Chapter 4: Circular Functions Lesson 4.. 4-. a.! b.! c. i. 0!! " radians 80! " 6 radians 4-. a. and b. ii. iii. 45!! " radians 80! " 4 radians 60!! " radians 80! " radians 4-. Possible patterns that can be seen: In Quadrant I: all - and y-values are positive. In Quadrant II: all -values are negative, all y-values are positive. In Quadrant III: all - and y-values are negative In Quadrant IV: all -values are positive, all y-values are negative. Notice there is symmetry across the -ais and y-ais with respect to the coordinates (without regarding the signs). 4-4. a. The coordinates are ±, ± 4-5. a.! " has coordinates!, c.! 7" 6 4-6. 7! 5! b. i.! has coordinates!,! a.! " # 4 has coordinates,! c.! " 5# 6!!! ", 4!! ", ( )! 6!! ", 4! 6!! ", " 5! 4!! ", "!! ", "!!! ( 0, ) has coordinates!,!, ( )! 4!!, ( )!!!! 6!!,!!!(", 0) 0!! ( 0,)!!! ( 0, " )! 6!! 7! 4!! 5!!! (, " ), ", " (, ) ii. (,! ) b.! 7" 4 has coordinates (,! ) d.! " has coordinates! 4, ( ) b.! 7" has coordinates, ( ) " d.! has coordinates! 4, ( )
Review and Preview 4.. 4-7. a. b. c.! 5" 4!!!, 8!!! ", 5!!!, " 4-8. a. c + s s + c b. s + c! s " c c. s + c! c " s d. Cannot be simplified. e. s + c! c " s so 4(! s ) 4c f. 4! 4s 4(! s ) 4c 4-9. 500 revolutions minute! " radians revolution! minute 60 seconds 50" radians per second 4-0. a. log (9) because 9. b. log c. log 5 (65) 4 because 5 4 65. d. log (0) is impossible because there is no -value that makes 0.! because!. e. log 5 (?) means? 5 5. f. log 5 (?) means? 5 5. 4-. a. b.! +! (+)!! + +!(+)! + (+) (+)!+ (+) (!)(+) (+)! 4-. a. hours! 45 miles 5miles, hours! 55 miles 0miles, 0 + 5 45 miles hour hour b. See graph at right. c. Area under curve 45! + 55! 45 d. They are the same because the area under the curve is the distance traveled. speed (mph) 55 45 time (hours)
4-. a. 48!! " radians 80! 4" 5 b.!500! " # radians 80!! 5# 9! c. " 80!! radians 95! d.! 0" 9 # 80! " radians!400! 4-4. a. y 6()! 6( "! ) 6( ) (! ) 6( ) 4 4(4) b. y (4) + (4) (4) 6(4) 6(4 ) 6( 4 ) 6() c. y 60 ( ) + 60 ( ) ( ) 0! ( ) " # $ 0 ( 8 ) d. y 8( 4 9 ) + 8( 4 9 ) 4 ( 9 ) 8( 4 9 ) 6 ( 8 ) 6! 4 ( 9 ) " # $ 6 ( ) Lesson 4.. 4-5. a. Because P(, y) is a point on the unit circle, the length from the origin to the circle is. Thus, the hypotenuse of the triangle has length. b. sin! opp hyp y y c. cos! adj hyp d. A circle with center (h, k) and radius r is described by: (! h) + (y! k) r. The unit circle has center (0, 0) and radius r so the equation of the circle is: + y. e. Since cos! and y sin!, the equation of the circle can be written: cos! + sin!. f. Given the triangle, the Pythagorean theorem says + y. This triangle in the unit circle generated the equation cos! + sin!, thus they named it the Pythagorean Identity. 4-6. Domain: all angles; Range: [, ] 4-7. a. tan! y b. tan! sin! cos! 4-8. a. The y-value sine is positive in Quadrant I and II. b. The -value cosine is positive in Quadrants I and IV. c. Tangent is positive when sine and cosine have the same sign, i.e., in quadrants I and III. d. The mnemonic All Students Take Control is used to say that All three trigonometric functions are positive in Quadrant I, the Sine function the only function that is positive in Quadrant II, the Tangent function is the only function that is positive in Quadrant III, and the Cosine function is the only function that is positive in Quadrant IV.
4-9. The coordinates are the same ecept for the signs. The reference angle for 4! is!. 4-0. a. The coordinates for! on the unit circle are b. 5! c. 7! 4 has a reference angle of! 4 which has coordinates III, the cosine function is negative. Thus, cos( 5! 4 ) ". 6 has a reference angle of! 6 which has coordinates III, the sine function is negative. Thus, sin 7! 6, ( ), so sin (! )., ( ). Since 5! 4 ". d.! " has a reference angle of! which has coordinates IV, the cosine function is positive. Thus, cos(! " ). 4-. To find sin!, find the sin of!, then square your result. a. sin! b. cos! c. tan! d. sin! 4. so sin! ( ) so ( cos! ) ( ) 4 + cos (! ) 4 + 4 4 4. sin(! ) cos(! ) " so tan! (, ). Since 7! 6, ( ). Since! " ( ). e. Remember: sin! + cos!. Here,! " so sin! + cos!. 4-. a. sin! is the y-coordinate of the point on the unit circle, so sin! 0.8 in this case. b. Using sin! + cos! and sin! 0.8 we get: (0.8) + cos! cos! " (0.8) is in Quadrant is in Quadrant is in Quadrant cos! ± " (0.8) ± " 0.64 ± 0.6 ±0.6 Since the point P is in Quadrant II, we know that the cosine function is negative, so cos! "0.6.
