Trigonometric Functions

Size: px

Start display at page:

Download "Trigonometric Functions"

May Lindsey
5 years ago
Views:

1 Exercise. Degrees nd Rdins Chpter Trigonometric Functions EXERCISE. Degrees nd Rdins 4. Since 45 corresponds to rdin mesure of π/4 rd, we hve: 90 = 45 corresponds to π/4 or π/ rd. 5 = 7 45 corresponds to 7 π/4 or 7π/4 rd. 5 = 45 corresponds to π/4 or π/4 rd. 60 = 8 45 corresponds to 8 π/4 or π rd. 80 = 4 45 corresponds to 4 π/4 or π rd. 5 = 5 45 corresponds to 5 π/4 or 5π/4 rd. 70 = 6 45 corresponds to 6 π/4 or π/ rd. 6. The ngle of degree mesure 0 is smller, since which is pproximtely 8.6. rdin corresponds to degree mesure of Coterminl ngles hve mesures = 495 nd 5 60 = 5 Coterminl ngles hve mesures = 40 nd = Coterminl ngles hve mesures = 0 nd = 500 Coterminl ngles hve mesures + π = 7 rd nd π = 5 rd 8

2 Chpter Trigonometric Functions Coterminl ngles hve mesures π = rd nd π = 4 4 rd Coterminl ngles hve mesures π = rd nd π = 4 5 rd. corresponds to corresponds to corresponds to rd; 5 corresponds to 5 80 = 6 80 rd = rd = 5 rd.6 rd rd.885 rd rd rd 80& 8. rd corresponds to % ( ; rd corresponds to 80 ' & ) = ' % ( % 540 & ( 7.9 ' (.4)(80) & 0..4 rd corresponds to % ( = ' % 5 & ( 80. '. 9 rd corresponds to 9 80 ' & ) = 40 (Exct) % ( 4. Set clcultor in degree mode. (A) (0.750) rd is 4.97 (B) (.5) rd is (C) (.80) rd is 7.74 (D) ( 7.) rd is Set clcultor in rdin mode. (A) 5 is 0.6 rd (C) 47 is 7.67 rd (B) 87 is.64 rd (D) 75. is.060 rd 8. Since θ = s, we hve r (A) θ = 5cm 65cm = 0.75 rd;.5 (B) θ = =.65 rd; cm 40 cm (C) θ =.5cm 40 cm =.0875 rd; 76.9 (D) θ = 0 cm 40 cm = 5.5 rd;

3 Exercise. Degrees nd Rdins 40. (A) s = rθ (B) s = rθ = (5)(0.8) =.4 in ' = (5) & 7 ) =. in % 80 ( (C) s = rθ (D) s = rθ = (5)(.57) = in ' = (5) & 75 ) = 5.7 in % 80 ( 4. Since θ r = 80 θ d, if oth sides of this eqution re cut in hlf, then θ r = 80 θ d Thus if θ d is cut in hlf, θ r is lso cut in hlf. 44. Since θ r = s r, nd if s (the numertor) is douled while r is held constnt, then θ r will e douled. 46. (A) A = r θ (B) A = r θ = (0.5) (0.50) 8.7 ft = ' (0.5) & 5.0 ) 4.4 ft % 80 ( (C) A = r θ (D) A = r θ = (0.5) (.74) 95.9 ft = ' (0.5) & 05 ) 0.0 ft % 80 ( 48. Since θ = s r, θ in rdin mesure, θ = s = s = π + 4. So 9 4 is coterminl with π + 4 π = 4. Since 4 is etween 0 nd, its terminl side lies in qudrnt I is coterminl with (696 60) = 6. Since 6 is etween 70 nd 60, its terminl side lies in qudrnt IV. 54. is coterminl with + π = 5. Since 5 is etween lies in qudrnt I. nd π, its terminl side is coterminl with ( ) =. Since is etween 70 nd 60, its terminl side lies in qudrnt I rd is pproximtely 46 which is coterminl with = 786 nd with = 46 nd with = 66. Since 66 is etween 0 nd 90, its terminl side lies in qudrnt IV = (6.985) rd = (.56) 80 = rd =.048 rd 0

