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CHAPTER 2 2.1 RADIAN AND DEGREE MEASURE 2.2 T HE TRIGONOMETRIC FUNCTIONS AND THE UNIT CIRCLE 2.3 TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES 2.4 T RIGONOMETRIC FUNCTIONS OF ANY ANGLE 2.5 G RAPHS OF SINE AND COSINE FUNCTIONS 2.6 OTHER TRIGONOMETRIC GRAPHS 2.7 INVERSE TRIGONOMETRIC FUNCTIONS 2.8 TRIGONOMETRY ApPLICATIONS OF TRIGONOMETRY 2.1 Angles RADIAN AND DEGREE MEASURE I Radian Measure / Degree Measure I Applications Angles FIGURE 2.1 As derived from the Greek language, the word trigonometry means "measurement of triangles." Initially, trigonometry dealt with relationships among the sides and angles of triangles. As such, it was used in the develop ment of astronomy. navigation, and surveying. With the advent of calculus in the 17th century and a resulting expansion of knowledge in the physical sciences, a different perspective arose--one that viewed the classic trigonometric relationships as functions with the set of real numbers as their domains. Consequently, the applications of trigonometry expanded to include a vast number of physical phenomena involving rotations or vibrations. These include sound waves, light rays, planetary orbits, vibrating strings, pendulums. and orbits of atomic particles. Our approach to trigonome try incorporates both perspectives, starting with angles and their measure. An angle is determined by rotating a ray (half-line) about its endpoint. The starting position of the ray is called the initial side of the angle, and the position after rotation is called the terminal side, as shown in Figure 2.1. The endpoint of the ray is called the vertex of the angle. 105 106 CHAPTER 2 TRIGONOMETRY In a coordinate system, an angle is in standard position if its vertex is the origin and its initial side coincides with the positive x-axis, as shown in Figure 2.2. Positive angles are generated by counterclockwise rotation, and negative angles by clockwise rotation, as shown in Figures 2.3 and 2.4. To label angles in trigonometry, we use the Greek letters a (alpha), f3 (beta), and (J (theta), as well as uppercase letters A, B, and C. In Figure 2.4, note that the angles a and f3 have the same initial and terminal sides. Such angles are coterminal. y y )' Positive angle (counterclockwise) ----.:....-4------..;~:c --....lIo:-~--:c Initial side --,.-".I----:C Negative angle (clockwise) Coterrninal Angles FIGURE 2.2 FIGURE 2.3 FIGURE 2.4 Radian Measure The measure of an angle is determined by the amount of rotation from the initial to the terminal side. One way to measure angles is in radians. This type of measure is needed in calculus. To define a radian we use a central angle of a circle, one whose vertex is the center of the circle, as shown in Figure 2.5. Central Angle FIGURE 2.5 e DEFINITION OF A RADIAN One radian is the measure of a central angle (J that subtends (intercepts) an arc s equal in length to the radius r of the circle (see figure). arc length '" radius when e 1 radian SECTION 2.1 2 radians 3 radians -f...J===:::::::..~:---+L- RADIAN AND DEGREE MEASURE 107 Because the circumference of a circle is 21Tr, it follows that a central angle of one full revolution (counterclockwise) corresponds to an arc length of s = 21Tr. Moreover, because each radian intercepts an arc of length r, we conclude that one full revolution corresponds to an angle of 21Tr r 6 radians 21T radians. Note that since 21T = 6.28, there are a little more than six radius lengths in a full circle, as shown in Figure 2.6. In general, the radian measure of a central angle () is obtained by dividing the arc length s by r. That is, 5 radians FIGURE 2.6 s - = (), Radian measure r where () is measured in radians. Because the units of measure for sand r are the same, this ratio is unitless--it is simply a real number. The radian measures of several common angles are shown in Figure 2.7. e 12I (21T) rr e 2 t Quadrant II fJ I 4 (21T) 6 1T =2 I 8 (21T) I e = :2 (21T) 1T 4 = 1T I 1T 6 (21T) = 3 fJ = 21T Quadrant I Radian Measure for Several Common Angles 1T 1T = 1T 0<9<2 2<9<1T FIGURE 2.7 e= rr - - - - - - i - - - - - < - fJ Quadrant III Quadrant IV 31T 1T<9<2 31T 2 < 9 < 21T 3rr 8= 2 FIGURE 2.8 = 0 Recall that the four quadrants in a coordinate system are numbered coun terclockwise as 1, II, III, and IV. Figure 2.8 shows which angles between 0 and 2'1T lie in each of the four quadrants. You can find an angle that is coterminal to a given angle ()by adding or subtracting 2'1T (one revolution), as demonstrated in Example I (Note that a given angle has many coterminal angles. For instance, () 1T/6 is coterrninal with both I 31T/6 and -111T/6.) 108 CHAPTER 2 TRIGONOMETRY Sketching and Finding Coterminal Angles EXAMPLE 1 A. To find an angle that is cotenninal to the positive angle (} = 13'TT/6. you can subtract 2'TT to obtain 13 'TT 6 Thus, the terminal side of (} lies in Quadrant l. Its sketch is shown in Figure 2.9(a). B. To find an angle that is cotenninal to the positive angle (} 3'TT/4, you can subtract 2'TT to obtain 5'TT 4' as shown in Figure 2.9(b). REMARK The phrase "the terminal side of fJ lies in a quadrant" can be abbreviated by simply saying that "0 lies in a quadrant." The terminal sides of the quadrant angles 0, 'TT/2, 'TT, and 3'TT/2 do not lie within any of the four quadrants. -27T/3, you C. To find an angle that is coterminal to the negative angle (} can add 27T to obtain 27T 4'TT (} = - 3 + 27T = 4' as shown in Figure 2.9(c). 1317 e 6 6 1'--'---4:""'-'-'_0 417 e =317 4 17 'TI-L----=:>i--_- 0 3 l ' - - - - J - -_ _ O 517 4 311 (a) FIGURE 31' }1! T (b) 2 (el T ......... ... 2.9 Figure 2.10 shows several common angles with their radian measures. Note that we classify angles between 0 and 7T/2 radians as acute and angles between 7T/2 and 7T as obtuse. ~ ~ Acute angle: between 0 and 17 2 FIGURE 2.10 17 8= 2 Right angle: quarter revolution 8 Obtuse angle: 17 between 2 and 17 Straight angle: half revolution 217 Full revolution SECTION 2.1 109 RADIAN AND DEGREE MEASURE Two positive angles a and f3 are said to be complementary (or comple ments of each other) if their sum is 1T/2, as shown in Figure 2.II(a). For example, 1T/6 and 1T/3 are complementary angles because 1T 1T -+-= 1T 632 (a) Complementary angles Two positive angles are supplementary (or supplements of each other) if their sum is 1T, as shown in Figure 2.] ](b). For example, 21T/3 and 1T/3 are supplementary angles because 21T - 3 + 1T == 1T. 3 EXAMPLE 2 (b) Supplementary angles FIGURE Complementary. Supplementary. and Coterminal Angles A. The complement of (J = 1T/12 is 2.11 1T ]2 1T 2 61T ]2 1T 51T ]2 = 12' as shown in Figure 2.l2(a). B. The supplement of (J = 51T/6 is 51T 1T 1T 6=6' as shown in Figure 2.12(b) C. In radian measure, a coterminal angle is found by adding or subtracting 21T. For 171T/6, you subtract 21Tto obtain (J 171T 6 as shown in Figure 2.12(c). Thus, 171T/6 and 51T/6 are coterminal. 7T 7T 7T 2 - 2 2 57T 12 7T +"':"''''::''1--'--0 7T 12 6 37T (b) (a) FIGURE 2.12 (c) 2 ... " .. " ... ". 110 CHAPTER Z y 90° = .--,---- TRIGONOMETRY Degree Measure ~ (360°) ~(3600) 60° 45° = ~(360o) 30° = 12 (360°) 0° --t-----l<---'----i-''---x 360" 330° 315 300° 0 A second way to measure angles is in terms of degrees. A measure of one degree (10) is equivalent to 1/360 of a complete revolution about the vertex. To measure angles in degrees, it is convenient to mark degrees on the circum ference of a circle as shown in Figure 2.13. Thus, a full revolution (counter clockwise) corresponds to 360°, a half revolution to 180°, and a quarter revolution to 90°. Because 21T radians is the measure of an angle of one complete revolution, degrees and radians are related by equations 3600 21T rad and 180 = 0 1T rad. From the latter equation, we obtain FIGURE 2.13 o I = - 1T 180 rad and I rad which lead to the following conversion rules. CONVERSIONS: DEGREES +-+ RADIANS REMARK Note that when no units of angle measure are specified, radian measure is implied. For instance, if we write e = 1T or e = 2. we mean e 1T radians or e 2 radians. 1T rad I. To convert degrees to radians, multiply degrees by 1800 ' 2. To convert radians to degrees, multiply radians by 1800 1T rad . To apply these two conversion rules, use the basic relationship 180°. EXAMPLE 3 rad = Converting from Degrees to Radians (l35~)C;or~) A. 135° 1T Multiply by 1T/180 31T = -rad 4 B. 540° = (540.deg-)( 1T rad ) 180.Geg Multiply by 1T/180 = 31T rad C. - 270° (-270 .deg-)( 31T --rad 2 1T rad ) 180.eeg Multiply by 1T/180 ....... .. SECTION 2.1 Techno{091J Note _ _ _ _ _ __ Calculators and graphing utilities have both and radian modes, \Vhen you are using a calculator or graphing utility, be sure you use the correct mode, Most graphing utilities allow you to convert directly from degrees to radians and from radians to degrees, Try using a graphing utility to convert 0 135 to radians and TI/2 radians to degrees, EXA M PLE 4 Converting from Radians to Degrees ~rad = (-~~)C~~) A. lIt RADIAN AND DEGREE MEASURE Mulrip!\' !J.,' J 80!" = -90° B. 9; rad = (9; ~)C~O~) = 810° c. 2 rad = (2fttdJC~~) 360 -deg Mulripl,' b,' 180/" 1T ........ .. "" 144.59° With calculators it is convenient to use decimal degrees to denote fractional parts of degrees. Historically, however, fractional parts of degrees were ex pressed in minutes and seconds, using the prime (') and double prime (") notations, respectively, That is, I' = one minute I (1°) 60 I I" = one second = -(I ') 60 Consequently, an angle of 64 degrees, 32 minutes, and 47 seconds is repre sented by 8 64° 32' 47", Many calculators have special keys for converting an angle in degrees, minutes, and seconds (DO M' S") into decimal degree form, and conversely, If your calculator does not have these special keys, you can use the techniques demonstrated in the next example to make the conversions, EXAMPLE 5 Converting an Angle from DO :W S" to Decimal Form Convert 152° 15' 29" to decimal degree form, SOLUTION 152° 15' 29" 1520 + ( -15)0 + (29)° - 60 3600 "" 1520 + 0,25 0 + 0,00806° 152,25806°, & •••••• II • 112 CHAPTER 2 TRIGONOMETRY Applications The radian measure formula, e = sir, can be used to measure arc length along a circle. Specifically, for a circle of radius r, a central angle e subtends an arc of length s given by s = where re, Lel1gTh IIf c;rdi/or <iII e is measured in radians. EXA M PLE 6 Finding Arc Length A circle has a radius of 4 inches. Find the length of the arc cut off (subtended) by a central angle of 240°, as shown in Figure 2.14. SOLUTION re, To use the formula s = 240° we must first convert 240° to radian measure. = (240 deg) ( 7T rad) = -47T rad 180 deg 3 Then, using a radius of r = 4 inches, we find the arc length to be s = re 167T 3 = 16.76 inches. FIGURE 2.14 Note that the units for units. re are determined by the units for r because e has no ........ . The formula for the length of a circular arc can be used to analyze the motion of a particle moving at a constant speed along a circular path. Assume the particle is moving at a constant speed along a circular path (of radius r). If s is the length of the arc traveled in time t, then we say that the speed of the particle is distance Speed = - - time s Moreover, if e is the angle (in radian measure) corresponding to the arc length s, then the angular speed of the particle is e Angular speed = -. t SECTION 2.1 113 RADIAN AND DEGREE MEASURE Finding the Speed of an Object EXAMPLE 7 The second hand on a clock is 4 inches long, as shown in Figure 2.] 5. Find the speed of the tip of this second hand. SOLUTION The time required for the second hand to make one full revolution is 60 seconds I minute. The distance traveled by the tip of the second hand in one revolution is s = 211'(radius) 211'(4) = 811' inches. Therefore, the speed of the tip of the second hand is FIGURE 2.15 s 811' inches . = 0.419 m./sec. Speed = - = r 60 seconds ....... .. Finding Angular Speed and Linear Speed EXAMPLE 8 A lawn roller that is 30 inches in diameter makes one revolution every ~ second, as shown in Figure 2.16. A. Find the angular speed of the roller in radians per second. B. How fast is the roller moving across the lawn? SOLUTION A. Because there are 211' radians in one revolution. it follows that the angular speed is e Angular speed = r 211' rad = 2.411' rad/sec. B. Because the diameter is 30 inches. r = 15 and s = 211'r = 3011' inches. Thus, Speed = s t 3011' in. FIGURE 2.16 = 3611' in./sec = 113.1 in./sec. ....... .. 114 CHAPTER 2 TRIGONOMETRY If you have a programmable calculator, try entering an "angle-drawing program." Then use the program to draw several different angles. The following program is for the TI-81. Programs for other calculators are given in Appendix B. AN ANGLE DRAWING PROGRAM :Disp "ENTER,MODE" :Disp "C,RADIAN" :Disp "I,DEGREE" :Inp'..lt M :Disp "ENTER,ANGLE" :Input T : O--'I'rnin : abs T--Trr,ax : .15-+Tstep : cos T--A : sin '-.;:'--B : Para;-:-, : 1--S :If M=l : fi,!: /2.80-+':' :Rad : If T<C :-1--+5 : " ( .25-. 04T') cos "--Xn : ,. S (.25+. 04T) sin T"-+Y l l :Cl~Draw :AII-Off -1.sl---'---+-+-J."'+--+--'----ll.5 : 1. 5--Xmin : 1. 5--Xmax : l--Xscl :-l--Y;-:-,in : l--Ymax : l--Yscl : DispGraph :=-.ine(C,O,A,B) :Pause : F:mction :End To run this program, enter 0 for radian mode or 1 for degree mode. Then enter any angle (the angle can be negative or larger than 360°). After the angle is dis played, press ENTER to clear the display. The figure shows an angle of -750°, as drawn by this program. WARM-UP The following warm-up exercises involve skills that were covered in earlier ~tions,; You will use these skills in the exercise set for ,this section. In Exercises I-to, solve for x. 3. 51T 1T;: - 6 45 x 180 2. 790 +x 4.2'17" - 1T 5.-=- 7. x 1T 180 x 9. - 60 20 3 4 = 720 + = 180 + 135 1. x 8. 5'17" X = 3" 240 180 180 330 'IT x 6.-= 10. _x_ 3600 X x 1T 0.0125 SECTION 2. I 115 RADIAN AND DEGREE MEASURE SECTION 2.1 . EXERCISES ........................................................................................................................ In Exercises 1-4. determine the quadrant in which the terminal side of the angle lies. (The angle is given in radians.) L (a) 7r. 5 r. 5' (b) - 13. (a) (b) ~_450 II r. 9 2. (a) r. 12 (b) 3. (a) I (b) -2 4. (a) 5.63 In Exercises 13-16. determine two coterminal angles (one pos itive and one negative) for the angle. Give your results in degrees. (b) -2.25 14. (a) / ~o ,~e" -I," (bJe=L In Exercises 5 and 6, detennine the quadrant in which the terminal side of the angle lies. S. (a) 1300 6. (a) .- 260" 15. (a) (b) 28SO e= (b) 740 c /~ (b) -3.4 0 . , In Exercises 7-10. sketch the angle in standard position. 7. 5r. (a) (b) (b) , 5[J . 7r. 1 " \ 4 5r. 2 In Exercises 17-20, find (if possible) the positive angle comple ment and the positive angle supplement of the angle. 9. (a) 30° (b) 150" 10. (a) 405 0 (b) -480° In Exercises 11 and 12, determine two coterminal angles (one positive and one negative) for the angle. Give your results in radians. 11. (a) (b) 12. (a) -420° .(~'.\ (b) 3 8. (a) e= 16. (a) 4 2r. (b) f"" 21' e=- 15 ~. r. 17. (a) 3 18. (a) J 3r. (b) 19. (a) 18° (b) 115 0 20. (a) 79° (b) 150c 4 (b) 2 In Exercises 21-24. express the angle in degree measure. (Do not use a calculator.) 3r. 21. (a) 2 22. (a) 23. (a) (b) 7r. 12 7r. :3 II r. 24. ( a ) 6 (b) 7r. 6 r. 9 II r. (b) 30 34r. 15 (b) 116 CHAPTER 2 TRIGONOMETRY In Exercises 25-28, express the angle in radian measure as a multiple of 71'. (Do not use a calculator.) 25. (a) 26. (al 27. (a) 28. (a) 30° 315 0 -20° - 2700 (b) 1500 (b) 1200 (b) -240° (b) 144° In Exercises 29-32, convert the angle from degrees to radian measure. Express your result to three decimal places. 29. (a) 115° 30. (al - 216.35° 0 31. (al 532 32. (al -0.83° (b) 87.40 (b) -48.27° (b) 0.54° (b) 345 0 In Exercises 43-46, find the radian measure of the central angle of a circle of radius r that intercepts an arc of length s. Radius 43. 15 inches 44. 16 feet 45. 14.5 centimeters 46. 80 kilometers Arc 4 inches 10 feet 25 centimeters 160 kilometers In Exercises 47-50, on the circle of radius r find the length of the arc intercepted by the central angle e. Radius 47. 15 inches 48. 9 feet 49. 6 meters 50. 40 centimeters Central Angle 180° 60° 2 radians 371' 4' radians In Exercises 33-36, convert the angle from radian to degree measure. Express your result to three decimal places. 33. (al 71' 7 1571' 34. ( a ) 8 35. (aJ -4.271' 36. (al -2 571' (b) II (b) 6.571' (b) 4.8 (b) -0.S7 In Exercises 37 and 38, convert the angle measurement to decimal form. 37. (a) 245° 10' 38. (a) -13SO 36" (b) 2° 12' (b) -408° 16' 2S" In Exercises 39-42. convert the angle measurement to DO M' S" form. 39. (aJ 240.6° 40. (aJ - 345.12° 41. (a) 2.S 42. (a) ~-0.35S (bJ 145.8° (b) 0.45 (b) -3.58 (b) 0.7865 Distance Between Cities In Exercises 51-54, find the distance between the two cities. Assume that the earth is a sphere of radius 4000 miles and that the cities are on the same meridian (one city is due north of the other). Latitude --_._ 51. Dallas Omaha 52. San Francisco Seattle 53. Miami Erie 54. Johannesburg, South Africa Jerusalem, Israel 32° 41° 37° 47° 25° 42° 26° 31° 47' 9" N 15' 42" N 46' 39" N 36' 32" N 46' 37" N 7' IS" N 10' S 47' N 55. Difference in Latitudes Assuming that the earth is a sphere of radius 4000 miles, what is the difference in lati tude of two cities, one of which is 325 miles due north of the other? 56. Difference in Latitudes Assuming that the earth is a sphere of radius 4000 miles, what is the difference in lati tude of two cities, one of which is 500 miles due north of the other? SECTION 2.1 57. Instrumentation The pointer on a voltmeter is 2 inches long (see figure). Find the angle through which the pointer rotates when it moves inch on the scale. 1 RADIAN AND DEGREE MEASURE 117 59. Angular Speed A car is moving at the rate of 50 miles per hour. and the diameter of each wheel is 2.5 feel. (a) Find the number of revolutions per minute of the rotating wheels. (b) Find the angular speed of the wheels in radians per minute. 60. Angular Speed A truck is moving at the rate of 50 miles per hour. and the diameter of each wheel is 3 feel. (a) Find the number of revolutions per minute the wheels are rotat ing. (b) Find the angular speed of the wheels in radians per minute. 