4-. a. Label the missing side A. Using the Pythagorean theorem: b. sin! opp hyp 5 c. Quadrant IV has values of! : " <! < ". A + 4 5 d. If! is in Quadrant IV, the sine function is negative, thus sin! " 5. A 5! 6 9 A 9 Review and Preview 4.. 4-4. a. Using a 0!! 60!! 90! triangle, we know that sin 0!. Since 0!! is the reference angle for 4! which is in Quadrant III, we know the sine function is negative and sin 4! ". b. Using a 0!! 60!! 90! triangle, we know that cos 0!. Since 0!! is the reference angle for! which is in Quadrant IV, we know the cosine function is positive and cos!. c. Using a 0!! 60!! 90! triangle, we know that sin 60!. Since 60!! is the reference 6 angle for! 5" 6! sin! 5" 6 which is in Quadrant III, we know the sine function is negative and d. Using a 0!! 60!! 90! triangle, we know cos 0!. Since 0!! is the reference angle for! " which is in Quadrant III, we know the cosine function is negative and!. cos! " 4-5. a. and b. See digrams at right. c. sin! + cos!. If cos! 5, cos! 5 Then, sin! " cos! " 9 5 6 5 sin! ± 6 5 ± 4 5 d. tan! sin! cos! ± 4 5 5 ± 4 5 " 5 ± 4 9 5. 5! 4-6. a. log M + log N log M + log N log(mn ) b. (log P! log Q) log P Q P log Q
4-7. log ( ) log (! )! log() 4-8. a. The amplitude is the height from peak to trough, i.e.! 4. b. The period is the length from peak to peak, or 4. c. Domain: all real numbers. Range:! " y ". d. Roots, or -intercepts, are points where the graph crosses the -ais. This happens at all odd integers: r n +, n any integer. 4-9.! is the reference angle for!, ",#. a. sin! is in Quadrant II where sine is positive, so sin!. sin! is in Quadrant III where sine is negative, so sin! ". sin! is in Quadrant IV where sine is negative, so sin! ". b. sin! + cos!. If sin!, sin! Here cos! is positive because it is in Quadrant I. 9. Then, cos! " sin! " 9 8 9 cos! 8 9 c. cos! is in quadrant II where cosine is negative, so cos! ". cos! is in Quadrant III where cosine is negative, so cos! ". cos! is in Quadrant IV where cosine is positive, so cos!. 4-0. a. y! y y! y y!y y b. y + y y + y y y c. y! y! " y!y y!y d. (! ) ( y! y ) y! y! y + y y! y! y!y
4-. a. b. c. d. Lesson 4.. 4-5. a. The graphs have the same shape, and are shifted versions of each other. b. Each completes one full cycle in! c. The maimum value is, minimum is. 4-6. a. Domain of y is!" # # ". The range is! " y ". b. Domain of y cos is!" # # ". The range is! " y ". 4-7. a. -min: 6.5; -ma: 6.5; y-min: 4; y-ma: 4. There are approimately cycles on the screen.! 6.8 is close to 6.5. b. -min: 5.5; -ma: 5.5; y-min: 4; y-ma: 4. There are approimately cycles on the screen. The screen range is close to 60 to 60. Review and Preview 4.. 4-8. a. The period for both is!. b. The amplitude is. c. The y-intercept for y is (0, 0). The -intercepts for y are (n!, 0), n any integer.! d. The y-intercept for y cos is (0, ). The -intercepts for y cos are + n!, 0 any integer., n
4-9. a. (sin!) (0.6) 0.6 b. sin! (sin!) (0.6) 0.6 c. Impossible, to calculate sin!, we must know the value of! first. 4-40. cos! + sin! cos! " sin! cos! " 5 44 69 cos! ± Since! " # "!, the cosine function is negative, so cos! " sin!. tan! cos! 5 " " 5 # " 5. 4-4.! a. 4 has a reference angle of! 4 which has cos! 4. Since! is in Quadrant II, the 4 cosine function is negative, thus cos! 4 ".! b. 6 has a reference angle of! 6 which has sin! 6!. Since is in Quadrant IV, the sine 6 function is negative, thus sin! 6 ". c.! " has a reference angle of! which has sin!. Since! " is in Quadrant III, the sine function is negative, thus sin! "!. 5! d. 4 has a reference angle of! 4 which has tan! 4 sin! 4 cos! 4. Since 5! is in 4 Quadrant III, the tangent function is positive, thus tan 5! 4. 4-4. a. 5 + 7! + 4 7 5! + 7 + 4 7 (5! ) + ( + 4) 7 + 7 7 c. 0 + 45!! 5 +!! 5 5 + 5 ( + ) 5 5 5 b. 4 + 5! 7 ( ) +! 7 ( ) + 5 + + (5 + ) + 6 4 (4! 7)! d.! + 7 + 8 " "! " " " " + " " + " " "! " + + " 6 + 9
