones & Bartlett Learning Jones & Bartlett Learning, Jones & Bartlett 5Learning, LL (a) Horizontal compression (b) Vertical compression and reflection in -ais Jones & Bartlett FIGURE Learning, Graph of function in Eample 5 (c) Horizontal shift (d) Vertical shift to the right p/6 units, we see in Figure 4.4.9(c) two ccles of 5 sec(3 p /) on the intervals [ p/, p/6] and [p/6, 5p/6]. The vertical asmptotes 5p/, 5p/6, 5p/6, and 5p/shown in Figure 4.4.9(b) are shifted to 5p/3, Jones 5& 0, Bartlett 5p/3, Learning, and 5 p/3. Observe that the -intercept Jones A0, B in & Figure Bartlett 4.4.9(b) Learning, LL FOR is now SALE moved OR to Ap/6, B in Figure 4.4.9(c). (iv) Finall, we obtain the graph 5 sec(3 p/) in Figure 4.4.9(d) b shifting the graph of 5 sec(3 p/) in Figure 4.4.9(c) upward units. Eercises 4.4 FOR Answers SALE OR to selected odd-numbered problems begin on page ANS 6.. tan cot In Problems and, complete the given table. p 3 3p 4 5p 6 7p 6 5p 4 4p 3 3p 5p 3 p 7p 4 p 6 p. p 3p 5p p sec csc 7p 6 5p 4 4p 3 3p 5p 7p p p In Problems 3 8, find the indicated value without the use of a calculator. Jones & Bartlett Learning, 3. cot 3p 3p 9p 4. csca 5. tan 6. sec 7p b 6 FOR SALE OR 5p 7. csca p 3p tan 3p cota 0. tana 3 b 3 b 4 6 b 0p 7p 9p. sec. cot 3. csc 5p 4. sec 3 Jones 6 & Bartlett Learning, 4 5. sec(0 ) 6. tan FOR 405 SALE OR 7. csc cot(70 ) 46 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

42 Jones & Bartlett Learning.. In Problems 9 6, use the given information to find the values of the remaining five trigonometric functions. 9. tan 5, p /,,p 0. cot 5, p,, 3p/. csc 5 4 3, 0,,p/. sec 55, p/,,p 3. sin 5 3, p/,,p 4. cos 5/"5, p,, 3p/ Jones & Bartlett 5. cos Learning, 5 3p /,, p 6. sin 5 4 3, Jones 5, 0 &, Bartlett,p/Learning, LL 7. If 3 cos 5 sin, find all values of tan, cot, sec, and csc. 8. If csc 5 sec, find all values of tan, cot, sin, and cos. In Problems 9 36, find the period, -intercepts, and the vertical asmptotes of the Jones & Bartlett Learning, given function. Sketch at least one ccle Jones of the & Bartlett graph. Learning, 9. 5 tan p tan p 3. 5 cot 3. 5cot tana Jones cota p p & Bartlett Learning, 4 b FOR SALE OR b cot p tana 5p 6 b Jones & Bartlett In Problems Learning, 37 44, find the period and the vertical Jones asmptotes & Bartlett of the given Learning, function. LL FOR SALE Sketch OR at least one ccle of the graph. p 37. 5sec sec Jones & Bartlett Learning, csc p Jones 40. & Bartlett 5 csc Learning, seca3 p 4. 5 csc (4 p) b csca p sec( p) b In Problems 45 and 46, FOR use the SALE graphs OR of 5 tan and 5 sec to find numbers A and C for which the given equalit is true. 45. cot 5 A tan( C) 46. csc 5 A sec( C) Jones & Bartlett Calculus-Related Learning, Problems FOR SALE In Problems OR 47 56, use the Pthagorean identities (8) (0) FOR and SALE the indicated OR DISTRIBUT substitution to rewrite the given algebraic epression as a trigonometric epression without radicals. Assume that a "a ; 5 a cos u, 0 #u#p 48. "a ; 5 a tan u, p/,u,p/ 49. " a ; 5 a sec u, 0 #u,p/ 50. "6 5 ; sin u, p / #u#p/ 5. "9 ; 5 3 sin u, p /,u,p/ 5. " 4; 5FOR sec u, SALE 0 #u,p/ OR 4.4 Other Trigonometric Functions 47

43 Jones & Bartlett Learning.. " ; 5!3 sec Jones u, 0 #u,p/ & Bartlett Learning, 54. (36 ) 3 / ; 5 6 tan u, p/,u,p/ 55. "7 ; 5!7 tan u, p /,u,p/ Jones & "5 Bartlett Learning, FOR 56. SALE ; 5!5 cos u, 0 #u#p OR For Discussion Jones & Bartlett Learning, 57. Use a calculator in radian mode to compare Jones the values & Bartlett of tan(.57) Learning, and tan(.58). Eplain the difference in these values. 58. Use a calculator in radian mode to compare the values of cot(3.4) and cot(3.5). 59. Eplain wh there are no real numbers satisfing the equation 9 csc For which real numbers is (a) sin # csc? (b) sin, csc? 6. For which real numbers Jones is (a)& sec Bartlett # cos? Learning, (b) sec, cos? 6. Discuss and then sketch the FOR graphs SALE of 5OR 0sec 0 and 5 0csc Sum and Difference Formulas FOR INTRODUCTION SALE OR There are man identities involving trigonometric FOR SALE functions. OR DISTRIBUT A trigonometric identit is an equation or formula involving onl trigonometric functions that is valid for all angles measured in degrees or radians or for real numbers for which both sides of the equalit are defined. The most basic of all these identities the Jones & Bartlett Learning, Pthagorean identities, the Quotient identities, Jones and & Reciprocal Bartlett identities were Learning, dis- cussed in Section 4.4. In this section we develop FOR some SALE additional OR identities that are of FOR SALE OR particular importance in courses in mathematics and science. Sum and Difference Formulas The sum and difference formulas for the cosine and sine functions Jones are identities & Bartlett that Learning, reduce cos( ), cos( ), sin( ), and sin( FOR ) to epressions SALE OR that involve cos, cos, sin, and sin. We will derive the formula for cos( ) first, and then we will use that result to obtain the others. For convenience, let us suppose that and represent angles measured in radians. As shown in FIGURE Jones & Bartlett Learning, 4.5.(a), let d denote the distance between Jones P( & ) Bartlett and P( ). Learning, If LL we place the angle in standard position as shown in Figure 4.5.(b), then d is P( ) = (cos, sin ) Jones & Bartlett P( ) Learning, = (cos, sin ) d 48 CHAPTER 4 TRIGONOMETRIC FUNCTIONS P( ) = (cos ( ), sin( )) d P(0) = (, 0) (a) Jones & Bartlett Learning, (b) FIGURE 4.5. The difference of two angles

44 Jones & Bartlett Learning.. also the distance between P( ) and P(0). Equating the squares of these distances gives (cos cos ) (sin sin ) 5 (cos( ) ) sin ( ) or cos cos cos cos sin sin sin sin 5 cos ( ) cos( ) sin ( ). FOR SALE In view OR of the Pthagorean identit (8) of Section 4.4, FOR SALE OR DISTRIBUT cos sin 5, cos sin 5, cos ( ) sin ( ) 5, and so the preceding equation simplifies to cos( ) 5Jones cos & cos Bartlett sin Learning, sin. This last result can be put to work immediatel to find the cosine of the sum of two angles. Since can be rewritten as the difference ( ), cos( ) 5 cos( ( )) Jones & Bartlett 5 cos cos( Learning, ) sin sin( ). B the even odd identities, cos( ) 5 cos and sin( ) 5sin, it follows that the last line is the same as cos( ) 5 cos cos sin sin. FOR SALE The OR two results just obtained are summarized in the net FOR theorem. SALE OR DISTRIBUT THEOREM 4.5. Sum and Difference Formulas for the Cosine Jones & Bartlett Learning, For all real numbers and, cos( ) 5FOR cos SALE cos OR sin sin cos( ) 5 cos cos sin sin () () EXAMPLE FOR Cosine SALE of OR a Sum Evaluate cos(7p/). Solution We have no wa of evaluating cos(7p/) directl. However, observe that Jones & Bartlett Learning, 7p radians Jones 5 p 3 & Bartlett p 4. Learning, LL Because 7p/ radians is a second-quadrant angle, we know that the value of cos(7p/) is negative. Proceeding, the sum formula () of the Theorem 4.5. gives 7p cos 5 cosap 3 Jones p 4 b 5 & cos Bartlett p 3 cos p 4 Learning, sin p 3 sin p 4 5!!3! 5! (!3). 4 Using!!3 5!6 Jones, this result & Bartlett can also Learning, be written as cos(7p/) 5 (!!6 )/4. Since!6.!, we see FOR that SALE cos(7p/) OR, 0, as epected. 4.5 Sum and Difference Formulas 49

45 Jones & Bartlett Learning.. To obtain the corresponding sum/difference identities for the sine function we will make use of two identities: See (5) in Section 4.3. cosa p b 5 sin and sina p b 5cos. These identities were first discovered in Section 4.3 b shifting the graphs of the cosine Jones and & sine. Bartlett However, Learning, both results in (3) can now be proved using Jones (): & Bartlett Learning, LL FOR SALE OR cosa p b 5 cos cos p sin sin p 5 cos # 0 sin # 5 sin, zero b () of Theorem 4.5. T T cos 5 cosa p p b 5 cosap ap Jones bb 5 & sinap Bartlett b Learning, 5sina p b. Now from the first equation in (3), the sine of the sum can be written (3) sin( ) 5 cosa( ) p Jones & Bartlett b Learning, 5 cosa FOR a SALE p bb OR 5 cos cosa p b sin sina p b d b () of Theorem 4.5. Jones & Bartlett Learning, 5 cos sin sin (cos ). FOR The SALE last line OR is traditionall written as d Jones b &(3) Bartlett Learning, LL sin( ) 5 sin cos cos sin. To obtain the sine of the difference, we use again cos( ) 5 cos and sin( ) 5sin : sin( ) 5 sin( ( )) 5 sin FOR cos( SALE ) OR cos sin( ) 5 sin cos cos sin. THEOREM 4.5. Sum and Difference Formulas for the Sine For all real numbers and Jones, & Bartlett Learning, sin( ) 5 sin cos cos sin (4) sin( ) 5 sin cos cos sin (5) FOR EXAMPLE SALE OR Sine of a Sum Evaluate sin(7p/). Solution We proceed as in Eample, ecept we use the sum formula (4) of Theorem 4.5.: 7p sin 5 sinap 3 p 4 b 5 sin pfor 3 cos psale 4 cos OR p 3 sin p 4 5!3!! 5! A!3B. 4 As in Eample, the result can FOR be rewritten SALE as sin(7p/) OR 5 (!!6)/4. 50 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

46 Jones & Bartlett Learning.. Since we know Jones the value & of Bartlett cos(7p/) Learning, from Eample, we can also compute the value of sin(7p/) using the Pthagorean identit (8) of Section 4.4: sin 7p cos 7p 5. Jones & Bartlett We solve Learning, for sin(7p/) and take the positive square Jones root: & Bartlett Learning, LL 7p sin 5 7p cos Å 5 Ç c! 4 A!3B d (6) 4!3 #!3 5 Jones 5 & Bartlett. Learning, Å 8 FOR SALE OR Although the number in (6) does not look like the result obtained in Eample, the values are the same. See Problem 77 in Eercises 4.5. There are sum and difference formulas for the tangent function as well. We can derive the sum formula Jones using & the Bartlett sum formulas Learning, for the sine and cosine as follows: tan( ) 5 sin( ) (7) cos( ) 5 sin cos cos sin. cos cos sin sin Jones & Bartlett We now Learning, divide the numerator and denominator of Jones (7) b cos & Bartlett cos (assuming Learning, that LL and are such that cos cos 0), sin cos cos sin cos cos cos cos tan tan tan( ) 5 cos cos sin 5. sin tan tan cos cos cos cos FOR SALE OR The derivation of the difference formula for tan( ) is obtained in a similar manner. We summarize the two results. (8) THEOREM 4.5.3FOR Sum SALE and OR Difference Formulas for the Tangent For real numbers and for which the functions are defined, tan( ) 5 tan tan (9) tan tan Jones & Bartlett Learning, LL tan( ) 5 tan tan FOR SALE OR DISTRIBUT (0) tan tan EXAMPLE 3 Tangent of FOR a Difference SALE OR Evaluate tan(p/). Solution If we think of p/ as an angle in radians, then pjones radians & 5 Bartlett Learning, 30 5 p 4 p 6 radians. 4.5 Sum and Difference Formulas 5

47 Jones & Bartlett Learning.. It follows from formula (0) of Theorem 4.5.3: p p tan 5 tanap 4 p tan 6 b 5 4 tan p 6 You should rework this eample using p tan 4 tan p p/ 5p/3 p/4 6 to see that the result is the same. Jones & Bartlett 5 Learning,!3 # 5!3!3 5!3 #!3!3!3 A!3 B 5 5 4!3 5!3. Strictl speaking, we reall do not need the identities for tan( 6 ), since we can alwas compute sin( 6 ) and cos( 6 ) using (), (), (4), (5) and then proceed as in (7), that is, form the quotient sin( 6 )/cos( 6 ). Double-Angle Formulas Man useful trigonometric formulas can be derived from the sum and difference formulas. The double-angle formulas for the cosine and sine functions epress the cosine and sine of in terms of the cosine and sine of. If we set 5 5 in () and use cos( ) 5 cos, then cos 5 cos cos sin sin 5 cos sin. Similarl, b setting 5 5 in (4) and using sin( ) 5 sin, then these two terms are Jones equal & Bartlett Learning, T TFOR SALE OR sin 5 sin cos cos sin 5 sin cos. We summarize the last two results along with the double-angle formula for the tangent function. THEOREM Double-Angle FOR SALE Formulas OR For a real number for which the functions are defined, tan 5!3 d rationalizing the denominator cos 5 cos sin sin 5 sin cos tan tan () Jones & Bartlett () Learning, LL FOR SALE OR (3) DISTRIBUT EXAMPLE 4 Using the Double-Angle Formulas FOR SALE OR If sin 54and p,, 3p/, find the eact FOR values SALE of cos OR and sin. Solution First, we compute cos using sin cos 5. Since p,, 3p/, cos, 0, and so we choose the negative square root: Jones & cos 5 " sin Bartlett 5 Å Learning, a 4 b! CHAPTER 4 TRIGONOMETRIC FUNCTIONS

48 Jones & Bartlett Learning.. From the double-angle formula () of Theorem 4.5.4, FOR cos SALE 5 cos OR sin!5 5 a 4 b a 4 b Jones 8. & Bartlett Learning, LL Finall, from the double-angle formula (), sin 5 sin cos 5 a 4 ba!5 4 b 5!5 8. The formula in () has two useful FOR alternative SALE forms. OR B (8) of Section 4.4, we know that sin 5 cos. Substituting the last epression into () ields cos 5 cos ( cos ) or cos 5 cos. On the other hand, Jones if we substitute & Bartlett cos Learning, 5 sin into () we get cos 5 sin. Half-Angle Formulas The alternative forms (4) and (5) of the double-angle formula () are the source of two half-angle formulas. Solving (4) and (5) for cos Jones & Bartlett and sin Learning, gives, respectivel, cos 5 FOR ( cos ) (6) and sin 5 SALE OR DISTRIBUT ( cos ). B replacing the smbol in (6) b / and using (/) 5, we obtain the half-angle formulas for the cosine and sine functions. THEOREM Half-Angle Formulas For a real number for which the functions are defined, (4) (5) cos 5 ( cos ) Jones & Bartlett Learning, FOR SALE OR sin 5 ( cos ) (7) (8) tan (9) 5 cos cos EXAMPLE 5 Using the Half-Angle Formulas Find the eact values of cos(5p/8) and sin(5p/8). Jones & Bartlett Learning, Solution If we let 5 5p/4, then Jones / 5& 5p/8 Bartlett and Learning, formulas (7) and (8) of Theorem ield, respectivel, and cos 5p 8 5 a cos 5p 4 b 5 c a!! bd 5, 4 sin 5p Jones 8 5 a & Bartlett cos 5p 4 b 5 Learning, c a!! bd Sum and Difference Formulas 53

49 Jones & Bartlett Learning.. Because 5p/8 radians is a second-quadrant angle, cos(5p/8), 0 and sin(5p/8). 0. Therefore, we take the negative square root for the value of the cosine, 5p cos 8 5! "! 5, Å 4 and the positive square root for the value of the sine, 5p sin 8 5! "! 5. Å 4 If want the eact value of, sa, tan(5p/8) we can use the results of Eample 5 or formula (9) with 5 5p/4. Either wa, the result is the same 5p! FOR SALE tan 5" 5!. d e rationalizing OR the denominator 8 "! Reducing Powers of Sine and Cosine As discussed in Section 3.7, the princi- topics of stud in calculus Jones are derivatives & Bartlett and Learning, integrals of functions. Trigonometric pal identities are especiall useful FOR in integral SALE calculus. OR To evaluate integrals of powers of the sine and cosine, specificall cos n and sin n, where n $ is an even positive integer, ou would use the half-angle formulas in the form given in (6). The idea is to epress cos n and sin n in terms of one or more of cosine functions raised to the first power. The net eample illustrates the idea. FOR EXAMPLE SALE OR 6 Reducing a Power Epress sin 4 in terms of first-powers of cosine functions. Solution We first use the laws of eponents to rewrite sin 4 as (sin ) and then use Jones & Bartlett Learning, the second identit in (6) to replace sin Jones in terms & of Bartlett cos : Learning, d e replace sin using sin 4 5 (sin ) 5 c ( cos ) d second formula in (6) d epand 5 4 cos 4 c ( cos 4) d now replace cos using d the first formula in (6) with replaced b 5Jones 4 cos & Bartlett Learning, 4 cos cos cos 4. 8 As required, the last line contains onl first-powers of cosine functions of multiples of the original independent variable. ES FROM THE CLASSROOM (i) Should ou memorize FOR all the SALE identities OR presented in this section? You should consult our instructor about this, but in the opinion of the authors, ou should at the ver least memorize formulas (), (), (4), (5), (), and (). (ii) When ou enroll in a calculus course, check the title of our tet. If it has the words Earl Transcendentals in its chlorophlle/shutterstock, Inc. 54 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

50 Jones & Bartlett Learning.. title, then Jones our knowledge & Bartlett of the Learning, graphs and properties of the trigonometric functions FOR will come SALE into OR pla almost immediatel. (iii) At some point in our stud of integral calculus ou ma be required to evaluate integrals of products such as sin sin 5 and sin 0 cos 4. Jones & Bartlett Learning, One wa of doing this is to use the sum/difference Jones & formulas Bartlett to Learning, devise an LL identit that converts these products into either FOR a sum SALE of sines OR or DISTRIBUT a sum of cosines. This will be the topic of discussion in the net section. Eercises 4.5 Answers to selected odd-numbered problems begin on page ANS 6. In Problems, use a sum or difference formula to find the eact value of the given trigonometric function. Do not use a calculator. p p. cos. sin Jones & Bartlett Learning, 3. sin cos 75 7p p 5. sin 6. cos 7. tan 5p 8. cosa 5p Jones & Bartlett Learning, p 9. sina p Jones b & Bartlett Learning, LL 0. tan b p 7p. sin. tan 3. cos sin 65 Jones & Bartlett Learning, 5. tan 65 Jones 6. & Bartlett cos 95 Learning, 7. sin 95 FOR 8. SALE tan 95 OR 9. cos sin 345 3p 7p. cos. tan In Problems 3 8, Jones use a double-angle & Bartlett formula Learning, to write the given epression as a single trigonometric function FOR of twice SALE the angle. OR 3. cos b sin b 4. cos t sin t 5. p 6. cos 9 sin 5 Jones & Bartlett tan Learning, 3t Jones sin cos & Bartlett Learning, LL FOR SALE OR tan 3t FOR SALE OR DISTRIBUT In Problems 9 34, use the given information to find (a) cos, (b) sin, and (c) tan. 9. sin 5!/3, p/,,p 30. cos 5!3/5, 3p/,, p 3. tan 5, p,, 3p/ 3. csc 53, p,, 3p/ Jones & Bartlett Learning, 33. sec cot 5 4 5, p/,,p Jones & Bartlett Learning, 3, 0,,p/ In Problems 35 44, use a half-angle formula to find the eact value of the given trigonometric function. Do not use a calculator. 35. cos(p/) 36. sin(p/8) 37. sin(3p/8) Jones & Bartlett Learning, 38. tan(p/) 39. cos sin5 4.5 Sum and Difference Formulas 55

51 Jones & Bartlett Learning.. 4. tan05 4. cot csc(3p/) 44. sec(3p/8) In Problems 45 50, use the given information to find (a) cos(/), (b) sin(/), and (c) tan(/). 45. sin 5 3, p/,,p 46. cos 5 4 5, 3p/,, p Jones 47. & tan Bartlett 5, Learning, p,, 3p/ 48. csc 5 9, 0 Jones,,p/ & Bartlett Learning, LL FOR 49. SALE sec 5OR 3, 0,, cot 5 4, 90 FOR,, SALE 80 OR DISTRIBUT In Problems 5 60, verif the given identit. 5. sin cos (sin sin 3 ) 5. cos cos 3 3 cos Jones & Bartlett Learning, 53. (sin cos ) 5 sin 54. Jones cos & 5Bartlett cos 4 Learning, sin 4 FOR SALE OR 55. cot 5 (cot tan ) 56. sec 5 cos tan cot tan sin cos tan cot tan 59. tan 60. tan 5 sin 5 cos sin cos 56 CHAPTER 4 TRIGONOMETRIC FUNCTIONS In Problems 6 64, proceed as in Eample 6 and use (6) to rewrite the given function in terms of first powers of cosine functions. 6. sin 5 6. cos 63. cos sin 4 3 In Problems 65 and 66, proceed as in Problems 6 64 to rewrite the given function in terms of first powers of cosine functions. You will also need (7) of Section.5 and the identit in Problem 5 written in the form cos cos Jones & Bartlett Learning, FOR cos 3. 4 SALE OR 65. sin cos 6 In Problems 67 70, rewrite the given function as a single trigonometric function involving no products or squares. Give Jones the amplitude & Bartlett and period Learning, of the function cos FOR SALE 68. OR 5 sin(/)cos(/) sin cos cos 4 5 sin 4 7. If P( ) and P( ) are points in quadrant II on the terminal side of the angles and respectivel, with cos and sin 5 5, 3 3, find (a) sin( ), Jones &(b) Bartlett cos( Learning, ), (c) sin( ), and (d) cos( Jones ). & Bartlett Learning, LL FOR 7. SALE If is a quadrant II angle, is a quadrant III angle, sin and tan , 8 OR 4, find (a) sin( ), (b) sin( ), (c) cos( ), and (d) cos( ). Applications Jones & Bartlett Learning, 73. Mach Number The ratio of the speed Jones of an airplane & Bartlett to the speed Learning, of sound is θ FOR SALE OR called the Mach number M of the plane. If FOR M., SALE the plane OR makes sound waves that form a (moving) cone, as shown in FIGURE A sonic boom is heard at the intersection of the cone with the ground. If the verte angle of the cone is u, then u sin 5 Jones M. FIGURE 4.5. & Bartlett Airplane in Learning, Problem FOR 73SALE OR If u5p/6, find the eact FOR value of SALE the Mach OR number.