61. Angular Speed A 2-inch-diameter pulley on an electric • • FtGURE FOR 57 58. Electric Hoist An electric hoist is used to lift a piece of equipment (see figure). The diameter of the drum on the hoist is 8 inches and the equipment must be raised I foot. Find the number of degrees through which the drum must rotate. motor that runs at 1700 revolutions per minute is connected by a belt to a 4-inch-diameter pulley on a saw arbor. (a) Find the angular speed (in radians per minute) of each pulley. (b) Find the rotational speed (in rpm) of the saw. 62. Angular Speed How long will it take a pulley rotating at 12 radians per second to make 100 revolutions? 63. Circular Saw Speed The circular blade on a saw has a diameter of 7.5 inches and the blade rotates at 2400 revolu tions per minute (see figure). (a) Find the angular speed in radians per second. (b) Find the speed of the saw teeth (in feet per second) as they contact the wood being cuI. l-- 7.5 in. FIGURE FOR 63 64. Speed of a Bicycle The radii of the sprocket assemblies and the wheel of the bicycle in the figure are 4 inches. 2 inches. and 13 inches. respectively. If the cyclist is pedal ing at the rate of I revolution per second. find the speed of the bicycle in (a) feet per second and (b) miles per hour. 1 ft I T FIGURE FOR 58 -1 1--2 in. FIGURE FOR 64 118 CHAPTER 2 TRIGONOMETRY ClRCL.E Tdgurwmetric Functions ! Domain and Period of Td~ fU!ictiom with a Calculator The Crill Circle The Unit Circle }' The two historical perspectives of trigonometry incorporate different methods for introducing the trigonometric functions. Our first introduction to these functions is based on the unit circle. Consider the unit circle given by (0, 1) (~1, 0) (1, 0) ---<t----T------t--x (0, Unit Circle: x~ FIGURE 1) as shown in Figure 2.17. Imagine that the real number line is wrapped around this circle, with positive numbers corresponding to a counterclockwise wrap ping and negative numbers corresponding to a clockwise wrapping, as shown in Figure 2.18. + y' 2.17 y (J, 0) -j-----..----+- x -I -2 (a) Positive numbers FIGURE -2 (b) Negative numbers 2.18 As the real number line is wrapped around the unit circle, each real number t will correspond with a point (x, y) on the circle. For example, the real number 0 corresponds to the point (I, 0). Moreover, because the unit circle has a circumference of 21T, the real number 21T will also correspond to the point (1, 0), SECTION 2.2 THE TRIGONOMETRIC FUNCTIONS AND THE UNIT CIRCLE 119 The Trigonometric Functions From the preceding discussion, it follows that the coordinates x and yare two functions of the real variable t. These coordinates are used to define the six trigonometric functions of t. sine cosine tangent cosecant secant cotangent These six functions are normally abbreviated as sin, esc, cos, sec, tan, and cot, respectively. DEFINITION OF TRIGONOMETRIC FUNCTIONS Let t be a real number at (x, y) the point on the unit circle corresponding to t. sin t = Y esc t cost=x I sect=-,x*-O x y (0, I) v tan t = '--, x (I, 0) x*-O cot t = y*-O y x , y*-O Y ~------~----~~_x (-I, 0) 'I As an aid to memorizing these definitions, note that the functions in the second column are the reciprocals of the corresponding functions in the first column. (0, Unit Circle Divided into 8 Equal Arcs FIGURE 2.19 ° y ( _~2' v'23) (0, 1) ~--.,.---~ (~ 3 ~) (_v'2'2 2' v'23) (-;,D (1, O) ~~----~------~~X (0, -I) Unit Circle Divided into 12 Equal Arcs FIGURE 2.20 In the definition of the trigonometric functions note that the tangent or secant is not defined when x = 0. For instance, because t = n/2 corresponds (0, I), it follows that tan( n/2) and sec( n/2) are undefined. Simi to (x, y) O. For instance, lar/y, the cotangent or cosecant is not defined when y because t = 0 comesponds to (x, y) = (1,0), cot and esc 0 are undefined. In Figure 2.19, the unit circle has been divided into eight equal arcs, corresponding to t-values of n n 3n 0, 4' 2' 4' 5n 3n 7n n, 4' 2' 4' and 2n. Similarly, in Figure 2.20, the unit circle has been divided into 12 equal arcs, corresponding to t-values of n n n 2n 5n 7n 4n 3n 5n I In 0, 6' 3' 2' 3' 6' n, 6' 3' 2' 3' 6' and 2n. Using the (x, y) coordinates in Figures 2.19 and 2.20, you can easily evaluate the trigonometric functions for common t-values. This procedure is demonstrated in Examples I, 2, and 3. 120 CHAPTER 2 TRIGONOMETRY E\aJuating Trigonometric Functions of Real Xumbers EXAMPLE 1 DISCOVEr{y Use a graphing utility to draw the unit circle as follows. Evaluate the six trigonometric functions at the following real numbers. A. t I. Set the utility 10 radian and parametric mode. 2. Let 0 T :$ 6.3. with = 0.314 3. Let -1.5 :$ X :$ 1.5 and -- 1.5 :$ Y :$ 1.5. 4. Let XIT cos T and Yn = sin T. 5. Activate the graphing utility. The screen should look like that shown in the figure. Use the trace key to move the cursor around the unit circle. The T-values represent the measure of in radians. What do the X-and Y-values represent? Explain your reasoning. e 1T ="6 51T = '4 B. t SOLUTION = 1T/6 corresponds to the first-quadrant point (V3)2, 1/2), you can write the following. A. Because t 1T sin 6 1T cos - = x = 6 1T tan - 6 1T csc 6 y=- Y x 2 v3 2 =2 2 v3 1T sec 6 1/2 1T cot 6 v3/2 (x, y) = 2v3 3 v3 51T/4 corresponds to the third-quadrant point (x, y) you can write the following. B. Because t ( - \12/2, - \12/2), \12 51T sin - = y 4 51T cos 4 1--+---+---+---11.5 51T tan - 4 x 2 = \12 2 \12/2 y =:.. x 51T csc- = 4 sec cot 2 -\12 51T 4 51T 4 ....... .. Evaluating Trigonometric Functions of Real Numbers EXAMPLE 2 Evaluate the six trigonometric functions at the following real numbers. A. t = 0 B. t = 7T SOLUTION A. t 0 corresponds to the point (x, y) sin 0 y cos 0 x v tan 0 = =x 0 = 1 =0 = (L 0) on the unit circle. csc 0 is undefined sec 0 cot 0 is undefined SECTION 2.2 B. t = 1T THE TRIGONOMETRIC FUNCTIONS AND THE UNIT CIRCLE corresponds to the point (x, .v) sin 1T Y cos 1T = X tan 1T =0 v -I =- = - 0 x Ex AMPLE 3 = 0 1 = (- I, 121 0) on the unit circle. CSC 1T is undefined sec 1T = -I cot 1T is undefined .. " " ... " " Evaluating Trigonometric Functions of Real "iumbers Evaluate the six trigonometric functions at the following real numbers. 1T A. 1= B. t = 3 51T SOLUTION A. Moving clockwise around the unit circle, you can see that corresponds to the point (x, y) (1/2, v3/2) Sin( -~) cos( -~) =x 2 tan ( -~) = ~ = -V3 v3 y= 2 sec( r ~) csc( -~) r./3 I 2 =2 1T) = cot ( - '3 X Y B. Moving counterclockwise around the unit circle one and a quarter revolu 51T/2 corresponds to the point (x. y) = (0. I). tions, you can see that f . sm tan (0. -1) -1 ;::; x ;::; 1 FIGURE 2.21 2 y= 5r. cos -~ = x 2 y ~---+---+--- 51T x 0 5r. v = '- is undefined 2 x 51T 2 5r. csc ~ sec 2 cot ~ is undefined 5r. 2 x - = 0 y .. " . " ... " -1;::; y ;::; 1 Domain and Period of Sine and Cosine The domain of the sine and cosine functions is the set of all real numbers. To determine the range of these two functions. consider the unit circle shown in y and cos f = x. Moreover. Figure 2.21. Because r = 1. it follows that sin t 122 CHAPTER 2 TRIGONOMETRY because (x, y) is on the unit circle, you know that -] S Y S J and -1 S x S 1, and it follows that the values of the sine and cosine also range between I and I. That is, -ls),SI I S sin t S -Isxsi S cos t S I. and Suppose you add 211' to each value of t in the interval [0, 211' J. thus completing a second revolution around the unit circle, as shown in Figure 2.22. The values of sin(t + 211') and cos(t + 211') correspond to those of sin t and cos t. Similar results can be obtained for repeated revolutions (positive or negative) on the unit circle, This leads to the general result sin(t + 21Tn) In Figure 2.22, positive multiples of 21T were added to the t values. You could just as well have added negative multiples. For instance, 1T/4 21T and 1T/4 41T are also coterrninal to 1T/4. = sin t and cos(t + 21Tn) = cos t for any integer n and real number t. Functions that behave in such a repetitive (or cyclic) manner are called periodic. t := y 37T 37T t := t 4' 4 + := 7T 7T 2 2' 27T, ... + 7T 27T, 2 + 47T, 7T 7T 4' 4 + 7T 27T, 57T 57T := 4+ 47T, ... 7T, 37T, •.. -----+-----+-----+--~x t := 0, t ... 4' 4 + 21T, ... 77T 77T 4' 4 + 27T, 47T, ... 77T 21T, 4 + 47T, ... Repeated Revolutions on the Unit Circle FIGURE 2.22 DEFINITION OF A PERIODIC FUNCTION A functionJis periodic if there exists a positive real number c such that J(t + c) = J(t) for all t in the domain off The least number c for which J is periodic is called the period off From this definition it follows that the sine and cosine functions are periodic and have a period of 211'. The other four trigonometric functions are also periodic, and we will say more about this in Section 2.6. SECTION 2.2 THE TRIGONOMETRIC FUNCTIONS AND THE UNIT CIRCLE Csing thE: hTioct In EXAMPLE 4- A. Because 2.luate S~ne 123 and 1317 17 6 = 217 + 6' you have 1317 = sin ( 217 sin (5 B. Because cos( - 717 2 -417 7;) = . 17 17) + "6 Sin - 6 = I 2 . 17 + -, you have 2 cos( -417 + ~) = cos ~= 0, Recall from Section 1.7 that a function f is even if f( -t) = f(t) and is odd if f( -t) = - f(t). Of the six trigonometric functions, two are even and four are odd, as stated in the following theorem. EVEN AND 000 TRIGONOMETRIC FUNCTIONS The cosine and secant functions are even. cos( -t) = cos t sec( -t) sec t The sine, cosecant, tangent, and cotangent functions are odd. sine - t) -sin t tan( - t) = -tan t csc( -t) = -csc t cot( -t) = -cot t Evaluating Trig Functions with a Calculator DISCOVERY Set your graphing utility in degree mode and enter tan 9(), What happens" Why" Now set your graphing utility to radian mode and enter lan(3.14159; 2!. Explain the calculator's anS\\er. From the arc length formula s = r8, with r = I, you can see that each real number t measures a central angle (in radians). That is, t = r8 1(8) = 8 radians. Thus, when you are evaluating trigonometric functions. it doesn'r make any difference whether you consider t to be a real number or an angle given in radians. A scientific calculator can be used to obtain decimal approximations of the values of the trigonometric functions, Before doing this, however. you must be sure that the calculator is set to the correct mode: degrees or radians. Here are two examples. 124 CHAPTER 2 TRIGONOMETRY Degree mode: cos 28 Radian mode: tan 12 j)i.\pjrl'" 11' Most calculators do not have keys for the cosecant, secant, and cotangent functions. To evaluate these functions, use the reciprocal key with the func tions sine, cosine, and tangent. Here is an example. 11' esc 8 \ = . ( / ) = 2.613. Sin 11' EXA[v'!PLE 5 8 Using a Calculator to Evaluate Trigonometric Functions Use a calculator to evaluate the following. (Round to three decimal places.) A. sin(-76.4)" B. cot 1.5 c. sec(5° 40' 12") SOLUTION Rounded to 3 FunCTion Mode Degree Radian Degree A. sine 76.4t B. cot 1.5 c. sec(5° 40' 12") Note that 5° 40' 12" = 5.67°. Decimal Places -0.97196\ 0006 0.0709148443 1.004916618 -0.972 0.071 1.005 ......... w Suppose you are tutoring a student who is just learning trigonometry. Your stu dent was asked to evaluate the cosine of 2 radians and, using a calculator, ob tained the following. You BE THE INSTRUCTOR Keyscrokes 0.999390827 You know that 2 radians lies in the second quadrant. You also know that this implies that the cosine of 2 radians should be negative. What did your student do wrong? SECTION 2.2 WARM-UP THE TRIGONOMETRIC FUNCTIONS AND THE UNIT CIRCLE 125 .The following wann-up exercises involve skills that were covered in earlier sections. You ,Will use theSe skills in the exercise set for this section. .fu EXerclses,1 and 2; simplify the expression. • 1. I 2· ~r: -V3 V2 2. 2 ~ r: -v2 2 2 In Exercises 3 and 4, find an angle () in the interval [0, 21TJ that is coterminal with the given angle. 3 81T • 3 4. 4 In Exercises 5 and 6, convert the angle to radian measure. 5. 30· 6. 13Y In Exercises 7 and 8, convert the angle to degree measure. 7. 7T 3" radians 8. - 31T 2 radians 9. Determine the circumference of a circle with radius I. 10. Determine the arc length of a semicircle with radius I. SECTION 2.2 . EXERCISES ...•.•••.....••....................................... .....••.......•••..................................... .....•...... In Exercises 1-8, find the point (x, y) on the unit circle that corresponds to the real number I (see Figures 2.19 and 2.20). 1. [ == 3. t 1T 4 51T 6 41T 5. 1==3 37T 7. 1 = 2 1T 2. [ 4. 6. 3 51T I 4 1= 8. [ = 111T 6 1T In Exercises 9-16. evaluate the sine. cosine. and tangent of the real number. 9. 11. 13. 1T 4 51T t = 4 I IT. I 15. [ 6 41T 3 1T 4 10. t == [ 12. 14. 16. 5'77 I = 6 21T I 3 7rr I 4 126 CHAPTER 2 TRIGONOMETRY In Exercises 17-22, evaluate (if possible) the six trigonometric functions of the real number. 37T 27T 17. 1 = - IS. 1= 7T 19. 1=- 37T 20. 1 = 4 2 2 47T 21. 1= 3 117T 22. 1= 3 6 In Exercises 45 and 46, use the accompanying figure and a straightedge to approximate the values of the trigonometric functions. In Exercises 47 and 48, approximate the solutions of the equations. Use 0 :5 I :5 27T. 45. (a) sin 5 (b) cos 2 46. (a) sin 0.75 (b) cos 2.5 47. (a) sin I = 0.25 48. (a) sin I = -0.75 (b) cos I = 0.75 (b) cos I = -0.25 1.75 2.00 In Exercises 23-30, evaluate the trigonometric function using its period as an aid. 23. sin 37T 24. cos 37T 87T 25. cos3 26. sin \ 2.25 0.8 \ 2.50 97T .1.0 o. 0.4 2.75 4 137T) 2S. sin ( -6 29. sin( _ 9;) 30. cos( _ 8;) 1.25 1 1.00 ' / 1 t 0.75 / 0.50 0.25 0.2 3.00 197T 27. cos - 6 1.50 1 I 3.25 I I 0.2 ! ! 0.4 0.6 I:::J 6.25 0.8 1.0 6.00 3.50 1 1 / In Exercises 31-36, use the value of the trigonometric function to evaluate the indicated functions. 31. sin t = / 4.00 t 4.50 (a) sin I -i (a) cos (-I) (b) sec (-I) 49. Harmonic MOlioll (b) sec(-I) 35. sin t = ~ (a) sin( 7T - I) (a) cos( 7T - r) (b) sin(1 + 7T) (b) cos(r + 7T) In Exercises 37-44, use a calculator to evaluate the trigonomet ric function. (Set your calculator in the correct mode and round your answer to four decimal places.) 41. tan I 10.5° 43. csc 0.8 The displacement from equilibrium of an oscillating weight suspended by a spring is y(l) = I = ~ 4 39. cos 34.2° 5.00 FIGURE FOR 45-48 34. cos I = - ~ 7T 4.75 (b) csc I (a) cos I 37. sin - 5.50 (b) CSc(-I) 32. sin( -I) = ~ 33. COS(-I) = 5.75 5.25 4.25 (a) sin(-I) 36. co, 3.75 3S. tan 7T where v is the displacement in feet and I is the time in seconds. Find the displacement when (a) I = 0, (b) I = j, and (c) I = ~. 50. Harmonic Malian The displacement from equilibrium of an oscillating weight suspended by a spring and subject to the damping effect of friction is y(t) = 40. cot I 42. sec 54.9° 44. sin( -0.9) j cos 61, je I cos 61, where y is the displacement in feet and I is the time in seconds. Find the displacement when (a) I = 0, (b) I = j, and (c) I = ~. 5 E C TI 0 N 2.3 51. Electric Circuits The initial current and charge in the electrical circuit shown in the accompanying figure is zero. The current when J 00 volts is applied to the circuit is given by TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES 127 52. Use the unit circle to verify that the cosine and secant functions are even and the sine. cosecant tangent, and cotangent functions are odd. 5e- 21 sin t I if the resistance, inductance, and capacitance are 80 ohms, 20 henrys, and 0.01 farads, respectively. Approximate the 0.7 seconds after the voltage is applied. current t c R o------c E(t) FIGURE FOR 51 2.3 TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES Trigonometric Functions of an Acute Angle I Trigonometric Identities I Applications Involving Right Triangles Trigonometric Functions of an Acute Angle second look at the trigonometric functions is from a right triangle per spective. In Figure 2.23, the three sides of the triangle are the hypotenuse, the opposite side (the side opposite the angle 8). and the adjacent side (the side adjacent to the angle 8). Using the lengths of these three sides. you can form six ratios that define the six trigonometric functions of the acute angle 8. In the following definition it is important to see that 0° < 8 < 90° and that for such angJes the value of each of the six trigonometric functions is positive. OUf .8 Adjacent side FIGURE 2.23 128 CHAPTER 2 TRIGONOMETRY RIGHT TRIANG.LE DEFINITION OF TRIGONOMETRIC FUNCTIONS Let e be an acute angle of a right triangle. Then the six trigonometric functions of the angle e are defined as follows. sin e csc hyp cos 8 hyp sec8= adj hyp tan 8 cot adj hyp e =opp e =adj opp The abbreviations opp, adj, and hyp represent the lengths of the three sides of a right triangle. the length of the side opposite 8 the length of the side adjacent to the length of the hypotenuse opp adj hyp e Find the values of the six trigonometric functions of shown in Figure 2.24. e in the right triangle SOLUTION By the Pythagorean Theorem, (hyp)2 = (opp)! \125 hyp Thus, adj SJ ;y §' C' .:$>"<. 4 sin 8 cos '\8 e \ 3 FIGURE 2.24 tan 8 = 3, opp = 4, 4 hyp - 5 3 hyp adj 5 (adj)2, it follows that = 5. and hyp = 5. csc e= hyp opp = 5 4 sec e = hyp adj = 5 3 cot 8 + - adj opp 3 4 "' ,. " :; .. ". " '" ~ !:!,;;""'~~'r:::£iiJ.£ In Example I, you were given the lengths of the sides of the right triangle, but not the angle 8. A much more common problem in trigonometry is to be asked to find the trigonometric functions for a given acute angle e. To do this, construct a right triangle having e as one of its angles. 5 E C T JON 2.3 129 TR tGONOMETR IC FUNCTIONS AND RIGHT TRIANGLES E\aluating Trigonometric Functions of 45° EXAMPLE 2 Find the values of sin 45°, cos 45°, and tan 45°. SOLUTION Construct a right triangle having 45° as one of its acute angles, as shown in Figure 2.25. Choose the length of the adjacent side to be 1. From geometry, you know that the other acute angle is also 45°. Hence, the triangle is isosceles and the length of the opposlle side is also I. Using the Pythagorean Theorem, you can find the length of the hypotenuse to be VI" + hyp = 12 vi Finally, you can write the following. opp hyp cos 45° 45° tan 450 V2 1 = V2 = 2 adj 1 V2 hyp V2 2 = opp = adj FIGURE 2.25 Ex AMPLE 3 ,. ,. ,. II II ,. .. ,. /I Evaluating Trigonometric Functions of 30° and 60° Use the equilateral triangle shown in Figure 2.26 to find the values of sin 60°, cos 60°, sin 30°, and cos 30°. SOLUTION Try using the Pythagorean Theorem to verify the lengtlE> of the sides given in Figure 2.26. For (J = 60°, you have adj I, opp = V3, and hyp = 2, which implies that FIGURE 2.26 opp -= V3 hyp 2 For (J DISCQVERY Set your graphing utility in degree mode and choose un anglc \. Now evaluate costr) and sinl90 xL Whut do you oh,cn c·: Repeat this experiment II ith ,inC,) and COS\I.)O .11. = 30°, you have adj . 