4-4. 4! a. is in Quadrant II, so the sine function (y-value) is positive and the cosine function (value) is negative. The signs of the coordinates are (!,+). 7 b.! " is in Quadrant III, so the sine and cosine functions are negative. The signs of the 5 coordinates are (!,!). is in Quadrant I, so the sine and cosine functions are positive. The signs of the coordinates are (+,+). d.!" on the unit circle is the point (!,0), so the sign of the coordinates are (!,0). 7! c. 9 4-44. See graph at right. No, f () is not continuous because you have to lift your pencil at 0. 4-45. a. y + y + c.!y + +y (+y)+(!y)!y b. y! y! y "! y " y d.!y + +y!y + +y (!y++y)!y y 4-46. a + y 7 Add the equations together: ( + a) 7 + b Solve for : 7+b +a! y b Substitute this value for into any of the original two equations:! y b becomes ( 7+b +a )! y b. Solve this equation for y: y 7+b +a! b +b +b!b(+a)!ab +a +a +a! b +b +a! b(+a) +a
Lesson 4.. 4-47. a. See graph at right. b. To shift y! units to the right and units up, the function becomes +. y! " c. See graph at right. d. To shift y h units to the right and k units up, we write: y sin(! h) + k. 4-48. a. The amplitude and period of and cos are the same. The shape is the same, but one is a shifted version of the other. b. To get from cos, shift cos to the left or right by!, for eample.. c. y can be written as y cos! " d. y cos can be written as y sin ( +! ). e. cos! " and cos sin ( +! ). 4-49. a. To give y an amplitude of, write: y 0.5. b. To reflect y across the -ais, write: y!. c. A sine function with an amplitude a and shifted up by k units is: y a + k. 4-50. From problem 4-47, we have y sin(! h) + k. To add the vertical stretch factor, write: y a sin(! h) + k. 4-5. a. y starts in the middle and goes up. b. y! cos starts at the bottom and goes up. c. y! starts in the middle and goes down. 4-5. a. The graph of y sin()! starts in the middle and goes up, has an amplitude of, and is shifted down units. Solution continues on net page.
4-5. Solution continued from previous page. b. The graph of y!cos()! starts at the bottom and goes up, has an amplitude of, and is shifted down unit. c. The graph of y! sin ( + " 4 ) + starts in the middle and goes down, is shifted! units to the left, and shifted 4 up unit. d. The graph of y cos ( +! ) " starts at the top and goes down, has an amplitude of, is shifted to the left by units, and is shifted down units.! Review and Preview 4.. 4-5. a. From y, the graph has an amplitude of and is shifted down units: y!. From y cos, the graph has an amplitude of, is shifted down units, and shifted to the!. right! 4 units: y cos! " 4 b. From y, the graph has an amplitude of, is shifted up unit, and is shifted to the +. right! 4 units: y! " 4 From y cos, the graph has an amplitude of, is shifted up unit, and is reflected across the -ais: y! cos +. 4-54. a. A radian is an angle measure found by wrapping a radius around a circle. b. 40!! " radians 80! 4" c. 5! 6 " 80!! radians 50!
4-55. a. We know the point W is in Quadrant III, so the sine function is negative. Furthermore, we know sin! is the y-value in the coordinates for W. Since we know W has coordinates (?,!0.6), we know sin! "0.6. b. Let the origin be labeled with O. Because this is a unit circle, we know WO. We are given VW 0.6. Then, VO + VW WO Now, cos! VO WO VO + (0.6) VO + 0.84 VO! 0.84 0.66 VO 0.785 0.785. But, since we are in Quadrant III, the cosine function is negative so cos! "0.785. c. The sine function is also negative in Quadrant IV, so! is in Quadrant IV. d. The cosine function is also negative in Quadrant II, so! is in Quadrant II. 4-56. a. b.!! c. The reciprocal of 0 is an integer divided by 0, which is undefined. 4-57. a. See graph at right above. The graphs are the same. b. See graph at right below. y cos( +! ) is equivalent to y! cos. c. It reverses the sign because it moves the angle quadrants ahead. 4-58. Area! 0 sin! 0 k 9 " k0 4-59. ( + )! ( + )! 0 0. Let a + then we get: a! a! 0 0 which has solutions: a ± 9!4()(!0) ± 49 ±7!, 5 () Using a! in a +, we get:! + "! 5. Using a 5 in a + we get: 5 +!.