52 Jones & Bartlett Learning Cardiovascular Branching A mathematical model for blood flow in a large blood vessel predicts that the optimal values of the angles u and u, which represent the (positive) FOR angles SALE of OR the smaller daughter branches (vessels) with respect to the ais of the parent branch, are given b and cos u 5 A 0 A A cos u 5 A 0 A A, A 0 A Jones & Bartlett A 0 A Learning, LL FOR SALE OR where A 0 is the cross-sectional area of the parent branch FOR and SALE A and OR A are DISTRIBUT the cross-sectional areas of the daughter branches. See FIGURE Let c5u u be the junction angle, as shown in the figure. A 0 Blood flow θ A θ ψ = θ + θ A FIGURE FOR SALE Branching OR of a large blood vessel in Problem 74 (a) Show that cos c5 A 0 A Jones A & Bartlett Learning, LL. A A FOR SALE OR DISTRIBUT (b) Show that for the optimal values of and u, the cross-sectional area of the daughter branches, A A, is greater than or equal to that of the parent branch. Therefore, the blood must slow down in the daughter branches. FOR SALE OR 75. Range of a Projectile We saw in FOR Problem SALE 56 of OR Eercises v 0 4. that if a projectile, such as a shot put, is released from a height h, upward at an angle u with θ velocit v 0, the range R at which it strikes the ground is given b Jones & Bartlett Learning, h Ground R R 5 v 0 cosu Jones & gbartlett av 0 sinu Learning, "v 0 sin ughb, where g is the acceleration due to gravit. See FIGURE (a) Show that when h 5 0 the range of the projectile is FIGURE Projectile in R 5 v 0 sin u Problem 75. g (b) It can be shown that the maimum range R ma is achieved when the angle u satisfies the equation cos u 5 gh Jones & v 0 Bartlett gh. Learning, Show that maimum range is FOR SALE OR R ma 5 v 0"v 0 gh g b using the Jones epressions & Bartlett for R and Learning, cosu and the half-angle formulas for the sine and the cosine FOR with SALE t 5OR u. 4.5 Sum and Difference Formulas 57 u

53 Jones & Bartlett Learning.. For Discussion 76. Discuss: Wh would ou epect FOR SALE to get an OR error message from our calculator when ou tr to evaluate tan 35 tan 55 tan 35 tan 55? 77. In Eample we showed that sin (7p/) 5 4(!!6). Following the eample, we then showed that sin (7p/) 5 "!3. Demonstrate that these two answers are equivalent. 78. Discuss: How would ou find a formula that epresses sin 3u in terms of sin u? Carr out our ideas. 79. In Problem 7, in what quadrants do P( ) and P( ) lie? FOR SALE OR 80. In Problem 7, in which quadrant does the FOR terminal SALE side of OR lie? The terminal side of? 4.6 Product-to-Sum and Sum-to-Product Formulas INTRODUCTION There are instances, especiall in integral calculus, where it is necessar to convert a product of sine and cosine functions to a sum of these functions. Moreover, in solving trigonometric equations we ma find it convenient to convert a sum of sine and cosine functions into a product of these functions. In the discussion that follows & Bartlett we establish Learning, trigonometric identities or formulas that do Jones the job. & Bartlett Learning, LL Jones Reduction of a Product to a Sum The product-to-sum formulas given in the net theorem are direct consequences of the sum and difference formulas for the cosine and sine functions in Section 4.5. Jones & Bartlett Learning, THEOREM 4.6. Product-to-Sum Jones Formulas & Bartlett Learning, For all real numbers and, sin sin 5 [cos( ) cos( )] () cos cos 5 [cos( ) cos( )] () sin cos Jones 5 [sin( & Bartlett ) Learning, sin( )] (3) PROOF: To prove (), we use () and () of Theorem 4.5.: cos( ) 5 cos cos sin sin cos( ) 5 cos cos sin sin. (4) (5) Jones Subtracting & Bartlett (5) Learning, from (4) ields cos( ) cos( ) 5 sin sin. And so, sin sin 5 [cos( ) cos( )] Jones & Bartlett Learning, which is (). Similarl, b adding (4) and (5) Jones we get& Bartlett Learning, cos( ) cos( ) 5 cos cos, which in turn, ields formula (): cos cos 5 [cos( ) cos( )]. Formula (3) follows analogousl Jones & Bartlett b adding Learning, the sum and difference formulas for the sine, (4) and (5) in Theorem FOR 4.5. SALE of Section OR CHAPTER 4 TRIGONOMETRIC FUNCTIONS

54 Jones & Bartlett Learning.. Although we do not feel that it is necessar to memorize () (3) in Theorem 4.6., ou should listen to what our instructor requires. B remembering the procedure just illustrated in the proof of Theorem 4.6. each of these formulas can be derived on the spot. Jones & Bartlett EXAMPLE Learning, Using () of Theorem Jones 4.6.& Bartlett Learning, LL FOR SALE Use OR a product-to-sum formula to rewrite the product cos FOR u cos SALE 3u as a OR sum. DISTRIBUT Solution From formula () of Theorem 4.6. with the identifications 5 u and 5 3u, we obtain cos u cos 3u 5 [cos (u Jones 3u) & Bartlett cos (ulearning, 3u)] 5 [cos (u) FOR cos SALE 5u] OR d cos(u) 5 cos u 5 [cos ucos 5u]. EXAMPLE Jones & Using Bartlett (3) of Learning, Theorem 4.6. Use a product-to-sum formula to find the eact value of the product sin 45 cos 5. Solution Using formula (3) of Theorem 4.6. with 5 45 and 5 5, we have Jones & Bartlett Learning, sin 45 cos [sin(45 5 ) Jones sin(45 & Bartlett 5 )] Learning, LL [sin 60 sin 30 ]. FOR SALE OR DISTRIBUT sin 45 cos 5 5 Because sin 60 5!3 and sin 30 5 we observe that the eact value of the given product is [sin 60 sin 30 ] 5 (!3 ) 5 4(!3 ). Reduction of a Sum to a Product The results in Theorem 4.6. can now be used to derive the sum-to-product formulas. THEOREM 4.6. Sum-to-Product Formulas For all real numbers and, sin sin 5 sin cos (6) Jones & Bartlett Learning, LL sin sin 5 cos sin FOR SALE OR DISTRIBUT (7) cos cos 5 cos cos (8) cos cos Jones 5 & sin Bartlett sin Learning, (9) PROOF: B replacing, in turn, the smbols and in () of Theorem 4.6. b Jones & Bartlett Learning, and, FOR SALE OR (0) 4.6 Product-to-Sum and Sum-to-Product Formulas 59

55 Jones & Bartlett Learning.. we get sin sin 5 ccosa FOR SALE OR b cosa 5 [cos cos ]. Multipling the last epression b ields sin sin 5 cos cos, which is formula (9). In a similar manner, each of the remaining product-to-sum formulas together with the substitutions in (0) ields one of the sum-to-product formulas. bd EXAMPLE 3 Using (9) of Theorem 4.6. Use a sum-to-product formula Jones to rewrite & Bartlett the sum Learning, cos t cos 5t as a product. Solution We use formula (9) of FOR Theorem SALE 4.6. OR with 5 t and 5 5t: cos t cos 5t 5 sina t 5t b sina t 5t b 5 sin 3t sin(t) d sin(t) 5sin t 5 sin 3t sin t EXAMPLE 4 Using (6) of Theorem 4.6. Jones & Bartlett Learning, Use a sum-to-product formula to find the eact Jones value & of Bartlett the sum sin Learning, 75 sin 5. Solution In this case we use formula (6) of Theorem 4.6. with 5 75 and 5 5 : sin 75 sin 5 5 sina b cos a b 5 sin 45 cos 30. Because sin 45 5! and cos FOR 30 SALE 5!3 OR the eact value of the given sum sin 75 sin 5 5 sin 45 cos 30 5 (!)(!3) 5!6. Eercises 4.6 Answers to selected odd-numbered problems begin on page ANS 7. Jones & Bartlett Learning, In Problems, use a product-to-sum formula Jones in & Theorem Bartlett 4.6. Learning, to write the given product as a sum of cosines or a sum of sines. 3t.. sin cos t cos 4u cos 3u 3. sin sin 5 4. sin 0 cos cos 6. sin t sin t 3 cos 3 60 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

56 Jones & Bartlett Learning.. 7. sin 8 cos 8. sin pu cos 7pu 9. cos 3b sin b 0. 6 sin a sin 4a. sina p FOR. sinat p b cosat p 4 b sina p SALE OR 4 b b In Problems 3 8, use a product-to-sum formula in Theorem 4.6. to find the eact value Jones & Bartlett of the given Learning, epression. Do not use a calculator sin 5p FOR SALE OR DISTRIBUT 5p cos sin p 8 cosp 8 5. sin 75 sin 5 6. cos 5 cos sin 97.5 sin sin 05 cos 95 In Problems 9 30, use a sum-to-product-formula FOR SALE OR in Theorem 4.6. to write the given sum as a product of cosines, a product of sines, or a product of a sine and a cosine. 9. sin sin 5 0. cos 3u cos u 9.. sin sin 3 cos cos Jones & Bartlett Learning, 3. cos cos 6 FOR SALE OR 4. sin 5t sin 3t 5. sin v t sin v t 6. (cos a cos b) 7. cos t 8. sin(u p) sin(u p) 9. sinat p cos 5t 30. cosat p b cosat p b sinat p b Jones & Bartlett Learning, b LL In Problems 3 36, use a sum-to-product-formula in Theorem 4.6. to find the eact value of the given epression. Do not use a calculator. 3p p sin sin 5p! sin! sin 5p Jones & Bartlett Learning, 33. cos 05 cos 5 FOR 34. SALE cos 5 OR cos sin 95 sin cos 95 cos05 Applications 37. Sound Wave A note produced b a certain musical instrument results in a sound wave described b FOR SALE OR f (t) sin 500pt 0.03 sin 000pt, where f (t) is the difference between atmospheric pressure and air pressure in dnes per square centimeter at the eardrum after t seconds. Epress f as the product of a sine and a cosine function. 38. Beats If two piano wires struck b the same ke are slightl out of tune, the difference between the atmospheric pressure and air pressure at the eardrum can be represented b the function f (t) 5 a cos pb t a cos pb t, where the value of the constant b is close to the value of constant b. The FOR SALE OR variations in loudness that occur are called beats. See FIGURE The two strings can be tuned to the same frequenc b tightening one of them while sounding both until the beats disappear. (a) Use a sum formula to write f (t) as a product. (b) Show that Jones f (t) can & be Bartlett considered Learning, a cosine function with period /(b b ) and variable amplitude FOR SALE a cos OR p(b b )t. 4.6 Product-to-Sum and Sum-to-Product Formulas 6

57 Jones & Bartlett Learning.. = a cos pb t t = a cos pb t t (c) Use a graphing utilit to obtain the graph of f in the case pb 5 5, pb 5 4, and a Alternating Current The term sin vt sin(vt f) is encountered in the derivation of an epression for Jones the power & Bartlett in an alternating-current Learning, circuit. Show that this term can be written as FOR [cos fsale cos(vt OR f)]. For Discussion = a cos pb t + a cos pb t FIGURE 4.6. Graph for Problem If 3 5p, then show that Jones & Bartlett Learning, sin sin sin sin sin Jones sin 3. & Bartlett Learning, LL FOR 4. SALE Write as OR a product of cosines: cos t cos 4t cos 6t. FOR SALE OR DISTRIBUT 4. Simplif: cos t cos t cos 3t. 4.7 Inverse Jones & Trigonometric Bartlett Learning, Functions Recall, a function f is one-to-one if ever in its range corresponds to eactl one in its domain. INTRODUCTION Although we can find the values of the trigonometric functions of real numbers or angles, in man applications we must do the reverse: Given the value of a trigonometric function, find a corresponding angle or number. This suggests we consider inverse trigonometric functions. Before we define the inverse trigonometric functions, let s recall from Section.8 Jones some of & the Bartlett properties Learning, of a one-to-one function f and its inverse f. Properties of Inverse Functions If 5 f () is a one-to-one function, then there is a unique inverse function f with the following properties: FOR Properties SALE OR of Inverse Functions The domain of f 5 range of f. The range of f 5 domain of f. 5 f () is equivalent to 5 f (). Jones & Bartlett Learning, The graphs of f and f are reflections Jones in the & line Bartlett 5. Learning, f ( f ()) 5 for in the domain of f FOR. SALE OR f ( f ()) 5 for in the domain of f. Inspection of the graphs of the various trigonometric functions clearl shows that none of these functions are one-to-one. Jones In Section & Bartlett.8 we discussed Learning, the fact that if a function f is not FOR SALE See Eample OR 8 in Section.8. one-to-one, it ma be possible to FOR restrict SALE the function OR to a portion of its domain where it is 6 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

58 Jones & Bartlett Learning.. one-to-one. Then we can define an inverse for f on that restricted domain. Normall, when we restrict the domain, we make sure to preserve the entire range of the original function. Arcsine Function From FIGURE 4.7.(a) we see that the function 5 sin on the closed interval [p/, p/] takes on all values in its range [, ]. Notice that an horizontal line drawn to intersect the red portion of the graph can do so at most once. Jones & Bartlett Thus the Learning, sine function on this restricted domain is Jones one-to-one & Bartlett and has Learning, an inverse. LL FOR SALE There OR are two commonl used notations to denote the inverse FOR of SALE the function OR DISTRIBUT shown in Figure 4.7.(b): arcsin or sin, and are read arcsine of and inverse sine of, respectivel. = sin on (, ) = sin on [ /, /] (a) Not a one-to-one function 4.7 Inverse Trigonometric Functions 63 (b) A one-to-one function FIGURE 4.7. Restricting the domain of 5 sin to produce a one-to-one function In FIGURE 4.7.(a) we have reflected the portion of the graph of 5 sin on the interval Jones & Bartlett [p/, Learning, p/] (the red graph in Figure 4.7.(b)) about Jones the line & 5 Bartlett to obtain Learning, the graph LL of 5 arcsin (in blue). For clarit, we have reproduced this blue graph in Figure 4.7.(b). As this curve shows, the domain of the arcsine function is [, ] and the range is [p/, p/]. = arcsin (a) = (b) FIGURE 4.7. Graph of 5 arcsin is the blue curve = arcsin = sin Proceeding as in (6) of Section.8, the inverse of 5 sin, p/ # #p/, Jones & Bartlett is obtained Learning, b interchanging the smbols and, Jones that is, & Bartlett 5 arcsin Learning, is defined LL FOR SALE implicitl OR b 5 sin, p # # p. The following definition summarizes the discussion. DEFINITION 4.7. Arcsine FOR Function SALE OR The arcsine function, or inverse sine function, is defined b 5 arcsin if and onl if 5 sin where # # Jones and p/ & Bartlett # #p/. Learning, ()

59 Jones & Bartlett Learning.. In words: The arcsine of the number FOR is that SALE number OR (or radian-measured angle) satisfing p/ # # p/ whose sine is. When using the notation sin it is important to realize that is not an eponent; rather, it denotes an inverse function. The notation arcsin has an advantage Jones over & Bartlett the notation Learning, sin that there is no and hence Jones no potential & Bartlett for misinterpretation; SALE moreover, OR the prefi arc refers to an angle the angle FOR whose SALE sine is OR. But DISTRIBUT Learning, LL FOR since 5 arcsin and 5 sin are used interchangeabl in calculus and in applications, we will continue to alternate their use so that ou become comfortable with both notations. EXAMPLE Evaluating the Inverse FOR SALE Sine Function OR Note of Caution: (sin ) 5 sin sin Find (a) arcsin, (b) sin ( ), and (c) sin (). Solution (a) If we let 5 arcsin, then b () we must find the number (or radian- angle) that satisfies Jones sin & 5Bartlett and Since sin (p/6) 5 p/ Learning, # #p/. measured and p/6 satisfies the inequalit FOR p/ SALE # #p/ OR it follows that 5p/6. (b) If we let 5 sin ( ), then sin 5. Since we must choose such that p/ # #p/, we find that 5p/6. (c) Letting 5 sin (), we have that sin 5and p/ # #p/. FOR Hence SALE 5p/. OR In parts (b) and (c) of Eample we were careful to choose so that p/ # #p/. For eample, it is a common error to think that because Jones & Bartlett Learning, sin(3p/) 5, then necessaril sin () Jones can be & taken Bartlett to be Learning, 3p/. Remember: If FOR SALE OR 5 sin, then is subject to the restriction FOR p/ SALE # #p/ OR and 3p/ does not satisf this inequalit. Read this paragraph several times. EXAMPLE Evaluating a Composition Jones & Bartlett Without using a calculator, find tan(sin Learning, FOR SALE 4). OR Solution We must find the tangent of the angle of t radians with sine equal to 4, that is, tan t, where t 5 sin 4. The angle t is shown in FIGURE Since tan t 5 sin t cos t 5 4 cos t, t 4 FOR we want SALE to determine OR the value of cos t. From Figure and the FOR Pthagorean SALE identit OR DISTRIBUT cos t sin t cos t 5, we see that a 4 b cos t 5 or cos t 5!5 4. FIGURE The Jones angle t 5& sin Bartlett 4 in Learning, Hence we have Eample tan t 5 /4!5/4 5!5 5!5 5, and so Jones tan asin & Bartlett!5 b 5 tan tlearning, CHAPTER 4 TRIGONOMETRIC FUNCTIONS

60 Jones & Bartlett Learning.. Arccosine Function Jones & If we Bartlett restrict the Learning, domain of the cosine function to the closed interval [0, p], the resulting FOR SALE function OR is one-to-one and thus has an inverse. We denote this inverse b arccos or cos. B interchanging the smbols and in 5 cos, 0 # #p, the inverse function 5Learning, arccos is defined implicitl b Jones & Bartlett 5 cos, 0 # #p. DEFINITION 4.7. Arccosine Function Jones & Bartlett Learning, The arccosine function, orinverse Jones cosine & function, Bartlett is defined Learning, b 5 arccos FOR if and onl SALE if OR 5 cos where # # and 0 # #p. () The graphs shown Jones in FIGURE & Bartlett illustrate Learning, how the function 5 cos restricted to the interval [0, p] becomes FOR a SALE one-to-one OR function. The inverse of the function shown in Figure 4.7.4(b) is 5 arccos. = cos on (, ) Jones & Bartlett = cos Learning, LL FOR SALE on [0, OR ] DISTRIBUT 0 (a) Not a one-to-one function (b) A one-to-one function FIGURE Restricting the domain of 5 cos to produce a one-to-one function = arccos Jones & Bartlett Learning, B reflecting Jones the graph & of Bartlett the one-to-one Learning, function in Figure 4.7.4(b) in the line 5 we obtain the graph FOR of SALE 5 arccos OR shown in FIGURE Note that the figure clearl shows that the domain and range of 5 arccos are [, ] and [0, p], respectivel. EXAMPLE 3 Evaluating the Inverse Cosine Function FIGURE Graph of 5 arccos Jones & Bartlett Find (a) Learning, arccos (!/) (b) cos (!3/). Solution (a) If we let 5 arccos(!/), then cos 5!/ and 0 # #p. Thus 5p/4. (b) Letting 5 cos (!3/), we have that cos 5!3/, and we must find such that 0 # #p. Therefore, 5 5p/6 since cos (5p/6) 5!3/. Jones & Bartlett sin t t Learning, EXAMPLE 4 Evaluating FOR the SALE = cos t Compositions OR of Functions Write sin(cos ) as an algebraic epression in. Solution In FIGURE we have constructed an angle of t radians with cosine equal to Jones & Bartlett FIGURE Learning, The angle. Then t 5 cos Jones, or 5& cos Bartlett t, where Learning, 0 # t #p. Now to find sin (cos ) 5 sin t, FOR SALE t 5 cos OR in Eample 4 we use the identit sinfor t cos SALE t 5. OR Thus 4.7 Inverse Trigonometric Functions 65

61 Jones & Bartlett Learning.. Jones sin & t Bartlett 5 Learning, FOR SALE sin t 5OR sin t 5 " sin (cos ) 5 ". We use the positive square root of, since the range of cos is [0, p], and the Jones sine & of Bartlett an angle Learning, t the first or second quadrant is positive. Arctangent Function If we restrict the domain of tan to the open interval (p/, p/), then the resulting function is one-to-one and thus has an inverse. This inverse is denoted b arctan or Jones tan &. Bartlett Learning, DEFINITION Arctangent Function The arctangent, or inverse Jones tangent, & Bartlett function is Learning, defined b 5 arctan FOR SALE if and OR onl if 5 tan where `,,`and p/,,p/. (3) The graphs shown in FIGURE illustrate how the function 5 tan restricted to the open interval (p/, p/) becomes a one-to-one function. = tan = tan on ( /, /) (a) Not a one-to-one function (b) A one-to-one function FIGURE Restricting the domain of 5 tan to produce a one-to-one function = arctan B reflecting the graph of the one-to-one function in Figure 4.7.7(b) in the line 5 we obtain the graph of 5 arctan shown in FIGURE We see in the figure that the domain and range of 5 arctan are, in turn, the intervals (`, `) and FIGURE Jones Graph of& Bartlett Learning, (p/, p/). 5 arctan EXAMPLE 5 Evaluating the Inverse Tangent Function Find tan (). Solution If tan () 5 Jones, then tan & Bartlett 5, where Learning, p/,,p/. It follows that tan () 5 5p/4. 66 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

62 Jones & Bartlett Learning.. EXAMPLE 6Jones & Evaluating Bartlett Learning, the Compositions of Functions Without using a calculator, FOR find SALE OR Solution If we let t 5 arctan( 5 3), then tan t The Pthagorean identit tan t 5 sec t can be used to find sec t: 5 a 3 b 5 sec t sec t 5 5 Å Å 9 5 "34 3. Jones & Bartlett Learning, In the preceding line we take the positive Jones square & Bartlett root because Learning, t 5 arctan( 5 3) is in the interval (p/, p/) (the range of the FOR arctangent SALE function) OR and the secant of an angle t in the first or fourth quadrant is positive. Also, from sec t 5!34/3 we find the value of cos t from the reciprocal identit: cos t 5 sec t 5!34/3 5 3!34. Finall, we can use the FOR identit SALE tan OR t 5 sin t/cos t in the form sin t 5 tan t cos t to compute It follows that sin (arctan ( 5 3)). sin (arctan ( 5 3)). sin t 5 tan t cos t 5 a 5 3 ba 3!34 b 5 5!34. Properties of the Inverses Recall from Section.8 that f ( f ()) 5 and f ( f ()) 5 hold for an function f and its inverse under suitable restrictions on. Thus for the inverse trigonometric functions, we have the following properties. THEOREM 4.7. Properties FOR of Inverse SALE OR Trigonometric Functions (i) arcsin(sin ) 5 sin (sin ) 5 if p/ # #p/ (ii) sin(arcsin ) 5 sin(sin ) 5 if # # (iii) arccos(cos ) 5 cos (cos ) 5 if 0 # #p (iv) cos(arccos Jones ) 5 cos(cos & Bartlett ) 5Learning, if # # (v) arctan(tan ) FOR 5 tan SALE (tan ) OR 5 if p/,,p/ (vi) tan(arctan ) 5 tan(tan ) 5 if `,,` Jones & Bartlett EXAMPLE Learning, 7 Using the Inverse Properties Jones & Bartlett Learning, LL Without using a calculator, evaluate: p (a) (b) cos (c) tan 3p sin asin atan b acos 3 b 4 b. Jones & Bartlett Learning, Solution In each case we use the Jones properties & Bartlett of the inverse Learning, trigonometric functions given in Theorem (a) Because p/ satisfies p/ # #p/it follows from propert (i) that sin p asin b 5 p. (b) B propert (iv), cos FOR (cossale 3) 5OR Inverse Trigonometric Functions 67

63 Jones & Bartlett Learning.. (c) In this case we cannot appl propert (v), because 3p/4 is not in the interval (p/, p/). If we first evaluate tan(3p/4) 5, then we have see Eample 5 T tan 3p atan 4 b 5 tan () 5 p 4. FOR SALE In the net OR section we illustrate how inverse trigonometric functions FOR SALE can be OR used DISTRIBUT to solve trigonometric equations. Postscript The Other Inverse Trig Functions The functions cot, sec, and Jones & Bartlett Learning, csc also have inverses when their domains are Jones suitabl & Bartlett restricted. Learning, See Problems 49 5 FOR SALE OR in Eercises 4.7. Because these functions are not FOR used SALE as often OR as arctan, arccos, and arcsin, most scientific calculators do not have kes for them. However, an calculator that computes arcsin, arccos, and arctan can be used to obtain values for arccsc, arcsec, and arccot. Unlike the fact that sec 5 /cos, we note that sec /cos ; rather, sec 5 cos (/) for 0 0 $. Similar relationships hold for csc and cot. See Problems in Eercises 4.7. Eercises 4.7 Answers Jones to selected & Bartlett odd-numbered Learning, problems begin on page ANS 7. In Problems 4, find the eact value of the given trigonometric epression. Do not use a calculator.. sin 0. tan!3 Jones & Bartlett Learning,!3 3. arccos() Jones 4. arcsin & Bartlett Learning, FOR SALE OR 5. arccos 6. arctan (!3)!3 7. sin a!3 8. cos b! 9. tan Jones & Bartlett 0. sin Learning,. arctan a!3. arccos( 3 b ) 3. sin a! 4. arctan 0 Jones & Bartlett Learning, b In Problems 5 3, find the eact value of the given trigonometric epression. Do not use a calculator. 5. sin(cos 3 5) 6. cos(sin 3) Jones & Bartlett Learning, 7. tan(arccos ( 3)) 8. Jones sin(arctan & Bartlett 4) Learning, 9. cos(arctan ()) 0. FOR tan(sinsale ( 6)) OR. csc(sin 3 5). sec(tan 4) 3. sin(sin 5) 4. cos(cos ( 4 5)) 5. tan(tan.) 6. sin(arcsin 0.75) p 7. arcsinasin p 8. arccos acos 6 b 3 b 68 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

