= = 6 radians 15. θ = s 30 feet (2π) = 13 radians. rad θ R, we have: 180 = 60

Transcription

1 SECTION - 7 CHAPTER Section. A positive ngle is produced y counterclockwise rottion from the initil side to the terminl side, negtive ngle y clockwise rottion.. Answers will vry. 5. Answers will vry. 7. Since rottion corresponds to 0, 9 rottion corresponds to (0 ) = Since rottion corresponds to 0, rottion corresponds to (0 ) = 70. Since rottion corresponds to 0, 9 8 rottion corresponds to 9 (0 ) = θ = s centimeters = = rdins 5. θ = s 0 feet = =.5 rdins r centimeters r feet 7. Since rottion corresponds to π rdins, 8 rottion corresponds to 8 (π) = rdins 9. As in prolem 7, rottion corresponds to (π) = rdins. As in prolem 7, rottion corresponds to (π) = rdins rd. Using the reltion θ R = 80 θ D, rd 5. Using the reltion θ R = 80 θ D, rd we hve: 80 0 = rd rd we hve: 80 ( 5 ) = rd rd 80 0 = rd rd 80 ( 90 ) = rd rd = rd rd 80 ( 5 ) = rd rd 80 0 = rd rd ( 80 ) = π rd 80 rd = 5 rd rd 80 = π rd Using the reltion θ D = 80 rd θ R, we hve: 80 = 0 80 = = 0 = π = π = 0 9. Using the reltion θ D = 80 rd θ R, 80 we hve: = ( π) = = ( π) = 0. True. "Stndrd position" mens tht oth ngles hve the positive x xis s their initil sides. If they hve the sme initil side nd the sme mesure, they must hve the sme terminl side s well.

2 8 CHAPTER TRIGONOMETRIC FUNCTIONS. True. Angles tht re complementry hve mesures tht dd up to 90, so if oth re positive nd dd to 90 the mesure of ech hs to e etween 0 nd 90. Tht's the definition of cute. 5. Flse. The terminl side of the ngle 5 (or 7 ) is in qudrnt I '" = = '9" = = = (0.0 0)' =.5' = '(0.5 0)" = '". 0. = 0 (0. 0)' = 0.8' = 0 '(0.8 0)" = 0 '" rd 5. Using the reltion θ R = 80 θ rd D, we hve =.7 rd 80 rd 7. Using the reltion θ R = 80 θ rd D, we hve 08. =.89 rd First we convert 5'" to deciml degrees: 5 5'" = 0 00 =. rd Then we use the reltion θ R = 80 θ D to find θ R = rd (. ) = 0. rdins Using the reltion θ D = 80 θ rd R, we hve 80 (0.9) = Using the reltion θ D = 80 θ rd R, we hve 80 (.) = Using the reltion θ D = 80 θ rd R, we hve 80 (.5) =.5 For Prolems 57-8 the following sketch is useful: 57. From the sketch, we find, since 80 < 50 < 70, 50 is III qudrnt ngle. 59. Since 70 < 75 < 0, 75 is IV qudrnt ngle.. Since < < π, is II qudrnt ngle.. is qudrntl ngle. 5. Since 0 < 0 < 70, 0 is I qudrnt ngle. 7. Since.57 <.5 < 0,.5 is IV qudrnt ngle.