4-60. a. Ryan traveled 60 miles both ways, so 0 miles. b. 60miles! hour 60 miles + 60miles! hour hours + hours hours 0 miles c. Total distance: 0 miles. Total time: hours. 0 miles Average speed: 40 miles per hour hours 4-6. a. 50 00 ( ) ().88 00 ( ) log ( ) log log c. 50 log log( )!.70 b. log ( + ) + log () log log (! ( + )) log! ( + ) + " 0 "± "4()("),"4 "± 49 "±7 log () is defined, log ("4) is undefined. Answer: only d. log ( + )! log () log log + log + + Lesson 4..4 4-6. a. csc A sin A a c c a c. sec A cos A b c c b b. cot B tan B b a a b 4-6. b. + c. i. tan B opposite! 9 adjacent ii. csc A sin A hypotenuse opposite iii. sin B opposite hypotenuse
4-64. a. tan! sin! cos! y c. csc! sin! y b. sec! cos! d. cot! cos! sin! y 4-65. a. sec! cos(! ) " c. cot 5! 6 tan 5! 6 cos 5! ( 6 ) " sin 5! 6 4-66. a. csc() sin() 0.4 7.086 " b. csc! 4 b. cot(0.857) tan(0.857) 0.9768.404 c sec(.54) cos(.54)!0.80906!.6 4-67. " d. tan! " sin(! 4 ) ( sin! " )! cos! " Let!ABC contain!"!b. Then, because sin! 0.6 opp, we know that hyp AC 0.6 and BC. Let AB then + (0.6) + 0.6! 0.6 0.64 ± 0.64 ±0.8 Because! " # "!, we are in Quadrant II where the cosine function is negative, so we want to let!0.8. Then, cos! AB BC "0.8. 4-68. a. Since (, y) is on the unit circle, we know + y. b. We substitute cos! and y sin! to get + y cos! + sin!. c. i. From sin! + cos! we get cos! " sin!. ii. From sin! + cos! we get sin! " cos!. iii. From sin! + cos! we get sin! + cos! " 0. cos! " " sin!!
4-69. a. Let!ABC be a right triangle with!b 90! and!a ". Then, b. Since! < " <!, we are in Quadrant III so the sine and cosine functions are negative, and the tangent function is positive. sin! " BC AC " 4 cos! " AB 4 AC " 5 4 sec! cos! " 5 4 cot! tan! 4 5 5 4 " 4 5 csc! sin! " 4 4 " 4 4 AC AB + BC AC 5 + 4 5 + 6 AC 4 AC 4 Review and Preview 4..4 4-70. a. sec! sec " c. tan! tan " cos ( " ) ( sin " ) cos " b. csc! csc " d. cot! cot " sin " tan " 4-7. a. sin!. The sine function is positive in Quadrants I and II. In Quadrant I this angle corresponds to! " 6. In Quadrant II this angle corresponds to! 5" 6. b. cos! ". The cosine function is negative in Quadrants II and III. The reference angle for is! ". In Quadrant II this reference angle corresponds to! ". In Quadrant III this reference angle corresponds to! 4". c. sin! ". The sine function is negative in Quadrants III and IV. The reference angle for which sin! is! ". In Quadrant III this corresponds to! 4" and in Quadrant IV this corresponds to! 5". d. cos! 0. The cosine function, which corresponds to the values of, is zero on the y-ais. These angles correspond to! " and! ". 4-7. a. cos! + sin! b. cos! + sin! c. cos 5 + sin 5 sin 5! cos 5 e. cos (! ) + sin (! ) cos (! ) " sin (! ) sin! " cos! d. cos! + sin! cos! " sin! f. 4![cos (" ) + sin (" ) ] 4 cos (" ) + 4 sin (" ) 4 4 cos (" ) 4 # 4 sin (" )
4-7. a. y! y y! y y!y y ( + ) b.! c. y! " # $ y d.!y + +y 4-74. +!! y! % & ' y! y! 4!4 +y (!y) ( +y) +!y (!y) ( +y)!y a. c! s c! (! c ) c! b. c +s + c 4-75. a. b. c.!s c!s!s + c +s!s c c c f () f ( ) f (!) 4-76. a. A 60! sector is 60! 60! 6 of the circle which would weigh 0g! 6 0g. b. A!r. The area of the circle with radius 0cm is: A 00!. This area weighs 0g. The area of the circle with radius 5cm is: A (5)! 5!. This area weighs: 5! 0 00! " 0 # 5! 70 g. 00! 4-77. a. y 7 log y log 7 log y log 7 log y log 7 log 7 y b. y 7 (y) 7 ( 7 ) 7 (y) 7 c. y log 5! 5 y 4-78. a. +.5 + 4 + 4.5 +... + 9.5 +0.5k k0! b. 5 + 8 +...+ 0! k + 5 c. cos() + cos(4) + cos(6) +... + 5 cos(0) k cos(k) 5! k 99 k0