64 Jones & Bartlett Learning sin 5p tan (tan p) Jones & Bartlett Learning, asin 6 b p 3. cos acos a p FOR SALE OR 3. arctan atan 4 bb 7 b Jones & Bartlett In Problems Learning, 33 40, write the given epression as an Jones algebraic & epression Bartlett Learning, in. LL 33. sin(tan ) 34. cos(tan ) 35. tan(arcsin ) 36. sec(arccos ) 37. cot(sin ) 38. cos(sin ) 39. csc(arctan ) 40. tan(arccos ) In Problems 4 48, sketch the graph of the given function arctan p arctan arcsin Jones 0 & Bartlett Learning, sin ( ) cos FOR SALE OR cos arccos( ) cos(arcsin ) 49. The arccotangent function can be defined b 5 arccot (or 5 cot ) if and onl if 5 cot, where 0,,p. Graph 5 arccot, and give the Jones & Bartlett domain Learning, and the range of this function. FOR SALE 50. OR The arccosecant function can be defined b 5 arccsc FOR SALE (or 5OR csc DISTRIBUT ) if and onl if 5 csc, where p/ # #p/and 0. Graph 5 arccsc, and give the domain and the range of this function. 5. One definition of the arcsecant function is 5 arcsec (or 5 sec ) if and onl if 5 sec, where 0 Jones # #p& and Bartlett p/. Learning, (See Problem 5 for an alternative definition.) Graph 5 FOR arcsec SALE, and give OR the domain and the range of this function. 5. An alternative definition of the arcsecant function can be made b restricting the domain of the secant function to [0, p/) [p, 3p/). Under this restriction, define the arcsecant function. Graph 5 arcsec, and give the domain and the range of this function. Jones & Bartlett Learning, 53. Using the definition FOR of the SALE arccotangent OR function from Problem 49, for what values of is it true that (a) cot(arccot ) 5 and (b) arccot (cot ) 5? 54. Using the definition of the arccosecant function from Problem 50, for what values of is it true that (a) csc(arccsc ) 5 and (b) arccsc(csc ) 5? 55. Using the definition of the arcsecant function from Problem 5, for what values of is it true that (a) sec(arcsec ) 5 and (b) arcsec(sec ) 5? 56. Verif that arccot 5 p arctan, for all real numbers. 57. Verif that arccsc 5 arcsin (/) for 0 0 $. 58. Verif that arcsec 5 arccos (/) for 0 0 $. In Problems 59 64, use the results of Problems and a calculator to find the indicated value. 59. cot csc (.3) 6. arccsc(.5) Jones & Bartlett Learning, 6. arccot(0.3) 63. arcsec(.) 64. sec Inverse Trigonometric Functions 69

65 Jones & Bartlett Learning.. Applications 65. Projectile Motion The angle FOR of SALE elevation OR u for a gun firing a bullet to hit a target v 0 at a horizontal distance R (assuming that the target and the muzzle of the gun are θ Target at the same height) satisfies R v 0 sin u R 5, g Gun FOR SALE where vor 0 is the muzzle velocit and g is the acceleration due FOR to gravit. SALE See OR DISTRIBUT FIGURE Angle of elevation in FIGURE Find the angle of elevation u if the target is 800 ft from the gun and Problem 65 the muzzle velocit is 00 ft/s. Use g 5 3 ft/s. [Hint: There are two solutions.] 66. Olmpic Sports For the Olmpic event, the hammer throw, it can be shown that Jones & Bartlett Learning, the maimum distance is achieved for the Jones release & angle Bartlett u (measured Learning, from the horizontal) that satisfies cos u 5 gh FOR SALE OR v 0 gh, where h is the height of the hammer above the ground at release, v 0 is the initial velocit, and g is the acceleration due to gravit. For v m/s and h 5.5 m, find the optimal Jones release & Bartlett angle. Use Learning, g m/s. 67. Highwa Design In the design FOR of SALE highwas OR and railroads, curves are banked to provide centripetal force for safet. The optimal banking angle u is given b tan u5v /Rg, where v is the speed of the vehicle, R is the radius of the curve, and g is the acceleration due to gravit. See FIGURE As the formula indicates, Bartlett for a Learning, given radius there is no one correct angle for Jones all speeds. & Consequentl, Bartlett Learning, LL θ Jones & FOR SALE curves are OR banked for the average speed of the traffic over them. FOR Find SALE the correct OR DISTRIBUT R banking angle for a curve of radius 600 ft on a countr road where speeds FIGURE Banked curve in average 30 mi/h. Use g 5 3 ft/s. [Hint: Use consistent units.] Problem Highwa Design Continued If m is the coefficient of friction between the car and the road, then the maimum velocit v m that a car can travel around a curve Jones & Bartlett Learning, without slipping is given b v m 5 gr tan Jones (u tan & Bartlett m), where Learning, u is the banking angle of the curve. Find v m for the countr FOR road in SALE Problem OR 67 if m Ladder About to Slip Consider a ladder of length L leaning against a house with a load at point P a distance measured from the bottom of the ladder. See P FIGURE L The angle u at which the ladder is at the verge of slipping can be shown to be related to Jones and the & coefficient Bartlett of Learning, friction c between the ladder and Ladder the ground b L 5 c c (c tan u). θ FIGURE 4.7. Ladder in Problem 69 (a) Find u when c 5 and the load is at the top of the ladder. 3 Jones &(b) Bartlett Find ulearning, when c and the load is 4 of the wa up Jones the ladder. & Bartlett Learning, LL For Discussion 70. Using a calculator set in radian mode, evaluate arctan (tan.8), arccos (cos.8), and arcsin (sin.8). Eplain the results. Jones & Bartlett Learning, 7. Using a calculator set in radian mode, evaluate Jones tan& Bartlett (tan ()), Learning, cos (cos()), and sin (sin()). Eplain the results. 7. In Section 4.3 we saw that the graphs of 5 sin and 5 cos are related b shifting and reflecting. Justif the identit 70 CHAPTER 4 TRIGONOMETRIC FUNCTIONS arcsin arccos 5 p, for all in [, ], b Jones finding & a similar Bartlett relationship Learning, between the graphs of 5 arcsin and 5 arccos FOR. SALE OR

66 Jones & Bartlett Learning With a calculator set in radian mode determine which of the following inverse trigonometric evaluations result in an error message: (a) sin (), (b) cos (), FOR (c) tan SALE (). OR Eplain. 74. Discuss: Can an periodic function be one-to-one? 75. Show that arcsin 3 5 arcsin arcsin [Hint: See (4) of Section 4.5.] 76. The angle of inclination of a line is defined to be the angle u measured counterclockwise Learning, between the line and the -ais 0 #u,p Jones (or & 0 Bartlett #u,80 ). Learning, See LL Jones & Bartlett FOR SALE OR FIGURE The angle of inclination of a horizontal FOR line is SALE 0 and the OR angle DISTRIBUT of inclination of a vertical line is p/. If L is an line with slope m (that is, not vertical), use FIGURE to show that m 5 tan u. L L L (cos θ, sin θ) θ θ (a) θ is acute (b) θ is obtuse FIGURE 4.7. Angles of inclination in Problem 76 θ = m FIGURE Line passing through the origin and a unit circle in Problem 76 Jones & Bartlett 77. Use Learning, Problem 76 to find the angle of inclination Jones of the given & Bartlett line. Learning, LL L FOR SALE OR (a) (b) 5 FOR SALE 5 OR DISTRIBUT θ L 78. Suppose that u is the acute angle between the two nonperpendicular intersecting lines L and L in FIGURE If L and L have slopes m and m, respectivel, then show that Jones m& θ θ Bartlett m Learning, tan u5 P P. FOR SALE m m OR FIGURE Two lines in Problem Use Problem 78 to find the acute angle u between the given pair of lines. (a) 5 0, 55 (b) 5 6, (a) Show that tan(tan tan tan 3) 5 0. (b) Discuss: How does the result in part (a) prove that tan tan tan 3 5p? 4.8 Trigonometric Equations INTRODUCTION In Sections 4.5 and 4.6 we considered trigonometric identities, which are equations involving trigonometric functions that are satisfied b all values of the variable for which both sides of the equalit are defined. In this section we eamine conditional trigonometric equations, that is, equations that are true for onl certain values of the variable. We discuss techniques for finding those values of the variable (if an) that satisf the equation. We begin b considering Jones & Bartlett the problem Learning, of finding all real numbers that satisf sin 5!/. Interpreted FOR as SALE the -coordinates OR of the points of intersection of the graphs 4.8 Trigonometric Equations 7

67 Jones & Bartlett Learning = sin FOR SALE FIGURE 4.8. OR Graphs of 5 sin and 5! of 5 sin and 5!/, FIGURE 4.8. shows that there eists infinitel man solutions of the equation sin 5!/:..., 7p () 4, p 4, Jones 9p 4, 7p & 4,. Bartlett.. Learning,..., 5p () 4, 3p 4, p 4, 9p 4,.... Note that in each of the lists () and (), two successive solutions differ b p 58p/4. This is a consequence of the Jones periodicit & Bartlett of the sine Learning, function. It is common for trigonometric equations to have an infinite FOR number SALE of OR solutions because of the periodicit of the trigonometric functions. In general, to obtain solutions of an equation such as sin 5!/, it is more convenient to use a unit circle and reference angles rather than a graph of the trigonometric function. We illustrate this approach in the following & eample. Bartlett Learning, Jones EXAMPLE Using the Unit Circle Find all real numbers satisfing sin 5!/. Solution If sin 5!/, the reference angle for is p/4 radian. Since the value of Jones & Bartlett Learning, sin is positive, the terminal side of the angle Jones lies & in Bartlett either the first Learning, or second quadrant. Thus, as shown in FIGURE 4.8., the onl solutions FOR SALE between OR 0 and p FOR SALE OR are 3 5 p and 5 3p 4. Since the sine function is periodic with period p, all of the remaining solutions can be obtained b adding integer multiples of p to these solutions. The two solutions are 5 p np and 4 5 3p 4 np, (3) FIGURE 4.8. Unit circle in Eample where n is an integer. The numbers that ou see in () and () correspond, respectivel, Jones to letting & Bartlett n 5, Learning, n 5 0, n 5, and n 5 in the first and second Jones formulas & Bartlett in (3). Learning, LL When we are faced with a more complicated equation, such as 4 sin 8 sin 3 5 0, the basic approach is to solve for a single trigonometric function (in this case, it would Jones & Bartlett Learning, be sin ) b using methods similar to those Jones for solving & Bartlett algebraic equations. Learning, EXAMPLE Solving a Trigonometric Equation b Factoring Find all solutions of 4 sin 8 sin Solution We first observe Jones that this is & a Bartlett quadratic Learning, equation sin, and that it factors as ( sin FOR SALE 3)( sin OR ) CHAPTER 4 TRIGONOMETRIC FUNCTIONS Jones & Bartlett = Learning,

68 Jones & Bartlett Learning This implies that either FOR sin SALE 5 3 OR or sin 5. The first equation has no solution since 0 sin 0 #. As we see in FIGURE the two 6 angles between 0 and p for which sin equals are 5 p Jones and 5 5p & Bartlett Learning, LL 6 6 FOR. SALE OR DISTRIBUT Therefore, b the periodicit of the sine function, the solutions are FIGURE Unit circle in 5 p np and 6 5 5p Eample Jones & Bartlett 6 Learning, np, where n is an integer. cos 0 5, cos p5 cos 3p 5, and so on. In general, cos np 5() n, where n is an integer. EXAMPLE 3 Checking for Lost Solutions Find all solutions of Jones & Bartlett Learning, FOR SALE sin OR 5 cos. Solution In order to work with a single trigonometric function, we divide both sides of the equation b cos to obtain tan 5. (5), cos p 5, Equation (5) is equivalent to (4) provided that cos 0. We observe that if cos 5 0, FOR SALE then OR as we have seen in () of Section 4.3, 5 (n )p/ FOR SALE 5 p/or np, DISTRIBUT for n an integer. B the sum formula for the sine, see (4) of Section 4.5 () n 0 T T T sin a p npb 5 sin p cos np cos p Jones & Bartlett Learning, Jones & Bartlett sin np Learning, 5()n 0, = 4 5 we see that these values of do not satisf the original equation. Thus we will find all 4 the solutions to (4) b solving equation (5). 4 Now tan 5 implies that the reference angle for is p/4 radian. Since tan 5. 0, the terminal side of the angle of radians can lie either in the first or in the third quadrant, Jones as shown & in Bartlett FIGURE Learning, Thus the solutions are FOR SALE 5 OR FOR = 5 p SALE OR 4 np and 5 5p 4 4 np, FIGURE Unit circle in Eample 3 where n is an integer. We can see from Figure that these two sets of numbers can be written more compactl as 5 p FOR tan is SALE OR 4 np, This also follows from the fact that p-periodic. where n is an integer. Losing Solutions When solving an equation, if ou divide b an epression containing a variable, ou ma lose some solutions of the original equation. For eample, in algebra a common mistake in solving equations such as 5 is to divide b to obtain 5. But b writing 5 as 5 0 or ( ) 5 0 we see that in fact 5 0 or 5. To prevent the loss of a solution ou must determine the values that make the epression zero and check to see whether the are solutions of the original equation. Note that in Eample Jones & 3, when Bartlett we divided Learning, b cos, we took care to check that no solutions were lost. 4.8 Trigonometric Equations 73 (4)

69 Jones & Bartlett Learning.. Whenever possible, it is preferable to avoid dividing b a variable epression. As illustrated with the algebraic equation 5, this can frequentl be accomplished b collecting all nonzero terms on FOR one side SALE of the OR equation and then factoring (something we could not do in Eample 3). Eample 4 illustrates this technique. Jones & EXAMPLE Bartlett Learning, 4 Solving a Trigonometric Equation Jones b & Factoring Bartlett Learning, LL Solve sin cos 5!3 cos. (6) Solution To avoid dividing b cos, we write the equation as Jones sin cos!3 & Bartlett Learning, FOR cos SALE 5 0 OR and factor: Thus either cos a sin cos!3 b 5 0. cos 5 0 or sin cos! Jones Since & Bartlett the cosine Learning, is zero for all odd multiples of p/, the solutions Jones of & cos Bartlett 5 0 arelearning, LL 5 (n ) p 5 p np, where n is an integer. See () in Section 4.5. Jones & Bartlett Learning, In the second equation we replace sin Jones cos & b Bartlett sin from Learning, the double-angle formula for the sine function to obtain an equation with a single trigonometric function: sin!3 5 0 or sin 5!3. Thus the reference angle for is p/3. Since the sine is negative, the angle must be Jones & Bartlett 4 Learning, 3 in either the third quadrant or the fourth quadrant. As FIGURE illustrates, either FOR SALE 5 OR 3 5 4p 3 np or 5 5p 3 np. 4 = 3 5 = 3 FIGURE Unit circle in Eample 4 Dividing b gives 5 p 3 np or 5 5p 6 np. FOR SALE OR DISTRIBUT Therefore, all solutions of (6) are 5 p np, 5 p 3 Jones np, and & Bartlett 5 5p 6 Learning, np, where n is an integer. In Eample 4 had we simplified the equation b dividing b cos and not checked to see whether the values of Jones for which & Bartlett cos 5 0Learning, satisfied equation (6), we would have lost the solutions 5p/ np FOR, where SALE n is an OR integer. 74 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

70 Jones & Bartlett Learning.. EXAMPLE 5Jones & Using Bartlett a Trigonometric Learning, Identit Solve 3 cos cos FOR 5. SALE OR Solution We observe that the given equation involves both the cosine of and the cosine of. Consequentl, we use the double-angle formula for the cosine in the form cos 5 cos FOR SALE to replace OR the equation b an equivalent equation that involves FOR SALE cos onl. OR We DISTRIBUT find that Jones d& See Bartlett (4) of Section Learning, 4.5 LL Therefore, cos 5 0, and the solutions are Jones p& Bartlett 5 (n ) 5 p Learning, FOR SALE OR np, where n is an integer. We are often interested in finding roots of an equation onl in a specified interval. EXAMPLE 6 FOR Using SALE a OR Trigonometric Identit Find all solutions of the equation cos t cos 5t 5 0 in the interval [0, ). Solution In this case it is helpful to use a sum-to-product formula. In Eample 3 of Section 4.6 we saw that b identifing 5 t and 5 5t, (9) of Theorem 4.6. gives FOR sin( t) SALE sint OR DISTRIBUT t 5t t 5t T cos t cos 5t 5 sin sin 5 sin 3t sin(t) 5 sin 3t sin t Replacing the sum cos t cos 5t in the given equation b the product sin 3t sin t Jones & Bartlett Learning, gives the equivalent equation sin 3t Jones sin t 5& 0, Bartlett or Learning, sin 3t FOR sin tsale 5 0. OR The last equation is satisfied if either sin 3t 5 0 or sin t 5 0. Then from () in Section 4.3 we see that sin 3t 5 0 implies whereas 3 cos ( cos ) 5 becomes cos t 5 np, n 5 0,,, 3,... sin t 5 0 implies FOR SALE OR t 5 np, n 5 0,,, 3,.... If ou think in terms of angles measured in radians and the unit circle, then the onl angles satisfing the condition that t be in the interval [0, p) correspond to n 0,, Jones & Bartlett, 3, 4, Learning, 5 in (7) and n,, 3 in (8): t 5 0, p/3, p/3, p, 4p/3, 5p/3, (9) or t 5p/, p, 3p/. (0) Jones & Bartlett Learning, The solution set of the original equation Jones is then & Bartlett the union Learning, of the two sets defined b the numbers in (9) and (0), that is 50, p/3, p/, p/3, p, 4p/3, 3p/, 5p/3 6. (7) (8) So far in this section we have viewed the variable in the trigonometric equation as representing either Jones a real number & Bartlett or an angle Learning, measured in radians. If the variable represents an angle measured FOR in SALE degrees, OR the technique for solving is the same. 4.8 Trigonometric Equations 75

71 Jones & Bartlett Learning.. Jones EXAMPLE 7 Equation When the Angle Is in Degrees θ = & 0 Bartlett Learning, Solve cos u 5, where u is FOR an angle SALE measured OR in degrees. 40 θ = 40 0 Solution Since cos u 5, the reference angle for u is 60 and the angle u must be in either the second or the third quadrant. FIGURE illustrates that either u 50 or u 540. An angle that is coterminal with one of these angles will also satisf Jones cos & u Bartlett 5. These Learning, angles are obtained b adding an integer Jones multiple & of Bartlett 360 to 0 Learning, LL FOR or to SALE 40 : OR FIGURE Unit circle in where n is an integer. Dividing b the last line ields the two solutions Eample 7 u n and u50 80 n. Etraneous Solutions When an equation is squared or multiplied b a variable epression, the resulting equation ma not be equivalent to the original. In other words, the new equation ma possess Jones solutions & Bartlett that are Learning, not solutions of the given equation. Such numbers are called etraneous FOR solutions. SALE OR For eample, the simple equation 5 has onl one solution but b squaring both sides the resulting equation 5 is now satisfied b 5 and 5. Thus number 5is an etraneous solution of the original equation. The net eample illustrates the same idea for trigonometric equations. FOR EXAMPLE SALE OR 8 Etraneous Roots 76 CHAPTER 4 TRIGONOMETRIC FUNCTIONS u n or u n, Find all solutions of tan a5sec a, where a is an angle measured in degrees. Solution The equation does not factor, but we see that if we square both sides, we can use a fundamental identit to obtain an equation involving a single trigonometric function: ( tan a) 5 (sec a) tan atan a5sec a d now use (9) of Section 4.4 tan atan a5 tan a tan a50 Jones & Bartlett tan a50. Learning, The values of a for 0 #a,360 at which tan a50are a50 and a580. Since we squared each side of the original equation, we ma have introduced etraneous Jones solutions. & Bartlett Therefore, Learning, it is important that we check all solutions Jones in the original & Bartlett equation. Learning, LL FOR Substituting SALE OR a50 into tan a5sec a, we obtain the true FOR statement SALE OR 0 5DISTRIBUT. But after substituting a580, we obtain the false statement 0 5. Therefore, 80 is an etraneous solution and a50 is the onl solution satisfing 0 #a,360. Thus, all the solutions of the equation are given b a n n, where n is an integer. For n 0, these are the angles that are coterminal with 0. Recall from Section. that to find the -intercepts of the graph of a function 5 f () we find the zeros Jones of f, that & Bartlett is, we must Learning, solve the equation f () 5 0. The following eample makes use of FOR this SALE fact. OR

72 Jones & Bartlett Learning.. EXAMPLE 9Jones & Intercepts Bartlett Learning, of a Graph Find the first three -intercepts FOR SALE of the graph OR of f () 5 sin cos on the positive -ais. Solution We must solve f () 5 0, that is, sin cos 5 0. It follows that either sin 5 0 or cos 5 0. From sin 5 0, we obtain 5 np, where n is an integer, or 5 np/, where n Jones & Bartlett is an integer. Learning, From cos 5 0, we find 5p/ np, Jones where & nbartlett is an integer. Learning, Then for LL FOR SALE n 5OR, 5 np/ gives 5p, whereas for n 5 0 and FOR n 5, SALE 5p/ OR DISTRIBUT np gives 5p/ and 5 3p/, respectivel. Thus the first three -intercepts on the positive -ais are (p/, 0), (p, 0), and (3p/, 0). Using Inverse Functions So Jones far all & of Bartlett the trigonometric Learning, equations have had solutions that were related b reference FOR angles SALE to the special OR angles 0, p/6, p/4, p/3, or p/. If this is not the case, we will see in the net eample how to use inverse trigonometric functions and a calculator to find solutions. EXAMPLE 0 FOR Solving SALE OR Equations Using Inverse Functions Find the solutions of 4 cos 3 cos 5 0 in the interval [0, p]. Solution We recognize that this is a quadratic equation in cos. Since the left-hand side of the equation does not readil factor, we appl the quadratic formula to obtain cos 5 3 6!4. FOR SALE OR DISTRIBUT 8 At this point we can discard the value (3!4)/8 <.8, because cos cannot be greater than. We then use the inverse cosine function and a calculator to solve the Jones & Bartlett Learning, remaining equation: cos 5 3!4 which implies 5 cos a 3!4 b < Of course in Eample Jones 0, & had Bartlett we attempted Learning, to compute cos [(3!4 )/8] with a calculator, we would FOR have received SALE OR an error message. Eercises 4.8 Answers to selected odd-numbered problems begin on page ANS 7. In Problems 6, find all solutions of the given trigonometric equation if represents an angle measured in radians.. sin 5!3/. cos 5!/ 3. sec 5! 4. tan 5 Jones & Bartlett Learning, 5. cot 5!3 Jones & 6. Bartlett csc 5 Learning, In Problems 7, find all solutions of the given trigonometric equation if represents a real number. 7. cos 5 8. sin 5 9. tan 5 0 Jones & Bartlett Learning, 0.!3 sec 5. csc 5.!3 cot Trigonometric Equations 77

73 Jones & Bartlett Learning.. In Problems 3 8, find all solutions of the given trigonometric equation if u represents an angle measured in degrees. 3. csc u5!3/3 4. sin u5! 5. cot u50 6.!3 sin u5cos u 7. sec u5 8. cos u! 5 0 Jones In & Problems Bartlett 9 46, Learning, find all solutions of the given trigonometric Jones equation & Bartlett if is a Learning, real LL FOR number SALE and OR u is an angle measured in degrees. 9. cos sin 3 sin sec 5 sec. tan (!3 )tan! cos u3 cos u50 4. sin usin u50 Jones & Bartlett Learning, 5. cot ucot u Jones sin & u( Bartlett!3)sin Learning, u!3 5 0 FOR SALE OR 7. cos 5 8. FOR sec SALE 5 OR 9. sin 3u tan 4u 5 3. cot (/) 5 3. csc (u/3) sin sin cos sin cos u 5sin u 36. sin u sin u cos u5 37. sin 4 sin Jones & Bartlett Learning, tan 4 u sec u350 FOR SALE OR cos u 39. sec sin 5 tan cos u 4. sin ucos u5 4. sin cos 5 0 sin Jones 43. & Bartlett Learning, sin!sin Jones 5 0 & Bartlett Learning, LL Ä 45. cos u!cos u cos u" tan u 5 In Problems 47 5, use a sum-to-product formula (as in Eample 6) to solve the given equation on the indicated interval. Jones & Bartlett Learning, 47. sin 6t sin 4t 5 0, [0, p) 48. Jones cos t & Bartlett cos 3t 5Learning, 0, [0, p) FOR SALE OR 49. cos ucos 4u 50, [p, p) 50. FOR sin 5aSALE sin 3aOR 50, [0, p) 5. sin 7 sin sin 3 5 0, (p, p) 5. sin cos sin 3 5 0, [0, 3p) In Problems 53 60, find the first three -intercepts of the graph of the given function on the positive -ais. 53. f () 55 sin (3 p) FOR SALE 54. OR f () 5 cos a p 4 b p 55. f () 5 sec 56. f () 5 cos p Jones 57. & f Bartlett () 5 sin Learning, tan 58. f () 5 Jones cos a & Bartlett p 3 b Learning, LL FOR 59. SALE f () 5 OR sin sin 60. f () 5 cos cos FOR 3 SALE OR DISTRIBUT [Hint: Write 3 5.] In Problems 6 64, b graphing determine whether the given equation has an solutions. Jones & Bartlett Learning, 6. tan 5 [Hint: Graph 5 tan and Jones 5 the & Bartlett same coordinate Learning, aes.] 6. sin cot cos 5 0 In Problems 65 70, using an inverse trigonometric function find the solutions of the given equation in the indicated interval. Round our answers to two decimal places cos cos Jones 5 0, [0, & Bartlett p] Learning, sin 8 sin 4 5 0, FOR [p/, SALE p/] OR 78 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