4 Chpter Trigonometric Functions '45 = % ' = 6.65 = (6.65),600 & 80 = rd rd = 4 80 = Since θ = s r, we hve = s r = 5 = 0.65 rd (4) 70. Since θ = 80 s ' & ), we hve % r ( = 80 s ' & ) = % r ( 80 4 ' & ) = 65 % 5 ( ' 7. s = rθ, where r = 5 km nd θ = (7 ) = 54, s = (5) &(54 ) ) 49 km % 80 ( 74. θ = (0.5) = 0.5 r = s = 6 = 5 cm At 4:0, the minute hnd hs moved of circumference from its position t the top of the clock. The hour hnd hs moved 4 twelfths of circumference from the sme position. Therefore, they form n ngle of C 4 C rdins, where C = totl circumference or π rdins. Thus, the desired ngle is (π) 4 9 (π) = π = = rd 0.79 rd We re to find centrl ngle θ, in circle of rdius cm, when the rc length is 9.5 cm. 80s s = rθ; θ = = 80(9.5) 5 80 r () 80. Since s = rθ we hve r = s = in 8. Since s = rθ, we hve s = (0 cm) 4 cm 84. Since s = rθ, we hve ' dimeter = s = (5.000 km) &.44 ) m % 80 ( 86. Since s = rθ, we hve 80 width of field s = (865 ft)(.5) 8 ft 80

5 Exercise. Degrees nd Rdins = 5 0 rd r = 55 mi s Since s = rθ, we hve width of oject s = (55 mi)(5 0 7 rd) = mi mi = ( mi)(,609 m/mi) 0.5 m mi = (0.5 m)(9.7 in) 4.9 in 90. Assuming tht n ngle corresponding to n entire circumference is swept out in yer (5 weeks), nd tht the mount swept out in weeks is proportionl to the time, we cn write ngle time = ngle time = 5 θ = 5 (π) = rd.57 rd error in distnce error in time 9. We use the proportion time, yer = (65 dys) 4 hours &,600 seconds &. dy % hour % x Thus, 65 seconds = r 65 4,600 seconds = ctul distnce ctul time. Let x = error in distnce. The ctul x = 65 ( miles) 65 4,600 = ( miles) 4,600 4,900 miles 94. Since A = r θ nd P = s + r = rθ + r, we cn eliminte θ etween the two equtions nd write A = r θ, θ = A r. P = r A % r ' + r = A & r Thus, P = (45.7) r. + (8.4) 5 cm 96. (A) The rdius of the pulley is 5 cm nd u meters of rope is the sme s n rc length of s = u meters. Use θ = s, where s nd r re in the sme unit. r (B) Yes. In the formul θ = s, r chnges when the dimeter chnges. r (C) Since s = 575 cm nd r = 5 cm, then θ = 575 = 5 rd. 5

6 Chpter Trigonometric Functions 98. Since the two wheels re coupled together, the distnce (rc length) tht the drive wheel turns is equl to the distnce tht the shft turns. Thus, s = r θ s = r θ r θ = r θ 00. In view of the explntion in Prolem 98 ove, θ = r θ r = 0 () = 0 rd. r = r = 7 () = 8 mm 0. The minimum distnce is chieved when the smllest front ger nd the lrgest rer ger re used. For one complete revolution t the pedls, the lrge rer ger will hve 0 revolution. Thus: 0 % the minimum distnce = (8π) ' 55 in. & EXERCISE. Liner nd Angulr Velocity 6. V = rω = 5(0.07) = 8.75 mm/sec 8. V = rω = 0.567(45) = 8.5 km/hr 0. ω = V r = 90 8 =.5 rd/sec. ω = V r = 55 =.85 rd/sec 5 4. ω = t = 5.0 = 0.74 rd/sec 6. ω = t = = 0.68 rd/sec revolutions per second = 500 π rd/sec =,000π rd/sec V = rω = % 6 & ((,000π)cm/sec =,000π cm/sec ' =,000π cm sec m = 94 m/sec 00 cm. ω = V r = 700 cm/sec cm rd/sec 4. ω = V r = rd/sec 6. The erth rottes once every.9 hours or π rdin in.9 hours. Thus, ω = θ t = rd.9 hr rd% V = rω = (,964 mi) ',04 mi/hr.9 hr &