Lesson 4.. 4-79. tan! opposite opposite hypotenuse adjacent adjacent hypotenuse sin! cos! cot! tan! cos! sin! 4-80. a. tan! " cos! sin! " cos! sin! cos! b. cot! "sec! cos! sin! " cos! sin! csc! c. sec! csc! cos! sin! sin! cos! tan! sin! cos! d. " cot! " csc! cos! " tan! sin! " sin! cos! sin! cot! 4-8. a. tan A! sec A sin A cos A! cos A + b. (tan A) sec A c. +cos A sin A + sin A +cos A sin A+! cos A+cos A!cos A sin A!cos A sin A!cos A sin A "sin A d.!cos A + +cos A+!cos A +cos A!cos A sin A sin A sin A cos! A cos A + sin A sin A + sin A!cos A +!cos sin A A!cos A +!cos A sin A (!cos A) (!cos A) sin A sin A sin A sin A+ sin A 4-8. a. The sine function is positive in Quadrants I and II. In Quadrant I,! " 6. In Quadrant II,! 5" 6. b. csc! sin! c. sec! cos! "! # 6 or 5# 6 cos! "! # 4 or 7# 4 4-8. a. csc! sin! " " sin! " #! 7$ 6 c. csc! sin! " or $ 6 sin! " #! 4$ or 5$ b. sec! cos! cos! "! # 6 or # 6 d. sec! is undefined when cos! 0 cos! 0 "! # or #
4-84. a. csc! sin! sin! 5 5 5 c. cos! 5 " sin! 5 # 5 4 5 b. tan! cot! 0.8 cot! 0.8.5 d. csc! sin! 5 sin! 5 cos! 5"4 5 5 5 5 4-85. a. Per measure from peak to peak!. Amp y measure from peak to center b. Per measure from peak to peak 6. Amp y measure from peak to center 5 4-86. a. log + log + + 9 8 4! log b. 50 (.5)! 65 (.5)! 7 log.5 7! log.5 7 + 4.9 4-87. a. cot! 7 cos! " 7 +7 " 7 4 b. csc! hypotenuse opposite " 4 4-88. Sine: shifted up, Amp! y " sin ( ) + Cosine: shifted up, shifted to left!, Amp! y cos ( + " ) + 4-89. a. 5 radians! 60! " radians 86.5! b. c. 00! " meters! 60! 0.944 meters d. No, 0º is just short of the shaded region. 4-90. a. a b 5 ab 5/ ab b b. 4 c. 5 y 6 y P! d. 5 5 y 6 y5 y
4-9. a + b + 7 c c! 7 a + b c!7 a +b Lesson 4.. 4-9. a. See graph at right, top. Period:!! b. See graph at right, middle. Period:! c. See graph at right, bottom. Period:! 0.5!! " 4! / d. The period is! b. e. In part (a), periods occur in!. In part (b), periods occur in!. In part (c), 0.5 periods occur in!. f. p! b or pb!. 4-9. a. g() is horizontally compressed b. The amplitude and shift are the same c. b stretches the graph of y horizontally. The period is! b. 4-94. a. y! sin(4) + 5 has period! 4!. b. y 5 cos c. y sin! 4-95. has period! Angular frequency is! 0! 5.! 4. d. y 4 cos! 7! 0 has period! "! has period 6!.! 7 7. 4-96. a. A cosine curve with period! is graph. b. A sine curve with amplitude and shifted down by unit is graph 4. c. A cosine curve with period!! and shifted up by unit is graph 5. d. A sine curve with period! 4! is graph. e. A sine curve with period! and shifted to the right by! is graph 8. 4 f. A cosine curve with period! and shifted down by units is graph. g. A sine curve with period!!, with amplitude, reflected across the -ais is graph 6. h. A cosine curve with period!, with amplitude, and shifted to the left by! is graph 7.
Review and Preview 4.. 4-98. a. cos sec + csc 4-99. cos cos + cos + sin +! + +! + + b. tan +cot sec csc cos + cos sec csc sin +cos cos cos cos cos 4-00. cos! + sin! tan! sec! cos! + sin! " sin! cos! cos! " cos! sin! + sin! " cos! cos! cos! cos! + sin! () sec! cos! 4-0. The radius of the circular cross section, r, is: 0 4!!!!!r 4 r 0 5 The area of the circular cross section is: A 4-0. See graph at right. By counting squares, or using the area program we can estimate the area as.! 4 5 5! y 4-0. a. Any angle in Quadrant 4. b. Any angle in Quadrant.