74 Jones & Bartlett Learning tan tan Jones 5 0, & Bartlett (p/, p/) Learning, 3 sin cos 5 0, [p/, p/] 5 cos 3 3 cos cos 5 0, [0, p] tan 4 3 tan 5 0, (p/, p/) Applications Jones & Bartlett 7. Isosceles Learning, Triangle B methods that will be discussed Jones in & Section Bartlett 4.0 Learning, it can be LL FOR SALE OR shown that the area of the isosceles triangle shown FOR in SALE FIGURE is OR given DISTRIBUT b θ A 5 sin u. If the length of side is 4, what value of u will give a triangle with area 4? 7. Circular Motion An object travels in a circular path centered at the origin with constant angular speed. The -coordinate Jones & of Bartlett the object Learning, at an time t seconds is given b 5 8 cos (pt p/). At FOR what SALE time(s) OR does the object cross the -ais? 73. Mach Number Use Problem 73 in Eercises 4.5 to find the verte angle of the cone of sound waves made b an airplane fling at Mach. FIGURE Isosceles triangle in 74. Alternating Current An electric generator produces a 60-ccle alternating current given b I(t) 5 30 sin 0p(t 36), 7 where I(t) is the current in amperes at t Problem 7 seconds. Find Jones the smallest & Bartlett positive value Learning, of t for which the current is 5 amperes. 75. Electrical Circuits FOR If the SALE voltage OR given b V 5 V 0 sin (vt a) is impressed on a series circuit, an alternating current is produced. If V volts, v50p radians per second, and a5p/6, when is the voltage equal to zero? 76. Refraction of Light Consider a ra of light passing from one medium Incident Air Jones & Bartlett (such Learning, as air) into another medium (such as a Jones crstal). & Let Bartlett f be the Learning, angle LL ra φ FOR SALE OR of incidence and u the angle of refraction. As shown FOR in SALE FIGURE OR 4.8.8, DISTRIBUT these angles are measured from a vertical line. According to Snell s law, there is a constant c that depends on the two mediums, such that Crstal Refracted sin f θ ra Assume that for light passing from air into a crstal, c sin u 5 c. Find f and u such that the angle Jones of incidence & Bartlett is twice Learning, the angle FIGURE Light ras in of refraction. Problem FOR 76 SALE OR 77. Snow Cover On the basis of data FOR collected SALE from OR 966 to 980, the etent of snow cover S in the northern hemisphere, measured in millions of square kilometers, can be modeled b the function p S(w) 5 5 cos (w 5), Jones & Bartlett Learning, 6 where w is the number of weeks past Januar. (a) How much snow FOR cover SALE does OR this formula predict for April Fool s Da? (Round w to the nearest integer.) (b) In which week does the formula predict the least amount of snow cover? (c) What month does this fall in? 4.9 Simple Harmonic Motion Jones & Bartlett Learning, INTRODUCTION Man phsical objects Jones vibrate & Bartlett or oscillate Learning, a regular manner, repeatedl moving back and forth over a definite time interval. Some eamples are clock pendulums, a mass on a spring, sound waves, strings on a guitar when plucked, the human heart, tides, and alternating current. In this section we will focus on mathematical models of the undamped oscillator motion of a mass on a spring. Before proceeding with the main discussion we need to discuss the graph of the sum of constant multiples Jones of & cos Bartlett B and sin Learning, B, that is, 5 c cos B c sin B, where and are constants. c c 4.9 Simple Harmonic Motion 79

75 Jones & Bartlett Learning.. Addition of Two Sinusoidal Jones & Functions Bartlett In Learning, Section 4.3 we eamined the graphs of horizontall shifted sine and cosine graphs. It turns out that an linear combination of a sine function and a cosine function of the form 5 c cos B c sin B, () where c and c are constants, can be epressed either as a shifted sine function Jones 5& A Bartlett sin(b f), Learning, B. 0, or as a shifted cosine function Jones 5 A cos(b & Bartlett f). Note Learning, LL FOR that SALE in () the OR sine and cosine functions sin B and cos B have the FOR same SALE period OR p/b. DISTRIBUT EXAMPLE Addition of a Sine and a Cosine Jones & Bartlett Learning, Graph the function 5 cos!3 sin. Jones & Bartlett Learning, FOR SALE OR Solution Using a graphing utilit we have shown FOR in FIGURE SALE 4.9.OR four ccles of the graphs of 5 cos (in red) and 5!3 sin (in green). It is apparent in FIGURE 4.9. that the period of the sum of these two functions is p, the common period of cos and sin. Also apparent is that the blue graph is a horizontall shifted sine (or cosine) function. Although Figure 4.9. suggests that the Jones amplitude & Bartlett of the function Learning, 5 cos!3 sin is, the eact phase shift of the graph is FOR certainl SALE is notor apparent. Jones & Bartlett = Learning, 3 sin The sine form 5 A sin(b f) is slightl easier to use than the cosine form 5 A cos(b f). FIGURE 4.9. Superimposed graphs of FIGURE 4.9. Graph of the sum 5 cos and 5!3sin in 5 cos!3sin in Eample Eample Reduction to a Sine Function We eamine onl the reduction of () to the form 5 A sin(b f), B. 0. FOR THEOREM SALE 4.9. Reduction of () to () For real numbers c, c, B,and, Jones & Bartlett Learning, where A and f are defined b = cos c cos B c sin B 5 A sin(b f) FOR SALE OR = cos 3 sin () A 5 "c FOR c SALE OR (3) and sin f5 c A tan f5 c c cos f5 c Jones & Bartlett A Learning, (4) 80 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

76 Jones & Bartlett Learning.. PROOF: To prove Jones (), we use & Bartlett the sum formula Learning, (4) of Section 4.5: A sin(b FOR f) SALE 5 A sin OR B cos fa cos B sin f 5 (A sin f) cos B (A cos f) sin B 5 c cos B c sin B and identif A sin f5c Thus, sin f5c /A 5 c /"c c, A cos f5c. and Jones & Bartlett cos f5c Learning, /A 5 c /"c c. EXAMPLE Eample Revisited Epress 5 cos!3 sin as a single sine function. Solution With the identifications c 5, c 5!3, and B 5, we have from (3) and (4), A 5 "c c 5 " (!3) 5!4 5, sin f5 Jones & Bartlett Learning, cos f5!3 tan f5!3. Although tan f5/!3 we cannot blindl assume that f5tan (/!3). The value we take for f must be consistent with the algebraic signs of sin f and cos f. Jones & Bartlett Thinking Learning, in terms of radian-measured angles, because Jones sin f &.0 Bartlett and cos Learning, f,0 the LL FOR SALE terminal OR side of the angle f lies in the second quadrant. FOR But SALE since the OR range DISTRIBUT of the inverse tangent function is the interval (p/, p/), tan (/!3) 5p/6is a fourth-quadrant angle. The correct angle is found b using the reference angle p/6 for tan (/!3) to find the second-quadrant angle Jones f5p p 6 5 & 5p Bartlett 6 radians. Learning, Therefore 5 cos!3 sin can be rewritten as 5 sin a 5p 5p b or 5 sin a 6 b. Hence the graph of FOR 5 cos SALE!3 OR sin is the graph of 5 sin, which has amplitude, period p/ 5p, and is shifted 5p/ units to the left. EXAMPLE 3 Eample Revisited FOR SALE Find OR the first two -intercepts on the positive -ais of FOR the graph SALE of OR the function DISTRIBUT in Eample. Solution The first alternative form 5 sin( 5p/6) of the function in Eample can be used to find the -intercepts of its graph. Recall, sin 5 0 when Jones & Bartlett Learning, 5 np, n 5 0, 6, 6,.... So b Jones replacing & Bartlett the smbol Learning, b 5p/6, we see sina 5p 6 b 5 0 FOR implies SALE OR 5p 6 5 np. Solving Jones & 5p Bartlett 6 5p Learning, and 5p 6 5 p 4.9 Simple Harmonic Motion 8

77 Jones & Bartlett Learning.. ields 5p/ and 7p/. Thus the first two intercepts on the positive -ais are Jones = & sin Bartlett Learning, (p/, 0) and (7p/, 0). The blue graph in FIGURE is the portion of the graph in Figure 4.9. on the interval [0, p]. We note that the -coordinates of the -intercepts of 5 sin( 5p/6) can 7 also be obtained b subtracting the phase shift 5p/ from the -coordinates of the -intercepts of the red graph of 5 sin in Figure = sin ( + ) Simple Harmonic Motion Consider the motion of a mass on a spring as shown FIGURE Graph of function in in FIGURE In the absence of frictional or damping forces, a mathematical model for Eample the displacement (or directed distance) of the mass measured from a position called the equilibrium position is given b the function (t) 5 0 cos vt FOR v 0 sin SALE vt, OR (5) v where t is time, 0 and v 0 are, respectivel, the initial (t 5 0) displacement and velocit of the mass. Oscillator motion modeled Jones b & the Bartlett function Learning, (5) is said to be simple harmonic FOR SALE OR A motion. Equilibrium More precisel, we have the following definition. A FIGURE An undamped DEFINITION 4.9. Simple Harmonic Motion spring/mass sstem ehibits A point moving on a coordinate line whose position at time t is given b simple harmonic motion (t) 5 A sin(vt f) or (t) 5 A cos(vt f) (6) where A, v.0, and f are constants, is said to ehibit simple harmonic motion. Special cases of the trigonometric functions in (6) are (t) 5 A sin vt, (t) 5 A cos vt, and (t) 5 c cos vt c sin vt. Terminolog The function (5) is said to be the equation of motion of the mass. Also, in (5), v5!k/m, where k is the spring constant (an indicator of the stiffness of the spring), m is the mass attached FOR SALE to the spring OR (measured in slugs or kilograms), 0 is the initial displacement of the mass (measured above or below the equilibrium position), v 0 is the initial velocit of the mass, t is time measured in seconds, and the period p of motion is p 5 p/v seconds. The number f 5 /p 5 /(p/v) 5v/p is called Jones the & frequenc Bartlett of Learning, motion. The frequenc indicates the number Jones of ccles & Bartlett completed Learning, b LL FOR the graph SALE per OR unit time. For eample, if the period of (5) is, sa, pfor 5 seconds, SALE then OR we DISTRIBUT know that one ccle of the function is complete in seconds. The frequenc f 5 /p 5 means one-half of a ccle is complete in second. In the stud of simple harmonic motion it is convenient to recast the equation of motion (5) as a single epression involving onl the sine function: (t) 5 A sin(vt FOR f). SALE OR (7) The reduction of (5) to the sine function (7) can be done in eactl the same manner as illustrated in Eample. In this situation we make the following identifications between () and (5): c 5 0, c 5FOR v 0/v, SALE A 5 "c OR c, and B 5v. 8 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

78 Jones & Bartlett Learning.. EXAMPLE 4Jones & Equation Bartlett of Learning, Motion (a) Find the equation of FOR simple SALE harmonic OR motion (5) for a spring mass sstem if m 5 slug, ft, k 5 4 lb/ft, and v ft/s. (b) Find the period and frequenc of motion. Solution (a) We begin with the simple harmonic motion equation (5). Since Jones & Bartlett k/m 5Learning, 4/( 6) 5 64, v5!k/m 5 8, and v 0/v 5( Jones 4 3) /8 5& 6, Bartlett therefore (5) Learning, becomes LL (t) 5 3 cos 8t 6 sin 8t. (8) (b) The period of motion is p/8 5p/4second; the frequenc is 4/p <.7 ccles per second. EXAMPLE 5 Eample 4 FOR Continued SALE OR FOR SALE OR Epress the equation of motion (8) as a single sine function (7). Solution With c 5 3, c 5 6, we find the amplitude of motion is Jones A& 5 Bartlett "( 3) Learning, ( 6) 5 6!7 ft. Then from sin f 5 3/!7 6, 0 cos f 5 6/! f tan f54 t 3 4 Jones 4 & Bartlett Learning, we can see from algebraic signs sin f,0and cos f.0that the terminal side of the angle = 7 sin(8t.358) 6 FOR SALE lies in the fourth quadrant. Hence the correct value of is tan OR DISTRIBUT f f (4) <.358. FIGURE Graph of the equation of The equation of motion is then (t) 5 6!7 sin(8t.358). As shown in FIGURE 4.9.5, motion in Eample 5 the amplitude of motion is A 5!7/6 < Since we are assuming that that is no resistance to the motion, once the spring/mass sstem is set in motion the model Jones & Bartlett Learning, indicates it stas in motion bouncing Jones back and & Bartlett forth between Learning, its maimum displacement!7/6 feet above the equilibrium FOR position SALE and a minimum OR of!7/6 feet below the equilibrium position. Onl in the two cases, c. 0, c. 0 or c, 0, c. 0, can we use tan f in (4) to write f5tan (c /c ).(Wh?) Correspondingl, f is a first or a fourth-quadrant angle. Eercises 4.9 Answers to selected odd-numbered problems begin on page ANS 8. In Problems 6, proceed as in Eample and write the given trigonometric function Jones & Bartlett in the form Learning, 5 A sin(b f). Sketch the graph and Jones give the & amplitude, Bartlett the Learning, period, and LL FOR SALE the phase OR shift. p.. 5 sin!3cos p 5 cos p sin p 3. 5!3 sin cos 4. 5!3cos 4 sin ! (sin cos ) 6. 5 sin cos In Problems 7 and 8, proceed as in Eample 3 and find the first two -intercepts on the positive -ais of the graph of the function. 7. 5cos p Jones sin p & Bartlett Learning, 8. 5 cos p sin p!3 4.9 Simple Harmonic Motion 83

79 Jones & Bartlett Learning.. In Problems 9 and 0, use (), (3), and (4) to write the left-hand side of the given equation in the form A sin(b f). Then find the solutions of the equation in the indicated interval cos!3 sin cos sin 5 ; [0, p] 5 ; [0, 4] In Problems 4, proceed as in Eamples 4 and 5 and use the given information to Jones epress & Bartlett the equation Learning, of simple harmonic motion (5) for a spring/mass Jones & Bartlett sstem in Learning, the LL FOR trigonometric SALE OR form (7). Give the amplitude, period, and frequenc FOR of motion. SALE OR DISTRIBUT. m 5 slug, ft, and v k 5 lb/ft,. slug, ft, and v ft/s m k 5 40 lb/ft, 4 ft/s 3. m 5 slug, 0 5 ft, k 5 6 lb/ft, and v 0 5 ft/s 4. m 5 slug, ft, k 5 00 lb/ft, and v ft/s 5. The equation of simple harmonic motion of a spring/mass sstem is 5 (t) 5 sin(t p/3). Determine the initial displacement 0 and initial velocit v 0 of the mass. [Hint: Use (5).] 6. Use the equation of simple Jones harmonic & Bartlett motion Learning, of the spring/mass sstem given in Problem 5 to find the times FOR for SALE which the OR mass passes through the equilibrium position 5 0. Applications Jones 7. & Current Bartlett Under Learning, certain conditions, the current I(t) measured Jones in amperes & Bartlett at time Learning, t LL FOR SALE in an electrical OR circuit is given b I(t) 5 I 0 [sin(vt u)cosf cos(vt u)sinf]. Epress I(t) as a single sine function of the form given in (7). [Hint: Review the sum formula in (4) of Theorem 4.5..] Jones & Bartlett Learning, 8. More Current In a certain kind of electrical Jones circuit, & Bartlett the current Learning, I(t) measured in amperes at time t seconds is given b FOR SALE OR (a) Change the form of Jones the function & Bartlett I(t) to the Learning, sine function form given in (7). (b) Give the period of the sine function in part (a) and the phase shift. Use the sine function to sketch two ccles of the graph of I(t). 9. Pendulum Motion An object that swings back and forth is called a phsical pendulum. A simple pendulum is a special case of the phsical pendulum and consists of a rod of length L with a mass m attached at one end. See FIGURE Jones & If Bartlett the motion Learning, of a simple pendulum takes place in a vertical Jones plane & and Bartlett it is Learning, LL FOR SALE assumed OR that the mass of the rod is negligible and no damping FOR forces SALE act on OR thedistribut Clock pendulum sstem, then it can be shown that the displacement angle u of the pendulum as a function of time t, measured from the vertical, is given b Kg Kua/Dreamstime.com u(t) 5u 0 cos vt sin vt. Jones v & Bartlett Learning, FOR Here u and v are the initial displacement and velocit and v5!g/l, and L SALE OR 0 0 θ g 5 3 ft/s is the acceleration due to gravit. (a) Find u(t) if u 0. 0 and v at t 5 0. (b) Show that the period (in seconds) of motion of a simple pendulum is m Jones L FIGURE & Bartlett Pendulum Learning, Jones & T Bartlett 5 p Learning, FOR Problem SALE 9 OR FOR SALE Ä OR g 84 CHAPTER 4 TRIGONOMETRIC FUNCTIONS I(t) 5 0 cosa0pt v 0 p 3 b.

80 Jones & Bartlett Learning.. 0. Pendulum Motion on the Moon Use Problem 9 to describe the motion of a simple pendulum: (a) On the Moon where the acceleration of due to gravit is 6 that of the Earth. (b) If its length is increased to 4L. (c) If its length is decreased to 4L. Jones & Bartlett Calculator/Computer Learning, Problems FOR SALE In Problems OR and, use a graphing utilit to obtain the FOR graph SALE of the OR given DISTRIBUT function f on the interval [0, p]. Use (), (3), and (4) to write f in the form f () 5 A sin(b f). Then find approimate solutions of indicated equation in the interval.. f () 5 3 cos 4 sin ; f () 5 5 Jones & Bartlett Learning,. f () 5 5 cos 3 sin 3; Jones f () 53 & Bartlett Learning, 4.0 Right Triangle Trigonometr INTRODUCTION The word trigonometr (from the Greek trigonon meaning triangle and metria meaning measurement ) refers to the measurement of triangles. In Section 4. we defined the trigonometric functions using coordinates of points on the unit circle and b using radian measure we were able to define the trigonometric functions of an angle. In this section we will show that the trigonometric functions of an acute angle in a right triangle have an equivalent definition in terms of the lengths of the Jones & Bartlett sides oflearning, the triangle. Terminolog In FIGURE 4.0.(a) we have drawn a right triangle with sides labeled a, b, and c (indicating their respective lengths) and one of the acute angles denoted b u. From the Pthagorean theorem we know that a b 5 c. The side opposite the right angle is called the hpotenuse; the remaining sides are referred to as the legs of the triangle. The legs labeled a and b are, Jones in turn, & said Bartlett to be thelearning, side adjacent to the angle u Jones & Bartlett Learning, and the side opposite the angle u. We will FOR also SALE use the OR abbreviations hp, adj, and opp to denote the lengths of these sides. c hpotenuse Jones & Bartlett Learning, Unit circle c FOR SALE oppositeor b b side θ sinθ θ a cosθ adjacent side Jones & Bartlett a Learning, LL (a) A right triangle with an (b) The unit circle with angle θ acute angle θ placed in standard position FIGURE 4.0. In (a) and (b) the right triangles are the same If we place u in standard position FOR and draw SALE a unit OR circle centered at the origin, we see from Figure 4.0.(b) that there are two similar right triangles containing the same angle u. Since corresponding sides of similar triangles are proportional, it follows that sin Jones u 5 b & c 5 Bartlett opp Learning, cos u and 5 a hp c 5 adj hp. 4.0 Right Triangle Trigonometr 85

81 Jones & Bartlett Learning.. Also, we have Jones & Bartlett hp Learning, opp tan ufor 5 sin SALE u cos u 5 b /c OR a/c 5 b a 5 opp adj. θ adj Then, appling the reciprocal identities () in Section 4.4, each trigonometric function FIGURE 4.0. Defining the Jones of & u can Bartlett be written Learning, as the ratio of the lengths of the sides of a right triangle as follows. trigonometric functions of u See FIGURE DEFINITION 4.0. Trigonometric Functions of u in a Right Triangle For an acute angle u in a right triangle as shown FOR in Figure SALE 4.0., OR sin u5 opp cos u5 adj hp hp tan u5 opp cot u5 adj () Jones & adj Bartlett Learning, opp FOR SALE sec u5 hp OR csc u5 hp adj opp Jones & EXAMPLE Bartlett Learning, Values of the Si Trigonometric Jones Functions & Bartlett Learning, LL Find the eact values of the si trigonometric functions of the angle u in the right triangle hp shown in FIGURE Solution From Figure we see that the side opposite u has length 8 and the side θ Jones & Bartlett Learning, adjacent has length 5. From the Pthagorean Jones theorem & Bartlett the hpotenuse Learning, c is 5 FOR SALE OR c and so FOR csale 5!89 OR 5 7. FIGURE Right triangle in Eample Thus from () the values of the si trigonometric functions are EXAMPLE Using a Right Triangle Sketch If u is an acute angle and sin u5 7, find the values of the other trigonometric functions Jones & Bartlett Learning, of u. FOR SALE OR Solution We sketch a right triangle with an FOR acute angle SALE u OR satisfing sin u5 7, b 7 making opp 5 and hp 5 7 as shown in FIGURE From the Pthagorean theorem we have θ adj Jones FIGURE & Bartlett Right Learning, triangle (adj) 5 7 so that (adj) FOR Eample SALE OR Thus, FOR adj 5 SALE!45 5OR 3!5. 86 CHAPTER 4 TRIGONOMETRIC FUNCTIONS sin u5 opp hp 5 8 adj, cos u5 7 hp 5 5 Jones & Bartlett Learning, 7, tan u5 opp FOR adj 5 SALE 8 OR adj, cot u5 5 opp 5 5 8, sec u5 hp adj 5 7, csc u5hp 5 opp

82 Jones & Bartlett Learning.. The values of the remaining five trigonometric functions are obtained from the definitions in (): cos u5 adj hp 5 3!5,sec u5hp 7 adj 5 7 3!5 5 7!5 5, tan u5 opp adj 5 3!5 5!5 adj, cot u5 5 opp 5 3!5 Jones & Bartlett, Learning, LL csc u5 hp opp 5 7. FOR SALE OR DISTRIBUT A Solving Right Triangles Applications of right triangle trigonometr in fields Jones & Bartlett Learning, such as surveing and navigation involve Jones solving & Bartlett right triangles. Learning, The epression to c α FOR SALE OR b solve a triangle means that we wish to FOR find the SALE length OR of each side and the measure of each angle in the triangle. We can solve an right triangle if we know either two sides β or one acute angle and one side. As the following eamples will show, sketching and labeling the triangle is an essential part of the solution process. It will be our general prac- B C a FIGURE Standard labeling tice to label a right triangle as shown in FIGURE The three vertices will be denoted Jones & Bartlett for a right triangle Learning, b A, B, and C, with Jones C at the & verte Bartlett of the Learning, right angle. We denote the angles at A and B b a and b and the lengths FOR of SALE the sides OR opposite these angles b a and b, respectivel. The length of the side opposite the right angle at C is denoted b c. EXAMPLE 3 Solving a Right Triangle FOR SALE Solve OR the right triangle having a hpotenuse of length 4 FOR!3 and SALE one 60 OR angle. DISTRIBUT A Solution First we make a sketch of the triangle and label it as shown in FIGURE We wish to find a, b, and b. Since a and b are complementar angles, ab590 ields 4 3 α = 60 b b590 a590 Jones & Bartlett Learning, FOR β SALE OR We are given the length of the hpotenuse, FOR namel, SALE hp OR 5 4!3. To find a, the length B C a of the side opposite the angle a560, we select the sine function. From FIGURE Right triangle in sin a5opp/hp, we obtain Eample 3 sin 60 5 a Jones & Bartlett or a 5 4!3 sin 60. 4!3 Learning, Since sin 60 5!3/, we have a 5 4!3 sin !3 a!3 b 5 6. FOR SALE Finall, OR to find the length b of the side adjacent to the 60 FOR angle, SALE we select OR DISTRIBUT the cosine function. From cos a5adj/hp, we obtain cos 60 5 b or b 5 4!3 cos 60. 4!3 Because cos 60 5, we find b 5 4!3 cos !3 A B 5!3. In Eample 3 once we determined a, we could have found b b using either the Pthagorean theorem Jones or the & tangent Bartlett function. Learning, In general, there are usuall several was to solve a triangle. 4.0 Right Triangle Trigonometr 87