7 Exercise. Liner nd Angulr Velocity 8. (A) The sun mkes full rottion, or π rdin in 7 dys, or 4.7 hours. Thus, ω = θ t = rd/sec (B) Note tht r = dimeter. Thus, & V = rω = % 865,400 miles( ( rd/sec) 4,00 mph ' 0. The Neptune trvels revolution, or π rdin, in 64 yers, or hours. Thus, ω = θ t = rd hr V = rω = ( rd' mi) & ),00 mph % hr (. Using suscript M to denote quntities ssocited with the moon of Jupiter, nd suscript J to denote quntities ssocited with Jupiter, we cn write: ω M = M t ω J = J t θ M = ω M t Using the hint (given in Prolem ), π = θ J θ M, hence, π = ω J t ω M t = (ω J ω M )t t = J M Thus, t = hr 4.5 (9 hr 55 min 9.9 hr; 4 hr 0 min = 4.5 hr) θ J = ω J t 4. (A) Using right-tringle trigonometry, c = 5 sec θ. Since rps corresponds to n ngulr velocity of π rd/sec, t the end of t sec, θ = πt. Sustituting the ltter into the eqution for c, we otin c = 5 sec πt. (B) The rte of chnge of the length of the light em seems to e incresing s time increses from 0.00 to 0.4. When t = 0.5, c = 5 sec(π/), which is not defined. The light hs mde one qurter turn nd the spot is no longer on the wll. TABLE t sec c ft In view of the ove formul (Prolem 5), we hve (8) 5 & % ((r)(60) ' = 6.5, 6,60 where r is the numer of revolutions per minute. Solving for r we otin r = (6.5)(6,60) (8) 5 66 rpm. & % ((60) ' 4

8 Chpter Trigonometric Functions EXERCISE. Trigonometric Functions 5. Q(, ) =, % ', r = + = & Q 5, % ' is on the unit circle. & =, so sin x =, cos x = 5, tn x = 5, cot x = 5, sec x =, csc x = Q, % ' = &, % ', r = + = & =, so Q, % ' is on the unit circle. & sin x =, cos x =, tn x =, cot x =, sec x =, csc x =. 6. Q, & % ' ( =, & % (, r = + = ' circle. + =, so Q, & % ( is on the unit ' sin x =, cos x =, tn x =, cot x =, sec x =, csc x =. 8. P(, 9), r = = 5 nd Q 4 5, % '. Thus 5& sin x = 5, cos x = 4 5, tn x = 4, cot x = 4, sec x = 5 4, csc x = P( 5, ), r = 5+44 = ; Q 5 %, '. Thus, & sin x =, cos x = 5, tn x = 5, cot x = 5, sec x =, csc x = 5.. P(4, 7), r = = 5; Q 4 5, 7 % '. Thus, 5& 7 4 sin x =, cos x = 5 5, tn x = 7 4, cot x = 4 5 5, sec x =, csc x = We sketch reference tringle nd lel wht we know. Since cos θ = r =, we know tht = nd r =. Use the Pythgoren theorem to find : + = = = 5 = 5 is positive since P(, ) is in qudrnt I. We cn now find the other five functions using Definition : 5 sin θ =, tn θ = 5, cot θ =, sec θ = 5, csc θ = terminl side of P(, ) 5 0 5

9 Exercise. Trigonometric Functions 6. We sketch reference tringle nd lel wht we know. Since sin θ = r = 4, we know tht = 4 nd r = 5. 5 Use the Pythgoren theorem to find : + ( 4) = 5 = 5 6 = 9 = is negtive since P(, ) is in qudrnt III. We cn now find the other five functions using Definition : cos θ = 5, tn θ = 4, cot θ = 4, sec θ = 5, csc θ = 5 4. terminl side of P(, ) 5 8. cot x = 4 nd x is Qudrnt I ngle: = 4, = nd r = (4) + () = 7, so sin x = r = 7 cos x = r = 4 7 tn x = = 4 sec x = r = 7 4 csc x = r = 7 0. tn x = nd x is Qudrnt IV ngle: =, = nd r = 0, so sin x = r = 0 cos x = r = 0 cot x = = = sec x = r = 0 csc x = r = 0 = 0 6