30° = -opp Sin and hyp I 2 = - cos 60° adj hyp I 2 = V3,opp = 1, and hyp = 2, which implies that and ° adj cos 30 = hyp V3 =. 2 .. ,. .. II .. ,. .. /I .. Because the angles 30°, 45°, and 60° (7T/6, 7T/4, and 7T/3) occur fre quently in trigonometry, we suggest that you learn to construct the triangles shown in Figures 2.25 and 2.26. 130 CHAPTER 2 TRIGONOMETRY SINE, COSINE, AND TANGENT OF SPECIAL ANGLES 7T 0 sin 30 = sin 6 • Sin 45 7T 2' cos 30° = Sin -7T4 = V2 2' 0 • sin 60" . 7T Sin - 3 cos 6 cos 45° = cos V3 = - V3 Ti' = 2' 7T V2 4 2 tan - = 6 3 tan 7T I cos - = -. 2' 3 4 7T tan 60° = tan - 2 3 ,r:; v3 Trigonometric Identities In the preceding list, note that sin 30° = ~ = cos 60", This occurs because 30° and 60° are complementary angles, In genera!, it can be shown that cofunctions of complementary angles are equal. That is, if () is an acute angle. then the following relationships are true. sin(90° - ()) tan(90° - 8) sec(90° - 8) = cos(90" 8) ()) cot (90° csc(90° - 8) cos 8 cot (J csc (J = sin 8 tan () sec 8 For instance, because 10° and 80° are complementary angles, it follows that cot 80°. sin 10c = cos 80° and tan 10° FUNDAMENTAL TRIGONOMETRIC IDENTITIES Reciprocal Identities sin 8 csc Tecnno(OiJlJ Note _ _ _ _ _ __ Use your calculator to confiml several of the trigonometric identities at the right for various values of 8, For instance, calculate (sin (cos 0,))' and observe that the vulue is l. (J csc sec 8 (J J = - sin cos 8 (J tan 8 cos 8 I cot =- (J I J cot 8 = - tan 8 =- sec 8 Quotient Identities sin () tan 8 = - cos (J CO! 8= cos 8 sin 8 Pythagorean Identities sin" 8 REMARK + cos 2 8 = We use sin' (J J + tan:: 8 = sec:: 8 to represent (sin 8)', cos' (J I + cot2 8 to represent (cos (J)', csc" 8 - and so on, SECTION 2.3 TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES 4 EXAMPLE 131 Applying Trigonometric Identities Let 6 be the acute angle such that sin 6 = 0.6. Use trigonometric identities find the values of the following. 10 B. tan 6 A. cos 6 SOLUTION + cos" 6 (0.6)2 + cos" 6 A. sin 2 6 cos" () = I - (0.6)" cos 6 = YO.64 = = 0.64 0.8. B. Now, knowing the values of sin () and cos (), you can find the value of tan (). tan () 0.8 sin 6 = 0.6 FIGURE 2.27 sin 6 0.6 cos 6 0.8 0.75. Try using the definitions of cos () and tan 6, and the triangle shown in Figure 2.27, to check these results. • .••.•••• EXAMPLE 5 Applying Trigonometric Identities Let () be an acute angle such that tan () the values of the following. A. cot 6 3. Use trigonometric identities to find B. sec () SOLUTION A. cot 6 3 tan () Rt'Ciprocai /dell IiI.' 3 B. sec 2 tan" 6 + sec 2 () 31 + I sec" () = 10 sec () = () Pyrlwgoreon idcntifY VIo tan 6 =3 FIGURE 2.28 Try using the definitions of cot () and sec 6, and the triangle shown in Figure 2.28, to check these results. .. ....... 132 CHAPTER 2 TRIGONOMETRY Applications Involving Right Triangles Many applications of trigonometry involve a process called solving right triangles. In this type of application, you are usually given two sides of a right triangle and asked to find one of its acute angles. or you are given one side and one of the acute angles and asked to find one of the other sides. EXAMPLE 6 Using Trigonometr) to Solve a Right Triangle A surveyor is standing 50 feet from the base of a large tree, as shown in Figure 2.29. The surveyor measures the angle of elevation to the top of the tree as 71.5°. How tall is the tree? SOLUTION From Figure 2.29, you can see that tan 71S = opp ad] y x where x 50 and y is the height of the tree. Thus. you can determine the height of the tree to be y x tan 71.5° = 50(2.9&&68) = 149.4 feet. EXAMPLE 7 ......... Using Trigonometry to Solve a Right Triangle A person is standing 200 yards from a river. Rather than walking directly to the river, the person walks 400 yards along a straight path to the river's edge. Find the acute angle 0 between this path and the river's edge, as indicated in Figure 2.30. SOLUTION From Figure 2.30, you can see that the sine of the angle 0 is FIGURE 2.30 sin 0 = opp hyp 200 400 2 Now, you can recognize that 0 = 30°. ...... .. . SECTION 2.3 TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES 133 In Example 7, you were able to recognize that the acute angle that satisfies the equation sin e = ~ is e = 30°. Suppose, however, that you were given the equation sin e == 0.6 and asked to find the acute angle e. Because . sm -30° I = -2 = 0 .5000 and . sm 45° = I v2 = 0.7071, you can figure that e lies somewhere between 30° and 45°. A more precise value of can be found using the inverse key on a calculator. (Consult your calculator manual to see how this key works on your own calculator.) For most calculators, one of the following keystroke sequences will work. e Degree mode: .61IN~ )sinl Degree mode: ~di [sir!) .6 ~~:ri~J Thus, you can conclude that if sin EXAMPLE 8 DI51'/0I' 3686989765 e Disl'/a\' 3686989765 0,6, then e Using Trigonometry to Solve a Right Triangle I 40 ft 1 A 40-foot flagpole casts a 30-foot shadow, as shown in Figure 2,31. Find the angle of elevation of the sun. Angle of elevation ~\. e FIGURE SOLUTION I / 1 .... · - - - 30 ft e, Figure 2.31 shows that the opposite and adjacent sides are known, Thus, tan ---1 40 30' = With a calculator in degree mode you can obtain 2.31 COMPARING DEFINITIONS OF TRIGONOMETRIC FUNCTIONS e = -opp adj e= 53.13°. ....... .. In Section 2.2 and in this section, we presented two different definitions of trigonometric functions. One was the "unit circle" definition and the other was the "right triangle" definition. Write a short paper that compares the two definitions. Then use both definitions to find the values of the six trigonometric functions at 30°. For this value of which definition do you prefer" For = 37T, which do you prefer and why? e e e, 134 CHAPTER 2 TRIGONOMETRY WARM-UP The following warm-up exercises involve skills that were covered in earlier sections. You will use these skills in the exercise set for this section. In Exercises 1-4, find the distance between the points. 1. (3, 8), 0, 4) 2. (5, 2), (2, -7) 3. (-4,0), (2, 8) 4. (-3, -3). (0, 0) In Exercises 5-10, perfonn the indicated operation(s). (Round your result to two decimal places.) 5. 0.300 x 4.125 7. 6. 7.30 2.40 9. 19,500 0.007 SECTION 2.3 . ....................... ~II X· 43.50 3740 8. 28.0 151.5 10. (10.5)(3401) 1240 EXERCISES ••• "."''' ..... \ln'''.'' .. ''~ll' .. ~.Cl .... O ••• ,,~ 0."1:"" . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . In Exercises 1-8. find the exact values of the six trigonometric functions of the angle e. (Use the Pythagorean Theorem to find the third side of the triangle.) 5. 1. 15 2 8 8. 7. 4. 3. 5 4 8 5 5 SECTION 2.3 In Exercises 9-16. sketch a right triangle corresponding to the trigonometric function of the acute angle (J Use the Pythago rean Theorem to determine the third side and then find the other five trigonometric functions of O. 9. sin 11. sec (J == ~ e 10. cot (J == 5 12. cos 0 ~ 14. csc 0 = J:} :2 13. tan (J == 3 IS. cot 0 16. sin e ;;, TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES In Exercises :25-34. use a calculator to evaluate each function. Round your result to four decimal places. (Be sure the calcula tor is in the correct mode.) (b) cos 80 0 25. (a) sin 10° 26. (a) tan 23.5° 27. (a) sin 16.35° (b) cot 66S (b) csc 16.35° 28. (a) cos 16° 18' 29. (a) sec 42° 12' (b) sin 73° 56' (b) 17. sin 60° == 2- cos 60° 1 3. (J (J == 5. 33. (a) csc I (b) sec 34. (a) 1 tan 2 (b) (n.2 I) cot( '!!. .2 ~) sin 30° (d) cot 60° v3 tan 30° In Exercises 35-40. find the value of (J in degrees (0° < (J < 90°) and radians (0 < e < 7T/2) without using a calculator. 3 (b) cot 60° (d) cot 30° 3v2 sec (J == - 35. {al sin 4 (a) sin (J (c) tan (J 20. sec 16 (b) cos 0.75 (b) (a) csc 30° (c) cos 30e 19. esc 32. (a) sec 0.75 =2 (a) tan 60° (c) cos 30e 18. sin 30° == :2 r. (b) tan 16 V1 csc 48° 7' (b) sec 4° 50' IS" 30. (aJ cos 4° 50' IS" r. 31. (a) cot In Exercises 17-20. use the given values and the trigonometric identities to evaluate the trigonometric functions. 135 (b) cos e (d) sec(90° tan 0 (a) cos e (c) cot(90° - 0) e) 2\16 cot e (d) sin (J 37. (b) esc (J (a) sec 39. (a) (b) I 36. (a) cos 2 2 I esc e sin v2 e 2 (b) 40. (a) cot 3 2 e I tan e = v3 cos e 38. (a) 2V3 e v2 tan (b) e= 2 e (bl cot (b) e sec (b) e v3 3 e v2 In Exercises 21-24. evaluate the trigonometric functions. 21. (a) cos 60° 22. (a) csc 30° 23. (a) cot 45° 7T 24. (al sin - 3 (b) Lan (b) sin 7T In Exercises 4 (00 < e < 90°) and radians 7T verse key on a calculator. 4 (b) cos 4Y 41. (b) esc 45° 43. (a) (b) (a) (b) 41--44. sin e == 0,8191 cos (j 0.0175 tan tan e= e find the (0 < value of e in degrees < 7T/2) by using the in (J 42. (a) cos (b) cos U920 44. (a) sin 0.4663 (b) cos = (J e (J 0.9848 0,8746 == 0.3746 e 0.3746 136 CHAPTER 2 TRIGONOMETRY 46. Solve for x. 45. Solve for y. 53. Height A 6-foot person standing 12 feet from a streetlight casts an 8-foot shadow (see figure). What is the height of the streetlight~ y 10 T 100 x 48. Solve for 47. Solve for x. r. FIGURE FOR 53 25 r 30 x 50. Solve for x. 49. Solve for r. 54. Height A 6-foot man walked from the base of a broad casting tower directly toward the tip of the shadow cast by the tower. When he was J 32 feet from the tower, his shadow started to appear beyond the tower's shadow. What is the height of the tower if the man's shadow is 3 feet long 0 ~30 10 x 55. Length A 20-foot ladder leaning against the side of a house makes a 75° angle with the ground (see figure). How far up the side of the house does the ladder reach') 51. Solve for y. 12 52. Solve for r. y 50" 20 FIGURE FOR 55 SECTION 2.3 56. Width of a River A biologist wants to know the width w of a river in order to properly set instruments for studying the pollutants in the water. From point A. the biologist walks downstream J00 feet and sights to point C. From this sighting. it is determined that SOc (see figure). How wide is the river? e 137 TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES 59. Machine Shop Calculations A steel plate has the form of a quarter circle with a radius of 24 inches. Two ·inch holes are to be drilled in the plate positioned as shown in the figure. Find the coordinates of the center of each hole. i y 24 20 FIGURE FOR 56 20 24 x FIGURE FOR 59 57. Distance From a J50·foot observation tower on the coast, a Coast Guard officer sights a boat in difficulty. The angle of depression of the boat is 4° (see figure). How far is the boat from the shoreline 0 FIGURE FOR 57 60. Machine Shop Calculations A tapered shaft has a diame ter of 2 inches at the small end and is 6 inches long (see figure). If the taper is 3°. find the diameter d of the large end of the shaft. 58. Angle of ElevatIOn A ramp 17 ~ feet in length rises to a loading platform that is 3 feet off the ground (see figure). Find the angle that the ramp makes with the ground. e * t-------.---~ FIGURE FOR 58 FIGURE FOR 60 6 in. ~-.~------i 138 CHAPTER 2 TRIGONOMETRY 61. Trigollomerric Funcrions by Acrual Measuremenr Use a compass to sketch a quarter of a circle of radius 10 centime ters. Using a protractor, construct an angle of 25° in stan dard position (see figure). Drop a perpendicular line from the point of intersection of the tenninal side of the angle and the arc of the circle. By actual measurement. calculate the coordinates (x, y) of the point of intersection and use these measurements to approximate the six trigonometric functions of a 25° angle. )' i 69. A 3D-foot ladder leaning against the side of a house is 4 feet from the house at the base (see figure). (a) How far up the side of the house does the ladder reach? Express your answer accurate to two decimal places. (b) Use the fact that cos e 4/30 to find the angle that the ladder makes with the ground. Express your answer in degrees accurate to two decimal places. (e) Use the tangent of the angle found in part (b) to answer part (a) again. (d) Why do your answers to parts (a) and (c) differ slightly? 10 FIGURE FOR 61 62. Trigonometric Functions by Actual Measurement Repeat Exercise 61 using an angle of 75°. In Exercises 63-68, determine whether the statement is true or false, and gi ve a reason for your answer. 63. sin 60° csc 60° = 1 64. sec 30° csc 60° 65. sin 4SO + cos 45° = 66. coe 10° 10° 67. sin 60° sin 30° = sin 2° 68. tan[ (0.8)2J tan 2(0.8) FIGURE FOR 69 SECTION 2.4 TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 139 2.4 TRIGONOMETRIC FUNCTIONS OF ANY ANGLE .......................................................................................... Trigonometric Functions of Any Angle ! Reference Angles Trigonometric Functions of Any Angle In Section 2.3, you learned to evaluate trigonometric functions of an acute angle. In this section you wi1llearn to evaluate trigonometric functions of any angle. DEFINITION OF TRIGONOMETRIC FUNCTIONS OF ANY ANGLE Let e be an angle in standard position with (x, y) any point (except the origin) on the terminal side of e and r = Vx" + y" (see figure). e y cos e= x - tan e= '-, x Sin csc r r V x*,O e= r -, y*,O \' sec e = xr cot e x y x*,O y*,O )' r = vx? + / REMARK Because r = cannol be zero, it follows that the sine and cosine functions are defined for any real value of 8. However, if x = 0, the tangent and secant of (] are undefined. For example, the tangent of 90° is undefined. Try calculating the tangent of 90° with your calculator. Similarly, if y 0, the cotangent and cosecant of 8 are undefined. - 140 CHAPTER 2 TRIGONOMETRY y (x, y) y If 0 is an acute angle, then these definitions coincide with those given in the previous section. To see this, note in Figure 2.32 that for an acute angle x adj, y := opp, and r = hypo e, Evaluating Trigonometric Functions EXAMPLE 1 Let ( 3, 4) be a point on the terminal side of tangent of O. \6 x e. Find the sine, cosine, and SOLUTION FIGURE 2.32 Referring to Figure 2.33, you can see that x y = Y(-3)" r = + 4" := - \I2s 3. y = 4. and S. Thus, you can write the following. e Y 4 r 5 cos 0 x r sin tan -3 -2 FIGURE 2.33 -] e= ~ := x -3 3 5 5 4 -3 - 4 ....... .. 3 The signs of trigonometric function values in the four quadrants can be determined easily from the definitions of the functions. For instance, because cos 0 = x r' it follows that cos () is positive where x > 0, which is in Quadrants I and IV. (Remember, r is always positive.) In a similar manner you can verify the results shown in Figure 2.34. y y rT i<9<1T Quadrant II sin 8: cos 8; tan 8; 1T 0<9<2 Quadrant I x<o y> 0., sin 8: ..... cos 8: -+ x>o >0 ,_ Y tan 8: + ;c Quadrant III sin e: cos 8: lan e: ~. Quadrant IV sin e: cos 8: + tan e: ;c x < o. y<o " ~ '. x > 0 y<O 3rT 1T<8<T Signs of Trigonometric Functions FIGURE 2.34 3rT T<8<21T SECTION 2.4 141 Evaluating Trigonometric Functions EXAMPLE 2 Given tan TRIGONOMETRIC FUNCTIONS OF ANY ANGLE e = 1and cos 0 > 0, find sin e and sec 0. SOLUTION Note that 0 lies in Quadrant IV because that is the only quadrant in which the tangent is negative and the cosine is positive. Moreover, using v 5 4 tane=~=-- x and the fact that y is negative in Quadrant IV, you can let y = - 5 and x = 4. Hence, r V 16 + 25 v4l and you have . SIn -5 e = ~v - - - = -0.7809 v4l r EXA M PLE 3 and sece= r - - = 1.6008. 4 x ....... .. Trigonometric Functions of Quadrant Angles Evaluate the sine function at the four quadrant angles 0, rr/2, rr, and 3rr/2. SOLUTION To begin, choose a point on the terminal side of each angle, as shown in Figure 2.35. For each of the four given points, r I, and you have y 1T 2 (0, I) sin 0 x (-1, 0) (l, 0) 0=0 r I . rr y =2 r SIn - 0 1T y . y sm rr = r 31T 2 FIGURE 2.35 (0, -I) . 3rr sm2 ~ = r IX. \ I -=1 i ix. I) 1 o - = 0 I I. Ol 10. I) It.r; =! I _I I = 1. Ix. I) = 10. 1.0\ -II Trying using Figure 2.35 to evaluate some of the other trigonometric functions at the four quadrant angles and check them on your calculator. ••••••••• Reference Angles The values of the trigonometric functions of angles greater than 90° (or less than 0°) can be determined from their values at corresponding acute angles called reference angles. 142 CHAPTER 2 TRIGONOMETRY Quadrant II DEFINITION OF REFERENCE ANGLES Reference angle: e' e' == e' == Let (J be an angle in standard position. Its reference angle is the acute angle (J' formed by the terminal side of (J and the horizontal axis. e , ! 11 - 180 0 e (radians) e (degrees) Figure 2.36 shows the reference angles for Find the reference angle Quadrant III e' e' ee- in Quadrants II, III, and IV. Finding Reference Angles EXAMPLE 4 Reference angle: e' (J 11 (radians) 0 180 (degrees) for each of the following. (J' B. (J = 2.3 C. (J = 135° SOLUTION A. Because axis is //' (e Reference angle: e' Quadrant IV e' == e' == 211 360° (J = 300° lies in Quadrant IV, the angle it makes with the x B. Because (J 2.3 lies between 1T/2 = Ic5708 and 1T = 3.1416, it follows that (J is in Quadrant II and its reference angle is e (radians) - e (degrees) (J' = 1T - 2.3 0.8416. Rat/wns C. Because (J = 135 is coterminal with 225°, it lies in Quadrant III. Hence, the reference angle is 0 FIGURE 2.36 (J' Degree.",. Figure 2.37 shows each angle (J and its reference angle (J'. y y y 0 e' == (a) e in Quadrant IV FIGURE 2.37 225 and -135 are coterminal. 11 (b) e in Quadrant II (c) e in Quadrant III ....... .. 0 SECTION 2.4 y TRIGONOMETRIC FUNCTIONS Of ANY ANGLE 143 To see how a reference angle is used to evaluate a trigonometric function, consider the point (x, y) on the terminal side of 8. as shown in Figure 2.38. By definition, you know that (.x; y) f ='~ sin 8 r v tan 8 and = :... x For the right triangle with acute angle 8' and sides of lengths Ix i and Iy I. you have opp sin 8' = opp = ------~~~-~_4--------X opp FIGURE 2.38 1)'1, adj Ixl r hyp tan 8' and _ opp _ Iyl - adj - Ix I Thus. it follows that sin 8 and sin 8' are equal, except possibly in sign. The same is true for tan 8 and tan 8' and for the other four trigonometric functions. In all cases, the sign of the function value can be determined by the quadrant in which lies. e EVALUATING TRIGONOMETRIC FUNCTIONS OF ANY ANGLE To find the value of a trigonometric function of any angle e. 1. Determine the function value for the associated reference angle e'. 