83 Jones & Bartlett Learning.. Use of a Calculator Jones If angles & other Bartlett than 30, Learning, 45, or 60 are involved in a problem, we can obtain approimations of the desired trigonometric function values with a calculator. For the remainder of this chapter, whenever an approimation is used, we will round the final results to the nearest hundredth unless the problem specifies otherwise. To take full advantage of the calculator s accurac, store the computed values of the trigonometric functions in the calculator for subsequent calculations. If, instead, a Jones rounded & Bartlett version Learning, of a displaed value is written down and then Jones later keed & Bartlett back into Learning, the LL FOR calculator, SALE the OR accurac of the final result ma be diminished. EXAMPLE 4 88 CHAPTER 4 TRIGONOMETRIC FUNCTIONS Solving a Right Triangle Solve the right triangle with legs of length 4 and 5. A Solution After sketching and labeling the triangle as shown in FIGURE 4.0.7, we see that we need to find c, a, and b. From the Pthagorean theorem, the hpotenuse c is given b α c 4 c 5 "5 4 5!4 < To find b, we use tan b5opp/adj. (B choosing to work with the given quantities, we Jones & βbartlett Learning, avoid error due to previous Jones B C approimations.) & Bartlett Thus Learning, we have FOR SALE 5 OR FIGURE Right triangle in tan b Eample 4 From a calculator set in degree mode, we find b < Since a590 b, we obtain a < Eercises 4.0 Answers to selected odd-numbered problems begin on page ANS 8. In Problems 0, find the values of the si trigonometric functions of the angle u in the given triangle. Jones & Bartlett Learning,. Jones. & Bartlett 3 Learning, 5 FOR 5 SALE OR 4 θ θ FIGURE Triangle for 3 Problem FIGURE Triangle for Problem θ θ FIGURE Triangle for Jones & Problem Bartlett 3 Learning, FIGURE 4.0. Jones Triangle & for Bartlett Learning, LL Problem θ θ FIGURE 4.0. Triangle for FIGURE Jones & Bartlett Learning, Triangle for Problem 5 Problem θ 3. θ FIGURE Triangle for Jones & Bartlett FIGURE Learning, Triangle for Problem 7 FOR SALE OR Problem 8

84 Jones & Bartlett Learning s θ θ FIGURE Triangle for s FIGURE Problem 9 Triangle for Problem Jones 0 & Bartlett Learning, LL A In Problems, find the indicated unknowns. Each problem refers to the triangle shown in FIGURE α. a 5 4, b57 ; b, c. c 5 0, b549 ; a, b Jones c & Bartlett Learning, b 3. b 5 8, b ; a, c 4. c 5 5, a550 ; a, b 5. b 5.5, c 5 3; a, b, a FOR 6. SALE a 5 5, OR b 5 ; a, b, c β 7. a 5 4, b 5 0; a, b, c 8. b 5 4, a558 ; a, c B a C 9. a 5 9, c 5 ; a, b, b 0. b 5 3, c 5 6; a, b, a FIGURE Triangle for. b 5 0, a53 ; a, c. a 5, a533.5 ; b, c Jones & Bartlett Problems Learning, In Problems 3 and 4, FOR solve SALE for in OR the given triangle Jones & Bartlett Learning, FIGURE Triangle for Problem FIGURE Triangle for Problem 4 In Problems 5 and 6, the cube given in the figure has a side of length s. Find the angle u between the diagonal AC of its base and the diagonal of the cube AD (Problem 5) and the Jones angle & u between Bartlett the Learning, edge AB and the diagonal of the cube AD (Problem 6). 5. D 6. D s θ C s A s C FIGURE 4.0. Cube in b Problem 5 Jones a & Bartlett A Learning, B Jones & Bartlett FIGURE Learning, Inscribed right FOR SALE triangle OR in Problem 7 c θ FOR SALE OR DISTRIBUT s A s B FIGURE 4.0. Cube in Problem 6 7. Inscribed Right Triangle If a right FOR triangle SALE is inscribed OR in a circle, then its hpotenuse is a diameter of the circle. In FIGURE the blue dot is the center of the circle. Find the area of the red right triangle with sides of length a, b,and c inscribed in a unit circle when the angle at verte A is Isosceles Triangle Jones An & isosceles Bartlett triangle Learning, is a triangle with two sides of the same length s. As a consequence FOR SALE of this OR definition, the angles opposite the equal sides 4.0 Right Triangle Trigonometr 89 s

85 Jones & Bartlett Learning.. have the same measure u. See FIGURE Epress the area A of an isosceles triangle in terms of s and u. 9. Equilateral Triangle An equilateral triangle is a triangle with all three sides of the same length s. As a consequence of this definition, the angles opposite the sides have the same measure 60. See FIGURE Epress the area A of an equilateral triangle as a function of s. 60 s s s s θ θ b FIGURE Isosceles triangle in Problem Jones & Bartlett s Learning, FIGURE FOR SALE Equilateral OR triangle in Problem Stacked Circles Suppose 0 circles of radius r are stacked within the red triangle shown in FIGURE Each circle is eternall tangent to its neighboring circles and each of the 9 outer circles in the stack is tangent to the red line(s). FIGURE Stacked circles in (a) Show that the triangle is equilateral. Problem 30 (b) Epress the area A of the equilateral triangle as a function of the radius r. 4. Applications of Right Triangles INTRODUCTION Right triangle trigonometr can be used to solve man practical problems, particularl those involving lengths, heights, Jones and & Bartlett distances. Learning, Jones & Bartlett Learning, EXAMPLE A kite is caught in the top branches of a tree. If the 90-ft kite string makes an angle of with the ground, estimate Jones the height & Bartlett of the tree Learning, b finding the distance from the kite to the ground. Solution Let h denote the height of the kite. From FIGURE 4.. we see that 90 ft h h 5 sin or h 5 90 sin. 90 FIGURE 4.. Tree in Eample FOR A calculator SALE OR set in degree mode gives h < 33.7 ft. 4 in. EXAMPLE 90 CHAPTER 4 TRIGONOMETRIC FUNCTIONS Finding the Height of a Tree Length of a Saw Cut A carpenter cuts the end of a 4-in.-wide board on a 5 bevel from the vertical, starting Jones & Bartlett Learning, 5 at a point in. from the end of the board. Find the lengths of the diagonal cut and the remaining side. See FIGURE 4... Solution Let,, and z be the (unknown) dimensions, as labeled in Figure 4... It 3 follows from the definition of the tangent function that in. Jones & Bartlett z Learning, 5 tan 5 so therefore 5 4 tan 5 <.87 in. FIGURE FOR 4.. SALE Saw OR cut in Eample 4

86 Jones & Bartlett Learning.. To find we observe that 4 FOR SALE OR 4 5 cos 5 so 5 < 4.4 in. cos 5 Since z 5 3 and <.87 in., we see that z <.5.87 < 3.37 in. FOR SALE OR Angles of Elevation and Depression The angle FOR between SALE an observer s OR DISTRIBUT line of sight to an object and the horizontal is given a special name. As FIGURE 4..3 illustrates, if the line of sight is to an object above the horizontal, the angle is called an angle of elevation, whereas if the line of sight is to an object below the horizontal, the angle is called an angle of depression. Object Line of sight Angle of elevation Horizontal Horizontal Angle of depression Line of sight Object FIGURE 4..3 Angles of elevation and depression FOR SALE EXAMPLE OR 3 Using Angles of Elevation A surveor uses an instrument called a theodolite to measure the angle of elevation between ground level and the top of a mountain. At one point the angle of elevation is measured to be 4. A half kilometer farther from the base of the mountain, the angle Jones & Bartlett Learning, of elevation is measured to be 37. Jones How high & is Bartlett the mountain? Learning, Solution Let h represent the height of FOR the mountain. SALE FIGURE OR 4..4 shows that there are two right triangles sharing the common side h, so we obtain two equations in two unknowns z and h: h h h Jones & Bartlett 37 4 Learning, Jones & 5Bartlett tan 37 Learning, and 5 tan 4. z 0.5 z FOR SALE 0.5 km OR z We can solve each of these for h, obtaining, respectivel, z FIGURE 4..4 Mountain in Eample 3 h 5 (z 0.5)tan 37 and h 5 z tan 4. Jones & Bartlett Equating Learning, the last two results gives an equation from Jones which & we Bartlett can determine Learning, the LL FOR SALE distance OR z: (z 0.5)tan 37 5 z tan 4. Solving for z gives us 0.5 tan 37 z 5 FOR SALE OR tan 37 tan 4. Using h 5 z tan 4 we find the height h of the mountain to be Jones & 0.5 Bartlett tan 37 Learning, tan 4 h 5 <.83 km. FOR SALE tan 37 OR tan 4 4. Applications of Right Triangles 9

87 Jones & Bartlett Learning.. EXAMPLE 4 Jones Glide Path & Bartlett Learning, Most airplanes approach San FOR Francisco SALE International OR Airport (SFO) on a straight 3 glide path starting at a point 5.5 mi from the field. A few ears ago, the FAA eperimented with a computerized two-segment approach where a plane approaches the field on a 6 glide path starting at a point 5.5 mi out and then switches to a 3 glide path.5 mi from Jones the & point Bartlett of touchdown. Learning, The point of this eperimental approach Jones was to & reduce Bartlett the noise Learning, LL FOR of the SALE planes OR over the outling residential areas. Compare the height FOR of SALE a plane Pr OR using DISTRIBUT the standard 3 approach with the height of a plane P using the eperimental approach when both planes are 5.5 mi from the airport. Solution For purposes of illustration, the angles and distances shown in FIGURE 4..5 Jones & Bartlett Learning, are eaggerated. P P P w FOR SALE z OR Q 6 Q 6 3 SFO 6 3 SFO.5 mi 4 mi Jones & Bartlett.5 mi Learning, LL 5.5 mi FOR SALE (b) (a) OR FIGURE 4..5 Glide paths in Eample 4 First, suppose is the height of plane Pr on the standard approach when it is 5.5 mi Jones & Bartlett Learning, out from the airport. As we see in Figure 4..5(a), Jones & Bartlett Learning, 5 tan 3 or tan Because distances from the airport are measured in miles, we convert to feet 5 5.5(580) tan 3 ft < 5 ft. Now, suppose z is the height FOR of plane SALE P on OR the eperimental approach when it is 5.5 mi out from the airport. As shown in Figure 4..5(b), z 5 w, so we use two right triangles to obtain 5 tan 3 or 5.5 tan 3 Jones & Bartlett Learning,.5 FOR w and SALE OR 5 tan 6 or w 5 4 tan 6. FOR SALE OR DISTRIBUT 4 Hence the approimate height of plane P at a point 5.5 mi out from the airport is z 5 w 5.5 tan 3 4 tan 6 Jones d mile & Bartlett 580 feet Learning, 5.5(580)tan 3 4(580)tan FOR SALE 6 < 635 OR ft. In other words, plane P is approimatel 3 ft higher than plane Pr. Building a Function Section.9 was devoted to setting up or constructing func- that were described or Jones epressed & Bartlett in words. Learning, As emphasized in that section, this tions is a task that ou will surel face FOR in a SALE course in OR calculus. Our final eample illustrates 9 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

88 Jones & Bartlett Learning.. a recommended procedure of sketching a figure and labeling quantities of interest with appropriate variables. EXAMPLE 5 Functions That Involve Trigonometr A plane fling horizontall at an altitude of miles approaches a radar station as shown Jones & Bartlett in FIGURE Learning, (a) Epress the distance d between the plane and the radar station as a function of the mi d angle of elevation u. (b) Epress the angle of elevation u of the plane as a function of the horizontal separation between the plane and the radar station. Ground Solution As shown in Figure 4..6, ufor is an acute SALE angle OR in a right triangle. Radar station (a) We can relate the distance d and the angle u b sin u5/d. Solving for d gives FIGURE 4..6 Plane in Eample 5 d(u) 5 sin u or d(u) 5 csc u, where 0,u#90 Jones. & Bartlett Learning, (b) The horizontal separation and u are related b tan u5/. We make use of the inverse tangent function to solve for u: where 0,,`. u() 5 tan, Eercises 4. Answers to selected odd-numbered problems begin on page ANS 8. Jones & Bartlett Learning,. A building casts a shadow 0 m Jones long. If & the Bartlett angle from Learning, the tip of the shadow to a point on top of the building is 69, FOR how high SALE is the OR building?. Two trees are on opposite sides of a river, as shown in FIGURE A baseline of T 00 ft is measured from tree T, and from that position the angle b to T is measured to be 9.7. If the baseline is perpendicular to the line segment between T River and T, find the distance between the two trees. T 00 ft 3. A 50-ft tower is located on the edge of a river. The angle of elevation between the opposite bank and the top of the tower is 37. How wide is the river? FIGURE 4..7 Trees and river in Problem 4. A surveor uses a geodometer to measure the straight-line distance from a point on the ground to a point on top of a mountain. Use the information given in FIGURE 4..8 to find the height of the mountain. Jones & Bartlett 5. An Learning, observer on the roof of building A measures Jones a 7 angle & Bartlett of depression Learning, LL 8000 ft h FOR SALE OR between the horizontal and the base of building B. The FOR angle SALE of elevation OR DISTRIBUT from the same point to the roof of the second building is 4.4. What is the height of.8 G building B if the height of building A is 50 ft? Assume buildings A and B are on the same horizontal plane. FIGURE 4..8 Mountain in Problem 4 6. Find the height h of a mountain using the information given in FIGURE The top of a 0-ft ladder is leaning against the edge of the roof of a house. If the angle of inclination of the ladder from the horizontal is 5, what is the approimate h height of the house and how far is the bottom of the ladder from the base of the house? An airplane fling horizontall at an altitude of 5,000 ft approaches a radar Jones & Bartlett Learning, km station located Jones a 000-ft-high & Bartlett hill. Learning, At one instant in time, the angle between FOR FIGURE SALE 4..9 OR Mountain in Problem 6 the radar dish pointed FOR at SALE the plane OR and the horizontal is 57. What is the 4. Applications of Right Triangles 93

89 Jones & Bartlett Learning.. straight-line distance in miles between the airplane and the radar station at that Jones P & Bartlett Learning, particular instant? 9. A 5-mi straight segment of FOR a road SALE climbs OR a 4000-ft hill. Determine the angle that the road makes with the horizontal A bo has dimensions as shown in FIGURE Find the length of the diagonal Q between the corners P and Q. What is the angle u formed between the diagonal 4 Jones & and Bartlett the bottom Learning, edge of the bo? 3 FOR. SALE Observers OR in two towns A and B on either side of a,000-ft FOR mountain SALE measure OR DISTRIBUT FIGURE 4..0 Bo in Problem 0 the angles of elevation between the ground and the top of the mountain. See FIGURE 4... Assuming that the towns and the mountaintop lie in the same vertical plane, find the horizontal distance between them.. A drawbridge* measures 7.5 m from shore to shore, and when completel open Jones,000 ft & Bartlett Learning, it makes an angle of 43 with the horizontal. See FIGURE 4..(a). When the 8 46 bridge is closed, the angle of depression from FOR the SALE shore to OR a point on the surface A B of the water below the opposite end is 7. See Figure 4..(b). When the FIGURE 4.. Mountain in Problem bridge is full open, what is the distance d between the highest point of the bridge and the water below? Bridge 7.5 m 7.5 m Jones & Bartlett Learning, 43 d 7 FOR SALE OR DISTRIBUT Jones & Bartlett Learning, (a) Open bridge Jones (b) Closed & bridge Bartlett Learning, FIGURE 4.. Drawbridge in Problem FOR SALE OR 3. A flagpole is located at the edge of a sheer 50-ft cliff at the bank of a river of width 40 ft. See FIGURE An observer on the opposite side of the river measures an angle of 9 between her line of sight to FOR the top SALE of the OR flagpole and her line of sight to the top of the cliff. Find the height of the 9 flagpole. 50 ft 4. From an observation site 000 ft from the base of Jones & Mt. Bartlett Rushmore Learning, the angle of elevation to the top of the FOR SALE sculpted OR head of George Washington is measured to Bust of George be 80.05, whereas the angle of elevation to the bottom of Washington on his head is Determine the height of George River Mt. Rushmore Washington s head. 5. The length of a Boeing 747 airplane is 3 ft. What is 40 Jones ft & Bartlett Learning, the plane s altitude if it subtends an angle of when it is FIGURE 4..3 Flagpole in Problem 3 directl above an observer on the ground? FOR See SALE FIGURE OR 6. The height of a gnomon (pin) of a sundial is 4 in. If it casts a 6-in. shadow, what is the angle of elevation of the Sun? Sundial *The drawbridge shown in Figure Jones 4.., & Bartlett where the Learning, span is continuousl balanced b a counterweight, is called a basculefor bridge. SALE OR 94 CHAPTER 4 TRIGONOMETRIC FUNCTIONS Paul Fries/ShutterStock, Inc. Peter Elvidge/ShutterStock, Inc.

90 Jones & Bartlett Learning.. 7. Weather radar is capable of measuring both the angle of elevation to the top of a thunderstorm and its range (the horizontal distance to the storm). If the range of a storm is 90 km and FOR the SALE angle of OR elevation is 4, can a passenger plane that is able to climb to 0 km fl over the storm? 8. Cloud ceiling is the lowest altitude at which solid cloud is present. The cloud ceiling at airports must be sufficientl high for safe takeoffs and landings. At Jones & Bartlett night Learning, the cloud ceiling can be determined b illuminating Jones & Bartlett the base of Learning, the clouds LL FOR SALE OR with a searchlight pointed verticall upward. If an FOR observer SALE is km OR from DISTRIBUT the searchlight and the angle of elevation to the base of the illuminated cloud is 8, find the cloud ceiling. See FIGURE (During the da cloud ceilings are generall estimated b sight. However, if an accurate reading is required, a balloon is inflated so that it will rise at a known constant rate. Then it is released Ground FIGURE 4..4 Airplane in Problem 5 and timed until it disappears into the cloud. The cloud ceiling is determined b multipling the rate b the time of FOR the ascent; SALE trigonometr OR is not required for this calculation.) Cloud cover 9. On a rescue flight, a U.S. Coast Guard helicopter approaches a container ship at an altitude of 800 ft as shown in FIGURE Measured from the front the helicopter, the Jones angle of & depression Bartlett of Learning, the ship s stern is 35.3 and the angle of depression of its bow FOR is 6.6. SALE Approimatel OR how long is the container ship? [Hint: Ignore the height of the bow and stern above sea level.] Observer 8 0. From a room on the top floor of a nearb hotel, the angle of elevation to the top of the Eiffel Tower is 7.5 ; from street level in front of the hotel the angle of Searchlight elevation to the top of the tower is.4. See FIGURE What is the distance km from the hotel to the midpoint M, shown in the figure, of the base of the tower? FIGURE 4..5 Searchlight in Problem FOR SALE 8 OR Approimatel how high is the hotel building? 800 ft M 063 ft FIGURE 4..6 Container ship in Problem 9 FIGURE 4..7 Eiffel Tower in Problem mi. The distance between the Earth and the Moon varies as the Moon revolves Observer around the Earth. At a particular time the geocentric paralla angle shown in Moon FIGURE 4..8 is measured to be. Calculate to the nearest hundred miles the d distance between the center of the Earth and the center of the Moon at this Earth instant. Assume that the radius of the Earth is 3963 miles. FOR SALE. OR The final length of a volcanic lava flow seems to FOR SALE OR DISTRIBUT FIGURE 4..8 Angle in Problem decrease as the elevation of the lava vent from which it originates increases. An empirical stud of Mt. Etna gives the final lava flow length L in terms of elevation h b the formula L h, Lava where L is measured in kilometers and h is Village Lava vent measured in meters. Suppose that a Sicilian 500 m village at elevation 750 m is on a 0 slope Mt. Etna m directl below Jones a lava vent & Bartlett at 500 m. Learning, See FIGURE According to the formula, how close will FOR the SALE lava flow OR get to the FOR FIGURE SALE 4..9 OR Lava flow in Problem village? 4. Applications of Right Triangles 95 Mar Lane/ShutterStock, Inc

91 _CH04_nd.qd 0/7/4 9:5 PM Page 96 Jones & Bartlett Learning.. 3. As shown in FIGURE 4..0, two tracking stations S and S sight a weather Weather balloon Jones & Bartlett Learning, balloon between them at elevation angles a and b, respectivel. Epress the height h FOR SALE OR FOR SALE h OR of the balloon in terms of a and b, and the distance c between the tracking stations. S Ground FIGURE 4..0 Problem 3 S 4. An entr in a soapbo derb rolls down a hill. Using the information given in FIGURE 4.., find the total distance d d that the soapbo travels. c 5. Find the height and area of the isosceles trapezoid shown in FIGURE 4... Weather balloon in Jones6.& An Bartlett Learning, Jones store & Bartlett Learning, escalator between the first and second floors of a department is 58 ft long 0 FIGURE Find the vertical and makes an angle of with the first floor. See distance between the floors. second floor d 00 ft 5 d FIGURE Soapbo in Problem ft 70 Jones4 & Bartlett Learning, FIGURE 4.. Problem 5 Lighthouse of Aleandria 0 first floor Finish FINISH 3 Trapezoid in FIGURE 4..3 Escalator in Problem 6 7. Recent Histor According to the online encclopedia Wikipedia, a French flown b Jean Boulet attained the world s of Jones & helicopter Bartlett Learning, record Jonesheight & Bartlett Learning, s,44 m in 97. What would the angle of elevation to the helicopter have FOR SALE OR FOR SALE OR h been from a point P on the ground 000 m from the point directl beneath the helicopter? 8. Ancient Histor In an article from the r r online encclopedia Wikipedia, the height h Jones & Bartlett Learning, & Bartlett Learning, of the Lighthouse of Aleandria, one ofjones the FOR SALE OR FOR SALE OR Seven Wonders of the Ancient World built θ between 80 and 47 B.C.E., is estimated to have been between 393 ft and 450 ft. The Center of article goes on to sa that there are ancient the Earth be seen on the Learning, Jones & Bartlett Learning, claims that the light could Jones & Bartlett FIGURE 4..4 Lighthouse in ocean up to 9 miles awa. Use the right Artist s rendering of the Lighthouse Problem FOR8SALE OR triangle in FIGURE 4..4 along with the two of Aleandria given heights h to determine the accurac of the 9 mile claim. Assume that the radius of the Earth is r mi and s is distance measured in miles on the ocean. [Hint: Use ft 5 /580 mi and (7) of Jones & Section Bartlett Learning, 4..] θ 9. SALE Reall Ancient Histor The Great FOR OR Pramid of Giza is believed to be the tomb 756 ft of the Pharaoh Khufu (AKA Cheops) and is the oldest of the Seven Wonders of the (a) Ancient World. When completed around Jones & Bartlett Learning, Ramp 560 B.C.E. the pramid had a square base with length 756 ft on a side and estimated height of 48 ft. φ Base (a) Find the angle of inclination u shown in FIGURE 4..5(a) that a side face of (b) Jones & Bartlett Jones & Bartlett Learning, of Giza the pramid makes with level The Great Pramid FIGURE 4..5 PramidLearning, in Problem 9 ground. PRILL/ShutterStock, Inc. George Baile/Dreamstime.com Beam of light 96 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