4-04. a. cos 7! 6 has a reference angle of! 6. cos (! 6 ) " cosine function is negative so cos 7! 6 b. Since! corresponds to the point (0,!), sin! ". c. sec 4! d. tan (! ) cos(4! ) " " sin(! ) cos(! ) " # " ".. Since 7! 6 is in Quadrant III, the 4-05. a. See graph at right above. Reflect the function over the -ais, halve the period, double the amplitude, shift the function up three units. b. See graph at right below. Shift the function to the right by! units, triple the amplitude, and shift the function down one unit. 4-06. First, log! because! and log 7 49 because 7 49. Then, log! log y! log 4 + log 4 y log y! log 4 + log 4 y " log 4 y Substituting in y we get: 6 y! 8 " " y! 4 Now, log y! "! y " y!!!!!!!!!!!!!!!and log 4 y " 4 y
Lesson 4..4 4-07. a. sin! + cos! sin! cos + cos!! cos! cos! tan! + sec! 4-08. a. tan! + sec! tan! sec! " c. sin! + cos! " cos! # sin! tan! + sec!. Then, ( # sin )( + tan ) cos $sec cos $ cos 4-09. a. (sec )(! cos ) tan ( cos )(! cos ) cos cos cos! cos! cos sin cos + sin 4-0. a. tan + cot csc sec cos + cos! cos sin +cos cos cos cos cos b. sin! + cos! sin! sin + cos!! sin! sin! + cot! csc! b. sin! + cos! " sin! # cos!!!!!sin! ( # cos!)( + cos!) d. + cot csc cot! csc! b. (cos A)(sec A! cos A) sin A sin A (cos A) cos A! cos A! cos A sin A cos A + sin A b. csc + cot csc!cot + cos! cos +cos!cos +cos!cos +cos +cos +cos!cos +cos
Review and Preview 4.. 4-. a. tan A! sec A sin A cos A! cos A sin A cos A + b. (tan A) sec A sin A sin A cos A sin A "sin A (cos A) + sin A sin A + c. +cos A sin A + sin A +cos A (+cos A) +sin A + cos A+cos A+sin A sin A(+cos A) sin A(+cos A) d.!cos A + +cos A 4-. +cos A (!cos A)(+cos A) +!cos A (!cos A)(+cos A) +cos A+!cos A (!cos A)(+cos A) (!cos A)(+cos A)!cos A sin A a. cot (sec! ) cos sin ( cos ) cos tan ( cos! ) cos cos cos! cos cos b. sin! " cos! " cos! " cos! " cos! c. tan! + 6 sin! cos + 6 sin!! cos + 6 cos!! cos sin!+6 cos!! cos! cos! + 5 cos! cos! cos! + 5 sec! + 5 d. cos y + sin y tan y cos y + sin y sin y cos y cos y + sin y cos y cos y cos y + sin y cos y 4-. cos y+sin y cos y cos y sec y a. csc! " sin! " sin! sin A sin A sin A + sin A sin A+ sin A + cos A (+cos A) sin A(+cos A) sin A(+cos A) sin A!cos ( cos )!cos ( cos ) "cos!+6 cos! cos +5 cos!! cos! The sine function is positive in Quadrants I and II. Therefore,! ", ". b. tan! " # sin! cos! " " sin!!!or!! cos! " /. / From the unit circle we know sin 5! 6 Also, sin! 6 Also, " and cos! 6 and cos ( 5! 6 ) ".. Then, sin(5! 6) cos(5! 6) " " # " ". sin(! 6) cos(! 6) " " # " ". Thus,! 5" 6, " 6. c. sec! cos! is undefined when cos! 0. cos! 0 when! ", ". d. cot! cos! sin! " when cos! " sin!. This is in Quadrants II and IV when! " 4, 7" 4.
4-4. a. Sinusoidal curves have either a sine or cosine base function. The amplitude is, and the curve is shifted down unit, and the period is!. Using cosine: y cos()!. ( ) ". If we use sine, we need to shift the curve left by! 4 units: y sin +! 4 b. Since the function starts at the origin, we can use sine. The amplitude is, the period is 4-5. 4!, and the function is reflected over the -ais: y!. a. + 4! 7 4 " + 4 "! 9 " 4 " + 4 "! 9 " " + "! " ( +! )! "! b. (! ) ( + ) ( ) + 4 "! "! " " + 6! " 4 + 6! 6! + 6 ( )! " + ( )! 6 + 5! 6 c.! 4-6. Using the two inner triangles as similar right triangles: 5!!!!! 45 5 4-7. a. See graph at right. b. The area under the curve can be broken up into a rectangle with coordinates (0, 0), (, 0), (0,8), (,8) and a triangle with coordinates (0,8), (,8), (, 78). The rectangle has an area of 8! 54. The height of the triangle is: 78! 8 60. The triangle has an area:!! 60 90. Adding these together we get a total area of: 90 + 54 44 units. c. The Gladiator traveled 44 miles in hours. 4-8. An eponential function with a horizontal asymptote at y 0 has the form: y A(B ) + 0 because as decreases without bound, y approaches 0, the horizontal asymptote. Since the function is increasing we know the eponent,, will be positive. Using the points (, 70) and (,45) we get two equations: 70 A(B ) + 0 AB + 045 A(B ) + 0 Solving the first equation for A we get: A 60 B. Substitute this in for to the second equation: 5 60 B (B) 60B. Solve for B: B 5 60 B.5 speed (mph) Solve for A: A 60 B 60.5 40. 60 40 0 The equation is: y 40(.5 ) + 0. time (hours)
Lesson 4.. 4-9. a. Let represent the number of days since birth. b. Let y represent how positive or negative the day is. c. The domain will be the lifespan: 0 days until death. d. The range will be twice the amplitude centered at zero: ( 00,00). The window settings should be [0, ] and y[!00,00]. 4-0. a. The team member with the highest physical cycle value P(). b. The team member with the lowest emotional cycle value E(). c. The team member with the highest intellectual cycle value I(). P() 00 sin! E() 00 sin! 8 I() 00 sin! 4-. Step : (007! 99) " 65 5840 Step : 5840 + 4 5844 Step : 5844 + ( + + + 0 + + ) 598 Michelle has been alive for 598 days. Her intellectual value is I(598) 00 sin! " 598 98.98, which is near a peak value. 4-. No, because P() 00 when ( +n 4 ), E() 00 when 8 ( +n 4 ), and I() 00 when ( +n 4 )., 8, and do not share a common factor and thus these three solutions will never coincide. Review and Preview 4.. 4-. sin! and sin! + cos!!!"!!cos! # sin! "! cos! # # sin! " " 4 9 " 5 9 " 5 cos! " " sec! cos! " " 5 " 5 5 5 tan! sin! cos! " " 5 # " " 5 5 5 5 cot! tan! " 5 " 5 " 5 5 5 csc! sin! 4-4. a. sin! ". The sine function is negative in Quadrants III and IV. sin! "! # 6 is the reference angle. sin! " #! 7$ 6, $ 6. b. cos! "! # is the reference angle. The cosine function is positive in Quadrants I 4 and IV, so! " 4, 7" 4. c. tan! "! # is the reference angle. The tangent function is positive in Quadrants I and III, so! ", 4". d. csc! " sin! " sin! "! # is the reference angle. The sine 4 function is positive in Quadrants I and II, so! " 4, " 4.