2. Depending on the quadrant in which e lies. prefix the appropriate sign to the function value. By using reference angles and the special angles discussed in the previous section. you can greatly extend your scope of exact trigonometric values. Table 2.1 lists the function values for selected reference angles. For instance, knowing the function values of 30° means that you know the function values of all angles for which 30° is a reference angle. TABLE 2.1 i (degrees) 0° 30° 0 - e 0 I 2 cos H J e 0 (j 45° I 60° 90° 180 0 270° ! H (radians) sin tan I _ .-~ i 1T 1T 6 4 1T 1T 3 2 - v2 v3 i ! v3 v2 I 2 -v3 3 31T 1T I v3 I 0 0 undef. I J 0 0 undef. 144 CHAPTER 2 TRIGONOMETRY 5 EXA M PLE Trigonometric Functions of ;\onacute Evaluate the following. A. cos 471' 1171' "3 C. csc- 4 SOLUTION A. Because e = 471'/3 lies in Quadrant III, the reference angle is e' = (471'/3) - 71' = 71'/3, as shown in Figure 2.39(a). Moreover, the cosine is negative in Quadrant III, so that 71' 1 (- )cos - = - . 3 2 471' cos - 3 B. Because -210° + 360° 150°, it follows that -2100 is coterminal with the second-quadrant angle 150°. Therefore, the reference angle is 1800 - 1500 30°, as shown in Figure 2.39(b). Finally, because the tangent is negative in Quadrant II, you have e V3 KCfi'reii. 3 C. Because (I 171'/4) - 271' 371'/4, it follows that 1171'/4 is coterminal with the second-quadrant angle 371'/4. Therefore, the reference angle is e' 71' - (371'/4) 71'/4. as shown in Figure 2.39(c). Because the cosecant is positive in Quadrant II, you have csc 1171' 4 = 71' (+ )csc - 4 1 sin 71'/4 = v2. Check the above values with your calculator. y y e =41T 3 e= e' 111T 4 e' =!!. 3 (a) (el (b) FIGURE 2.39 The fundamental trigonometric identities listed in the previous section (for an acute angle 8) are also valid when is any angle. e SECTION 2.4 TRIGONOMETRIC FUNCTIONS OF ANY ANGLE t'~ing EXAMPLE 6 145 Identities to Evaluate Trigonometric Functions *' Find the following. Let 8 be an angle in Quadrant II such that sin 8 B. tan 8 A. cos 8 SOLUTION A. Because sin 8 obtain *, use the Pythagorean Identity sin" 8 + cos" 8 = 1 to (:31)2 + cos" 8 = I 8 9 9 cos" 8 = 1 - - = -. < 0 in Quadrant II, use the negative root Because cos 8 cos 8 Vs 2\/2 = - ---;:: = - - - . Y9 3 B. Using the result from part A and the trigonometric identity tan 8 sin 81cos 8, you obtain tan 8 3 = --- = 2Y2 Y2 4 ......... 3 Scientific calculators can be used to approximate the values of trigono metric functions of any angle, as demonstrated in Example 7. EXA M PLE 7 Evaluating Trigonometric Functions with a Calculator Use a calculator to approximate the following values. (Round your answers to three decimal places.) TechnololJlJ Note _ _ _ _ _ __ utilities Remember that most do not have keys for the cosecant. ,ecant. and cotangent functions. To evaluate the,e functions. you need to with the 'ine. use the reciprocal cosine and u:.ngenl. For instance. In Example '",Al. you can lind the cotangent of .+10° by I tan-POlin A. cot 410 0 B. sin(-7) C. tan 141T 5 SOLUTION FunClion A. cot 4100 B. sin( -7) 141T C. tan 5 Mode Display Rounded 10 3 Decimal Places - Degree Radian 0.8390996312 -0.6569865987 0.839 -0.657 Radian -0.726542528 -0.727 ......... 146 CHAPTER 2 TRIGONOMETRY At this point, you have completed your introduction to basic trigonometry. You have measured angles in both degrees and radians. You have studied the definitions of the six trigonometric functions from a right triangle perspective and as functions of real numbers. In your remaining study of trigonometry. you will continue to rely on both perspectives. For your convenience we have included in the endpapers of this text a summary of basic trigonometry. There are many different techniques that people use to memorize (or reconstruct) trigonometric formu las. Here is one that we like to use for the sines and cosines of common angles. MEMORIZATION AIDS Ii sin fj cos fj OC 30" Va \./] 2 :: V3 Y4 2 45< 60° 90" vi] .. ") 2 2 V2 VI :: va 2 :: 2 Write a paragraph describing the pattern indicated by this table. Discuss other memory aids for trigonometric formulas. WARM-UP The following warm-up exercises involve skills that were covered in earlier sections. You will use these skills in the exercise set for this section. In Exercises 1-6. evaluate the trigonometric function from memory. 1. sin 30° 3. cos 7T 4 7T 5. sec6 2. tan 45° 7T 4. cot 3 7T 6. esc 4 In Exercises 7-10. use the given trigonometric function of an acute angJe (} to find the values of the remaining trigonometric functions. 7. tan e 9. sin (} 8. cos (} 10. sec e= 3 SECTION 2.4 SECTION 2.4 . 147 EXERCISES In Exercises J-4, determine the exact values of the six trigono metric functions of the angle e. 1. (a) TRIGONOMETRIC FUNCTIONS Of ANY ANGLE (b) y y • In Exercises 5-8, the point is on the terminal side of an angle in standard position. Determine the exact values of the SIX trigonometric functions of the angle. 5. (a) 0,24) (b) (7. 24) 6. 7. 8 (a) (8. 15) (-4. 10) (-5. -2) (b) (-9. -40) (a) (a) (b) (3. -5) (b) ( ~. 3) (8, -15) 2. (a) y (b) In Exercises 9-12. use the two similar triangles in the figure to find (a) the unknown sides of the triangles and (b) the six trlgonometric functions of the angles Cl'1 and 0'> y (-12, -5) (1, -I) FIGURE FOR 9-12 3. (a) (b) y (-\13, y ! . / ------~+_~--~x 9. B~~. . 1) ----+---..r-----_ x UI 10. b l = 3. b l 12.('1 l1.ul=l.c I 12.b, 4, = U, = 13.(', 9 = 26 2.b,=5 4.a2=4.b 2 10 (-2, -2) In Exercises 13-16. determine the quadrant in which (b) (-2, 4) e < 0 and cos e < 0 e > 0 and cos e < 0 sin e > 0 and cos e > 0 sin e < 0 and cos e > 0 SIn e > 0 and tan e < 0 cos e > 0 and tan e < 0 sec e > 0 and cot e < 0 esc e < 0 and tan e 0 13. (a) sin (b) sin J B -----~r_~----x 14. (a) (b) IS. (a) (b) 16. (a) (b) e lies. 148 CHAPTER 2 TRIGONOMETRY In Exercises 17-26. find the values (if possible) of the six trigonometric functions of e using the given functional value and constraint. 40. (a) 7T 41. (a) Functional Value 17. sin e 18. cos e= 4 5 19. tan 0 20. cos e == ~ 21. sec 0 = -2 tane<O sin e is undefined sin e == 0 tan e is undefined e> 44. (a) 0 7T 37T 2 2 22. cot -sos 23. sec 0 24. 0 -145 0 31. (a) 0 = 32. (a) e = 33. (a) 0 27T 3 77T 4 3.5 ll7T 34. (a) 0 == -3- e'. and sketch 0 (b) 0:= 127" (b) e:= 226° -72° (b) 0 (b) 0 = -239° 77T (b) 0 35. (a) 225 36. (a) 300° 37. (a) 750° 39. (a) :3 107T (b) 3 6 177T 3 45. (a) sin 10° 46. (a) sec 225 0 47. (a) cost - 110°) 48. (a) csc 330° 49. (a) tan 240 0 50. (a) cot 1.35 7T 51. (a) tan 9 52. (a) sine -0.65) (b) esc 10 0 (b) sec 135 0 (b) cos 250 0 (b) csc 150° (b) cot 210° (b) tan 1.35 107T (b) tan 9 (b) sin 5.63 6 (b) 0 9 (b) 0:= 5.8 77T (b) 0 10 0 (b) 330 (b) 510° 0 137T In Exercises 45-52, use a calculator to evaluate the trigonomet ric functions to four decimal places. (Be sure the calculator is set in the correct mode.) e In Exercises 53-58, find two values of that satisfy the equa tion. Give your answers in degrees (00 S (] < 360°) and radi ans (0 s e < 27T). Do not use a calculator. 53. (a) sin 0 = ~ 54. (a) cos (] (b) -225° 0 47T (b) 87T In Exercises 35-44, evaluate the sine, cosine, and tangent of each angle without using a calculator. 38. (a) -405 117T 4 2 -I In Exercises 27-34. find the reference angle and 0' in standard position. 27. (a) e 203° 28. (a) 0 = 309 29. (a) e:= - 245° (b) 2 6 7T OS27T 7T 25. The terminal side of 0 is in Quadrant III and lies on the line y == 2.x. 26. The terminal side of e is in Quadrant IV and lies on the line 4x + 3y := O. 30. (a) 0 42. (a) 43. (a) (b) 6 7T Constraint e lies in Quadrant II e lies in Quadrant III sin e < 0 ~ 57T (b) 4 57T 7T '4 (b) 120° 27T (b) - 3 55. (a) csc 0 = V2 2 2Y3 3 56. (a) sec (] 2 57. (a) tan (] = I 58. (a) sin 0 Y3 2 (b) sin (] == -~V2 (b) cos (] (b) cot (] 2 -I (b) sec 0 = -2 (b) cot (] (b) sin (] = - Y3 Y3 2 SECTION 2.4 In Exercises 59 and 60. use a caiculator to approximate two values of 0(0° 0 < 360°) that satisfy the equation, Round to two decimal places, 59. (a) sin 0 60. (a) cos 0 (b) sin 0,8191 = 0.8746 (b) 0 -0,2589 e= -02419 cos TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 149 72. Sales A company that produces a seasonal product fore casts monthly sales over the next two years to be 23, I + 0,4421 + 4.3 sin S 1TI '6' where S is measured in thousands of units and I IS the time in months, with I = I representing January 1991. Predict sales for the following months. (a) February 1991 (b) February 1992 (c) September 1991 (d) September 1992 In Exercises 61- 64, use a calculator to approximate two values of 0(0 ::; 0 < 21T) that satisfy the equation, Round to three decimal places, (b) cos 61. (al cos 0 = 0,9848 62. (a) sin 0 63. (a) tan e= 64. (a) cot 0 0.DI75 (b) sin 1.192 5,671 (b) tan e -05890 0 = -0,6691 e = -8,144 (b) cot 0 1.280 73. Dislance An airplane flying at an altitude of 5 miles is on a flight path that passes directly over an observer (see figure), If 0 is the angle of elevation from the observer to the plane. find the distance from the observer to the plane when (a) 0 = 30°. (b) 0 = 75 c • and (c) 0 = 90"' In Exercises 65-68, use the value of the given trigonometric function and trigonometric identities to find the required trigo nometric function of the angle in the specified quadrant e Given FUl1crioll e 5 66. tan e = ~ 67. csc e -2 68. sec e = :;9 65. sin .< Quadrant Find IV cos III IV III e sec e cot e tan e 5mi FIGURE FOR 73 In Exercises 69 and 70. evaluate the expression without using a calculator. 69. sin" 2 -" cos" 2 71. Average Temperature The average daily temperature T (in degrees Fahrenheit) for a city is T = 45 - 23 COS[~(I 365 74. Consider an angle in standard position with r = 10 cm. as shown in the figure. Write a short paragraph describing the changes in the magnitudes of T, Y. sin cos and tan as 0 increases continuously from Oc to 90°, e, y 32)J. where 1 is the time in days with I = I corresponding to January I, Find the average daily temperature on the fol lowing days, (b) July 4 (t = 185) (a) January I (C) October 18 (I = 291) (x, y) __ ~=- __ ______ x ~ FIGURE FOR 74 e. e. 150 CHAPTER 2 TRIGONOMETRY ,o... . . .:,;.,:;. GRAPHS OF SINE AND COSINE FUNCTIONS hsk Sine and Cosine Curves I Key Points on Basic Sine and Cosine ne~ Amplitude and Period of Sine and Cosine Curves / Translations .f' :;ine and Cosine Curves Basic Sine and Cosine Curves f, ~. In this section you will study techniques for sketching the graphs of the sine and cosine functions. The graph of the sine function is called a sine curve. In Figure 2.40, the solid portion of the graph represents one period of the func tion and is called one cycle of the sine curve. The gray portion of the graph indicates that the basic sine wave repeats indefinitely to the right and left. The graph of the cosine function is shown in Figure 2.41. To produce these graphs with a graphing utility, make sure you have set the mode to radians. Recall from Section 2.2 that the domain of the sine and cosine functions is the set of all real numbers. Moreover, the range of each function is the interval [-1. I]. and each function has a period of 27T. Do you see how this information is consistent with the basic graphs given in Figures 2.40 and 2.41 ? Note from Figures 2.40 and 2.41 that the sine graph is symmetric with respect to the origin, whereas the cosine graph is symmetric with respect to the y-axis. These properties of symmetry follow from the fact that the sine func tion is odd whereas the cosine function is even. Range: -1 y.:51 FIGURE 2.40 Period: 21T 2.41 Period: 21T FIGURE SECTION 2.5 GRAPHS OF SINE AND COSINE FUNCTIONS 151 Key Points on Basic Sine and Cosine Curves To construct the graphs of the basic sine and cosine functions by hand, it helps to note five key points in one period of each graph: the intercepts, maximum points. and minimum points. For the sine function, the key points are J ~ICr,'cpl \liJ,ximC:111 ~ I: l (' r,-' L' \' t \! j :)I1',';~lln (0,0), (~, 1). (7T. 0), e -I). \ 7T 2 ' (27T,0). For the cosine function. the key points are \!iJ,il11111ll \IlIli L1~1!li Inlc;'L'ef'! (~2' °), (0, I L (7T, - J), 1:-1 1 ;f'.. .-,:':.'! e;, Q). .1" (27T, J ). Note how the x-coordinates of these points divide the period of sin x and cos x into four equal parts, as indicated in Figure 2.42. }' }' I t Maximum Intercept Minimum Intercept Maximum Inte cept Maximum Intercept Minimum Intercept Full period Three-quarter Half period period Period: 211' Period: 211' FIGURE 2.42 EXAMPLE 1 Tecnnu(oiJlJ Note _______ Sketch the graph of y When using a graphing utility to graph trigonometric functions. pay special attention to the viewing rectangle you use. For instance. try graphing \' "" [sin(30x 1]/30 on the standard viewing rectangle. What do you observe') Use the zoom feature to find a rectangle that displavs a good view of the graph. 2 sin x on the interval l 7T, 47T]. SOLUTION The y-values for the key points of y = 2 sin x have twice the magnitude of those of y sin x. Thus, the key points for y 2 sin x are (0.0), (~'2). (7T.O), e;.-2). and (27T,O). 152 CHAPTER 2 TRIGONOMETRY By connecting these key points with a smooth curve and extending the curve in both directions over the interval [ 7T, 47T], you obtain the graph shown in Figure 2.43. Try using a graphing utility to check this result. FIGURE 2.43 Amplitude and Period of Sine and Cosine Curves DISCOVERY U~e a to graph r a ,in .\, where a 0.5. I. 2. and 3. ( a lie\\ in \\ hiel; -6.3 --+ } -+.) How doe, the I alue of a affect the graph '.' In the rest of this section, you will study the graphic effect of each of the constants a, b, c, and d in equations of the forms y = d + c) a sin(bx and y = d + a cos(bx A quick review of the transformations studied in Section 1.8 should help in this investigation. a sin x acts as a vertical stretch or vertical The constant factor a in y shrink of the basic sine curve, as shown in Example I. (If Ia I > I, the basic sine curve is stretched, and if •a 1 < I, the basic sine curve is shrunk.) The a sin x ranges between -a and a instead of result is that the graph of y between - 1 and 1. The absolute value of a is the amplitude of the function y = a sin x. DEFINITION OF AMPLITUDE OF SINE AND COSINE CURVES The amplitude of y and is given by Amplitude 1 a I· a sin x and y = a cos x is the largest value of y SECTION 2.5 EXAMPLE:2 153 GRAPHS OF SINE AND COSINE FUNCTIONS \erticai Shrink;ng <end Slretching Sketch the graphs of y = ~ cos x and y = 3 cos x. Because the amplitude of y = ~ cos x is minimum value is -!. For one cycle, 0 The amplitude of y (0,3), (~,o), 4, the maximum value is ~ and the :$ x :$ 277', the key points are 3 cos x is 3, and the key points are (77', 3), c;,o). and (277',3). The graphs of these two functions are shown in Figure 2.44. Try using a graphing utility to check this result. y 3 cos x Amplitude Determines Vertical Stretch or Shrink FIGURE 2.44 Reflection in the FIGURE 2.45 ,X- Axis "",.,,,, I;"D ~~-:-; You know from Section 1.8 that the graph of y -f(x) is a reHection (in the x-axis) of the graph of y = f(x). For instance, the graph of y = 3 cos x is a reflection of the graph of y = 3 cos x, as shown in Figure 2.45. Because y = a sin x completes one cycle from x 0 to X 277', it follows that y a sin bx completes one cycle from bx = 0 to bx 277'. This implies that y = a sin bx completes one cycle from x = 0 to x = 277'/b. 154 CHAPTER 2 TRIGONOMETR~ SCOVERY 10 draw th" " !i.5. i. "nd :: ; L~e a , c·,:.ngi;.' in \\hi;.;h 0 X s; 6.3 ,,,,J ~:: Y s; 2.1 Hov., does the •.. 'Lie:: of h affect the graph) PERIOD OF SINE AND COSINE FUNCTIONS Let b be a positive real number. The period of y y = a cos bx is 271'/b. a sin bx and Note that if 0 < b < I, the period of y = a sin bx is greater than 271' and represents a horizontal stretching of the graph of y a sin x. Similarly, if b > I, the period of y a sin bx is less than 271' and represents a horizontal shrinking of the graph of y = a sin x. If b is negative, use the identities sin( - x) = sin x and cos( - x) = cos x to rewrite the function. EXA M PLE:3 Horizontal Stretching and Shrinking Sketch the graph of y . X SIn 2' SOLUTION The amplitude is 1. Moreover, because b L the period is 271' /(4) 471'. By dividing the period-interval [0,471'] into four equal parts with the values 71',271', and 371', you obtain the following key points on the graph. (0,0), (71', I), (271', 0), (371', -I), and (471',0) The graph is shown in Figure 2.46. Use a graphing utility to check this result. y Y =\ SInX I - t ······r ~y_.sm. ~2" ....... " ............ "...' . ~ --rr X . , / ............ .. ;r. ". 17 217 ....... ... . _ _ _~ '( 141T ... . I J Period: 41T FIGURE 2.46 ••••••• • 111 SECTION 2.S EXAMPLE 4 GRAPHS OF SINE AND COSINE FUNCTIONS 155 Horizontal Stretching ,,:'d ''; sin 3x. Sketch the graph of y SOLUTION The amplitude is I, and because b = 3, the period is y= 27T 27T b 3 Dividing the period-interval [a, 27T /3] into four equal parts, you obtain the following key points on the graph. The graph is shown in Figure 2.47. Use a graphing utility to check this result. FIGURE 2.47 EXAMPLE 5 l'sing a Graphing Uiiit) Use a graphing utility to graph y = 2 cos O.4x and y 2 cos 4x on the same screen. Determine the period of each function from its graph. SOLUTION 3 '\ r~ OH!~~~~4Y~~~~44~~~18 I \ 27T v II 3 i y = 2 cos O.4x I FIGURE 2.48 The standard viewing rectangle is not appropriate for graphing most sine and cosine functions. You should choose a viewing rectangle that accommodates the amplitudes of the functions as well as any stretching or shrinking that may occur. For these graphs, a y-scale ranging between - 3 and 3 will accommo date the amplitude of 2. Figure 2.48 shows the two graphs and an appropriate 2 cos viewing rectangle. From the graph. you can estimate the period of y OAx to be about 16 and the period of )' = 2 cos 4x to be about 1.5. Alge braically, the period of y 2 cos O.4x is 0.4 i 57T 15.71, and the period of y 27T 7T 4 2 1.57. 