92 Jones & Bartlett Learning.. (b) In one of several theories on how the pramid was constructed, workers pulled massive stone blocks up an inclined ramp made out of mud bricks and stone rumple. See Figure 4..5(b). The height and width of the ramp had to be changed as the height of the pramid increased. Suppose at the point in time when the unfinished pramid was 300 ft high, the base of the ramp measured 984 ft. What was the angle of inclination f of the inclined ramp indicated in Jones & Bartlett Learning, Figure 4..5(b)? FOR SALE OR (c) What was the length of the ramp at the 300 ft stage? FOR [Hint: SALE Look OR at Figure DISTRIBUT 4..5(b) in cross section and use part (a).] 30. Medieval Histor The construction of the campanile, or bell tower, for the Cathedral of Pisa, Ital began in 73 C.E. After the construction of the second floor one side of its base started Jones sinking & into Bartlett the soft Learning, marsh ground. Because it continued to sink long after the completion of its construction in 37 C.E. the campanile came to be dubbed the Leaning Tower of Pisa. Over the centuries the tower has defied man ingenious attempts to correct its tilt, but after the most recent restoration in 00 the tower leans at angle 3.99 (corrected from 5.5 ) measured from the vertical. This angle is shown in FIGURE 4..6 along with the heights, measured Jones from & the Bartlett ground, Learning, of the high and low sides of tower belfr. Find the horizontal FOR displacement SALE OR d of the center P of the roof of the belfr from the vertical. [Hint: Keep in mind that the height of the point P measured from the ground is neither of the two heights given in the figure.] P d Belfr 86.0 ft ft Drozdowski/ShutterStock, Inc. Leaning Tower of Pisa Ground FIGURE 4..6 Displacement d of the Jones point P& on Bartlett the belfr roof Learning, in LL Problem FOR 30 SALE OR DISTRIBUT P d In Problems 3 34, proceed as in Eample 5 and translate the words into an appropriate Jones & Bartlett Learning, function. FOR SALE OR 3. A tracking telescope, located.5 km from the point of a rocket launch, follows a verticall ascending rocket. Epress the height h of the rocket as a function of Shore Lighthouse the angle of elevation u. 3. A searchlight one-half mile offshore illuminates a point P on the shore. Epress Jones & FIGURE Bartlett 4..7 Learning, Searchlight in the distance djones from the & searchlight Bartlett to Learning, the point of illumination P as a function of FOR Problem SALE 3 OR the angle u shown FOR in FIGURE SALE OR 4. Applications of Right Triangles 97

93 Jones & Bartlett Learning A statue is placed on a pedestal as shown in FIGURE Epress the viewing angle u as a function of the distance from the pedestal. 34. A woman on an island wishes FOR to SALE reach a OR point R on a straight shore on the mainland from a point P on the island. The point P is 9 mi from the shore and 5 mi from point R. See FIGURE If the woman rows a boat at a rate of 3 mi/h to a point Q on the mainland, then walks the rest of the wa at a rate of 5 mi/h, Jones & epress Bartlett the Learning, total time it takes the woman to reach point Jones R as a function & Bartlett of the Learning, LL FOR SALE indicated OR angle u. [Hint: Distance 5 rate 3 time. ] Shore Q R Ee level m 9 mi 5 mi θ Jones & m Bartlett Learning, P Island FIGURE 4..8 Viewing angle in Jones & Bartlett Problem 33 Learning, For Discussion FIGURE 4..9 Woman rowing to shore in Problem Consider the blue rectangle circumscribed around the red rectangle in FIGURE With the aid of calculus it can be shown that the area of the blue Jones & Bartlett Learning, rectangle is greatest when u5p/4. Find Jones this area & Bartlett in terms of Learning, a and b. FOR SALE OR 36. Home heating oil is often stored in a right circular FOR SALE clindrical OR tank of diameter D that rests horizontall. As FIGURE 4..3 shows, the depth of the oil can be measured b inserting a dipstick down a vertical diameter. If the dipstick indicates that the depth of the oil is d inches, then show that the volume V of the oil is given b V 5 V Jones 0 p ccos a d & Bartlett Learning, FOR SALE d b a OR D D b Ç a d D b d D d, where V 0 is the volume of the tank. Although not necessar, assume for simplicit that the tank is less than half full as shown in the figure. Dipstick 98 CHAPTER 4 TRIGONOMETRIC FUNCTIONS a θ b FIGURE Rectangles Jones in & Bartlett FIGURE Learning, 4..3 Oil tank in Problem 35 FOR SALE OR Problem 36 D d Oil

94 Jones & Bartlett Learning Regular Polgon Recall, in geometr a polgon is a closed plane figure whose sides are straight-line segments. If all sides of a polgon have equal lengths and if all angles have the FOR same SALE measure, OR then the polgon is said to be a regular polgon. A regular polgon with three sides is an equilateral triangle; a four-sided r R regular polgon is a square. The si-sided regular polgon shown FIGURE 4..3 is a heagon. In Figure 4..3 the perpendicular distance R from the center to a FIGURE 4..3 Heagon in Jones Problem 37 & Bartlett side Learning, of the polgon is called the inradius or the Jones apothem. & The Bartlett distance Learning, r from the LL FOR SALE OR center to a verte in Figure 4..3 is called the circumradius FOR SALE of the OR polgon. DISTRIBUT (a) Use trigonometr to epress the area A of a heagon as a function of the inradius R. (b) Epress the area A of a heagon as a function of the circumradius r. 38. Epress the area A of a eight-sided regular polgon, or octagon, shown in FIGURE as a function of the length S of one side. 39. Using trigonometric functions, generalize FOR SALE the three OR area formulas in Problems 37 S and 38 to an regular polgon having n sides. [Hint:You might want to keep track of the number 6 in Problem 37 and the number 8 in Problem 38.] FIGURE Octagon in Problem Intersection of Circles Again The centers of the red and blue circles in FIGURE are Jones B and & CBartlett and have coordinates Learning, ( 0, 0 ) and (, ), respectivel. Let d denote the distance FOR SALE between OR the centers. (a) Suppose each circle has radius r. Use a triangle in Figure 4..34(b) to epress the area A of the intersection of the circles, the ellow region in the Figure 4..34(a), in terms of r and u. (b) Show that the answer to Problem 95 in Eercises 4. is special case of the area formula in part (a). FOR SALE OR (c) Find area of the intersection of the circles ( 4) ( 4) 5 9 and ( 6) ( 8) 5 9. C D d θ/ r B r E 4. Law of Sines (a) FIGURE Intersecting circles in Problem 40 INTRODUCTION In Section 4.0 we saw how to solve right triangles. In this and the net section we consider two techniques for solving an oblique triangle, that is, a triangle Jones & Bartlett Learning, that has no right angle. An oblique triangle Jones has & either Bartlett three acute Learning, angles or two acute angles and an obtuse angle. (b) Area of an Oblique Triangle Before introducing the principal topic of this section, let us first eamine a familiar area problem. From geometr we know that the area A of an triangle is Jones A 5 (base & Bartlett 3 height). Learning, In the case of an oblique triangle we must generall use trigonometr FOR to SALE obtain the OR height h. As shown in FIGURE 4..(a) we begin 4. Law of Sines 99

95 Jones & Bartlett Learning.. b constructing an altitude with length in the blue triangle from verte B to side AC. Jones & Bartlett Learning, From the right triangle whose hpotenuse has Jones length & c Bartlett we see that Learning, h () c 5 sin a and so h 5 c sin a. Net, in Figure 4..(b) let denote the length of the altitude from verte C to side AB. Then from the right triangle Jones whose & Bartlett hpotenuse Learning, has length a we get h () a 5 sin FOR b SALE and so OR h 5 a sin b. Finall, b constructing an altitude from verte A to side BC we see from Figure 4..(c) that its length h 3 can be epressed in terms g: h 3 (3) b 5 sin g and so h 3 5 b sin g. FOR SALE OR DISTRIBUT Using A 5 (base 3 height) for the three heights h, h, and h 3 in (), (), and (3) along with the respective bases of lengths b, c, and a the area A of the oblique triangle can be Jones & Bartlett Learning, epressed in terms of the lengths of two sides Jones and the & included Bartlett angle: Learning, A 5 (4) bc sin a, A 5 FOR SALE ac sin b, A 5 OR ab sin g. Although we used an oblique triangle in which all angles were acute, the results in (4) are equall valid for a triangle with an obtuse angle. The formulas in (4) are certainl useful in their own right in finding area, but coincidentall we have proved a more FOR important SALE result. OR A Law of Sines Consider the oblique triangle ABC, shown in FIGURE 4.., with angles a, b, and g, sides BC, AC, and AB with corresponding lengths a, b, and c. If c Jones we & know Bartlett the length Learning, of one side and two other parts of the triangle, Jones that & Bartlett is, eitherlearning, LL b one side and two angles (SAA or ASA), or B a C two sides and an angle opposite one of the sides (SSA), FIGURE 4.. Oblique triangle then the remaining three parts can be found using the Law of Sines. THEOREM 4.. The Law of Sines FOR SALE OR 300 CHAPTER 4 TRIGONOMETRIC FUNCTIONS B c β A α h Jones & Bartlett a Learning, (a) b γ C FIGURE 4.. Oblique triangle with three acute angles B h c β A α h h a (b) Suppose angles a, b, and g, opposite sides of length a, b, and c are as shown in Figure 4... Then sin a Jones 5 sin b 5 sin g a& Bartlett b Learning, c b γ C B c β A α h 3 (c) b Jones & Bartlett a Learning, LL γ (5) C

96 Jones & Bartlett Learning.. Because the three formulas in (4) give the same area we see immediatel that FOR bc SALE sin a5 OR ac sin b5 ab sin g. B dividing each term in this double equalit b abc we obtain (5). FOR SALE EXAMPLE OR Solving an Oblique Triangle FOR (SAA) SALE OR DISTRIBUT A Find the remaining parts of the triangle shown in FIGURE c 6 Solution Let b50, a530, and b 5 6. Because the sum of the angles in a triangle 0 30 is 80 we have g and so g B Jones & Bartlett Learning, C a From (5) we then see that FIGURE 4..3 Triangle in Eample sin 30 sin 0 sin (6) a 6 c We use the first equalit in (6) to solve for a: sin 30 FOR SALE a 5 6 OR sin 0 < The second equalit in (6) gives c: EXAMPLE sin 30 c 5 6 sin 0 < Height of a Building (ASA) A building is situated on the side of a hill that slopes downward at an angle of 5. The Jones & Bartlett Learning, Sun is uphill from the building at an Jones angle of & elevation Bartlett of 4 Learning,. Find the building s height if it casts a shadow 36 ft long. Light beam 4 Q Solution Denote the height of the building on the downward slope b h and construct a right triangle QPS as shown in FIGURE Now a5 5 4 so that a57. Since DQPS is a right triangle, g gives g From the Jones & Law of Sines (5), Building Bartlett Learning, h sin 7 FOR sin 48 SALE OR sin 7 5 so h 5 36 <.99 ft. 4 h 36 sin 48 5 P S In Eamples and, where we were given two angles and a side opposite one of 36 ft shadow Jones & Bartlett these angles, Learning, each triangle had a unique solution. However, Jones this & ma Bartlett not alwas Learning, be true LL FIGURE 4..4 Triangle QPS in Eample for triangles where we know two sides and an angle opposite one of these sides. The net eample illustrates the latter situation. EXAMPLE 3 Two Triangles (SSA) Jones & Bartlett Learning, Find the remaining parts of the triangle Jones with & b550, Bartlett b Learning, 5 5, and c 5 6. Solution From the Law of Sines, we have sin 50 5 sin g or sin g5 6 sin 50 < From a calculator set Jones in degree & Bartlett mode, we Learning, obtain g < At this point it is essential to recall that the sine FOR function SALE is also OR positive for second quadrant angles. In other 4. Law of Sines 30

97 Jones & Bartlett Learning.. A words, there is another angle satisfing 0 #g#80 for which sin g < Using 66.8 as a reference angle we find the second quadrant angle to be Therefore, the two possibilities for g are g < and g < 3.8. Thus, as shown in FIGURE 4..5, there are two possible triangles 5 ABC and ABC satisfing the given three conditions. To complete the solution of triangle ABC (Figure 4..5(a)), we first find Jones a & 5Bartlett 80 glearning, b or a < To find the side opposite Jones this angle & Bartlett we use Learning, LL 50 B C FOR SALE sin OR 63.8 sin 50 sin FOR 63.8 SALE OR DISTRIBUT a 5 which gives a a a sin 50 b (a) or a < 5.8. A To complete the solution of triangle ABC (Figure 4..5(b)), we find a 5 Jones & Bartlett Learning, 80 g b or a < 6.8. Then from sin 6.8 sin 50 sin we find a a a sin 50 b 5 or a <.89. Jones & Bartlett Learning, 50 Ambiguous Case When solving triangles, the situation where two sides and an B FOR SALE OR C a angle opposite one of these sides are given (SSA) is called the ambiguous case. We have just seen in Eample 3 that the given information ma determine two different triangles. In the ambiguous case other complications can arise. For instance, suppose (b) FIGURE 4..5 Triangles in Eample 3 that the length of sides AB and AC (that is, c and b, respectivel) and the angle b in triangle ABC are specified. As shown in FIGURE 4..6, we draw the angle b and mark off FOR side SALE AB with OR length A c to locate the vertices A and B. The third FOR verte SALE C is located OR on DISTRIBUT the base b drawing an arc of a circle of radius b (the length of AC) with center A. As shown in FIGURE 4..7, there are four possible outcomes of this construction: c The arc does not intersect the base and no triangle is formed. Jones & Bartlett Learning, The arc intersects the base in two distinct Jones points & Bartlett C and C Learning, and two triangles are formed (as in Eample 3). B The arc intersects the base in one point and one triangle is formed. FIGURE 4..6 Horizontal base, the angle b, and side AB The arc is tangent to the base and a single right triangle is formed. b c B A 30 CHAPTER 4 TRIGONOMETRIC FUNCTIONS B c b A C C B A FOR SALE c OR b b c (a) No Jones triangle & Bartlett (b) Learning, Two triangles (c) Single triangle Jones (d) & Right Bartlett trianglelearning, LL FIGURE 4..7 Solution FOR SALE possibilities OR for the ambiguous case in the Law of Sines EXAMPLE 4 Determining the Parts of a Triangle (SSA) Find the remaining parts of the triangle with b540, b 5 5, and c 5 9. Solution From the Law of Sines (), we have sin 40 5 sin g and so sin g5 9 sin 40 < Since the sine of an angle must be between and, sin g <.570 is impossible. This means the triangle has no solution; Jones & the Bartlett side with Learning, length b is not long enough to reach the base. This is the case illustrated FOR in SALE Figure OR 4..7(a). C B A C b

98 Jones & Bartlett Learning.. Eercises 4. Answers to selected odd-numbered Jones & problems Bartlett begin Learning, on page ANS-8. In Problems 4, find the area A of the given triangle.. A. A 34 B FIGURE 4..9 Triangle for Problem C 49 B C 8.8 FIGURE 4..8 Triangle for Problem FOR SALE OR 3. A 4. A B C FIGURE 4..0 Triangle for Problem FIGURE 4.. Triangle for Problem 4 In Problems 5 0, refer to Figure 4... FOR SALE In Problems OR 5 0, use the Law of Sines to solve the triangle. FOR SALE OR DISTRIBUT 5. a580, b50, b a560, b55, c b537, g55, a a530, g575, a b57, b 5, c a50, a 5 9, c 5 4 Jones & Bartlett Learning,. g56, b 5 7, c 5 4 Jones &. Bartlett b50, Learning, g55, a g55, a 5 8, c 5 5 FOR 4. SALE a555, OR a 5 0, c g550, b 5 7, c a535, a 5 9, b 5 7. b530, a 5 0, b a540, g50, c 5 9. a50, a 5 8, c a575, g545, b 5 8 B C In Problems and Jones, solve & for Bartlett in the Learning, given triangle FIGURE FOR 4..3SALE Triangle OR for Problem DISTRIBUT FIGURE 4.. Triangle for Problem C FOR SALE 0 ftor Applications A Jones & Bartlett FIGURE Learning, 4..4 Pool in FOR SALE Problem OR 3 B Jones & Bartlett Learning, 3. Length of a Pool A 0-ft rope that is available to measure the length between two points A and B at opposite ends of a kidne-shaped swimming pool is not long enough. A third point C is found such that the distance from A to C is 0 ft. It is determined Jones that angle & Bartlett ACB is 5 Learning, and angle ABC is 35. Find the distance from A to B. See FIGURE FOR SALE OR 4. Law of Sines 303

99 Jones & Bartlett Learning.. 4. Width of a River Two points A and B lie on opposite sides of a river. Another point C is located on the same side of the river as B at a distance of 30 ft from B. If angle ABC is 05 and angle FOR ACB SALE is 0 OR, find the distance across the river from A to B. 5. Length of a Telephone Pole A telephone pole makes an angle of 8 with the level ground. As shown in FIGURE 4..5, the angle of elevation of the Sun is 76. Jones & Find Bartlett the length Learning, of the telephone pole if its shadow is 3.5 Jones m. (Assume & Bartlett that the Learning, tilt LL FOR SALE of the pole OR is awa from the Sun and in the same plane as the FOR pole SALE and the OR Sun.) DISTRIBUT 6. Not on the Level A man 5 ft 9 in. tall stands on a sidewalk that slopes down at a constant angle. A vertical street lamp directl behind him causes his shadow to be 5 ft long. The angle of depression from the top of the man to the tip of his 8 76 shadow is 3. Find the angle a, as shown in FIGURE 4..6, that the sidewalk 3.5 m makes with the horizontal. FIGURE 4..5 Telephone pole in Problem 5 7. How High? If the man in Problem 6 is 0 FOR ft down SALE the sidewalk OR from the street lamp, find the height of the light above the sidewalk. 8. Plane with an Altitude Angles of elevation to an airplane are measured from the top and the base of a building that is 0 m tall. The angle from the top of the building is 38, and the Jones angle from & Bartlett the base of Learning, the building is 40. Find the altitude of the airplane. 9. Angle of Drive The distance from the tee to the green on a particular golf hole is 370 d. A golfer hits his drive and paces its distance off at 0 d. From the point where the ball lies, he measures an angle of 60 between the tee and the green. Find the angle of his drive off the tee measured from the dashed line from the tee to the green shown in FIGURE Shadow Tee 60 Ball FIGURE 4..6 Sloping sidewalk in FIGURE 4..7 Angle of drive in Problem 6 Problem In Problem 9, what is the FOR distance SALE from OR the ball to the green? 3. Help! One Coast Guard vessel is located 4 nautical miles due south of a second Coast Guard vessel when the receive a distress signal from a sailboat. To offer assistance, the first vessel sails on a bearing of S50 E at 5 knots and the second vessel sails S0 E at 0 knots. Which one of the Coast Guard vessels reaches the Jones & sailboat Bartlett first? Learning, [Hint: The concept of bearing is reviewed Jones page & 306.] Bartlett Learning, LL 4.3 Law of Cosines A INTRODUCTION In the right triangle shown in FIGURE 4.3., the length c of the hpotenuse is related to the lengths a and b of the other two sides b the Pthagorean c theorem b Jones B & Bartlett a Learning, C In this section we will see that Jones () is & just Bartlett a special Learning, case of a general formula that relates FOR FIGURE SALE 4.3. OR Right triangle the lengths of the sides of an triangle. FOR SALE OR 304 CHAPTER 4 TRIGONOMETRIC FUNCTIONS c 5 a b. ()

100 Jones & Bartlett Learning.. A Law of Cosines An oblique triangle, such as that in FIGURE 4.3., for which we know either c three sides (SSS), or b two sides and the included angle (that is, the angle formed b the given sides) (SAS), B C a Jones & Bartlett cannot Learning, be solved directl using the Law of Sines. The Jones Law & of Bartlett Cosines that Learning, we consider OR net can be used to solve triangles in these two cases. FOR Like SALE the Law OR of DISTRIBUT Sines, (5) LL FIGURE 4.3. Oblique triangle FOR SALE of Section 4., the Law of Cosines is valid for an oblique triangle, but for convenience we prove the last two equations in () using a triangle in which the angles a, b, and g, are acute. THEOREM 4.3. The Law FOR of Cosines SALE OR Suppose angles a, b, and g, and opposite sides of length a, b,and c are as shown in Figure Then c A h PROOF: Let P denote the point where the altitude from the verte A intersects side BC. Then, since both ^BPA and ^CPA in FIGURE are right triangles we have from (), c 5 h (c cos b) FOR SALE OR DISTRIBUT (3) b and b 5 h (b cos g). (4) B Jones & Bartlett Learning, C c cos FOR P SALE b cos OR Moreover, from (4), FIGURE Acute triangle a 5 b c bc cos a Jones & Bartlett b 5 a Learning, c ac cos b FOR SALE c 5 aor b ab cos g Now the length of BC is a 5 c cos bb cos g so that c cos b5a b cos g. (5) h 5 b (b cos g). (6) Substituting (5) and (6) into (3) and simplifing ields the third equation in (): c 5 b (b cos g) (a b cos g) Jones 5 b & bbartlett cos ga Learning, ab cos gb cos g or c 5FOR a SALE b ab OR cos g. (7) Note that equation (7) reduces to the Pthagorean theorem () when g590. Similarl, if we use b cos g5a c cos b and h 5 c (c cos b) to eliminate b cos g and h in (4), we obtain the second equation in (). FOR SALE EXAMPLE OR Solving an Oblique Triangle FOR (SAS) SALE OR DISTRIBUT A Find the remaining parts of the triangle shown in FIGURE b Solution First, if we call the unknown side b and identif a 5, c 5 0, and b56, Jones & Bartlett Learning, then from the second equation in () Jones we can & write Bartlett Learning, 6 FOR SALE OR b 5 () FOR (0) SALE OR B C ()(0)cos 6. Therefore, b FIGURE Triangle in Eample < and so b < 5.3. Net, we use the Law of Cosines to determine the remaining angles in the triangle in Figure If g is the angle at the verte C, then the third equation in () gives 0 5 (5.3) FOR SALE ()(5.3) OR cos g or cos g < Law of Cosines 305 ()

101 Jones & Bartlett Learning.. With the aid of a calculator and the inverse cosine we find g < Note that since the cosine of an angle between 90 and 80 is negative, there is no need to consider two possibilities as we did in Eample 3 in Section 4.. Finall, the angle at the verte A is a580 bgor a < In Eample, observe that after b is found, we know two sides and an angle opposite & one Bartlett of these Learning, sides. Hence we could have used the Law of Jones Sines to & find Bartlett the angle Learning, g. LL Jones FOR SALE In the net OR eample we consider the case in which the lengths FOR of the SALE three sides OR DISTRIBUT of a triangle are given. EXAMPLE Determining the Angles in a Triangle (SSS) Jones & Bartlett Learning, Find the angles a, b, and g in the triangle shown Jones in & Bartlett FIGURE Learning, A Solution We use the Law of Cosines to find the angle opposite the longest side: (6)(7) cos g or cos g A calculator then gives g < Although we could use the Law of Cosines, we choose to find b b the Law Jones of Sines: & Bartlett Learning, B C sin b sin or sin b5 6 7 sin 87.7 < FIGURE Triangle in Eample Since g is the angle opposite the longest side it is the largest angle in the triangle, so b Jones must & Bartlett be an acute Learning, angle. Thus, sin b < ields b Jones < & Bartlett Finall, from Learning, LL FOR a580 SALE bg OR we find a < Bearing In navigation directions are given using bearings. A bearing designates the acute angle that a line makes with the north south line. For eample, FIGURE 4.3.6(a) illustrates a bearing of S40 W, meaning 40 degrees west of south. The bearings in Jones & Bartlett Learning, Figures 4.3.6(b) and 4.3.6(c) are N65 E and Jones S80 E, & Bartlett respectivel. Learning, 306 CHAPTER 4 TRIGONOMETRIC FUNCTIONS W N 40 S40 W S S S N (a) (b) (c) FIGURE Three eamples of bearings EXAMPLE 3 port Bearings of Two Ships (SAS) c Two ships leave a port at 7:00 AM, one traveling at knots (nautical miles per hour) Jones & Bartlett Learning, and the other at 0 knots. If the faster ship maintains Jones & a bearing Bartlett of N47 W Learning, and the 40 other FOR 0 SALE OR ship maintains a bearing of S0 W, what is their FOR separation SALE OR (to the nearest nautical mile) at :00 AM that da? S Solution Since the elapsed time is 4 hours, the faster ship has traveled 4 # 5 48 nautical miles from port and the slower ship 4 # nautical miles. Using these distances and the given bearings, Jones we & can Bartlett sketch the Learning, triangle (valid at :00 AM) shown FIGURE FOR SALE Ships OR in Eample 3 in FIGURE In the triangle, FOR c denotes SALE the distance OR separating the ships and g is the N 65 N65 E E W E W N 80 E S80 E