10 Chpter Trigonometric Functions. csc x = 5 7 nd x is Qudrnt III ngle: = 7, r = 5, so = (5) (7) = 4, nd sin x = r = 7 5 cos x = r = 4 5 tn x = = 7 4 = 7 4 cot x = = 4 7 = 4 7 sec x = r = 5 4 = sec x = nd x is Qudrnt II ngle: r =, =, so = =, nd sin x = r = cos x = r = tn x = = = cot x = = = csc x = r = = 6. No. For ll those vlues of θ for which oth re defined, tn θ = /(cot θ), hence oth re either positive or oth re negtive. 8. Rdin mode: cos(7 rd) = Rdin mode: cot (5 rd) = Degree mode: csc 9 = sin9 =.8 4. Degree mode: sin 98 = Degree mode: tn 48 = Rdin mode: sec 9 5 = cos Rdin mode: csc ( 9) = sin( 9) 44. Degree mode: tn( 8 ) = = Degree mode: cos ( 55 ) = =.6 5. (, ) = (, ), r = + = + = sin θ = r = cos θ = r = tn θ = = csc θ = r = sec θ = r = cot θ = = 7

11 Exercise. Trigonometric Functions 54. (, ) = (, ), r = + = ( ) + ( ) = sin θ = r = csc θ = r = cos θ = r = tn θ = = sec θ = r = cot θ = = 56. I, II 58. I, III 60. I, II 6. II, III 64. II, IV 66. II nd III 68. Given cos θ =, cot θ > 0, then the terminl side of θ is in qudrnt III. We sketch the reference tringle nd lel wht we know. Since cos θ = r =, we know tht = nd r =. Use the Pythgoren theorem to find : ( ) + = () = 4 = = is negtive since P(, ) is in qudrnt III. terminl side of - P(, ) The other five functions re: sin θ =, tn θ =, cot θ =, sec θ =, csc θ =. 70. Given csc θ = 5 (or sin θ = ), cos θ < 0, then the terminl side of 5 θ is in qudrnt II. We sketch the reference tringle nd lel wht we know. Since sin θ = r =, we know tht = nd r = 5. 5 Use the Pythgoren theorem to find : + () = ( 5 ) = 5 = 4 = is negtive since P(, ) is in qudrnt II. Now we hve: cos θ = 5, tn θ = 5, cot θ =, sec θ =. terminl side of P(, ) 8

Chpter Trigonometric Functions 7. tn x = nd cos x > 0, so x is Qudrnt IV ngle: =, = nd r = 5, so sin x = r = 5 cos x = r = 5 cot x = = = sec x = r = 5 csc x = r = 5 = 5 74. No.

12 Chpter Trigonometric Functions 7. tn x = nd cos x > 0, so x is Qudrnt IV ngle: =, = nd r = 5, so sin x = r = 5 cos x = r = 5 cot x = = = sec x = r = 5 csc x = r = 5 = No. For exmple cos & % ' ( = = cos & % (, ut α = ' nd β = re not coterminl. 76. Use the reciprocl identity sec x = 78. Degree mode: sin 7.85 = Use the reciprocl reltion sec θ = 8. Rdin mode: tn 4.78 = Use the reciprocl reltion csc θ = cos x. sec x = =.4098 cos. Degree mode: sec 07.5 = cos 07.5 =.0 sin. Rdin mode: csc( 0.408) = sin ( 0.408) = Degree mode: cot 5 5'55 = cot % ' = 7.05,600 & 88. Degree mode: cos( 4.8 ) = Degree mode: sin 605 4'75 = sin % ' = 0.96,600 & 9. Use the reciprocl reltion sec θ = cos. Rdin mode: sec.55 = cos.55 = Degree mode: cot( ) = Cotngent nd cosecnt. Since cot x = / nd csc x = r/, neither is defined when the terminl side of the ngle lies long the positive or negtive horizontl xis, ecuse will e 0, nd division y zero is not defined. 98. (A) Since θ = s r, nd s = 0, nd r, the mesure of CA, is 4, we hve θ = 0 4 =.5 rd. (B) Since cos θ = r nd sin θ =, we hve r = r cos θ = 4 cos.5 = r sin θ = 4 sin.5 Thus, (, ) = (4 cos.5, 4 sin.5) = (.0,.9) 9