2 cos 4x is 156 CHAPTER:2 TRIGONOMETRY Translations of Sine and Cosine Curves The constant c in the general equations y -= a sin(bx - c) and y a cos(bx c) creates a horizontal shift of the basic sine and cosine curves. The graph of y a sin(bx c) completes one cycle from bx - c 0 to bx - c = 27T. By solving for x in the inequality 0 :::; bx - c :::; 27T, you can find the interval for one cycle to be c b -'- C 27T - + b b' This implies that the period of y = a sin(bx - c) is 27T/b, and the graph of y = a sin bx is shifted by an amount c lb. The number c /b is the phase shift. ~---~------~-----.-.- .. , ~-~.-- 1Il ,Jnt\\ the \\ here (L a hid: 6.:;"nu--2::;Y !-l(I\\ Joe, lh~ \alllc: of ::.i affect GRAPHS OF THE SINE AND COSINE FUNCTIONS The graphs of y a sin(bx c) and y lowing characteristics. (Assume b > 0.) a cos(bx c) have the fol Amplitude = ia I Period = 27T /b The left and right endpoints corresponding to a one-cycle interval of the graphs can be determined by solving the equations bx - c = 0 and bx - c 27T. SECTION 2.5 157 GRAPHS OF SINE AND COSINE FUNCTIONS Horizontal Shift EXAMPLE 6 Describe the graph of y = ~ sin (x - ~). SOLUTION Because a = ~ and b the inequality o 17 -:s 3 17 X - - :s 3 <: x - J, the amplitude is ~ and the period is 217. By solving 217 717 3 ' you can see that the interval [17/3, 717/3] corresponds to one cycle of the graph. Dividing this interval into four equal parts produces the following key points. I) 1117 6 ,- 2' C;,o) and ....... .. The graph is shown in Figure 2.49. FIGURE 2.49 The final type of transformation is the vertical shift caused by the constant d in the equations y = d + a sin(bx - c) and y d + a cos(bx c). The shift is d units upward for d > 0 and downward for d < O. In other words, the graph oscillates about the horizontal line y = d instead of the x-axis. 7 EXAMPLE Vertical Shift Describe the graph of y =2+ 3 sin 2x. SOLUTION The amplitude is 3 and the period is 17. The key points over the interval [0, 17] are ··2 (0,2), Period: FIGURE 2.50 11' (~,5). (~'2). C;,-l). The graph is shown in Figure 2.50. and (17,2). ....... .. 158 CHAPTER 2 TRIGONOMETRY ~~:'.~:.7'::.~ Finding an Equation for a Given Graph Find the amplitude, period, and phase shift for the sine function whose graph is shown in Figure 2.51. Write an equation for this graph. SOLUTION The amplitude for this sine curve is 2. The period is 21T. and there is a right phase shift of 1T /2. Thus, you can write the following equation . A Shifted Sine Curve .\' 2 Sin( x - ~) FIGURE 2.51 .... '" .'" .. Try finding a cosine function with the same graph. A SINE SHOW If you have a programmable calculator, try running the following "sine-show program." This program simultaneously draws the unit circle and the correspond· ing points on the sine curve. After the circle and sine curve are drawn, you can connect points on the unit circle with their corresponding points on the sine curve by repeatedly pressing ENTER. After the program is complete, the screen should look like that shown in the figure. (This program is for the TI-81. Pro gram steps for other calculators are given in Appendix 8.) :Rad :ClrDraw :Line(-1.25, :DispGraph .19,-:.25, :O~N :Param :Simul ;-2.2S-Xmin ; 7T/2-X::,ax :3-Xscl . 9~Ymi:; :1.:"9-Ymax ::~Yscl :O-Tmin : 6 . 3-T::.ax :. 5-Tstep :"-1.2 J.ccs T"-X :"51::1 T"-Y i : "sin T"-Y , : Lbl 1 :N.L1-N : N7T/6. S-T : 1.25+cos T-A :sin ':'-8 : Line(A, B, C, 3) :Pat;se : f N=12 :Goto 2 :Goto 1 :Lbl 2 : Fc;nctiorl :Seqt;e:1ce :End .19) SECTION 2.5 WARM-UP GRAPHS OF SINE AND COSINE FUNCTIONS 159 The following warm-up exercises involve skills that were covered in earlier sections_ You will use these skills in the exercise set for this section. In Exercises I and 2, simplify the expression. 27T 1. 2. 27T In Exercise, 3-6, solve for x. 7T 3. 2.. 5. 3 + 6" 37TX 7T 4. 2.. 0 0 6. 27T 3 37TX + 6" 2" In Exercises 7-10. evaluate the trigonometric function without using a calculator. 7. sin " 2 8. sin 7T " 10. cos2 9. cos 0 SECTION 2.5 . EXERCISES ...........•.......................................................•..........................•......................... In Exercises 1-10. determine the period and amplitude of the function. Then describe the viewing rectangle for Exercises 1-6. 1. y 3. \' 2 sin 2x 3 x - cos 2 2 2. \ 4. \' 5. \' I ~ . SlO ITr 3 cos 3x 2 sin 3 7. ,\ 3 sin lOx 8. 5 .\ - cos 2 4 1/ 9. r :1 sin 47T.\ Tr\" 10. \ 3 cm, 10 6. y 5 - cos ') 7iX 2 160 CHAPTER 2 TRIGONOMETRY In Exercises II 16. describe the relationship between the graphs of f and g, In Exercises 29-42. sketch the graph of the function. (Include two full periods. and use a graphing utility to verify your result.) 11. fed 29. \' = -2 sin ger) 13. fix) !i'(x) 15. g(x) sin x sinCr 12. fix) = cos x g(r) = cos(x -'- 17) 17) 31. y = cos 14. j(x) = sin 3x g(rl = sin(-3xl cos 2" -cos 2x sin x 2 -'- sin x 16. fix) g(r) = cos 4x = _"! ... cos 35. .r 217X ="J - - 37. r In Exercises 17-24. sketch the graphs of the two functions on the same coordinate plane, (Include two full periods. and use a graphing utility to verify your resu)L) 17. fix) g(rl 18. fix) g(xl 19. fix) g(xl 20. fix) g(x) 2 22. g(x) 23. fix) g(x) 24. I(x) g(x) 217X sin3 (\x - "4rr) 39. v= 3 cos(x + ) 41. \' cos I 3 cos 4x \' 32. \' 34. \' 3 - sin 2 6 36. \ . 2 cos x 38. -2 SIIl(, . 3 I , \' 4 42. v 60171' l7X 10 cos 40. 17) l7X 17) COS(' 4 m I 5 cos 3 x In Exercises 43-54. use a graphing utility to graph the function, (Include two full periods,) .3 cos x cos x 2 cos 2, -cos 4x 21. g(x) Sin 2 sin x 4 sin x sin x sin 217X -sin3 33. y 4x 30. 6.1 sin 3 x 2 sin 2 4 sin l7X 4 sin l7X 2 cos x 2 cos (x -cos x -cos(x x :: .3 43. y = 3 cos(x 45. = -:: \' 3 17) - .3 + 44. cos (x- - -17) \2 4/ 47. y = 49. \' = cos( 2m 46. r -2 sin(4x + 71') - ~) \' + 4 cosC, + ~1 + 4/ 3 cos(6x ~ " C 48. \' -4 sin ~x - 50. \' 3 cos 52. \' 5 sin( 17 17) 54. \' 17) 'r3r) (rrx 2" + 2'17) \ 51. y '" -01 sin ( -17X + 171 , )0 / 53. \' = 5 cos( 17 - 2x) + 2 2x) 4 - 2 + 10 sin 120m 17) In Exercises 55-58. use the graph of the trigonometric function to find all real numbers x in the interval (- 217. 271'J that give the In Exercises 25-28. sketch the graphs of I and g on the same coordinate axes and show that ((xl = g(rl for all x, (Include two full periods. and use a graphing utility to verify your result,) 26. fix) = sin x sin x 25. ~) g(x) g(x) = -cos(, .:., cos x 27. g(x) f sin\ x \ 7r\ -I! specified function value, Function 55. sin x -~ 56. cos x -) 57. co, x 28. fix) g(x) cos x -cosc, 17) FIlIlC/lrll1 58. sin " 2 \/3 2 Value SECTION 2.5 In Exercises 59 and 60. find a and d for the function j(x) = a cos x -+- d so that the graph of j matches the GRAPHS OF SINE AND COSINE FUNCTIONS 161 69. Piallo TUlling When tuning a piano, a technician strikes a tuning fork for the A above middle C and sets up wave motion that can be approximated by v = 0.00 I sin 880m. where t is the time in seconds. (a) What is the period p of this function,) (b) The frequency j is by j = 1/p. What is the fre quency of this note') (c) Use a graphing utility to graph this function. In Exercises 61-64, find G, b. and c so that the graph of the function matches the graph in the 61. v = a sin{bx c) 62. y 70. Blood Pressure P = a cos(hx c) 64. v = 100 - 20 cos 5m 3 approximates the blood pressure P in millimeters of mer cury at time I in seconds for a person at rest. (a) Find the period of the function. (b) Find the number of heartbeats per minute. (c) Use a graphing utility to graph the pressure function. - 1.5 63. y The function a sin(hx - c) = a sin(bx c) Sales [n Exercises 71 and 72. use a graphing utility to graph the sales function over I year where S is sales in thousands of units and t is the time in months. with t 1 corresponding to January. Use the graph to determine the month of maximum sales and the month of minimum sales. 2e' and 5 cos x. Approximate any points of intersection of the graphs in the interval [-7T, 7T]. 65. Use a graphing utility to graph the functionsj(x) 71. S 22.3 - 3.4 cos g (xj 66. Use a graphing utility to determine the smallest illteger value of a such that the graphs of j(x) = 2 In x and g (x) = a cos x intersect more than once. 67. Respiratory Cvcle For a person at rest, the velocity v (in liters per second) of air flow during a respiratory cycle is 72. S = 74.50 'iT1 6 'iTt 43.75 sin - 6 In Exercises 73-76. describe the relationship between the graphs of the functions j and g. 73. 74. 75. 76. 7Tt \' = 0.85 sin-. 3 where t is the time in seconds. (Inhalation occurs when I' > O. and exhalation occurs when v < 0.) (a) Find the time for one full respiratory (bl Find the number of cycles per minute. (el Use a graphing utility to graph the velocity function. 68. Respirator>: Cvcle After exercising for a few minutes. a person has a respiratory cycle for which the velocity of air flow is approximated by \. ~ O~'2.57 7Tt 1.75 sin - . 2 Use this model to repeat Exercise 67. -2 -2 162 CHAPTER 2 TRIGONOMETRY 2.6 OTHER TRIGONOMETRIC GRAPHS •............................................................•.......................... Graphs of Tangent and Cotangent Functions I Graphs of the Reciprocal Functions I graphs of Combinations of Trigonometric Functions I Combinations of Algebraic and Trigonometric Functions I Damped Trigonometric Graphs Graphs of Tangent and Cotangent Functions Recall from Section 2.2 that the tangent functions is odd. That is, tan( - x) -tan x. Consequently, the graph of y = tan x is symmetric with respect to the origin. From the identity tan x sin x/cos x, you know that the tangent is undefined when cos x O. Two such values are x 1T/2 = ± 1.5708. I 7T x - tan x undef. 2 1.57 -1255.8 : 1.5 -14.1 -·1 o -1.56 0 I 1 1.5 1.57 1.56 14.1 1255.8 I~ ......... Period: 1T Domain: all 1T x ;/= "2 ± n1T Range: (-00, 00) Venical asymptotes: x = 1T "2 ± mT _ As indicated in the table, tan x increases without bound as x approaches 1T/2 from the left, and decreases without bound as x approaches 1T/2 from the right. Thus. the graph of y tan x has vertical asymptotes at x = 1T/2 and 1T/2. as shown in Figure 2.52. Moreover. because the period of the tangent function is 1T, vertical asymptotes also occur when x ± n1T. The do main of the tangent function is the set of all real numbers other than x = 1T/2 ± n1T. and the range is the set of all real numbers. Sketching the graph of a function with the form Y a tan(bx - c) is similar to sketching the graph of y = a sin(hx c) in that you locate key points which identify the intercepts and asymptotes. Two consecutive asymp totes can be found by solving the equations hx - c = FIGURE 2.52 I undef. 2 and bx 1T C =-. 2 The midpoint between two consecutive asymptotes is an x-intercept of the graph. After plotting the asymptotes and the x-intercept, plot a few additional points between the two asymptotes and sketch one cycle. Finally. sketch one or two additional cycles to the left and right. REMARK The period of the function r (I lan(iJx c) j, the distance between two •••••• consecutive asymptotes. The amplitude of a tangent function is not defined. - SECTION 2.6 OTHER TRIGONOMETRIC GRAPHS 163 Sketching the Graph of a Tangent Function EXAMPLE 1 x tan -, 2 Sketch the graph of y SOLUTION By solving the inequality -- < I < 7T 222 i---- l I X 7T I I 1 I I I I _7>-,,+-¥-t---H-I--+-++-;HH......)(· 3 I 1 I L I I '1 I - ....... _ [an 2. I", . . . .... y - X 1 1 -3 . you can see that two consecutive asymptotes occur at x = 7T and x = 7T. Between these two asymptotes, plot a few points including the x-intercept, as shown in the table. Three cycles of the graph are shown in Figure 2.53. Use a graphing utility to confirm this result. = - X .. . x tan 2 7r 7r 0 2 2 0 1 1 ......... ... FIGURE 2.53 EXAMPLE Sketching the Graph of a Tangent Function 2 -3 tan lx. Sketch the graph of y SOLUTION By solving the inequality 7T 7T --<lx< 2 2 7T 7T -<x< 4 4 you can see that two consecutive asymptotes occur at x = - 7T/4 and x = 7T/4. Between these two asymptotes, plot a few points as shown in the table, and complete one cycle. Four cycles of the graph are shown in Figure 2.54. Use a graphing utility to confirm this result. x FIGURE 2.54 -3 tan 2x 7r 7r -8 0 - 3 0 -3 8 ............. .. 164 CHAPTER 2 TRIGONOMETRY Period: 1T Domain: all x Range: (-00, 00) By comparing the graphs graph of *' n1T Vertical asymptotes: x n1T y y =a In Examples I and 2, you can see that the tan(bx + c) is increasing between consecutive vertical asymptotes if a > 0 and decreasing between consecutive vertical asymptotes if a < O. In other words, the graph for a < 0 is a reflection of the graph for a > 0, The graph of the cotangent function is similar to the graph of the tangent function. It also has a period of 11'. However, from the identity y . FIGURE 2.55 = cot x cos x sin x =- you can see that the cotangent function has vertical asymptotes at x = n11' (because sin x is zero at these x-values). The graph of the cotangent function is shown in Figure 2.55. Technofo91J Note _ _ _ _ _ __ You can use the tangent function on your graphing utility to obtain the graph of the cotangent function. For example. to graph the function \. = :: cot (xI3) from Ex<:mple 3. let \'1 ::/tanlxI3). If you select the viewing rectangle -9 :;; x :;; 18 and -6 :;; Y :;; 6, you should obtain a graph similar to that of Figure 2,56, EXAMPLE 3 Sketching the Graph of a Cotangent Function x Sketch the graph of y 2 cot '3' SOLUTION To locate two consecutive vertical asymptotes of the graph, you can solve the following inequality. x 0<-<11' 3 0< x < 311' Then, between these two asymptotes, plot the points shown in the following table, and complete one cycle of the graph. (Note that the period is 311', the distance between consecutive asymptotes.) Three cycles of the graph are shown in Figure 2.56. 311' X 2 cot FIGURE 2.56 i x 3 37r 2 4 911' 4 - I 2 0 i -2 " '" '" " " " " " " SECTION 2.6 OTHER TRIGONOMETRIC GRAPHS 165 Graphs of the Reciprocal Functions fJ SCOVERY Use a graphing utility to graph the functions}" = sin x and )'2 = csc.t "" l/sin.t on ttie same viewing rectangle. How are the graphs related? What happens to the graph of the cosecant function as x approaches the zeros of the sine function? Similarly, graphy, = cos.t and Jz = sec x = I/cos x on the same viewing rectangle. How are these functions related? Y The graphs of the two remaining trigonometric functions can be obtained from the graphs of the sine and cosine functions using the reciprocal identities esc x sin x and I sec x = - - . cos x For instance, at a given value for x, the y-coordinate for sec x is the reciprocal of the y-coordinate for cos x. Of course, when cos x = 0, the reciprocal does not exist. Near such values for x, the behavior of the secant function is similar to that of the tangent function. In other words, the graphs of tan x (sin x)/(cos x) and sec x = l/(cos x) have vertical asymptotes at x (?T/2) + n?T, because the cosine is zero at these x-values. Similarly, cot x (cos x)/(sin x) and csc x = I/{sin x) have vertical asymptotes where 0, that is, at x = n?T. sin x To sketch the graph of a secant or cosecant function, we suggest that you first make a sketch of its reciprocal function. For instance, to sketch the graph of y csc x, first sketch the graph of y = sin x. Then take reciprocals of the y-coordinates to obtain points on the graph of y = csc x. You can use this procedure to obtain the graphs shown in Figure 2.57. Cosecant: local minimum Period: 217 Domain: all x 1'117 Range: all y not in (-1, 1) Vertical asymptotes: x = 1'117 Symmetry: origin Period: 217 . 17 Domam: all x*-"2 + 1'117 Range: all y not in (-I, 1) FIGURE 2.57 Symmetry: y-axis *' Cosecant: local maximum FIGURE 2.58 Vertical asymptotes: x = 17 "2 + n17 In comparing the graphs of the secant and cosecant functions with those of the sine and cosine functions, note that the "hills" and "valleys" are inter changed. For example, a hill (or maximum point) on the sine curve corre sponds to a valley (a local minimum) on the cosecant curve. Similarly, a valley (or minimum point) on the sine curve corresponds to a hill (a local maximum) on the cosecant curve, as shown in Figure 2.58. EXAMPLE 4 Graphing a Cosecant Function Use a graphing utility y 2 Sin(x to +~) graph and \' = 2 csc(x +~). 166 CHAPTER Z TRIGONOMETRY SOLUTION The two graphs are shown in Figure 2.59. Note how the "hills" and "valleys" of each graph are related. For the function y 2 sin[x + (7T/4)], the ampli tude is 2 and the period is 27T. One cycle of the sine function corresponds to the interval from x -7T/4 to x 77T/4, Because the sine function is zero at the endpoints of this interval, the corresponding cosecant function y = 2 csc(x + ~) has vertical asymptotes at x = -7T/4. x FIGURE = 37T/4. and 77T/4. ....... .. 2.59 In Figure 2.60, we summarize the graphs, domains. ranges, and periods of the six basic trigonometric functions. Y 6 Domain: all Teals Range: [-I, 1] Period: 21T y Domain: all Teals Range: [-I. 1] Period: 21T 4 -+--J-.--f----:f--f---E-+_x *' -¥ :U I : -2' I " I -3 Penod: 21T 4 I ,~ I :u: W Ra~ge: (-00, , 2 '(\W¥': : : I :=cscx=~1 I + mr -I] and [I, 00) Domain: all x*'2 y . 3: : I ' : 1 217' 7T '-Ji~"'I! -2: : : -3 ; , I I \ • x 1 I ly=secx=~1 Graphs of the Six Trigonometric Functions FIGURE 2.60 *' 1T Domain: all x n1T Range: (-00. -1] and [I, 00) Period: 21T Y y Domain: all x n1T Range: (-<>0, 00) Period: 1T \ x SECTION 2.6 OTHER TRIGONOMETRIC GRAPHS 167 Graphs of Combinations of Trigonometric Functions Sums. differences. products. and quotients of periodic functions are also peri odic. The period of the combined function is the least common multiple of the periods of the component functions. This period is important for determining an appropriate viewing rectangle for a graphing utility. Finding the Period and Relative Extrema of a Function EXAMPLE 5 Graph y = sin x - cos 2x. Find the period and the relative minimums and maximums of this function. SOLUTION The period of sin x is 217 and the period of cos 2x is 17. Thus. the period of the given function is 217, because 217 is the least common multiple of 217 and 17. This conclusion is further reinforced by graphing the function, as shown in Figure 2.61. From the graph, it appears that the function has two relative maximums and two relative minimums in each complete cycle. For instance. between 0 and 217, the graph appears to have relative maximums when x 17/2 = 1.57 and when x = 317/2 = 4.71. Using the zoom and trace features of the graphing utility. you can find that the relative minimums occur when x = 3.40 and when x 6.03. = FIGURE 2.61 Relative maximums: (1.57,2.00) (4.71.0.00) Relative minimums: (3.40, (6.03, -l.J3) 1.13) ....... .. Finding the Period and Range of a Function EXAMPLE 6 Graph y = 2 sin 6x sin 4x. Find the period and range of this function. SOLUTION _ 3 14 ~-+-,-.,..f-M--+-'-+--f---.Jl-'-+--\-;f--'-+---I3.14 The period of 2 sin 6x is 17/3 and the period of sin 4x is 17/2. Thus. the period of the gi ven function is 17, because 17 is the least common multiple of 17/3 and 17/2. This conclusion is further reinforced by graphing the function. as shown in Figure 2.62. In the interval from 0 to 17, the maximum y-value occurs when x = 0.29 and y 2.89. The minimum value in this interval occurs when x = 2.86 and y = -2.89. Thus, the range of the function is approximated by = FIGURE 2.62 -2.89 :S Y :S 2.89. Range ....... .. 