102 Jones & Bartlett Learning.. angle opposite that side. Since 47 g we find g53. Finall, the Law of Cosines c (48)(40)cos 3 gives c < or c < Thus the distance between the ships (to the nearest nautical mile) is 74 nautical miles. Roman Sigaev/ShutterStock, Inc. ES FROM THE CLASSROOM (i) An important Jones first & Bartlett step in solving Learning, a triangle is determining which FOR of the SALE three approaches OR we have discussed to use: right triangle trigonometr, the Law of Sines, or the Law of Cosines. The following table describes the various tpes of problems and gives the most appropriate approach for each. Tpe of Triangle Information Given Technique Right Two sides or an angle and Basic definitions of sine, a side cosine, and tangent; Jones the Pthagorean & Bartlett theorem Learning, LL Oblique Two angles and a side Law of Sines (SAA or ASA) Oblique Two sides and an angle Law of Sines (if the opposite one of the sides given angle is acute, it is (SSA) an ambiguous case) Oblique Three sides (SSS) Law of Cosines Oblique Two sides and the Law of Cosines included angle (SAS) (ii) Here are some additional FOR SALE bits of OR advice for solving triangles. Students will frequentl use the Law of Sines when a right triangle trigonometric function could have been used. A right triangle approach is the simplest and most efficient. In appling the Law of Sines, if ou obtain a Jones value greater & Bartlett than for Learning, the sine LL FOR SALE OR of an angle, there is no solution. is less than 80. In the ambiguous case of the Law of Sines, when solving for the first unknown angle, ou must consider both the acute angle found from our calculator and its supplement as possible solutions. The supplement will be a solution if the sum of the supplement and the angle given in the triangle When three sides are given, check first to see whether the length of the longest side is greater than or equal to the sum of the lengths of the other two sides. If it is, there can be no solution (even though the given information indicates a Law of Cosines approach). This is because the shortest distance between two points Jones is the length & Bartlett of the line Learning, segment joining them. 4.3 Law of Cosines 307

103 Jones & Bartlett Learning.. Eercises 4.3 Answers to selected odd-numbered problems begin on page ANS 8. In Problems 6, refer to Figure In Problems 6, use the Law of Cosines to solve the triangle.. g565, a 5 5, b 5 8. b548, a 5 7, c a 5 8, b 5 0, c g53.5, a 5 4, b 5 8 Jones 5. & g597.33, Bartlett Learning, a 5 3, b a 5 7, b 5 9, Jones c 5 4 & Bartlett Learning, LL FOR 7. SALE a 5, OR b 5 9.5, c a56, b 5, FOR c 5SALE 8 OR DISTRIBUT 9. a 5 5, b 5 7, c a 5 6, b 5 5, c 5 7. a 5 3, b 5 4, c 5 5. a 5 5, b 5, c a 5 6, b 5 8, c 5 4. b530, a 5 4, c a5, b 5 3, c 5 9 Jones C & Bartlett Learning, 6. b500, a 5.3, b 5 6. FOR SALE OR Applications 7. How Far? A ship sails due west from a harbor for nautical miles. It 4. in. 3.5 in. then sails S6 W for another 5 nautical miles. How far is the ship from the harbor? 8. How Far Apart? Two Jones hikers leave & Bartlett their camp Learning, simultaneousl, taking bearings FOR A SALE B.5 in. OR of N4 W and S0 E, respectivel. FOR SALE If the OR each average a rate of 5 km/h, how FIGURE Triangle in Problem 9 far apart are the after h? 9. Bearings On a hiker s map point A is.5 in. due west of point B and point C is 3.5 in. from B and 4. in. from A, respectivel. See FIGURE l Find (a) the Jones & bearing Bartlett of ALearning, from C, and (b) the bearing of B from C. l FOR 0. SALE How Long OR Will It Take? Two ships leave port simultaneousl, FOR one SALE traveling OR at DISTRIBUT 5 knots and the other at knots. The maintain bearings of S4 W and S0 E, respectivel. After 3 h the first ship runs aground and the second ship immediatel goes to its aid. (a) How long will it take the second ship to reach the first ship if it travels at Jones & Bartlett Learning, 4 knots? FIGURE Robotic arm in Problem (b) What bearing should it take? Fire. A Robotic Arm A two-dimensional robot arm knows where it is b keeping track of a shoulder angle a and an elbow angle b. As shown in FIGURE 4.3.9, this arm has a fied point of rotation at the origin. The shoulder angle is measured counterclockwise from the -ais, and the elbow angle is measured counterclockwise from the upper to the lower arm. Suppose that the upper and lower arms are both of length and that the elbow angle b is prevented from hperetending beond 80. Find the angles a and b that will position the robot s hand A B? 3 at the point (, ). C. Which Wa? Two lookout towers are situated on mountain tops A and B,4 mi Jones & from Bartlett each other. Learning, A helicopter firefighting team is located Jones in a valle & Bartlett at point C, Learning, LL FIGURE Fire in Problem FOR SALE 3 mi from OR A and mi from B. Using the line between A and FOR B as a SALE reference, OR a DISTRIBUT lookout spots a fire at an angle of 40 from tower A and 8 from tower B. See 80 cm 80 cm FIGURE At what angle, measured from CB, should the helicopter fl in 70 order to head directl for the fire? 3. Making a Kite For the kite shown in FIGURE 4.3., use the Law of Cosines to find the lengths of the two dowels required for the diagonal supports. 0 cm FOR 0 cmsale OR 4. Bermuda Triangle The Bermuda Triangle, also known as the Devil s Triangle, is a triangular patch of the Atlantic Ocean where it is claimed b some to be a place of paranormal activities. This claim is bolstered b the fact that, over the ears, man planes and ships have disappeared without a trace within this region. The Jones three vertices & Bartlett of the Learning, triangle are generall taken to be FIGURE FOR 4.3. SALE Kite OR in Problem 3 at Miami (Florida), San Juan FOR (Puerto SALE Rico), OR and the islands of Bermuda. 308 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

104 Jones & Bartlett Learning.. Jones & Bartlett Learning, Bermuda Atlantic ocean Miami Bermuda triangle San Juan FIGURE 4.3. Jones Bermuda & Triangle Bartlett in Learning, Problem 4 Third base Second base Pitching rubber See FIGURE The distance from Miami to Bermuda is 035 mi, the distance from Bermuda to San Juan is 96 mi, and the distance from San Juan to Miami is 033 Jones mi. & Bartlett Learning, (a) Find the three acute angles in the triangle. (b) Determine the approimate area of the triangle. 5. Plaing Hardball (a) A professional baseball diamond is a square 90 feet on a side with a base at each verte. The rubber (or plate) on the pitcher s mound is 60.5 ft from home base on the line between home base and second base. See FIGURE What is the distance s between Jones the & pitching Bartlett rubber Learning, and LL 90 Jones ft & Bartlett Learning, first base? First base (b) In softball (where the ball is not soft), the dimensions of the diamond var according to age and gender of the plaers and whether it is fast pitch or slow pitch. What is the distance s for fast-pitch women s college softball where the diamond is 60 ft on a side and the distance from home base to the 60.5 ft 90 ft pitching rubber is 43 ft? Home 6. On the Clock Suppose the lengths of the minute and hour hands of an analog base clock are 6 inches and 4.5 inches, respectivel. Find the distance d between the tips of the minute and hour hands at 4 PM. FIGURE Baseball diamond in Problem 5 For Discussion 7. Heron s Formula FOR Use SALE the Law OR of Cosines to derive the formula A 5!s(s a)(s b)(s c) for the area of a triangle with sides a, b, c where s 5 (a b c). This Jones & Bartlett formula Learning, is named after the Greek mathematician Jones and inventor & Bartlett Heron Learning, of LL 59.3 ft FOR SALE OR Aleandria (c. 0 6 C.E.) but should actuall be credited FOR SALE to Archimedes. OR DISTRIBUT 8. Garden Plot Use Heron s formula in Problem 7 to find the area of a triangular garden plot if the lengths of the three sides are 5, 3, and 4 m, respectivel. 9. Corner Lot Find the area of the irregular corner lot shown in FIGURE ft [Hint: Divide the lot into two triangular Jones & lots Bartlett as shown Learning, and then find the area of each triangle. Use Heron s formula in Problem 7 for the area of the acute 88. ftfor SALE OR triangle.] 30. Use Heron s formula in Problem 7 to find the area of a triangle with vertices located at (3, ), (3, 6), and (0, 6) in a rectangular coordinate sstem. 50. ft 3. Blue Man The effort in climbing a flight of stairs depends largel on the fleing Jones & Bartlett FIGURE Learning, Corner lot in angle of the leading Jones knee. & Bartlett A simplified Learning, blue stick-figure model of a person FOR SALE Problem OR 9 walking up a staircase FOR indicates SALE OR that the maimum fleing of the knee occurs 4.3 Law of Cosines 309

105 Jones & Bartlett Learning.. when the back leg is straight and the hips are directl over the heel of the front foot. See FIGURE Show that cos u5a R a b Å 4 at a b (T /a) (R/a), a a where u is the knee joint angle, a is the length of the leg, R is the rise of a single h stair step, and T is the width of a step. [Hint: Let h be the vertical distance from a hip to heel of the leading leg, as shown in the figure. Set up two equations a involving h: one b appling the Pthagorean theorem to the right triangle R outlined in color and the other b using the Law of Cosines on the angle u. T Then eliminate h and solve for cos u. ] Jones & Bartlett Learning, 3. For a triangle with sides of lengths a, b, Jones and c and & gbartlett is the angle Learning, opposite c,we have seen on page 305 that when g is a right FOR angle SALE the Law OR of Cosines reduces to the Pthagorean theorem c 5 a b. How is c related to a b when (a) g is an acute angle (b) g is an obtuse angle? FIGURE Blue man in 33. Intersection of Circles Finale The centers of the red and blue circles in Problem 3 FIGURE are B and C and have coordinates ( 0, 0 ) and (, ), respectivel. Let d denote the distance between the centers. In Problem 95 in Eercises 4. and in Problem 40 in Eercises FOR 4. SALE ou OR were asked to find the area of the intersection of two circles when each circle had the same radius r. Now suppose that the radii of the circles are r 0 and r, r 0 r. (a) Use a triangle in Figure to find a general formula for the area of the Jones & Bartlett intersection Learning, of the circles in terms of r 0, r, u, and f. Jones & Bartlett Learning, LL FOR SALE (b) Find OR area of the intersection of the circles ( 4) ( 5) 5 4 and ( 6) ( 8) 5 9. D B r 0 f/ C d r q/ E FIGURE Intersecting circles in Problem Rhombus A rhombus is a parallelogram in which the four sides have the same length s. See FIGURE Since there is no agreement on whether a square is a A D rhombus, for purposes of this problem we will eclude it. In a rhombus opposite Jones & Bartlett Learning, angles are equal acute angles at vertices Jones B and & D Bartlett and obtuse Learning, angles at A and B. The acute and obtuse angles are supplementar. h s (a) Use geometr to show that the area of a rhombus is the product of the base times the height, that is, A 5 sh. B s C (b) Use trigonometr to show that the area of a rhombus is A 5 s sin u, where u Jones FIGURE & Bartlett Rhombus Learning, is an of the four interior Jones angles. & Bartlett Eplain Learning, wh this area formula is valid for FOR Problem SALE 34 OR an verte angle in the FOR rhombus. SALE OR 30 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

106 Jones & Bartlett Learning.. D (c) Show that Jones the area & of Bartlett a rhombus Learning, is A 5 d d, where d and d are the lengths of the line segments, FOR SALE or diagonals, OR AC and BD. (d) Use geometr to show the diagonals connecting opposite vertices bisect the A 8 verte angles. 4 (e) Use trigonometr to show that the diagonal lines are perpendicular [Hint: See Problem 76 in Eercises 4.7.] B C 34 Jones & Bartlett 35. Quadrilateral Learning, A quadrilateral (also called a quadrangle) is an polgon with FIGURE Quadrilateral FOR in SALE OR four sides. The rhombus discussed in Problem 34 is a quadrilateral. Discuss how Problem 35 to find the area of the quadrilateral given in FIGURE Carr out our ideas. 4.4 FOR The SALE Limit OR Concept Revisited INTRODUCTION As we saw in Section.0, the fundamental motivating problem of differential calculus, find a tangent line to the graph of the function, is answered b the concept of a limit. Jones In that & section Bartlett welearning, purposel kept the discussion about limits at an intuitive level; ourfor emphasis SALE wasor on reviewing the appropriate algebra, such as factoring and rationalization, necessar to be able to compute a limit analticall. In the stud of the calculus of the trigonometric functions ou will, of course, be epected to compute limits involving trigonometric functions. As the eamples in this section will illustrate, computation of trigonometric limits entail both algebraic manipulations and knowledge of basic trigonometric identities. We begin with a fundamental limit result for the sine FOR function. SALE OR DISTRIBUT An Important Trigonometric Limit To do the calculus of the trigonometric functions, sin, cos, tan, and so on, it is important to realize that the variable is a real number or an angle measured in radians. With that in mind, consider the numerical values of Jones & (sin )/ as approaches 0 from the right ( S Bartlett 0 Learning, ) given in the table that follows. S sin It is eas to see that the same results given in the table hold as S 0. Because sin is an odd function, for. 0 and, 0 we have sin() 5sin and as a consequence 5 sin In other words, when the value of is small in absolute value sin(). sin <. While numerical calculations such as this do not constitute a proof, the do suggest sin that as S 0. Using the limit smbol, we have motivated the following S result FOR sin SALE lim () S0 5. OR Problem 40 in Eercises 4.4 gives a guided tour through the basic steps of a proof of () that is usuall presented in calculus. Jones & Bartlett Learning, Important In this discussion Jones we make & Bartlett the same Learning, assumption that we did in Sections.5 and.0, namel, that all limits FOR under SALE consideration OR actuall eist. Everthing that we do, 4.4 The Limit Concept Revisited 3

107 Jones & Bartlett Learning.. algebraic manipulations, taking limits of products and quotients in the eamples in this section is predicated on this assumption. Other limits of importance FOR are SALE OR lim sin 5 sin a, () Sa lim cos 5 cos a. (3) Sa FOR The SALE results OR () and (3) are immediate consequences of the fact FOR that f() SALE 5 sin OR and DISTRIBUT g() 5 cos are continuous functions for all. As we have seen in Section 4.3 the graphs of sin and cos are smooth and unbroken. For eample, from (), lim sin 5 sin p Sp/6 Jones 6 5 & Bartlett Learning, (4) and sin 5 sin Also, from (3), Jones lim cos & Bartlett 5 cos 0Learning, 5. (5) S0 The results in (), (), and (3) are used often to compute other limits. As in Section.5 man of the limits considered in this section are limits of fractional epressions where both the numerator and the denominator are approaching 0. Recall, these kinds of limits are said to have the indeterminate form 0/0. Note that the limit () is of this indeterminate Jones & Bartlett form. Learning, EXAMPLE Using () 0 3 sin Find lim. S0 Solution We rewrite the fractional epression as two fractions with the same denominator : 0 3 sin lim 5 lim c 0 S0 S0 5 Jones 0 lim & d cancel the in the first epression S0 Bartlett 3 lim sin Learning, S0 FOR SALE OR sin 5 lim 0 3 lim d now use () S0 S # 5 7. FOR EXAMPLE SALE OR Using the Double-Angle Formula sin Find lim S0. Solution To evaluate the given limit, we make use of the double-angle formula sin 5 sin cos of Section 4.5 and the results in () and (5): 3 CHAPTER 4 TRIGONOMETRIC FUNCTIONS 3 sin d from (5) from () T T sin cos sin lim 5 lim 5 lim cos S0 S0 # sin S0 5 # # 5. Jones & Bartlett sin Learning, Thus, FOR limsale 5. (6) S0 OR lim S0

108 Jones & Bartlett Learning.. Using a Substitution Jones & We Bartlett are oftenlearning, interested in limits similar to that considered sin 5 in Eample. But iffor we wish SALE to find, OR sa, lim the procedure emploed in S0 Eample breaks down at a practical level since we have not developed a trigonometric identit for sin 5. There is an alternative procedure that allows us to quickl find sin k Jones & Bartlett lim Learning, where k 0 is an real constant, b simpl changing the variable b means S0, FOR SALE of a substitution. If we let t 5 k, then 5 t/k. Notice that as S OR DISTRIBUT 0 then necessaril t S 0. Thus we can write sin k lim S0 Thus we have proved the general result sin k lim 5 k. S0 sin 5 Hence lim 5 5. FOR See Problem SALE 5 OR in Eercises 4.4. S0 this limit is from () T 5 lim sin Jones t ts0 t/k 5 lim & sin Bartlett t # k ts0 t 5 k Learning, lim sin t 5 k. FOR SALE OR ts0 t (7) EXAMPLE 3 Trigonometric Limit tan FOR SALE Find OR lim S0. Solution Using the definition tan 5 sin /cos we can write sin tan lim S0 5 lim cos Jones & Bartlett 5 lim S0 S0 cos # sin 5 Learning, lim S0 cos # sin. From (5) and () we know that cos S and (sin )/ S as S 0, and so the preceding line becomes EXAMPLE 4 tan Jones & Bartlett lim S0 Learning, 5 # 5. Using a Pthagorean Identit cos Find lim. Jones & Bartlett S0 Learning, FOR SALE Solution OR To compute this limit we start with a bit of algebraic FOR SALE cleverness OR b DISTRIBUT multipling the numerator and denominator b the conjugate factor of the numerator. Net we use the fundamental Pthagorean identit sin cos 5 in the form cos 5 sin : cos cos lim 5 lim # cos S0 S0 cos cos 5 lim S0 ( cos ) Jones & Bartlett Learning, sin 5 lim FOR SALE OR S0 ( cos ). 4.4 The Limit Concept Revisited 33

109 Jones & Bartlett Learning.. For the net step we resort back to algebra to rewrite the fractional epression as a product, then use the results in (), (4), and (5): cos sin lim 5 lim S0 S0 ( cos ) sin 5 lim # sin S0 cos 5 # 0 (8) 5 0. That is, lim S0 cos Jones & Bartlett Learning, 5 0. FOR SALE OR From (8) we obtain a limit result that is used in calculus to find the derivatives of the sine and cosine functions. Since the limit in (8) is equal to 0, we can write lim S0 cos Jones (cos & Bartlett ) Learning, cos 5 lim 5 () lim 5 0. S0 FOR SALE OR S0 Dividing b then gives cos lim 5 0. (9) S0 The Calculus Connection In Section.0 we saw that the derivative of a function 5 f() is the function f r() defined b a limit of a difference quotient: f ( h) f () f r() 5 lim (0) hs0 Jones h & Bartlett. Learning, In computing this limit we shrink h to zero, but is held fied. Recall too, if a number 5 a is in the domains of f and fr, then f (a) is the -coordinate of the point of tangenc (a, f (a)) and f r(a) is the slope of the tangent line at that point. Derivatives of f() Jones sin & Bartlett and f() Learning, cos To find the derivative of f () 5 sin we use the four-step FOR process SALE illustrated OR in Eample 3 of Section.0. In the first step we use from Section 4.5 the sum formula for the sine function: Jones & Bartlett (i) With Learning, and h plaing the parts of and, we have Jones from (): & Bartlett Learning, LL f ( h) 5 sin( h) 5 sin cos h cos sin h. (ii) f ( h) f () 5 sin cos h cos sin h sin 5 sin (cos h ) cos sin h. Jones & Bartlett Learning, As we see in the net line, we cannot Jones cancel & the Bartlett h s in the Learning, difference quotient, but we can rewrite the epression FOR to make SALE use of OR the limit results in () and (9). f ( h) f () sin (cos h ) cos sin h (iii) 5 h h Jones 5 sin cos & Bartlett h Learning, cos sin h h h. 34 CHAPTER 4 TRIGONOMETRIC FUNCTIONS sin( ) 5 sin cos cos sin. ()

110 Jones & Bartlett Learning.. (iv) In this line, the smbol h plas the part of the smbol in () and (9): f (FOR h) SALE f () OR cos h sin h f r() 5 lim 5 sin lim cos lim hs0 h hs0 h hs0 h. From the limit results in () and (9), the last line is the same as f ( h) f () f r() 5 lim 5 sin # 0 cos # 5 cos. hs0 h In summar: The derivative of f () 5 sin Jones is f r() & 5Bartlett cos. Learning, () It is left to ou the student to show that: FOR SALE OR The derivative of f () 5 cos is f r() 5sin. (3) See Problems 3 and 4 in Eercises 4.4. EXAMPLE 5 FOR Equation SALE OR of a Tangent Line Find an equation of the tangent line to the graph of f () 5 sin at 5 4p/3. Solution We start b finding the point of tangenc. From f a 4p 3 b 5 sin 4p 3 5!3 we see that the point of tangenc is A4p/3,!3/B. The slope of the tangent line at that point is the derivative of f () 5 sin evaluated at the -coordinate. From () we Jones & Bartlett Learning, know that f r() 5 cos and so the Jones slope at A4p/3, & Bartlett!3/B Learning, is = sin FOR SALE OR fra 4p 3 b 5 cos 4p 3 5. Point of From the point-slope form of a line, an equation of the tangent line is Jones & Bartlett tangenc Learning, 4 (, 3 3 Slope is!3 Jones & Bartlett 5 a 4p Learning, 3 b or 5 p 3!3 f. ( 4 = 3 FIGURE 4.4. Tangent line in Eample 5 ( ( See FIGURE Eercises 4.4 Jones Answers & Bartlett to selected Learning, odd-numbered problems begin on page ANS 8. In Problems 8, use the results in (), (), (3), (7), and (9) to find the indicated limit. sin sin p lim S0. lim. S0 sin(u) Jones & Bartlett sin 3t Learning, 3. lim 4. lim us0 5. lim cos u FOR SALE ts0 6. lim sin 4t OR S5p/6 Sp/4 7. lim (cos 5 sin ) 8. lim cos sin Sp/ Sp/6 cos 8( cos u) 9. lim Jones & Bartlett Learning, 0. lim S0 0 us0 u 4.4 The Limit Concept Revisited 35

111 Jones & Bartlett Learning.. 4 sin sin 4 cos. lim Jones & Bartlett. limlearning, S0 S0 sin sin 3. lim 4. lim S0 S0 cos cos tan 5. lim 6. lim Sp/ cot S0 cos Jones & Bartlett Learning, LL FOR 7. SALE lim cot 8. lim S0 OR Sp/4 cos sin FOR SALE OR DISTRIBUT In Problems 9, proceed as in Eample 5 to find an equation of the tangent line to the graph of f () 5 sin at the indicated value of. Jones & Bartlett Learning, Jones 5p/ & Bartlett Learning, FOR SALE OR. 5p/6. FOR 5 p/3 SALE OR 3. Proceed as on pages and find the derivative of f () 5 cos. 4. Use the result of Problem 3 to find an equation of the tangent line to the graph of f () 5 cos at 5p/3. 5. Use the facts that cos 5 FOR SALE OR sin 5 lim 5 0 and lim 5 5 S0 S0 to find the derivative of f () 5 sin Use the result of Problem 5 to find an equation of the tangent line to the graph Jones & of Bartlett f () 5 sin Learning, 5 at 5p. Calculator/Computer Problems In Problems 7 and 8, use a calculator or computer to estimate the given limit b completing each table. Round the entries in each table to eight decimal places. cos FOR SALE OR 7. lim S0 S cos Eplain wh we do not have to consider S lim Jones & S Bartlett sin( Learning, ) S sin S FOR.9999 SALE OR sin 36 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

112 Jones & Bartlett Learning.. For Discussion In Problems 9 36, discuss FOR how SALE to use OR the result in () along with some clever algebra, trigonometr, or a substitution to find the given limit. sin 4 9. lim 30. lim S0 sin 3 S0 sin 5 Jones & Bartlett Learning, sin Jones sin & Bartlett Learning, LL FOR SALE 3. OR lim 3. lim S0 Sp p FOR SALE OR DISTRIBUT cosa 33. lim 34. lim pb S0 cos S0 sin( ) Jones & Bartlett Learning, 35. lim sin Jones 36. & Bartlett lim S0 S " Learning, Using what ou have learned in Problems FOR SALE 9 and OR 36, find the limit 4 lim S sin( ) without the aid of FOR the numerical SALE table OR in Problem (a) Use a calculator to complete the following table. S cos 4 cos (b) Find the limit lim using the method given in Eample 4. S0 4 Jones & Bartlett Learning, (c) Discuss an differences that ou FOR observe SALE between OR parts (a) and (b). 39. (a) A regular n-gon is an n-sided polgon inscribed in a circle; the polgon is formed b n equall spaced points on the circle. Suppose the polgon shown in FIGURE 4.4. represents a regular n-gon inscribed in a circle of r /n radius r. Use trigonometr to show that the area A(n) of the n-gon is given b A(n) 5 n r sina p n b. FIGURE 4.4. Inscribed Jones n-gon in & Bartlett Learning, See Problems in Eercises 4.. Problem 39 FOR SALE OR (b) It stands to reason that the area A(n) approaches FOR the area SALE of the OR circle DISTRIBUT as the number of sides of the n-gon increases. Compute A 00 and A 000. (c) Let 5 p/n in A(n) and note that as n S ` then S 0. Use () of this section to show that ns`a(n) lim 5pr. Jones & Bartlett Learning, 40. Consider a circle centered atjones the origin & Bartlett O with radius Learning,. As shown in FIGURE 4.4.3(a), let the shaded region FOR OPR SALE be aor sector of the circle with central angle t such that 0, t,p/. We see from Figures 4.4.3(a) 4.4.3(d) that area of ^OPR, area of sector OPR, area of ^OQR. () 4.4 The Limit Concept Revisited 37

113 Jones & Bartlett Learning.. t O R Jones & Bartlett Learning, P Q FOR SALE OR t t t 38 CHAPTER 4 TRIGONOMETRIC FUNCTIONS (a) Unit circle (b) Triangle OPR (c) Sector OPR (d) Right triangle OQR FIGURE Unit circle in Problem 40 (a) Show that the area of ^OPR is sin t FOR SALE OR and that the area of ^OQR is tan t. (b) Since the area of a sector of a circle is r u, where r is its radius and u is measured in radians, it follows that the area of sector OPR is t. Use this result, along with the areas in part (a), to show that the inequalit in () ields FOR SALE cos t, sin OR t,. t Chapter 4 Review Eercises (c) Discuss how the preceding inequalit proves () when we let t S 0. Answers to selected odd-numbered problems begin on page ANS-9. A. Fill in the Blanks Jones & Bartlett Learning, In Problems 0, fill in the blanks.. p/5 radians 5 degrees.. 0 degrees 5 radians. 3. The eact values of the coordinates of the point P(t) on the unit circle corresponding to t 5 5p/6 are. 4. The reference angle for Jones 4p/3 radians & Bartlett is Learning, radians. p 5. tan p 6. In standard position, the terminal side of the angle radians lies in the 5 quadrant. FOR 7. SALE If sin u5 OR 3 and u is in quadrant IV, then sec u5. 8. If tan t 5 and t is in quadrant III, then cos t The -intercept for the graph of the function 5 sec( p) is. 0. The values of t in the interval 30, p4 that satisf sin t 5 are. Jones & Bartlett Learning,. If sin u 5 3 5, 0, u,p/, and cos v 5 Jones /"5, & 3p/ Bartlett, v, Learning, p, then cos(u v) 5.. If cos t 53, p,t, 3p /, then cos t A sine function with period 4 and amplitude 6 is given b. 4. The first vertical asmptote for the graph of 5 tana p to the right of the 4 b -ais is.