4-5. a. cos A + tan A sin A cos A + sin A cos A sin A cos A cos A + sin A cos A cos A+sin A cos A cos A sec A b. sin! + tan! " cot! + cos! " sec! + csc! 4-6. sin! + sin! cos! " cos! sin! + cos! " cos! + sin! sin! + sin! " cos +! cos! + "cos! sin! sec + csc cos + sin (sec )(csc ) cos sin + " cos! cos! + sin! sin! " + sin cos sin + cos cos sin +cos cos sin cos sin 4-7. a. Shift the sine function to the left! units, the amplitude is, the function is reflected over 4 the -ais, and shifted up unit: y! sin ( + " 4 ) + b. Shift the cosine function to the right! units, the amplitude is, the function is reflected 4 over the -ais, and shifted up unit: y! sin (! " 4 ) + 4-8. a. The sine function is negative in Quadrant IV. The reference angle for sin 5! is! ". sin! Therefore, sin 5! ". b. The sine function is negative in Quadrant IV. The reference angle for cos! 4 is! " 4. cos! 4! Therefore, cos 4. c. cot! cos! Cosine is positive and sine is negative in Quadrant IV. Therefore cotangent sin! will be negative. The reference angle is! 6. Therefore cot! cos(! /6) ( 6 ) sin(! /6) / /, but we are in Quadrant IV so cot (! " 6 )!. d. sec! cos!,! corresponds to the point (0, ), therefore cos! 0 and sec! 0, which is undefined. 4-9. Let the distance from the tree to Rocky be d and the distance from Rocky to the tip of his shadow be, then tan! 0+t d+ and tan! 6. Solve the latter equation for : 6 tan!. Substitute this into the first equation: tan! 0+t d+ 6 tan! d 0+t tan! # 6 tan! 4+t tan! " d + 6 tan! 0+t tan!
4-0. a. If AT BT r, then!atb is a right triangle and AB! 90º. The portion of the string that goes around the circle is 90! "r.5r. Therefore the total length of the string is 60.5!r + r " 6.7r. b. Recall that tangent lines to circles are perpendicular to the radius. See diagram at right. The portion of the string that goes around the circle is 40 60! "r 4" r. Therefore the total length of the string is 4! r + r " 7.65r c. m!atb "0! 60! A r T r 60º 60º r B r Lesson 4.. 4-. a. and b. See graph at right. c. The parent graph can either be y or y cos. d. Yes, the period is!. e. We know g() a cos(! b) because g() + cos is a shift of the cosine function with a larger amplitude. The smallest positive value of b is! 4. f. a!.4 4-. Yes, the bold graph confirms the graph drawn without a calculator. 4-. It is also sinusoidal and it has the same period, but it has a greater amplitude. 4-4. It is also periodic, but it does not look sinusoidal. It has two small crests between each trough. 4-5. The graph looks like a diagonal sine curve. 4-6. a. The amplitude is greater for large absolute values of and decreases as gets closer to zero. b. y is bounded by y and y!. The graph is also symmetric about the y-ais.
Review and Preview 4.. 4-7. a. cos!5" 6 b. tan! 4 cos ( 5" 6 )! tan ( 7! 4 ) sin(7! 4) cos(7! 4) " " 4-8. a. cos! when! 0, ". b. tan! " when sin! cos! " # sin! " cos! #! $ 4, 7$ 4 c. csc! when sin! " sin! "! # 6, 5# 6 d. sec! when cos! or cos!. The cosine function is positive in Quadrants I and IV:! " 4, 7" 4. 4-9. a. csc! cot cos cos! " cos!sin cos cos cos b. sec + tan sec!tan cos + cos cos! cos + cos + cos + cos cos! cos +!sin + cos. 4-40. +tan + +cot +cot ++tan (+tan )(+cot ) +cot (+tan )(+cot ) + +tan (+tan )(+cot ) +cot +tan (+tan )(+cot +csc!+sec! ) (sec )(csc csc +sec ) (sec )(csc ) csc (sec )(csc + sec ) (sec )(csc ) sec + csc cos + sin 4-4. h 0 +5! h 0 +5 4-4. a. Starting from the cosine function, the amplitude is, the period is 4!, and the function is!. shifted down unit: y cos b. Starting from the sine function, the amplitude is, the period is!, the function is reflected over the -ais, and shifted up unit: y! +.