168 CHAPTER 2 TRIGONOMETRY Combinations of Algebraic and Trigonometric Functions Functions that are combinations of algebraic and trigonometric functions are not, in general. periodic. EXAMPLE 7 The Graph of a Nonperiodic Function Graph y x + cos x. Find the domain and range of the function. Approxi mate any zeros of the graph. SOLUTION The graph of the function is shown in Figure 2.63. Notice that even though the function is not periodic, it does have a pattern that repeats. Notice that the x. Both the domain and graph of y = x + cos x oscillates about the line y range of the function are the set of all real numbers. Using the zoom feature, x + cos x is approximately x = -0.739. you can find that the zero of y - 12.57 \'------'---'--.L-+t---'--'---'---'\ 12.57 FIGURE EXAMPLE 8 Graph y ....... .. 2.63 A :Function Involving Absolute Value Isin x I, and find the domain and range of the function. SOLUTION FIGURE 2.64 The domain of the function is the set of all real numbers. The graph of the function is shown in Figure 2.64. Notice that the minimum value of the function is 0 and the maximum value is 1. Thus, the range of the function is ....... .. given by 0 :s y :s 1. SECTION 2.6 OTHER TRIGONOMETRIC GRAPHS 169 Damped Trigonometric Graphs A product of two functions can be graphed using properties of the individual functions involved. For instance, consider the function J(x) = x sin x as the product of the functions y x and y sin x. Using properties of absolute value and the fact that Isin x I ::;; J. you obtain 0 ::;; Ix II sin x I ::;; Ix I. Consequently, I x I ::;; x sin x ::;; I x I' which means that the graph of J(x) and y x. Furthermore, since J(x) J(x) FIGURE 2.65 = x sin x =::::x x sin x = 0 at at x x sin x lies between the lines y 1T + 2 -x n7r x = n7r, the graph of J touches the line y - x or the line \ = x at x = (1T /2) + WTr and has x-intercepts at x n1T. The graph of J. together with y x and y - x, is shown in Figure 2.65. In the functionJ(x) = x sin x, the factor x is called the damping factor. By changing the damping factor, you can change the graph significantly. For example, look in Figure 2.66 at the graphs of .v = -x sin x and y e- x sin 3x. FIGURE 2.66 170 CHAPTER 2 TRIGONOMETRY Damped Cosine Wave EXAMPLE 9 Graph f(x) = T ,r;: cos x. SOLUTION A graph of f{x) = 2 -x12 cos x is shown in Figure 2.67. To analyze this fuction further, consider f(x) as the product of the two functions y = 2-,1" and cos x, y each of which has the set of real numbers as its domain. For any real number x, you know that 2 - ,/2 0 and cos x I :S I. Therefore, I 2 - xj2 I I cos x I :S 2-,12, which means that cos x :S 2 Furthermore, since f(x) = rx/2 cos x =2- at x = n1T and o 1T at X= the graph off touches the curve y = r and has intercepts at x (1T/2) + fl1T. FIGURE 2.67 2 + fl1T, ,/2 or the curve y ... ,. .. ... Use a graphing utility to graph the following functions for varying values of a, b, e, and d. GRAPHING UTILITIES y y y d d d + a sin(bx + e) + a tan(bx + e) + a sec(bx + e) y y y = d d d + + + a cos(bx a cot(bx a csc(bx + e) + c) + e) In a paper or a discussion, summarize the effects of the constants a. b, e, and d in these graphs. SECTION 2.6 WARM-UP 171 OTHER TRIGONOMETRIC GRAPHS ni~foilow4ngwarni-up exercises iravolVe ski,Dstliat)'f~co~ered in. earliersediODS' YO\!, will:~ tii~skms in the exercise setforthii'Sei:tl~~'. -.. - ; : - : . . ' c : In Exerdsesl-4. 1. f(x) • findtllex~values in the interVal (0; 2~] for~iCb f(x) sin x ,- 2. /{x) = cos x '< 3. f{x} is -l.O, or L = sin2x , In Exercises·S':'S, graph' the function. 5. Y j xl 7. y = sin TrX = cos 2x 8. y In Exercises 9 and 10, evaluate f(x) when x = 0, Tr/6, Tr/4, Tr/3, and Tr/2. 9. f(x) 2.6 . SECTION x cos x 10. f{x) x + sin x EXERCISES ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• e In Exercises 1-8, match the trigonometric function with its graph and describe the viewing rectangle. [The graphs are la beled (a), (b), (cl, (d), eel, (f), (g), and (h).] 1. Y sec 2x 'l 2. r "" tan 3x 3. \' = tan x - 2 4. Y 2 csc , ••••••••••••••••••••••• ~ •••••••••••••••••••••••• j j( x 2 5. \' cot 7TX 6. Y = ~ sec 7TX -sec x 7. Y 8. y = - 2 esc 2 = I (b) (a) . ., , ,-"--~---- r-~~~~--~~ II )1 t. . , -,I f ,, ,, 'n' 1 1 llli ,I , . , 172 CHAPTER Z TRIGONOMETRY In Exercises 9-30, sketch the graph of the function. (Include two full periods, and use a graphing utility to verify your result.) t 9. y tan x 11. y = tan 2x - ~ sec x ]5. y = -sec 7TX 13. Y 17. y = sec x 19. Y csc 2 21. y 23. Y 7TX - cot x x 3 7TX . 24. Y 7TX 2 = ~ tan x ~ cSC(X +~) 28. y = sec( 7T - x) 30. Y 2cot(X +~) In Exercises 31-40, use a graphing utility to graph the function. (Include two full periods.) x tan 3 31. Y 32. Y 33. y = - 2 sec 4x 35. \' == tan(x 39. ~) ~ cot ( x - ~) 37. Y v = 2 sec(2x - 7T) 34. -" = sec 36. Y 38. Y 49. v 2 51. Y 53. .\' =4 - 2 . x 2: SID 2 cos 50. y -3 + cos x 52. \' =5 .l sin 27TX 54.) = 2 + tan 7TX 7TX I ..;.. csc x 7TX -csc(4x - 7T) ~ sec( r.; + ~) 40. Y = 0, I tan( r; -'- ~) In Exercises 55-60, use a graphing utility to graph the func tion. Determine the period of the function and approximate any relative minimums and maximums of the function through one period. 55. y sin x 1- cos x 57. I(x) 2 sin x + sin 2x x 59. g(x) = cos x - cos '2 56. \' cos x - cos 2x 58. I(x) = 2 sin x + cos 2x I x 60. g(x) sin x - - SIn - 2 2 Function Value 41. tan x 1 42. cot x -\13 43. sec x 44. esc x -2 V2 In Exercises 45 and 46. use the graph of the function to deter· mine whether the function is even, odd, or neither. 45./(x) = sec x In Exercises 49-54, sketch the graph of the function. (Use a graphing utility to verify your result.) -tan 2x In Exercises 41-44, use the graph of the trigonometric function to find all real numbers x in the interval [-27T. 27T] that give the specified function value. Flmcrion 48. Consider the functions I(x) = tan(7Tx/2) and g(x) 1sec(7Tx/2) over the interval (- L I J. (a) Use a graphing utility to graph I and g in the same viewing rectangle, (b) Approximate the interval where I < g. 26. -" = sec(x + 7T) 4 27. Y = CSc(7T - x) 29. Y 2 22. \' = 3 cot 2 tan - esc 20. Y = ~ sec 2x 25. Y 10. \' ~ tan x 12. y = 3 tan 7TX 14. \' t sec x 16. y 2 sec 4x 18. \' = -2 sec 4x 47. Consider the functions/(x) = 2 sin x and = ~ esc x over the interval (0. 7Tl. (a) Use a graphing utility to graph I and g in the same viewing rectangle. (b) Approximate the interval where I > g, (c) Describe the behavior of each of the functions as x approaches 7T. How is the behavior of g related to the behavior of I as x approaches 7T~ 46. I(x} = tan x In Exercises 61-64, use a graphing utility to graph the function through three periods. 61. hex) sin x + ~ sin 5x 62. hex} = cos x ~ cos 2x 63.), - 3 + cos x + :; sin 2x 64. .\' = sin TTX .,- sin 7TX 2 SECTION 2.6 In Exercises 65-72, use a graphing utility (0 graph the given function and the algebraic component of the function in the same viewing rectangle. Notice that the graph of the function oscillates about the graph of the algebraic component. x + sin x 65. Y 67. f(x) 2 cos x 66. y tx - . 1Tl 69. get) 71. y 6S. f(x) 1+ sm x2 =- 8 2x - sin x 1TX + sin - 72. y 2 1TS - - sin 2 4 S = 70. h(s) 2 2 x2 4 - 16 = 2- x / 4 cos 1TX = e- r2 !2 sin x Function 92. f(x) (2 Interval sin t = [0. 41T ] ~ sin 3x + ~ sin 5x 93. f(x) = sin x 94. f(x) [0, 21T] sin x ~2 - ~(cos 1TX 1T + I cos 31TX) 9 [0, 1T] [0. 2] + 4 cos 1TX In Exercises 73-76, use a graphing utility to graph the function and the damping factor of the function in the same viewing rectangle. Describe the behavior of the function as x increases without bound. 73. f(x) 75. g(x) In Exercises 9\-94, use a graphing utility to graph the function over the specified interval. 91. f(t) = + cos x x 173 OTHER TRIGONOMETRIC GRAPHS 74. f(t) e-' cos t 76. h(l) 2-,2/ 4 sin I 95. Distance A plane flying at an altitude of 6 miles over level ground will pass directly over a radar antenna (see figure). Let d be the ground distance from the antenna to the point directly under the plane and let x be the angle of elevation to the plane from the antenna. Write d as a function of x, and graph the function over the interval 0< x < 1T. In Exercises 77 -80, use a graphing utility to graph the function and the equations }' = x and y - x in the same viewing rectangle. Describe the behavior of the given function as x ap proaches zero. 77. f(x) 79. g(x) = x cos x Ixl sin x I x sin x I 7S. f(x) SO. g(x) I x i cos x FIGURE FOR 95 In Exercises 81 -86, use a graphing utility to graph the function. Describe the behavior of the function as x approaches zero. 81. y 6 x > 0 82. y - + cos x, x sin 2x, x > 0 I cos x 84. f(x) = - - sin x 83. g(x) 4 x = + x I 85. f(x) == sin x 86. hex) x sin x 96. Television Coverage A television camera is on a review ing platform 100 feet from the street on which a parade will be passing from left to right (see figure). Express the distance d from the camera to a particular unit in the parade as a function of the angle x, and graph the func tion over the interval -1T/2 < x < 1T/2. (Consider x as negative when a unit in the parade approaches from the left. ) In Exercises 87-90, use a graphing utility to graph the func tions f and g in the same viewing rectangle. Use the graphs to determine the relationship between the functions. 87. f(x) = sin x + cos( x + 8S. f(x) 89. f(x) = sin _ 2 90. f(x) - cos 2 ~), == 1(I cos( x + sin x x, g(x) 1TX 2' ~), I g(x) = 0 g(x) 2 sin x cos 2x) () I( ) g x =="2 I + cos 1TX TJ( I Camera FIGURE FOR 96 174 CHAPTER 2 TRIGONOMETRY 97. Sales The projected monthly sales S (in thousands of units) of a seasonal product is modeled by S = 74 71'[ 3t ..,- 40 sin 6' where t is the time in months, with f = I corresponding to January. Graph this sales function over I year. 98. Sales The projected monthly sales S (in thousands of units) of a seasonal product is modeled by 101. Harmonic MOlion An object weighing W pounds is sus pended from the ceiling by a steel spring (see figure). The weight is pulled downward (positive direction) from its equilibrium position and released. The resulting motion of the weight is described by the function y = ~e-f/4 cos 41, I > 0, where y is the distance in feet and I is the time in seconds. Graph the function. 71'1 S 25 + 21 + 20 sin 6' where I is the time in months, with I I corresponding to January. Graph this sales function over I year. 99. Predator-Prey Problem Suppose the population of a certain predator at time I (in months) in a given region is estimated to be P = 10,000 Equilibrium 1 2m + 3000 sin 24' and the population of its primary food source (its prey) is estimated to be p 2m 15.000 + 5000 cos - . . 24 FIGURE FOR 101 Graph both of these functions in the same viewing rectan gle, and explain the oscillations in the size of each popu lation. 100. Normal Temperatures The normal monthly high tem perature for Erie, Pennsylvania, is approximated by H(I) 54.33 20.38 cos 71'1 6 71'1 - 15.69 sin - , 6 and the normal monthly low temperature is approxi mated by L(I) = 39.36 - 15.70 cos 71't 71'1 6 14.16 sin 6' where t is the time in months, with 1 I corresponding to January. (Source: NOAA) Use a graphing utility to graph the functions over a period of I year, and use the graphs to answer the following questions. (a) During what part of the year is the difference between the normal high and low temperatures greatest? When is it smallest? (b) The Sun is farthest north in the sky around June 21, but the graph shows the warmest temperatures at a later date. Approximate the lag time of the tempera tures relative to the position of the Sun. SECTION 2.7 2.7 INVERSE TRIGONOMETRIC FUNCTIONS 175 INVERSE TRIGONOMETRIC FUNCTIONS Inverse Sine Function I Other Inverse Trigonometric Functions I Compositions of Trigonometric and Inverse Trigonometric Functions Inverse Sine Function Recall that for a function to have an inverse. it must be one-to-one. From Figure 2.68 it is clear that y = sin x is not one-to-one because different values of x yield the same y-value. If, however, you restrict the domain to the interval -7T/2 :s x :s 7T/2 (corresponding to the darker portion of the graph in Fig ure 2.68), the following properties hold. \. On the interval l 7T/2, 7T/2], the function 2. On the interval 7T/2, 7T/2], Y -I :s sin x I. 3. On the interval [-7T/2, 7T/2], Y y = sin x is increasing. sin x takes on its full range of values, sin x is a one-to-one function. x:s 7T/2, y = sin x has a unique Thus, on the restricted domain -7T/2 inverse called the inverse sine function. It is denoted by y = arcsin x y = sin-I x. or y ---_~r.~----~----~f------+----~~-----~. x Sin x is one-to-one on this interval. FIGURE 2.68 The notation sin -I x is consistent with the inverse function notation I-I (x), which is used in Section 1.10. The arcsin x notation (read as "the arcsine of x ") comes from the association of a central angle with its subtended arc length on a unit circle. Thus, arcsin x means the angle (or arc) whose sine is x. Both notations, arcsin x and sin -I x, are commonly used in mathematics, so remem ber that sin -I x denotes the inverse sine function rather than 1/sin x. The values of arcsin x lie in the interval 7T/2, 7T/2]' The graph of y = arcsin x is shown in Example 2. Try producing this graph with a graphing utility. Note that the domain of y = arcsin x is -I :s x :s 1. 176 CHAPTER 2 TRIGONOMETRY .. - - . . ~ DEFINITfONOFINVER5.ESiNE·FUNCTION When evaluating the in verse sine function, it helps to remem ber the phrase "the arcsine of x is the angle (or number) whose sine is x." REMARK The inverse sine function is defined by y = arcsin sin y if and only if x x, where I 5 x 5 1 and 7T/2 5 Y 5 7T/2. The domain of \' arcsin x is [ I, l], and the range is [-rr/2, 7T/2l. As with trigonometric functions, much of the work with inverse trigono metric functions can be done by exact calculations rather than by calculator approximations. Exact calculations help to increase your understanding of the inverse functions by relating them to the triangle definitions of the trigonomet ric functions. EXAMPLE 1 Evaluating the Inverse Sine Function Find the values of the following (if possible). Try verifying the answers in Example I with your calculator. What happens when you try to calcu late sin -I 2? REMARK A. arcsin ( -~) • B. SIn ~I r V3 2 C. sin 12 SOLUTION arcsin( - ~) implies that A. By definition, y . I 2' . Because sin( arcSin( sin -" 5 \" 5 . rr 2 -. - 4, you can conclude that \' rr/6) -~) B. Bydefinition,y rr 2 for Sin \" = - - -rr/6 and : = sin-tV'3/2) implies that V3 for 2 ' rr 2 rr 2 y5-. . V3/2, you can conclude that y Because sin(rr/3) 7T/3 and rr 3 c. I t is not possible to eval uate v sin ~ I x when x 2, because there is no angIe whose sine is 2. Remember that the domain of the inverse sine function is [ 1, I]. • ....... . SEC T I ON 2.7 INVERSE TRIGONOMETRIC FUNCTIONS 177 From Section 1.10, you know that graphs of Inverse functions are x. reflections of each other in the line y EXAMPLE 2 Sketching the Graph of the Arcsine Function Sketch a graph of v arcsin x bv hand. SOLUTION By definition, the equations y = arcsin x sin y and x are equivalent for 7T/2 ~ Y ~ Hence, their graphs are the same. By assigning values to y in the second equation. you can construct the following table of values. ! 0 V x sin v -\ 2 2 0 7T 7T 7T 6 4 2 v2 2 2 The resulting graph of y arcsin x is shown in Figure 2.69. Note that it is the reflection (in the line y = x) of the darker part of Figure 2.68. Be sure you see that Figure 2.69 shows the entire graph of the inverse sine function. Remember arcsin x is the closed interval [ 7T/2. 7T/2]' You can that the range of y verify this graph with a graphing utility for the function y == arcsin x . FIGURE 2.69 ......... 178 CHAPTER 2 TRIGONOMETRY Other Inverse Trigonometric F'unctions .y The cosine function is decreasing on the interval 0 :5 x :5 1T, as shown in Figure 2.70. Consequently, on this interval the cosine has an inverse function, which is called the inverse cosine function and is denoted by y == arccos x Cos x is one-to-one on this interval. or y = cos - I x. Similarly, you can define an inverse tangent function by restricting the domain of y == tan x to the interval 1T /2, 1T /2). The following list summa rizes the definitions of the three most common inverse trigonometric functions. The remaining three are discussed in the exercise set. FIGURE 2.70 Domain Function y arcsin x if and only if sin y = x y arccos x if and only if cos y == x y == arctan x if and only if tan y == x 1T 1T -:5v< -}:5x:5 2 ,- 2 O:5y:51T 1T 1T - 2 <v< . 2 x<x<x The graphs of these three inverse trigonometric functions are shown in Figure 2.71. Domain: [-I, IJ 1T Range: 2 !!.]2 Domain: [-1. I J Range: [0, 1TJ Domain: (-00, 00) \' y ~~+----+----:if--'--+---+--- -2 -i11 FIGURE 2.71 ----------------- x 179 INVERSE TRIGONOMETRIC FUNCTIONS SECTION 2.7 Evaluating Inverse Trigonometric Functions EXAMPLE 3 Find the exact values of the following. v2 A. arccos 2 B. arccos(-I) C. arctan ° SOLUTION v2/2 and A. Because cos(17'/4) = 17'/4 lies in [0,17'], it follows that v2 17' arccos - - = -. 2 4 B. Because cos 17' = -1 and 17' lies in [0, 17'], it follows that arccos( - 1) C. Because tan ° ° ° arctan 0 REMARK In Example 4, had you set the calculator to the degree mode, the display would have been in degrees rather than radians. This convention is peculiar to calculators. By definition, the values of inverse trigonometric functions are always in radians. 