13 Exercise. Trigonometric Functions 00. (A) Since θ = s r, nd s = 4, nd r, the mesure of CA, is, we hve θ = 4 = 4 rd. (B) Since cos θ = r nd sin θ =, we hve r = r cos θ = cos 4 = r sin θ = sin 4 Thus, (, ) = ( cos 4, sin 4) = ( 0.654, 0.757) 0. From the figure, we note: s = rθ r = + = + 4 = 5 = 5 4 (,4) s tn θ = = 4 5 θ = tn 4 Thus, s = 5 tn units (clcultor in rdin mode). 04. We sketch the reference tringle nd lel wht we know. Since cot θ = = nd since the terminl side of θ is in qudrnt III, we know tht = nd =. Use the Pythgoren theorem to find r: r = + = ( ) + ( ) = + 4 = 5 or r = 5 Now we hve: sin θ = 5, cos θ = 5, tn θ =, sec θ = 5, csc θ = 5. terminl side of P(, ) 06. If θ = 60, I = k cos 60 = k(0.50) = 0.50k. Thus, t the ngle θ = 60 the light intensity I will e 50% of the verticl intensity. 08. If θ = 8 (summer solstice), E = k cos θ = k cos 8 = 0.99k. If θ = 55 (winter solstice), E = k cos θ = k cos 55 = 0.57 k. 0. (A) If n = 6, A = n tn 80 & = 6 tn 80 & = 6 tn 0 = n % 6 % If n = 0, A = 0 tn 80 & = 0 tn 8 = % If n = 00, A = 00 tn 80 & = 00 tn.8 = % If n =,000, A =,000 tn 80 & =,000 tn 0.8 =.460., 000% 80 If n = 0,000, A = 0,000 tn & = 0,000 tn 0.08 = , 000% n ,000 A n

14 Chpter Trigonometric Functions (B) The re of the circle is A = πr = π() = π, nd A n seem to pproch π, the re of the circle, s n increses. (C) No. An n sided polygon is lwys polygon, no mtter the size of n, ut circumscried polygon cn e mde s close to the circle s you like y tking n sufficiently lrge.. From Prolem, x = cos 0πt = 5 (sin 0t) Now for t = 0, we hve x = cos (sin 0) = + 5 = 6 For t = 0.0 sec, we hve x = cos 0.π + 5 (sin 0.) = = From Prolem, I = 5 sin(48πt π) Now for t = 0.0 sec, we hve I = 5 sin(4.88π π) = 5 sin(.88π) mp 6. (A) If the ngle of inclintion θ = 89., then m = tn θ = tn 89. = 7.6 If the ngle of inclintion θ = 79, then m = tn θ = tn 79 = 0.0 (B) If the ngle of inclintion θ = 0, then m = tn θ = tn 0 Then the eqution of the line is given y y + 4 = tn 0 (x 7) y = tn 0 (x 7) 4 y = x tn 0 7 tn 0 4 = 5.4x +.0 EXERCISE.4 Additionl Applictions 0. Use n = sin n sin,. Use n = sin n sin, where n =., n =.00, nd α = 4. where n =.66, n =., nd α = sin sin 45.0 Solve for β: = Solve for β: =.00 sin. sin sin 4.. sin 45.0 sin β = sin β =..66 sin 4. β = sin & β = sin. sin 45.0 &. %.66 % = 5 = The index of refrction for flint glss is n =.66 nd tht for ir is.00. Find the ngle of incidence α such tht the ngle of refrction β is 90. sin sin = n ; sin α = sin 90 ; α = sin n () & = 7 %