114 Jones & Bartlett Learning.. 5. sin t cos t 5Jones & sinat Bartlett p 4 b. Learning, 6. If sin t 5 then cosat p 6,. b 5 p 7. The amplitude of 50 cos is cosa p. 6 5p 4 b 5 9p 9. The eact value of arccosacos. 5 b 5 Jones p & Bartlett Learning, 0. The period of the function 5 sin FOR is. 3 t SALE OR B. True/False In Problems 0, answer true or false.. If tan t 5 3 4, then sin t 5 3 and cos t In a right triangle, If sin u5 then cot u5 60 6,. 3p 3. sec(p) 5 csc FOR SALE 4. OR There is no angle t such that sec t sin(p t) 5sin t 6. sec u5tan u 7. (5, 0) is an -intercept of the graph of 5 3 sin p. Jones & Bartlett Learning, 8. Ap/3, /!3B is a point on the Jones graph & of Bartlett 5 cot. Learning, 9. The range of the function 5 csc FOR is (`, SALE ] OR < [, `). 0. The graph of 5 csc does not intersect the -ais.. The line 5p/is a vertical asmptote for the graph of 5 tan.. If tan( p) 5 0.3, then tan For the sine function Jones & 5 Bartlett sin we Learning, have # #. 4. sin 6 5 sin 3 cos 3 5. The graph of 5 sina p is the graph of 5 sin shifted p/3 units to 3 b the right. Jones & Bartlett 6. Since Learning, tan(5p/4) 5, then arctan() 5 5p/4. Jones & Bartlett Learning, LL FOR SALE 7. OR arcsina B f () 5 arcsin is not periodic. 9. f () 5 sin is p periodic. 0. f () 5 sin(cos ) is an even function. C. Review Eercises In Problems 6, find all t in the interval 30, p4 that satisf the given equation.. cos t sin t cos t sin t 5 0. cos t sin t sin t 5Jones 0 & Bartlett Learning, 4. sin t 5 tan t 5. sin t cos t 5 FOR SALE OR 6. tan t 3 cot t 5 CHAPTER 4 Review Eercises 39

115 Jones & Bartlett Learning.. In Problems 7 0, solve the triangle satisfing the given conditions. 7. a530, b570, b 5 0FOR SALE 8. OR g545, a 5 5, c a55, b 5 0, c a 5 4, b 5 6, c CHAPTER 4 TRIGONOMETRIC FUNCTIONS In Problems 8, find the indicated value without using a calculator. Jones. & cos Bartlett A B Learning,. arcsin() FOR 3. SALE cotacos OR 3 4B 4. cosaarcsin 5B FOR SALE OR DISTRIBUT 5. sin (sin p) 6. cos(arccos 0.4) 7. sinaarccosa3bb 5 8. arctan(cos p) In Problems 9 and 0, write the given epression as an algebraic epression in. FOR SALE OR 9. sin(arccos) 0. sectan FOR SALE OR In Problems 4, the given graph can be interpreted as a rigid/nonrigid transformation of the graph of 5 sin and of the graph of 5 cos. Find an equation of the graph using the sine function. Then find an equation of the same graph using the cosine function... FIGURE 4.R. FOR SALE OR Graph for FOR SALE OR DISTRIBUT Problem FIGURE 4.R. Graph for Problem Jones & Bartlett Learning, FOR SALE OR FIGURE 4.R.4 Graph for Problem 4 FIGURE 4.R.3 Graph for Problem 3 h 5. A surveor 00 m from the base of an overhanging cliff measures a angle of elevation from that point to the top of the cliff. See FIGURE 4.R.5. If the cliff makes an angle of 65 with the horizontal ground, determine 00 m Jones & its Bartlett height h. Learning, FIGURE 4.R.5 Cliff in Problem 5 FOR 6. SALE Arocketislaunchedfromgroundlevelatanangleofelevationof OR FOR 43. SALE Iftherocket OR DISTRIBUT hits a drone target plane fling at 0,000 ft, find the horizontal distance between the R rocket launch site and the point directl beneath the plane.what is the straight-line Ramp S distance between the rocket launch site and the target plane? 7. A competition water-skier leaves a ramp at point R and lands at S. See FIGURE 4.R.6. A judge at point J measures an /RJS as 47. If the distance from the ramp to the 47 judge is 0 ft, find the length of the jump. Assume that /SRJ is The angle between two sides of a parallelogram is 40. If the lengths of the sides J are 5 and 0 cm, find the lengths of the two diagonals. 9. A weather satellite orbiting the equator of the Earth at a height of H 5 36,000 km spots Jones a thunderstorm & Bartlett to the Learning, north at P at an angle of FIGURE FOR 4.R.6 SALE Water-skier OR in Problem 7 u56.5 from its vertical. FOR See FIGURE SALE 4.R.7. OR

116 Jones & Bartlett Learning.. R P R FIGURE 4.R.7 Satellite in Problem 9 (a) Given that the Earth s radius is approimatel R km, find the latitude f of the thunderstorm. (b) Show that angles u and f are related b Jones R & sin Bartlett f tan u5 H R( cos f). Learning, 30. It can be shown that a basketball of diameter d approaching the basket from an angle u to the horizontal will pass through a hoop of diameter D if D sin u.d, where 0 #u#90. See FIGURE 4.R.8. If the basketball has diameter 4.6 cm Jones & FIGURE Bartlett 4.R.8Learning, Basketball in and the hoop has Jones diameter & Bartlett 45 cm, what Learning, range of approach angles u will result in FOR SALE Problem OR 30 a basket? 3. Each of the 4 NAVSTAR Global Positioning Sstem (GPS) satellites orbits GPS Satellite the Earth at an altitude of h 5 0,00 km. Using this network of satellites, an h inepensive hand-held GPS receiver can determine its position on the surface GPS Receiver s GPS Receiver Jones & Bartlett of the Learning, Earth to within 0 m. Find the greatest distance Jones s &(in Bartlett km) on the Learning, surface LL of the Earth that can be observed from a single GPS satellite. See FIGURE 4.R.9. Take the radius of the Earth to be 6370 km. [Hint: Find the central angle u subtended b s.] 3. An airplane fling horizontall at a speed of 400 miles per hour is climbing at an angle of 6 from the horizontal. When it passes directl over a car traveling 60 miles per hour, it is miles Jones above the & car. Bartlett Assuming Learning, that the airplane and the FIGURE 4.R.9FOR GPS satellite SALE in OR car remain in the same vertical plane, FOR find SALE the angle OR of elevation from the car to Problem 3 the airplane after 30 minutes. 33. A house measures 45 ft from front to back. The roof measures 3 ft from the front of the house to the peak and 8 ft from the peak to the back of the house. See FIGURE 4.R.0. Find the angles of elevation of the front and back parts of the roof. 34. A regular five-sided polgon is called a regular pentagon. See Figure Using radian measure, determine the sum of the verte angles in a regular pentagon. 35. In FIGURE 4.R. the blue, green, and red circles are of radii 3, 4, and 6, respectivel. Learning, The dots represent the centers of the circles. Jones Let d& denote Bartlett the distance Learning, LL Jones & Bartlett FOR SALE OR between the centers of the blue and red circles. Determine FOR SALE the angle OR u shown DISTRIBUT in the figure that corresponds to d 5 4. H 3 ft 8 ft θ d 45 ft FIGURE 4.R.0 House FOR in Problem SALE 33 OR FIGURE 4.R. Circles in Problem 35 CHAPTER 4 Review Eercises 3

117 _CH04_nd.qd 0/3/4 8:4 PM Page 3 Jones & Bartlett Learning Navigator s Error An airplane is supposed to fl 500 mi due west to a refueling & Bartlett Learning, rendezvous point. If ajones 5 error is made in the heading, how far is the plane from the rendezvous point afterfor flingsale 400 mi?or Through what angle must the airplane turn in order to correct its course at that point? 37. National Historic Landmark Completed in 90 on a triangular cit block, the Flatiron Building in New York Cit was declared a National Historic Jones & Landmark Bartlett in Learning, 4.R.. The original stor Jones Bartlett Learning, stone&clad steel-frame 989. See FIGURE building is considered to be one of the first skscrapers built in the cit. The sides of the building measure 73 ft along Fifth Avenue, 87 ft along East nd Street, and 90 ft along Broadwa. (a) Show that the base of the building is approimatel a right triangle. (b) Assuming that the base of the building is a right triangle, find the two acute angles in it. Songquan Deng/ShutterStock, Inc. New York, NY a adw Flatiron Building in Bro Fifth avenue Jones East nd street & Bartlett Learning, FOR FIGURE 4.R. Triangular SALE OR block in Problem Volcanic Cones Viewed from the side, a volcanic cinder cone usuall looks like. Studies&ofBartlett cinder cones that are less than an isosceles Jones & Bartlett Learning, trapezoid. See FIGURE 4.R.3 Jones Learning, W 50,000 ears old indicate that cone height and crater width are related to H FOR SALE OR co cr the cone width Wco b the equations Hco Wco and Wcr Wco. If Wco 5.00, use these equations to determine the base angle f of the trapezoid in Figure 4.R.3. Wcr Greg Vaughn/Alam Images. φ FOR SALE Volcanic OR cinder cones in Joe Ferrer/ShutterStock, Inc. Haleakala Crater, Maui, Hawaii Hco φ Jones & Bartlett Learning, Wco FIGURE 4.R.3 Volcanic cinder cone in Problem Angels Flight Claimed to be the world s shortest railwa, Angels Flight is a Jones & Bartlett Learning, Jones & Bartlett Learning, funicular railwa consisting of two cars (named Olivet and Sinai) that transports people up and down the steep hill between Hill Street and California Plaza in downtown Los Angeles, CA. The original railwa dates back to 90 and, in its present form, is onl 98 ft long. If the angle of elevation of the tracks at its base on Hill Street is 33, then how high is the hill? Jones Bartlett Learning, 40. Distance Across a Canon JonesFrom & Bartlett Learning, the floor of a canon it takes 6 ft of rope to Angels&Flight in downtown Los Angeles ft reach the top of one canon wall and 86 to reach the top of the opposite wall. 3 CHAPTER 4 TRIGONOMETRIC FUNCTIONS

118 _CH04_nd.qd 0/6/4 9:49 PM Page 33 Jones & Bartlett Learning.. See FIGURE 4.R.4. If the two ropes make an angle of 3, what is the distance d from the top of one canon wall to the other? 4. How Fast? An observer at a horizontal distance of mile (580 ft) watches an 86 ft F-5E Strike Eagle fighter jet go into a vertical climb. See FIGURE 4.R.5. If the 6 ft 3 angle of elevation of the jet at the observer changes from the initial measurement of to 75. in 30 seconds, then how fast (in feet per minute) is its rate of Jones & Bartlett Learning, climb? d FIGURE 4.R.4 FOR Canon in Problem 40 Courtes of Master Sgt. Lance Cheung, U.S. Air Force SALE OR Observer US Air Force F-5E Strike Eagle ft FIGURE 4.R.5 Climbing F-5E in Problem Jones4& Bartlett Learning, 4. Building Height Two buildings were constructed on a inclined lot as shown in FIGURE 4.R.6. The angle of elevation from the right side of the roof of the brown building to the left side of the roof of the gra building is 3. From the same building, Jones the & Bartlett Learning, spot on the roof of the brown angle of depression to the base of the gra building is 48. Use the additional the figure to determine the FORinformation SALE ORin heights of the facing sides of the buildings relative to the inclined lot ft Jupiterimages/Stockbte/Thinkstock FIGURE 4.R.6 Buildings in Problem Estimating Tree Height There are man sophisticated instruments, such as Jones & Bartlett One wa oflearning, finding the height of a tree: climb it FOR SALE OR range finders and inclinometers (or clinometer), that are invaluable in determining an accurate measurement of the height of an object. A nontechnical method for determining the height of, sa, a tree is to climb the tree and then drop the weighted end of a measuring-tape line to the ground. Assuming that ou have no Jones Bartlett measuring devices and& that climbinglearning, a tree is not practical (or even legal), eplain wh the following method gives an approimation to the height of a tree: FOR SALE OR CHAPTER 4 Review Eercises 33

119 Jones & Bartlett Learning.. Find a patch of level ground containing the tree. Guess the distance from the ground to our eelevel and the length of our walking stride. Find two straight sticks, back awa from the FOR tree (counting SALE OR our strides) holding the sticks at ee level like this / (one parallel to the ground and the other adjusted b following the top of the tree). Stop when ou think the angle between the sticks is 45. The height of the tree is approimatel: (number of strides) 3 (stride length) ee level height. 44. Gate of Europe The Puerta de Europa towers are twin office buildings in Madrid, Spain. To accommodate a required setback from the wide Paseo de la Castellana, the towers were built in 996 at an angle of 5 from the vertical. In the photo, note the vertical line on the side of each building. The sides of the buildings shown in FIGURE 4.R.7 are congruent parallelograms. Use the information in the figure to find the lengths s and s FOR of the SALE sides of OR the parallelogram and the distance d between the roofs of the towers. Gln Thomas Photograph/Alam Images Puerta de Europa towers and the Paseo de la Castellana in Madrid, Spain 5 d s 374 ft s 63 FOR ft SALE OR DISTRIBUT FIGURE 4.R.7 Towers in Problem Stairs in Homes In home construction, stairs are usuall constructed using two Jones & Bartlett Learning, stringers, which are boards that have been Jones notched & Bartlett to accommodate Learning, the treads (steps) and the risers. Suppose the stairs consists FOR of SALE 9 risers, OR the riser height is 7.75 inches, and the tread depth is 0 inches. See FIGURE 4.R.8. Use two different methods to find the approimate length L of a stringer. Stairs showing a stringer FIGURE Jones 4.R.8& Stringer Bartlett in Problem Learning, El Castillo The most prominent feature in the archaeological site of Chichén Itzá is a step pramid on whose flat top rests a temple to the Maan feathered-serpent god Kukulkán. The Kukulkán pramid, more commonl known b the Spanish name El Castillo, was built around 900 C.E. and is located in the Meican state of Yucátan. On each of the Jones four faces & Bartlett of the pramid Learning, there is a protruding stone-block stairwa rising to the 79 ft high FOR temple SALE level, OR although onl two of the staircases 34 CHAPTER 4 TRIGONOMETRIC FUNCTIONS Robert Ranson/ShutterStock, Inc. Stringer length Tread depth Jones Riser & Bartlett Learning, LL FOR heightsale OR DISTRIBUT

120 Jones & Bartlett Learning.. have been completel restored. The angle of inclination a stairwa relative to the ground is 45 whereas the angle of inclination of a face is Prior to 006, tourists were allowed FOR to climb SALE one OR of the steep stairwas aided b a rope positioned in the middle of the stairwa and anchored at the base of the pramid and at the temple level. Find the approimate length L of the rope. See FIGURE 4.R.9. Tourists are no longer allowed to climb El Castillo agustavop/istock/thinkstock Jones Temple & Bartlett Rope Learning, LL anchor FOR SALE OR Side DISTRIBUT of stairwa 79 ft Rope Rope 53.3 anchor Jones & Bartlett Learning, 45 FOR FIGURE SALE 4.R.9 OR Pramid in Problem 46 In Problems 47 60, translate the words into an appropriate function. 47. A 0-ft-long water Jones trough & Bartlett has ends Learning, the form of isosceles triangles with sides that are 4 ft long. See Figure.9.7 in Eercises.9. As shown in FIGURE θ 4.R.0, let u denote the angle between the vertical and one of the sides of a triangular 4 end. Epress the volume of the trough as a function of u. 48. A person driving a car approaches a freewa sign as shown in FIGURE 4.R.. Let u be her viewing angle of the sign and let represent her horizontal distance FIGURE 4.R.0 End Jones & Bartlett (measured Learning, feet) to that sign. Epress u as a function Jones of &. Bartlett Learning, LL of water trough in Problem 47 FOR SALE 49. OR As shown in FIGURE 4.R., a plank is supported b a FOR sawhorse SALE so that OR one DISTRIBUT end rests on the ground and the other end rests against a building. Epress the length of the plank as a function of the indicated angle u. Jones 4 ft& Bartlett Learning, HOLLYWOOD 8 ft FIGURE 4.R. Freewa sign in Problem 48 FIGURE 4.R. Plank in Problem 49 θ 3 ft 4 ft 50. A farmer wishes to enclose a pasture in the form of a right triangle using 000 ft Jones & Bartlett of fencing Learning, on hand. See FIGURE 4.R.3. Show that Jones the area & of Bartlett the pasture Learning, as a LL FOR SALE OR function of the indicated angle u is Au 5 cot u # 000 a cot ucsc u b. Jones & Bartlett θ Learning, FOR FIGURE SALE 4.R.3 OR Pasture in Problem 50 CHAPTER 4 Review Eercises 35

121 Jones & Bartlett Learning.. 5. Epress the volume of the bo shown in FIGURE 4.R.4 as a function of the indicated angle u. 5 ft 5. A corner of an 8.5 in. 3 FOR in. piece SALE of paper OR θ is folded over to the other edge of the paper as shown in FIGURE 4.R.5. Epress the length L of the crease as a 6 ft ft function of the angle u shown in the figure. 53. A gutter is to be made from a sheet of metal 30 cm wide b turning up the edges FIGURE 4.R.4 Bo in Problem 5 Jones & of Bartlett width 0 Learning, cm along each side so that the sides make equal Jones angles & Bartlett f with thelearning, LL FOR SALE vertical. OR See FIGURE 4.R.6. Epress the cross-sectional area of FOR the gutter SALE as OR a DISTRIBUT function of the angle f. 8.5 in. θ in. FIGURE 4.R.5 Folded paper in Problem 5 L 0 cm 0 cm 0 cm FIGURE 4.R.6 Gutter in Problem A metal pipe is to be carried horizontall around a right-angled corner from a 6 ft hallwa 8 feet wide into a hallwa that is 6 feet wide. See FIGURE 4.R.7. Epress the length L of the pipe as a function of the angle u shown in the figure. θ 55. Circle of Latitude A circle of latitude is circle that connects all locations on the Earth that have the same latitude f. A circle of latitude is also referred to as a Jones & Bartlett Learning, line of latitude, or a parallel, because if Jones the Earth & is Bartlett represented Learning, as a circle in a 8 ft two-dimensional coordinate sstem, then the FOR circles SALE of latitude OR appear as (parallel) horizontal lines. See FIGURE 4.R.8. FIGURE 4.R.7 Pipe in Problem 54 (a) If the radius R of the Earth is taken to be 3959 miles, epress the radius r of a circle of latitude as a function of f. (b) Find the radius of the Artic Circle if its latitude is 66 33r44sN. (c) A line of longitude, Jones or meridian, & Bartlett is one half Learning, of a great circle whose center is the center of the Earth. FOR Longitudes SALE are OR measured east/west from the prime meridian that runs through the Roal Observator at Greenwich, England. See Figure 4..4 in Eercises 4.. The longitudes of Boston, Massachusetts and Detroit, Michigan are, respectivel, 7 3r 37sW and 83 r44sw but the Jones & Bartlett latitude Learning, of both cities is approimatel the same. What Jones is the & distance Bartlett Learning, LL FOR SALE between OR Boston and Detroit measured on the circle of latitude FOR SALE at 4 0rN? OR DISTRIBUT (d) Measured on a meridian, what is the distance between two points that differ b one degree of latitude? Circle of latitude Line of longitude Line of latitude (a) (b) FIGURE 4.R.8 Planet FOR Earth SALE in Problem OR CHAPTER 4 TRIGONOMETRIC FUNCTIONS

122 Jones & Bartlett Learning More Latitude (a) Use Problem 55 to show that the circumference of a circle of latitude as a function of the latitude angle f is given b C f 5 C e cos f, where C is the circumference FOR of SALE the Earth OR at e the equator. Find C e. (b) Use part (a) to find the circumference of the Arctic Circle. (c) Use part (a) to find the distance around the world at the latitude 5 45r N. 57. High Dive A diver jumps from a high platform with an initial downward Jones & Bartlett velocit Learning, of ft/s toward the center of a large circular Jones tank & Bartlett of water. See Learning, LL 00 ft θ FOR SALE OR FIGURE 4.R.9. From phsics, the height of the diver above FOR ground SALE level OR is DISTRIBUT given s b s(t) 56t t 00, where s is measured in feet and t $ 0 is time in seconds. See (5) in Section.4. (a) Epress the angle u shown in the figure as a function of s. Jones & Bartlett (b) For the function in part (a), what value does u approach as s S 5? 5 ftlearning, 58. Circular Segment A circular segment is the region formed between a chord of a circle and its associated arc AB X AB Ground FOR SALE OR 5 ft. This is the light red region in FIGURE 4.R.30. For a circle of fied radius r, epress the area of a circular segment as a function of the central angle u, where 0,u,p. 59. Equilateral Arch An equilateral arch is obtained b constructing circular arcs on two sides of Jones an equilateral & Bartlett triangle. Learning, Let ABC be an equilateral triangle with sides of length s as FOR shown SALE in red OR in FIGURE 4.R.3. A circular arc CB X is drawn from verte C to verte B using a circle of radius s centered at A. In like manner, CA X is C an arc of the circle of radius s centered at B. Equilateral arches were used etensivel in Gothic architecture in the design of church windows and doorwas. Use r θ the concept of a circular segment discussed in Problem 58 to epress the area of A Jones B & Bartlett Learning, FOR SALE OR an equilateral arch as a function of the length s. FOR SALE OR DISTRIBUT FIGURE 4.R.9 Diver in Problem 57 FIGURE 4.R.30 Circular segment in Problem 58 C Circular arc Aristodis/ShutterStock, Inc. The equilateral arch is found throughout the Doumo di Milano, the largest Gothic cathedral in the world. Circular segment s A B s FIGURE 4.R.3 Equilateral arch in Problem Epress the perimeter of the equilateral arch in Problem 59 as a function of s. s CHAPTER 4 Review Eercises 37