17'. = and lies in 17' 12. 17'/2), it follows that ....... . . = O. In Example 3, you were able to find the exact values of the given inverse trigonometric functions without a calculator. In the next example, a calculator is necessary to approximate the function values. Tecbno(oepJ Note __.._ _ _ __ Most graphing utilities do not have keys for evaluating the inverse cotangent function, the inverse secant function. or the inverse cosecant function. Is it still possible to calculate arcsec 3.4? If you let x = arcsec 3.4, then sec x 3.4 and cos x = I/sec x 1/3.4. Hence. using the inverse cosine function key. x = 1.272. Similarly, to evaluate the other inverse trigonometric functions, use the following identities. arccsc x = arcsin arccot x = ; + x arctan (-x) EXAMPLE 4 Using a Calculator to Evaluate Inverse Trigonometric Functions Use a calculator to approximate the values (if possible). A. arctan ( -8.45) B. arcsin 0.2447 C. arccos 2 SOLUTION Function A. arctan( -8.45) B. arcsin 0.2447 C. arccos 2 Rounded 10 3 Decimal Places Mode Radian Radian Radian 1.45300 1005 0.2472102741 ERROR -1.453 0.247 Note that the error in part (C) occurs because the domain of the inverse cosine .. ...... . function is [- I, I]. 180 CHAPTER z TRIGONOMETRY Compositions of Trigonometric and Inverse Trigonometric Functions Recall from Section 1.10 that inverse functions possess the properties /(F1(x)) x I(/(X)) and = x. The inverse trigonometric versions of these properties are as follows. REMARK Keep in mind that these in verse properties do not apply for arbi trary values of x and y. For instance. arcSin(Sin 3;) = arcsin(-l) INVERSE PROPERTIES I s x s I and -n'/2 s y s n/2. then If sin(arcsin x) 7T 37T 2 2 - - if:. In other words. the property arcsin(sin y) = y is not valid for values of y outside the interval [- 7T /2. 7T /2]. If - I s x S I and 0 cos( arccos x) = x C OV. E R Y (a) Use a graphing utility to graph y =arcsin(sin x). What are the domairiand range of the function? Explain why arcsin(sin 4) dOes not equal 4. (b) Use a graphing utility to graph y sin(arcsinx). What are the domain and range of the function? Explain why sin(arcsin 4 lis not defined. arcsin(sin y) y. y S n, then and arccos( cos y) = v. n/2 < y < n/2, then If tan(arctan x) EXAMPLE 5 f) IS and x x arctan(tan y) and y. Lsing Inverse Properties If possible, find the exact values. A. tan[arctan( -5)] B. . (. 5n) arCSin c. SIn-:3 cos(cos-1n) SOLUTION A. Because - 5 lies in the domain of arctan x, the inverse property applies, and you have tan[arctan(-5)] = -5. B. In this case, /3 does not lie within the range of the arcsine function, 7T /2 S x S 7T /2. However, 57T /3 is coterminal with 5n which does lie in the range of the arcsine function, and you have arcsin( sin 5;) = arcsin [ sin( -~) ] 3 c. T he Try verifying these results with your calculator. REMARK expression cos( cos -I 7T) is not defined, because cos -I 7T is not defined. Remember that the domain of the inverse cosine function is [-1, lJ. .. ....... SECTION 2.7 181 INVERSE TRIGONOMETRIC FUNCTIONS Example 6 shows how to use right triangles to find exact values of functions of inverse functions. Example 7 shows how to use triangles to convert a trigonometric expression into an algebraic one. This conversion technique is used frequently in calculus. Evaluating Functions of Inverse Trigonometric Functions EXAMPLE 6 Find the exact values of the following. =Vs ~) A. tan( arccos SOLUTION 2 = arccos ~, then cos u = ~. Because cos u is positive, u is a first-quadrant angle. You can sketch and label angle u as shown in Fig ure 2.72. Consequently. A. If you let u FIGURE 2.72 tan( arccos~) = tan u opp Vs adj 2 B. If you let u = arcsin( - ~), then sin u == ~. Because sin u is negative, u is a fourth-quadrant angle. You can sketch and label u as shown in Figure 2.73. Consequently, cos[ arcsin( FIGURE 2.73 D] = cos u 4 5 = adj hyp ............ .... 7 Some Problems from Calculus EXA M PLE Write each of the following as an algebraic expression in x. A. sin(arccos 3x), n l , B. cot(arccos 3x), SOLUTION ./1 v - (3.xi \u 3x FIGURE 2.74 If you Jet u cos u = arccos 3x, then cos u Because 3x hyp' you can sketch a right triangle with acute angle u, as shown in Figure 2.74. From this triangle, you can convert each expression to algebraic form. A. sin(arccos 3x) REMARK In Example 7, a similar ar gument can be made for x-values ly ing in the interval [-1/3,0]. Why do we restrict x < ~ in part (B)') = 3x. sin u = B. cot(arccos 3x} = cot u hyp = adj opp = VI - 9x 2 , 3x o I X < -3 1 3 O~x<- .............. .... 182 CHAPTER 2 .......................... INVERSE FUNCTIONS TRIGONOMETRY You have studied inverses for several types of functions. Match each of the func tions in the left column with its inverse function in the right column. l. f(x) (a) (b) (c) (d) (e) x 2. f(x) O:5x 3. f(x) 4. f(x) 5. f(x) In x 6. f(x) sin x, 7. f(x) cos x, 8. f(x) tan x. eX 1T 1T 2 2 -:5x:5 0 x:5 1T 1T 1T 2 2 -:5x:5 FI(x) = arcsin x FI(x) = In x FI(x) = Vx FI(x) = arctan x F I (x) = arccos x (f) FI(x) = Vx (g) FI(x) := eX (h) FI(x) =x Provide reasons for your answers. WARM-UP The following warm-up exer~ises involVe skills that were covered in earlier sectiou.s. You will use these. skills in the exercise set for this section. In Exercises 1-4, evaluate the trigonometric function without using a calculator. 1. Sin( ~) 2. cos 3. tan ( -~) 4. sin 1T 7T 4 In Exercises 5 and 6, find a real number x in the interval [- 7T /2, 7T/2] that has the same sine value as the given value. 5. sin 27T 6 . 57T • sJn 6 In Exercises 7 and 8, find a real number x in the interval [0, 7T] that has the same cosine value as the given value. 7. cos 37T 8. cos( ~) In Exercises 9 and 10, find a real number x in the interval (-7T/2. 7T/2) that has the same tangent value as the given value. 9. tan 4rr 10. tan 37T 4 SECTION 2.7 183 INVERSE TRIGONOMETRIC FUNCTIONS SECTION 2.7 . EXERCISES .•.....•.......•......•.•............•...........................•...................................................... In Exercises 1-16. evaluate the expression without using a calculator. 1. arcsin ~ 3. arccos ~ 2. arcsin 0 4. arccos 0 In Exercises 37-44, find the exact value of the expression without using a calculator. (Hint: Make a sketch of a right triangle, as illustrated in Example 6.) 37. sin(arctan ~) 38. sec(arcsin ~) 39. cos(arctan 2) 40. sin(arccos r 5. arctan V3 3 v:) 7. arccos ( 9. arctan ( 11. arccos ( 6. arctan ( I ) ~) 8. arcsin ( yJ) 10. arctan( yJ) D 12. arcsin V2 2 yJ 2 14. arctan ( 15. arctan 0 16. arccos I 13. arcsin ~) 41. cos(arcsin 43. sec[arctan( - 17. 19. 21. 23. 25. 27. arccos 0.28 18, 20. 22. 24. 26. 28. arcsin( -0.75) arctan( - 2) arcsin 0.31 arccos(-O.4l) arctan 0.92 arcsin 0.45 arccos( -0.8) arctan 15 arccos 0.26 arcsin( -0.125) arctan 2.8 In Exercises 45 and 46, use a graphing utility to graph I and g in the same viewing rectangle to verify that the two are equal. Identify any asymptotes of the graphs. 45./(x) = sin (arctan 2x), 29. I(x) = tan x, 30./(x) sin x, ~), 2x g(x) == ---;=== g(x) = - - x - In Exercises 47-56, write an algebraic expression that is equiv alent to the given expression. (Hint: Sketch a right triangle, as demonstrated in Example 7.) 47. cot(arctan x) 49. cos(arcsin 2x) 48. sin(arctan xl 50. sec (arctan 3x) 51. sin (arccos xl 52. cot ( arctan; ) 53. tan ( arccos In Exercises 29 and 30, use a graphing utility to graph/, g, and y == x in the same viewing rectangle to verify geometrically that g is the inverse of f (Be sure to properly restrict the domain of f) 42. csc[arctan( - fi)) 44. tan[arcsin( -~)J t)J 46·/(x) == tan(arccos In Exercises 17-28, use a calculator to approximate the given value. (Round your result to two decimal places.) ~) 55. csc (arccos I' ~) 54. sec [arcsin(x - 1)) ~) -r-)h' X 56. cos ( arcsin In Exercises 57-60, fill in the blanks. g (x) == arctan x g(x) == arcsin x 9 57. arctan x ~ In Exercises 31-36, use the properties of inverse trigonometric functions to evaluate the expression. 58. arcsin 31. sin(arcsin 0.3) 33. cos[arccos( -0.1)] 32. tan(arctan 25) 34. sin[arcsin( -0.2)J 59. arccos 35. arcsin(sin 37T) 36. arccos (cos 7;) 60. arccos ). arcsin ( 6 x =I' 0 ). == arccos( 3 x - 2 ( 2 == arctan arcsin ( ), 05x 6 ) Ix - 21 2 184 CHAPTER 2 TRIGONOMETRY In Exercises 61-64. graph the function. 61. fLr) arcsin!x - I) 62. fLI I = -::; + arctan \ 63. fCd 64. f(r) 68. Pholograph1' A television camera at ground level is filming the lift-off of the space shuttle at a point 2000 feet from the launch pad (see figure). If (j is the angle of eleva tion to the shuttle and s is the height of the shuttle in feet. write (j as a function of s. Find (j when (a) s = 1000 and (b) s = 4000. arctan 2x x arccos it 4 In Exercises 65 and 66. write the given function in terms of the sine function by using the identity , A cos wI + B sin wI Sin(wl + arctan-). A, 65. f(1I 3 cos 21 + 3 sin 21 66. fUi = 4 7TI I s B' Verify your result by using a graphing utility to graph both forms of the function cos ; 3 sin 7TI 1-1------ 2000ft - - - - - 1 FIGURE FOR 68 67. PllOf()graphr A photographer is taking a picture of a 4-foot-square painting hung in an art gallery. The camera lens is I foot below the lower edge of the painting (see figure). The angle {3 subtended by the camera lens x feet from the painting is given by {3 = arctan 4x + 5 x > O. (a) Use a graphing utility to graph (3 as a function of x. (b) Move the cursor along the graph to approximate the distance from the picture when {3 is maximum. (c) Identify any asymptote of the graph and discuss its meaning in the context of the problem. FIGURE FOR 67 69. Docking a Baal A boat is pulled in by means of a winch located on a dock 12 feet above the deck of the boat (see figure). If (j is the angle of elevation from the boat to the winch and s is the length of the rope from the winch to the boat, write () as a function of s. Find (j when (a) s 48 feet and (b) s 24 feet. FIGURE FOR 69 SECTION 2.7 70. Area INVERSE TRIGONOMETRIC FUNCTIONS 185 In calculus, it is shown that the area of the region bounded by the graphs of y 0, r I/(x' + I L x (I, and x b is given by 73. Define the inverse cosecant function by restricting the do main of the cosecant to the intervals [- To /2. 0) and (0. 7T /2). and sketch its graph. Area = arctan b - arctan a 74. Use the results of Exercises 71-73 to evaluate (hc follo\'.· ing without using a calculator. (see figure), Find the area for the following values of a and b, (a) a = 0, b (b) a -l.b=1 (e) a = 0, b = 3 (d) a = I.b=3 (a) arcsecY2 (b) arc sec I (c) arceot - (d) arc sec 2 In Exercises 75-79. verify the identity with a calculator. then prove the identi ty. 75. arcsin( - -arcsin x x) 76. arctan( - x) = -arctan x 77. arccos( xl = 78. arctan x + arctan 79. arcsin x FIGURE FOR 70 + 72. Define the inverse secant function by restricting the domain of the secant to the intervals [0, 1T /2) and (1T /2. 1TJ, and sketch its graph. 1T 2. x x 0 To arccos x ') 80. The Chebyshev polynomial of degree formula T,,(x) 71. Define the inverse cotangent function by restricting the domain of the cotangent to the interval (0. 1T). and sketch its graph, arccos x 1T 11 11 cos(n arccos xl for is defined by the 1:0; x I and I. 2. 3.. (a) Show that T,,(x) (b) Show that T1(xj (C) Find = I. x. the quadratic polynomial T,Ln. (d) Graph T,(x) and 4x' 3x on the same viewing rectangle. What do you observe') 186 CHAPTER 2 TRIGONOMETRY 2.8 ApPLICATIONS OF TRIGONOMETRY Applications Involving Right Triangles Harmonic Motion / Trigonometry and Bearings / Applications Involving Right Triangles In keeping with our twofold perspective of trigonometry, this section includes both right triangle applications and applications that emphasize the periodic nature of the trigonometric functions. In this section, the three angles of a right triangle are denoted by the letters A, B, and C (where C is the right angle), and the lengths of the sides opposite these angles are denoted by the letters a, b, and c (where c is the hypotenuse). B &J , a 0 A)4.2 b =' 19.4 FIGURE 2.75 I EXAMPLE 1 Solving a Right Triangle, Given One Acute Angle and One Side Solve the right triangle having A = 34.20 and b ure 2.75. = 19.4, as shown in Fig C SOLUTION Because C = 90°, it follows that A + B = 90° and To solve for a, use the fact that tan A opp = ~ adj b -> a = b tan A. Thus, a = 19.4 tan 34.2° "'" 13.18. Similarly. to solve for c, use the fact that cos A adj hyp b c -- c b cos A Thus, c = 19.4 = 23.46. cos 34.2° ....... .. SEC TI 0 N 2.8 187 ApPLICATIONS OF TRIGONOMETRY Finding a Side of a Right Triangle EXAMPLE 2 B A safety regulation states that the maximum angle of elevation for a rescue ladder is n°. If a fire department's longest ladder is 110 feet what is the maximum safe rescue height? / r---+---'---'-i SOLUTION c=llOft A sketch is shown in Figure 2.76. From the equation a sin A = -. C it follows that a = c sin A = 1 10(sin 72°) = 104.6 feet. ......... FIGURE 2.76 EXAMPLE 3 Finding a Side of a Right Triangle At a point 200 feet from the base of a building, the angle of elevation to the bottom of a smokestack is 35°, whereas the angle of elevation to the lOp is 53°, as shown in Figure 2.77. Find the height s of the smokestack alone. SOLUTION T Note from Figure 2.77 that this problem involves two right triangles. In the smaller right triangle, use the fact that tan 35° a/200 to conclude that the height of the building is s r a = 200 tan 35°. Now. from the larger right triangle, use the equation a 1 tan 53° -"-'---:,,.,...~;..jJ a +s =- 200 to conclude that a +s s = 200 tan 53° FIGURE 2.77 Observer Horizontal '.,...:...;=:..:==..,. Object ,: Angle of Angle of depression '-. elevation Observer Object Horizontal FIGURE 2.78 200 tan 53°. Hence. the height of the smokestack is a 200 tan 53° - 200 tan 35° 125.4 feet. .... , .... Examples 2 and 3 used the term angle of elevation to represent the angle from the horizontal upward to an object. For objects that lie below the horizon tal. it is common to use the term angle of depression, as shown in Figure 2.78. In Examples I through 3. you found the lengths of the sides of a right triangle. given an acute angle and the length of one of the sides. You can also find the angles of a right triangle given only the lengths of two sides, as demonstrated in Example 4. 188 CHAPTER 2 TRIGONOMETRY EXAMPLE 4 Finding an Acute Angle of a Right Triangle A swimming pool is 20 meters long and 12 meters wide. The bottom of the pool is slanted so that the water depth is 1.3 meters at the shallow end and 4 meters at the deep end. as shown in Figure 2.79. Find the angle of depression of the bottom of the pool. FIGURE 2.79 REMARK Note that the width of the pool, 12 meters, is irrelevant to the problem. SOLUTION Using the tangent function, you can see that tan A = opp = 2.7 = 0.135. adj 20 Thus, the angle of depression is given by A = arctan 0.135 = 0.13419 radians = 7.69°. ........ . Trigonometry and Bearings In surveying and navigation, directions are generally given in terms of bear in~s. A bearing measures the acute angle a path or line of sight makes with a fixed north-south line. For instance, in Figure 2.BO(a), the bearing is S 35° E, meaning 35 degrees east a/south. Similarly, the bearings in Figure 2.BO(b) and (c) are N BO° W and N 45° E, respectively. N N i W·-=""""'I---E W ----,"'---..... E S s s (a) S 35° E FIGURE 2.80 0 (b) N80 W (el N 45° E SECTION 2.8 EXAMPLE 5 ApPLICATIONS OF TRIGONOMETRY 189 Finding Directions in Terms of Bearings A ship leaves port at noon and heads due west at 20 knots (nautical miles per hour). At 2 P.M., to avoid a stonn, the ship changes course to N 54° W, as shown in Figure 2.81. Find the ship's bearing and distance from the port of departure at 3 P.M. nm = 2(20 nm) _I A FIGURE 2.81 SOLUTION Because the ship travels west at 20 knots for two hours, the length of AB is 40. Similarly, BD is 20. In triangle BCD, you have B 90° 54° = 36°. The two sides of this triangle are detennined as follows. sin B = b ~ 20 20 sin 36" and d cos B = • 20 d = 20 cos 36° Now, in triangle ACD. you can detennine angle A as follows. b 20 sin 36° tan A = - - = = 0.2092494 d + 40 20 cos 36° + 40 A = arctan 0.2092494 "" 0.2062732 radians = 11.82° The angle with the north-south line is 90° bearing of the ship is N 78.18° W. Bearing Finally, from triangle ACD, you have sin A b 11.82° = 78.18°. Therefore, the b Ie, which yields 20 sin 36° c=--= sin A sin(l1.82) c = 57.4 nautical miles. Distance from pOri • • • • • • II • • til 190 CHAPTER 2 TRIGONOMETRY Harmonic Motion The periodic nature of the trigonometric functions is useful for describing the motion of a point on an object that vibrates, oscillates. rotates. or is moved by wave motion. For example. consider a ball that is bophing up and down on the end of a spring, as shown in Figure 2.82. Suppose that 10 centimeters is the maximum distance the ball moves vertically upward or downward from its equilibrium (at rest) position. Suppose further that the time it takes for the ball to move from its maximum displacement above zero to its maximum displacement below zero and back again is t = 4 seconds. Assuming the ideal conditions of perfect elasticity and no friction or air resistance, the ball would continue to move up and down in a uniform and regular manner. From this spring you can conclude that the period (time for one complete cycle) of the motion is Period = 4 seconds and that its amplitude (maximum displacement from equilibrium) is Amplitude = 10 centimeters. Motion of this nature can be described by a sine or cosine function. and is called simple harmonic motion. 10 em 10 em 10 em Oem o em o em -10 em -10 em -10 em Maximum negative displacement Equilibrium Maximum positive displacement Simple Harmonic Motion FIGURE 2.82 SECTION 2.8 ApPLICATIONS OF TRIGONOMETRY th~t mov~s'o[;iddordi~~te 191 .. A· point HiieisiIl:siillpJeJiarlllonic motion ifitSdistancedf~omthe.origt{~tjiIne byeith~r . . -: '-, - "', .