15 Exercise.4 Additionl Applictions 6. A diver looking stright up towrds the surfce receives cone of light rys, since ll of the light rys ove the surfce re refrcted downwrd. Outside of the cone will e drk. Within the cone the driver will see wht is ove the surfce in ll directions, including ots nd uoys on the wter, people stnding on the shore or dock, nd so on. The cone ngle θ is out Use n n = sin α sin β, where n =.00. α = 90 4 = 47 r sin β = = r in n Solve for n :.00 = sin in. We use sin = S w S, where θ = 54 nd S = 55 km/hr. Then we solve for S w : sin 54 = S w 55 ; S w = 55 sin 7 = 5 km/hr 4. We use sin = S s S, where S = Mch. =.S s. Then we solve for θ: sin = S s ; sin.s s =.. Thus = 8, θ = We use sin = S S p, where θ = 0 nd S = 0 0 cm/sec. Then we solve for S p : sin 0 = 00 S p ; S p = 00 sin 6 =.4 00 cm/sec 8. d = + sin 4θ =.8 + ( 4.) sin(4 40 ) =.8 4. sin 60 = 0. Let x = OA, then r = (x + 4) + (48) r = 9' =,04 nd (x + 4) = (,04) (48) =,0.50 or x =, =,6.50 t A 48 t We oserve tht: t 40.5+,6.50,0.50 = 48 = t 96 x r Solving for t we otin: (,6.50)(96) (,0.50)(90.5) t =, in. O

16 Chpter Trigonometric Functions EXERCISE.5 Exct Vlue for Specil Angles nd Rel Numers Note for Prolems 6 6: The reference ngle α is the ngle (lwys tken positive) etween the terminl side of θ nd the horizontl xis. 6. α = θ = 45 P α = 45 = P 0. α = π/4 = π/4-4 P. α = π 5 6 = 6 4. α = = 0 6. α = π 5 = P P 5 6 P (, ) = (, 0), r = cos 0 = r = = 0. (, ) = (, 0), r = cot 0 = which is not defined since = 0. Use the specil 0 60 tringle s the reference tringle. Use the sides of the reference tringle to determine P(, ) nd r. Then use Definition. (, ) 0 (, ) = (, ), r = sin 0 = r =

17 Exercise.5 Exct Vlue for Specil Angles nd Rel Numers 4. Use the specil 0 60 tringle s the reference tringle. Use the sides of the reference tringle to determine P(, ). Then use Definition. (, ) 0 (, ) = (, ), r = tn 0 = = or 6. Locte the 0 60 reference tringle, determine P(, ) nd r, then evlute. / (, ) (, ) = (, ), r = csc = r = or 8. (, ) = (0, ), r = cos & % ' ( = cos & % ( ' = r = 0 = 0 / / 0. In view of Prolem 8, tn = = 0 which is not defined.. Locte the 0 60 reference tringle, determine (, ) nd r, then evlute. (, ) sin & % ( = ' r = 8. Locte the 0 60 reference tringle, determine (, ) nd r, then evlute. -0 (, ) cot 6 = = = 4. Locte the 0 60 reference tringle, determine (, ) nd r, then evlute. -0 cot( 0 ) = = = 40. (, ) = (, 0), r = sin( π) = r = 0 = 0 csc( π) = r = 0 which is not defined. 6. Locte the 45 reference tringle, determine (, ) nd r, then evlute. 7/4 45 (, ) cos 7 4 = r = or 4. Locte the 0 60 reference tringle, determine (, ) nd r, then evlute (, ) sin( 00 ) = sin 60 = r = 4

18 Chpter Trigonometric Functions 44. Locte the 0 60 reference tringle, determine (, ) nd r, then evlute. (, ) Locte the 0 60 reference tringle, determine (, ) nd r, then evlute sec( 50 ) = r = = or tn 480 = = 48. The cotngent function is not defined t θ = 0, π, nd π, ecuse cot θ = / nd = 0 for ny point on the horizontl xis. 50. The secnt function is not defined t θ = π/ nd π/, ecuse sec θ = r/ nd = 0 for ny point on the verticl xis. 5. It is shown tht cos(90 ) = 0 nd sin & % ( = = , so tht the 6 ' clcultor displys exct vlues of these. Thus sin % 0 / & 60 ( is not ' given exctly. To find sin % & (, locte the 0 60 reference ' tringle, determine (, ) nd r, then evlute. sin & % ( = ' 54. It is shown tht sin( 50 ) = = nd tn 5 & % ( =.000, so tht the clcultor displys exct 4 ' vlues for these. Thus tn( 50 ) is not given exctly. To find tn( 50 ), locte the 0 60 reference tringle, determine (, ) nd r, then evlute. 56. Drw reference tringle in the first qudrnt with side djcent reference ngle nd hypotenuse. Oserve tht this is specil 45 tringle tn( 50 ) = = = or 58. Drw reference tringle in the third qudrnt with side opposite reference ngle nd hypotenuse. Oserve tht this is specil 0 60 tringle. - 0 (A) θ = 45 (B) θ = 4 (A) θ = 0 (B) θ = 7 6 5