4-4. The parachute will be deployed after 0 seconds. The height will be represented by the area under the curve v 0t between t 0 and t 0 seconds. The curve maps out a triangle with base 0 and height 0(0) 00. The area is:!0! 00 500 ft. Closure Merge Problems 4-44.. a. sin! BD BD b. cos! PD PD.!A and! are both complementary to!apb..!abp!!pdb because of angle-angle similarity. 4. a. sin A PA b. csc A sin A PA c. tan A sin A cos A PA AB PA AB d. cot A tan A AB 5. Both!PBC and!pdc have angles!, 90!, and! " 90!, therefore they are similar. 6. a. tan! BC BC b. cos! PC c. sec! cos! /PC PC sin! BD cos! PD tan! BC csc! PA sec! PC cot! AB 4-45. a. From!PDB we get: PD + BD sin! + cos! b. From!PBC we get: + BC PC + tan! sec! c. From!ABP we get: + AB AP + cot! csc!
4-46. (cot! + tan!) cot! + cot! tan! + tan! csc! " + cot! tan! + sec! " csc! " + tan! tan! + sec! " csc! " + + sec! " csc! + sec! This is the Pythagorean theorem applied to!pac. 4-47. a. Juan travels the distance of the circumference of his tire:!() 6! inches. This is the period. b. Because the reflector is farthest from the ground at time t 0, we use the cosine function. The amplitude is and the function is shifted up units (the radius of the tire). The period is 6!. h() cos +. c. After riding 0ft! in. 40 in. h(40) cos 40 ft. +.8 in. ft in d. Juan travels 9. feet per second or 9.! " 5 in sec ft sec e. After 5 minutes, or 00 seconds, h(00) cos 5! 00, so h(t) cos 5 t + 8.68 in. +. Closure Problems CL 4-48. a. sin ( 5! ) has a reference angle of sin (! ). 5! function is negative, so sin ( 5! ) " b. cos (! " 4 ) has a reference angle of cos (! 4 ).! " 4 function is positive, so cos (! " 4 ). c. csc ( 5! 6 ) has a reference angle of csc (! 6 ) sin(! 6) the sine and cosecant functions are positive, so csc ( 5! 6 ). d. tan! sin(! ) ( ) cos(! ). sin (! ) and cos (! ) 0 so tan (! ) 5! cos(5! ) e. cot ( ) sin(5! ) " " # " " f. sec ( 7! 6 ) has a reference angle of sec (! 6 ) is in Quadrant IV where the sine is in Quadrant IV where the cosine. 5! 6 is in Quadrant II where sin(! ) cos(! ) 0 cos(! 6) III where the cosine and secant functions are negative, so sec 7! 6. 7! 6 " is undefined.. is in Quadrant CL 4-49. a. Starting with sin!, the amplitude is, the function is shifted up units, and the function is +. shifted right! units: y sin! " # b. Starting with cos!, the amplitude is, the function is shifted up units, and the function is reflected over the -ais: y! cos " +.
CL 4-50. sec!cos!cos cos cos!cos " cos cos tan cos! cos CL 4-5. a. b. cos! cos cos! cos cos c. CL 4-5. a. tan A(csc A! sin A) tan A csc A! tan A sin A sin A cos A " sin A! sin A cos A " sin A cos A! sin A cos A!sin A cos A cos A cos A cos A b. (csc + cot )(! cos ) csc! cos csc + cot! cot cos! cos " + tan! tan " cos! cos + cos! cos! cos!cos CL 4-5. sec! 5 4 " cos! 4 5 csc! < 0 " sin! < 0 cot! cos! sin! 4 5 5 4 5 " 5 # 4 sin! + cos! " sin! + 4 5 sin! # ( 4 5 ) 9 5 sin! # 5
CL 4-54. sin ( + ) + cos ( + ) + Let u + then we have: sin (u) + cos (u) u u Substitute back in: +!! CL 4-55. cos! a. "sin! + cos! cos!(+sin!) cos!("sin!) cos!(+sin!)+cos!("sin!) + +sin! ("sin!)(+sin!) (+sin!)("sin!) ("sin!)(+sin!) b. cos!((+sin!)+("sin!)) ("sin!)(+sin!) cos! "sin! cos! cos! cos! sec! tan +cot cos + cos cos sin +cos cos CL 4-56. The period is 4 seconds, the amplitude is!0 5 and since the peak is at 8 feet, the function is shifted up units: h 5 cos! t CL 4-57. + y! + cos CL 4-58. a. b. # c. h() (! ) + 4, " " 4 $ %& 5, > 4
CL 4-59. 9 " "! + ( ) + $ $ ( # & # % % '.8 using left-endpoint rectangles. 0 0 " " $ $ + (! + # & # % % ' 0. using right-endpoint rectangles. CL 4-60. Midpoint rectangles.8+0..094 log b M N P log b M + log b N! log b P log b M + log b N! log b P (.)!.5!.4.5