-',. .. ',- .,,' -.' :..i$ , . . .... . ' .' ,- -~. - ". t given - " :', where aand·:w. ~ert!alnuinbers stich thatw >0. J11emotion has amplitudela j, periOd 2 'IT! w, and rr~c:jtiency'w /2 'IT. EXAMPLE Simple Harmonic Motion 6 Write the equation for the simple harmonic motion of the ball described in Figure 2.82, where the period is 4 seconds. What is the frequency of this motion? SOLUTION Because the spring is at equilibrium (d d = 0) when t = 0, use the equation a sin wt. Moreover, because the maximum displacement from zero is 10 and the period is 4, you have = Ia I = Amplitude 2 'IT Period = - w = 4 10 -> w 'IT 2 Consequently, the equation of motion is 'IT d 10 sin "2(' Note that the choice of a = 10 or a = 10 depends on whether the ball initially moves up or down. The frequency is given by Frequency w 'IT /2 1 = - - = - cycle per second. 2 'IT 2 'IT 4 ...... ... 192 CHAPTER 2 TRIGONOMETRY FIGURE 2.83 One illustration of the relation between sine waves and harmonic motion is seen in the wave motion resulting from dropping a stone into a calm pool of water. The waves move outward in roughly the shape of sine (or cosine) waves, as shown in Figure 2.83. As an example, suppose you are fishing and your fishing bob is attached so that it does not move horizontally. As the waves move outward from the dropped stone, your fishing bob will move up and down in simple harmonic motion, as sho'wn in Figure 2.84. y 1 I 1 Simple Harmonic Motion EXAMPLE 7 Given the equation for simple harmonic motion, 37T d = 6 cos-t 4 ' A fishing bob moves in a vertical direction as waves move to the right FIGURE 2.84 find the following. A. The maximum displacement B. The frequency c. The value of d when t = 4 D. The least positive value of t for which d = 0 SOLUTION The given equation has the form d = a cos wI, with a = 6 and w = 37T /4. A. The maximum displacement (from the point of equilibrium) is given by the amplitude. Thus, the maximum displacement is 6. w 37T/4 3 I . f' B. Frequency = 27T = ---z;:- = "8 cyc e per umt 0 time C. d = 6cos[3;(4)] 6 cos 37T = 6(-1) = -6 D. To find the least positive value of t for which d = 0, solve the equation d = 37T 6 cos - [ 4 TechnofolJlJ Note You can use your graphing utility to solve part D as follows. Graph the function YI = 6 cose:,) on the viewing rectangle 0 :5 X :5 6, -7 :5 Y :5 7, and observe that the first x -intercept is approximately x = 0.7. Using the zoom and trace features, you will see that x "" 0.667. 0 to obtain 37T 4 7T 37T 57T 2' 2 ' 2 -1=- 10 2 t = -3' 2, 3 ' .... Thus, the least positive value of t is [ = ~. ....... .. SECTION 2.8 APPLICATIONS OF TRIGONOMETRY 193 Many other physical phenomena can be characterized by wave motion. These include electromagnetic waves such as radio waves, television waves, and microwaves. Radio waves transmit sound in two different ways. For an AM station, the amplitude of the wave is modified to carry sound (AM stands for amplitude modulation). See Figure 2.85(a). An FM radio signal has its frequency modified in order to carry sound, hence the term frequency modu lation. See Figure 2.85(b). (a) AM: Amplitude modulation (b) FM: Frequency modulation FIGURE 2.85 Radio Waves Suppose you are teaching a class in trigonometry. Write two "right triangle prob l~ms" that you think would be reasonable to ask your students to solve. (Assume that your students have 5 minutes to solve each problem.) You BE THE INSTRUCTOR 194 CHAPTER 2 TRIGONOMETRY covered in earlier sections. You amplitude arid period of the function. 8. f(~) = .: 10. g (x) ! sin 1TX X = 0.2 cot '4 SECTION 2.8 . EXERCISES ...................•..................•...•.•...•..•.....•........•..................•....................•............• In Exercises I -10. solve the right triangle shown in the figure. (Round your result to two decimal places.) In Exercises II and 12. find the altitude of the isosceles triangle shown in the figure. (Round your result to two decimal places.) B c a c A b FIGURE FOR 1-10 1. A = b == 10 20°. 3. B = 71°. b 5. A 12° 15'. 7. a == 6, 9. b = 16, b= C 24 c = 430.5 10 52 (j 2. B == 54°, c == 15 4. A = 8Ao, a = 40.5 6. B = 65° 8. a 25, 12'. 10. b = 1.32. C a = b FIGURE FOR 11 AND 12 14.2 == 35 9.45 C f 11. () = 52°. b = 4 inches 12. () b 18°, 10 meters SECTION 2.8 13. Length oj a Shadow If the sun is 30° above the horizon, find the length of a shadow cast by a silo that is 70 feet high (see figure). FIGURE FOR 13 ApPLICATIONS OF" TRIGONOMETRY 195 16. Height The length of a shadow of a tree is 125 feet when the angle of elevation of the sun is 33° (see figure). Approx imate the height h of the tree. FIGURE FOR 16 14. Length oj a Shadow The sun is 20° above the horizon. Find the length of a shadow cast by a building that is 600 feet high (see figure). 17. Angle oj Elevation An amateur radio operator erects a 75-foot vertical tower for his antenna (see figure). Find the angle of elevation to the top of the tower at a point on level ground 50 feet from the base. /~ 1 600 ft FIGU RE FOR 14 15. Height A ladder of length 16 feet leans against the side of a house (see figure). Find the height h of the top of the ladder if the angle of elevation of the ladder is 74°. FIGURE FOR 15 FIGURE FOR 17 18. Angle oj Elevation The height of an outdoor basketball backboard is 124 feet, and the backboard casts a shadow 17 ~ feet long (see figure). Find the angle of elevation of the sun. FIGURE FOR 18 196 CHAPTER 2 TRIGONOMETRY 19. Angle ofDepression A spacecraft is traveling in a circular orbit 100 miles above the surface of the earth (see figure). Find the angle of depression from the spacecraft to the horizon. Assume that the radius of the earth is 4000 miles. 23. Height From a point 50 feet in front of a church. the angles of elevation to the base of the steeple and to the top of the steeple are 35° and 47° 40', respectively (see figure). Find the height of the steeple. Orbit FIGURE FOR 23 Angle of depression FIGURE FOR 19 24. Height From a point 100 feet in front of a public library, the angles of elevation to the base of the flagpole and to the top of the pole are 28° and 39° 45', respectively. The flagpole is mounted on the front of the library's roof (see figure). Find the height of the pole. 20. Angle of Depression Find the angle of depression from the top of a lighthouse 250 feet above water level to the water line of a ship 2 miles offshore. 21. Airplane Ascent When an airplane leaves the runway (see figl'.re), its angle of climb is 18° and its speed is 275 feet per second. Find the altitude of the plane after 1 minute. lOOft----l FIGURE FOR 24 25. Navigation An airplane flying at 550 miles per hour has a bearing of N 52° E. After flying for 1.5 hours, how far FIGURE FOR 21 22. Mountain Descent A sign on the roadway at the top of a mountain indicates that for the next 4 miles the grade is 12S (see figure). Find the change in elevation for a car descending the mountain. north and how far east has the plane traveled from its point of departure? 26. Navigation A ship leaves port at noon and has a bearing of S 27° W. If the ship is sailing at 20 knots, how many nautical miles south and how many nautical miles west has the ship traveled by 6:00 PM? 27. Navigation A ship is 45 miles east and 30 miles south of port. If the captain wants to sail directly to port, what bearing should be taken? 28. Navigation FIGURE FOR 22 A plane is 120 miles north and 85 miles east of an airport. If the pilot wants to fly directly to the airport, what bearing should be taken? SECTION 2.8 29. Surveying A surveyor wishes to find the distance across a swamp (see figure). The bearing from A to B is N 32° W. The surveyor walks 50 yards from A, and at point C the bearing to B is N 68° W. (a) Find the bearing from A to C. (b) Find the distance from A to B. APPLICATIONS OF TRIGONOMETRY 197 32. Distance Between Towns A passenger in an airplane flying at 30,000 feet sees two towns directly to the left of the airplane. The angles of depression to the towns are 28° and 55° (see figure). How far apart are the towns') 30,000 ft I 1 FIGURE FOR 32 FIGURE FOR 29 33. Altitude ofa Plane 30. Location of a Fire Two fire towers are 20 miles apart, tower A being due west of tower B. A fire is spotted from the towers, and the bearings from A and Bare N 76° E and N 56° W (see figure). Find the distance d of the fire from the line segment A.B. [Hint: Use the fact that d = 20/(cot 14° + cot 34°).] d FIGURE FOR 30 A plane is observed approaching your home, and you assume it is traveling at 550 miles per hour. If the angle of elevation of the plane is 16° at one time, and I minute later the angle is 57°, approximate the altitude of the plane. 34. Height of a Mountain In traveling across flat land, you notice a mountain directly in front of you. The angle of elevation (to the peak) is 3.5°. After you drive 13 miles closer to the mountain, the angle of elevation is 9°. Approx imate the height of the mountain. 35. Length A regular pentagon is inscribed in a circle of radius 25 inches. Find the length of the sides of the pentagon. 36. Length A regular hexagon is inscribed in a circle of ra dius 25 inches. Find the length of the sides of the hexagon. 37. Wrench Size Use the figure to find the distance J across the flat sides of the hexagonal nut as a function of r. 31. Distance Between Ships An observer in a lighthouse 300 feet above sea level spots two ships directly offshore. The angles of depression to the ships are 4° and 6.5" (see figure). How far apart are the ships? r-·······_-x FIGURE FOR 31 '1 FIGURE FOR 37 198 CHAPTER 2 TRIGONOMETRY 38. Bolt Circle The figure shows a circular sheet 25 cm in diameter containing 12 equally spaced bolt holes. Deter mine the straight-line distance between the centers of adja cent bolt holes. ----------------------- l' 40. 6 ____________ 1 ! ~ 6 ____~~__~~____~1 f--9~",,__4 1....-·--------36 -----------l·1 Harmonic Motion In Exercises 41-44, for the simple har monic motion described by the given trigonometric function, find (a) the maximum displacement, (b) the frequency, and (c) the least possible value of ( for which d == O. FIGURE FOR 38 Trusses In Exercises 39 and 40, find the lengths of all the pieces of the truss. 41. d = 4 cos 8m 43. d = k sin 120"t 42. d 44. d ~ cos 20"t = t!:. sin 792m 45. Tuning Fork A point on the end of a tuning fork moves in simple harmonic motion described by d = a sin wt, Find w given that the tuning fork for middle C has a frequency of 264 vibrations per second. 46. Wave Motion A buoy oscillates in simple harmonic mo tion as waves go past. At a given time it is noted that the buoy moves a total of 3.5 feet from its low point to its high point, and that it returns to its high point every 10 seconds, Write an equation that describes the motion of the buoy if, at t = 0, it is at its high point. CHAPTER 2 CHAPTER 111T 1. - 21T 2. 3. -110° 4. -.405° 9 In Exercises 5-8, convert the angle measurement to decimal form. (Round your result to two decimal places.) 5. 135 0 16' 45" 7. 5° 22' 53" 6. -234° 50" 8. 280 0 8' 50" In Exercises 9-12, convert the angle measurement to D° M' 5" form. 9. 135.2r ll. -85.15° 199 2 . REVIEW EXERCISES In Exercises 1-4, sketch the angle in standard position, and list one positive and one negative coterminal angle. 4 REVIEW EXERCISES 0 10. 25.1 12. -327.85° In Exercises 31-34. find the remaining five trigonometric func tions of satisfying the given conditions. (Him: Sketch a right triangle.) e e ~. tan e < 0 e= sin e > 0 sin e = ~. cos e < 0 cos e = - ~. sin e > 0 31. sec 32. tan 33. 34. In Exercises 35-40. evaluate the trigonometric function with out using a calculator. 1T rr 35. tan 3 36. sec 51T 37. sin 3 /5rr) 38. cot~ 6 J 40. esc 270° 39. cos 495° 4 In Exercises 13-16, convert the angle measurement from radi ans to degrees. (Round your result to two decimal places.) 51T 31T 13. - 14. 15. -3.5 16. 1.75 7 5 In Exercises 41 -44. use a calculator to evaluate the trigonomet ric function. (Round your result to two decimal places.) 41. tan 33° 42. csc lOse 121T 43. s e e - 44. 5 In Exercises 17-20, convert the angle measurement from de grees to radians. (Round your result to four decimal places.) 17. 480° 19. -33 0 45' 18. -16.5° 20. 84° 15' Sin( ~) e e In Exercises 21-24, find the reference angle for the given angle. 21. 252 0 23. 61T 5 22. 640° 17rr 24. 3 In Exercises 25-30, find the six trigonometric functions of the (in standard position) whose terminal side passes angle through the point. e 25. (12. 16) 27.(-7.2) 29. (-4. -6) 26. (x.4x) 28. (4. -8) 30. (~.~) e In Exercises 45-48. find two values of in degrees ~ < 360°) and in radians (0 ~ < 21T) without using a calculator. (0° 45. cos e= 47. esc (J 2 -2 46. sec 48. tan e is undefined V3 (J = 3 e In Exercises 49-52, find two values of in degrees (0 0 < 360°) and in radians (0 ~ < 2rr) by using a cal culator. e e sec e = 49. sin 51. e 0.8387 1.0353 50. cot 52. esc e= e= 1.5399 11.4737 200 C HAP T E R 2 TR IGONOMETRY In Exercises 53-56, write an algebraic expression for the given expression. 53. sec[ arcsin(x - I)] 54. tan ( arccos 85. Alritude of a Triangle the figure. ~) / 2 55. Sin(arccos 4 x x 2 ) 10m 56. csc(arcsin lOx) /\50 In Exercises 57 -70. sketch a graph of the function through two full periods. (Use a graphing utility to verify your resulL) 57. Y 3 cos 58. Y = -2 sin 27TX 59. fix) = 5 sin ~ 60. fix) 5 1 7TX - - cos- 61. fix) 4 4 66. h(r) 67. f(r) csc( 3r - , !!.)' 2 tower is 225 feet. and it casts a shadow of length 105 feet (see figure), Find the angle of elevation of the Sun. -1) 3 cos(r + 7T) = sec(r - ~) 68. fir) = 3 csc(2r + !!.1 4/ 7TO "8 I lOcos ( 120m 70. E(r) -7T3) In Exercises 7 J -80, use a graphing utility to graph the func tion. x 73. Y = . - - sm x 4 71. fix) x "7" 3 cos 7T X 75. h(O) = 0 sin 7TO 77. y = arcsin 86. Angle of Elevarion The height of a radio transmission 4 _ 7T4) tan (,t FIGURE FOR 85 7TX 64. g(r) 65. h(r) 8 cos( 0 62. fix) = -tan 63. g(r) = ~ sin(r - 7T) 69. f(O) = cot 7TX 2x Find the altitude of the triangle m x 2 7T 79. fix) = - + arctan x 2 72. g(x) = 3 ( sin 74. Y 76. fit) 78. y 4 - x 4 TTX "3 + cos + I) 7TX 2.5e·'/4 sin 2m 2 arccos x 80. fix) = arccos(x .. 7T) In Exercises 8 J -84, use a graphing utility to graph the func tion. Use the graph to determine if the function is periodic. If the function is periodic, find any relative maximum or mini mum points through one period. 81. fix) e""' 83. g(x) = 2 sin x cos2 x 82. g(x) 84. h(x) FIGURE FOR 86 87. Shuttle Height An observer 2.5 miles from the launch pad of a space shuttle measures the angle of elevation to the base of the vehicle to be 28° soon after lift- off (see figure). How high is the shuttle at that instant? (Assume the shuttle is still moving vertically.) = sin e' 4 sin 2 x cos 2 x T J r--__________ h. I ~l //~ / / / 2.5 mi 1~ FIGURE FOR 87 . Observer ------~- CHAPTER 2 From city A to city B. a plane flies 650 miles at a bearing of N 48° E, From city B to city C, the plane flies 810 miles at a bearing of S 65° E. Find the distance from A to C and the bearing from A to C. 88. Distance A train travels 2.5 miles on a straight track with a grade of 10 10' (see figure). What is the vertical rise of the train in that distance? 89. Railroad Grade FIGURE FOR 89 90. Distance Between Towns A passenger in an airplane flying at 35,000 feet sees two towns directly to the left of the airplane. The angles of depression to the towns are 32° and 76° (see figure). How far apart are the towns? Town 2 FtGURE FOR 90 REVIEW EXERCISES 201 91. Using calculus. it can be shown that the sine and cosine functions can be approximated by the polynomials x 3 Xl X7 sin X = x - - + 3! 5! 7! and cos x = 1 x 2 x· -+ 2! 4! x6 6" where x is in radians, (a) Use a graphing utility to graph the sine function and its polynomial approximation in the same viewing rec tangle. (b) Use a graphing utility to graph the cosine function and its polynomial approximation in the same viewing rectangle, (c) Study the patterns in the polynomial approximations of the sine and cosine functions and guess the next term in each, Then repeat parts (a) and (b). Do you think your guesses were correct? How did the accuracy of the approximations change when additional terms were added? 202 CHAPTER Z TRIGONOMETRY CHAPTERS 1 AND 2 . CUMULATIVE TEST ......•...............................•........•........................................................................ Take this test as you would take a test in class. After you are done. check your work against the answers in the back of the book. In Exercises I and 2, solve the equation. 1. I - 4(x - 3) =~ 2. 3x 2 + X - 10 = 0 In Exercises 3-6, sketch the graph of the equation without the aid of a graphing utility. 3. x 5.y 3y + 12 = 0 ~ 4. y 6.}' = x 2 4x + I 3-ix-2i 7. A line passes through the point (4, 9) with slope m ~. Find the coordinates of three additional points on the line. (There are many correct answers.) 8. Find an equation of the line passing through the points (- L I) and (3, 8). x/ex - 2) at the 9. Evaluate (ifpossihle) the functionf(x) specified values of the independent variable. (a) f(6) (b) f(2) (c) f(2r) (d) f(s + 2) 10. Express the area A of an equilateral triangle as a function of the length s of its sides. 11. Express the angle 41T/9 in degree measure and sketch the angle in standard position. 12. Express the angle 1200 in radian measure as a multiple of 1T and sketch the angle in standard position. 13. The terminal side of an angle (j in standard position passes through the point (12. 5). Evaluate the six trigonometric functions of the angle. 14. If cos r ~ where 1Tj2 < r < 'IT, find sin t and tan t. 15. Use a calculator to approximate two values of (j(0° ~ (j < 360°) such that sec (} = 2.125. Round your answer to two decimal places. 16. Sketch a graph of each of the functions through two periods. (a) y = 3 sin 2x (b) f(x) = 2 cos(x 1TX (c) g(x) = tan 2 (d) her) = sec .:) 2/ t 17. Write a sentence describing the relationship between the graphs of the functions f(x) = sin x and g. (a) g(x) 10 + sin x (c) g(x) = sin( x + ~) (b) g(x) = sin (d) g(x) 'lTX :2 -sin x ]8. Consider the function fix) sin 3x 2 cos x. (a) Use a graphing utility to graph the function. (b) Approximate (accurate to one decimal place) the zero of the function in the interval [0, 3 J. (c) Approximate (accurate to one decimal place) the max imum value of the function in the interval [0. 3J. 19. Evaluate the expression without the aid of a calculator. 4 V3 (a) arcsin (b) arctan 20. Write an expression that is equivalent to sin(arccos 2x). 21. From a point on the ground 600 feet from the foot of a cliff, the angle of elevation to the top of the cliff is 320 30'. How high is the cliff?