19 Exercise.5 Exct Vlue for Specil Angles nd Rel Numers 60. Drw reference tringle in the second qudrnt with side djcent reference ngle nd side opposite. Oserve tht this is specil 45 tringle. (A) θ = 5 (B) θ = All should equl 0.079, ecuse the cosine function is periodic with period π. 66. Clcultor in rdin mode: (A) cot( ) = 0.64 (B) cot 8.7 =. (C) cot(.64) = 4 cos( ) cos 8.7 = 0.64 sin( ) sin 8.7 =. cos(.64) = 4 sin(.64) 68. (A) cos 5 = 0.8 (B) cos(.4) = 0.67 (C) cos(,00) = 0.67 cos( 5) = 0.8 cos(.4) = 0.67 cos(,00) = (A) sin 4 = 0.0 (B) sin ( 6.) = 0.69 (C) sin 766 = 0.7 cos 4 = 0.0 cos ( 6.) = 0.69 cos 766 = 0.7 % 7. cos x sec x = cos x ' = Use Identity () cos x & 74. tn x csc x = sin x cos x sin x = cos x 76. cos x sin x = cos x cos x = cos x 78. tn( x) cos( x) = sin( x) cos( x) = sec x Use Identities (4) nd () = sec x Use Identity (9) cos( x) = sin( x) = sin x Use Identity (4) 80. We cn drw reference tringle in the second qudrnt with side opposite reference ngle nd side djcent. We cn lso drw reference tringle in the fourth qudrnt with side opposite reference ngle nd side djcent. Ech tringle is specil 0 60 tringle We cn drw reference tringle in the second qudrnt with side djcent reference ngle nd hypotenuse. We cn lso drw reference tringle in the third qudrnt with side djcent nd hypotenuse. Ech tringle is specil 0 60 tringle. 0 θ = 0 or θ = 00 θ = 5 6 or θ = 7 6 6

20 Chpter Trigonometric Functions 84. For x = 4, sin x =, which is the lest positive x in rdon mesure for which sin x =. 86. For x =, tn x = ; x = is the lest positive x. 88. (A) Since 4 x = cos 0 nd cos 0 = ; 4 x = ; x = 8 Since y 4 = tn 0 nd tn 0 = ; y 4 = ; y = 4 (B) Since 5 y = tn 45 nd tn 45 = ; 5 y = ; y = 5 Since 5 x = sin 45 nd sin 45 = ; 5 x = ; x = 5 (C) Since x = cos 60 nd cos 60 = ; x = ; x = Since y = sin 60 nd sin 60 = ; y = ; y = 90. Since (, ) is on unit circle with (, ) = ( , ), we cn solve cos s = or sin s = Then s = cos ( ) = 0.54 or s = sin (0.4940) = (A) Identity (5) (B) Identity (9) (C) Identity () 94. π or g(x) = 4 cos x % ' is periodic ecuse cosine function is periodic. If T is the period of g(x), then we & should hve: g(x) = g(x + T) g(x + T) = 4 cos x +T % & ' = 4 cos x + T % ' & ) = 4 cos x % & ' cos T % & '( sin x % & ' sin T %, * '- + &. The right-hnd-side will e equl to g(x) = 4 cos x % & ' if cos T % & ' = 0 nd sin T % ' = 0. & Thus, T = 6π is the period. 98. k(x) = x cos x is not periodic since cos x is periodic ut x is not. The product of two terms (cos x nd x) is periodic if oth fctors re periodic nd with the sme period. 00. S = 0.5 S = S + cos S = cos 0.5 =.7758 S = S + cos S = cos.7758 = S 4 = S + cos S = cos = S 5 = S 4 + cos S 4 = cos = =

at its center, then the measure of this angle in radians (abbreviated rad) is the length of the arc that subtends the angle.

at its center, then the measure of this angle in radians (abbreviated rad) is the length of the arc that subtends the angle. Notes 6 ngle Mesure Definition of Rdin If circle of rdius is drwn with the vertex of n ngle Mesure: t its center, then the mesure of this ngle in rdins (revited rd) is the length of the rc tht sutends