3 SECTION A centrl ngle of rdin mesure is n ngle sutended y n rc of the sme length s the rdius of the circle. 7. Of ll the coterminl ngles 0 + n0, = 570 is in the correct intervl. 7. Of ll the coterminl ngles 5 + n0, 5 0 = 5 is in the correct intervl Of ll the coterminl ngles + nπ, + π = nd + π = re in the correct intervl. 77. Of ll the coterminl ngles π + nπ, π π = π nd π π = π re in the correct intervl. s 79. We pply c = with s = 500 mi, nd θ = Then = c 0 or 500 = 7.5 c 0 0c 500 = 0c 7.5 c 0 80,000 = 7.5c c =,000 mi 8. The 7.5 ngle nd θ hve common side. (An extended verticl pole in Alexndri will pss through the center of the erth.) The sun's rys re essentilly prllel when they rrive t the erth. Thus, the other two sides of the ngles re prllel, since sun ry to the ottom of the well, when extended, will pss through the center of the erth. From geometry we know tht the lternte interior ngles mde y line intersecting two prllel lines re equl. Therefore, θ = To clculte ngulr speed, we need rdins nd seconds. 00 revolutions π rdins = 00π rdins minute = 0 sec. 00 rdins ngulr speed = = 0.9 rd 0sec sec To clculte liner speed, we need the circumference of the wheel. c = πd = π In 00 revolutions, the liner distnce is 00 π = 00π feet. 00 feet liner speed = =.8 ft/sec 0sec 85. From the figure, it should e cler tht the minute (lrger) hnd is displced π rdins from noon, while the hour (smller) hnd is displced θ = of full revolution, or π rdins from noon. Hence the lrger ngle etween the hnds hs mesure θ + π, or π + π = π + π = 7 π rdins. 87. We use θ = s r with r = θ = 000 = 00 rdins. 5 (0) = 5 centimeters nd s = 0 meters 00 centimeters meter = 000 centimeters. Then 89. In one yer the line sweeps out one full revolution, or π rdins. In one week the line sweeps out 5 of one full revolution, or 5 π = rdins. = 0. rdin to two deciml plces.

4 50 CHAPTER TRIGONOMETRIC FUNCTIONS 9. Following exmple, we reson tht points on the circumference of ech wheel trvel the sme distnce. Then s = r θ = r θ = s. The rdius of the front wheel is (0) or 0 centimeters. The rdius of the ck wheel is (0) or 0 centimeters. So 0 θ = 8(0) θ = 0 0 θ = rdins 9. Front wheel: d = 0 cm; c = πd = 0π cm. One complete revolution is π rdins. So π rd = 0π cm or cm = 0 rdin. In one hour, the liner distnce is 0 km which is,000,000 cm.,000,000 cm = 0,000,000 rdin = 50,000 rdins 50, 000 rdins hr =.9 rd/sec hr 00sec Bck wheel: d = 0 cm; c = 0π cm π rd = 0π cm or cm = 0 rdin,000,000 cm =,000,000 rdin =,. rdins 0,. rdins hr = 9. rd/sec hr 00sec 95. We use c s = r θ with r = mi nd θ = 9. 0 rd. Then c ( )(9. 0 ) = 85,000 mi 97. We use c s = r θ. r = 750 ft. θ must e converted to rdins to use the formul, thus, using rd θ R = 80.5 we hve θ R = rd. Then c 750 = ft. 7 7 Section. The unit circle is the circle with center the origin nd rdius. In the uv-coordinte system it hs the eqution u + v =.. Answers will vry. 5. Since tn x =, when = 0 the tngent is undefined. This occurs when x = 5,,,... For Prolems 7 8, the following sketch is useful: 7. W(π) = (, 0) 9. W = (0, ). W 7 =,. W =, 5. W =, 7. W( π) = (, 0)

5 SECTION - 5 It's helpful to rememer these rules for finding the coordintes of the circulr points when x is multiple of π: ) If x hs denomintor of, like 5, the coordintes re,. Locte the qudrnt to decide on the pproprite signs. ) If x hs denomintor of, like, the coordintes re,. Locte the qudrnt to decide on the pproprite signs. ) If x hs denomintor of, like 7, the coordintes re,. Locte the qudrnt to decide on the pproprite signs. ) If the denomintor is, or there is no denomintor, like or π, x is on one of the coordinte xes nd you should e le to find the coordintes y just locting x on the unit circle. IMPORTANT: These rules re only vlid if x is reduced frction! For exmple, hs denomintor ut cn e reduced to. 9. W(π) = (, 0) so sin π = 0..W 7. W =, so tn 7 = = (0, ) so sec = is not defined. 0 = 5. W =, so cos = 7. W =, csc so = = 9. W( π) = (, 0) so cot( π) = is not defined. 0. Since cos x =, it is negtive in qudrnts II nd III.. Since sin x =, it is positive in qudrnts I nd II. 5. Since cot x =, it is negtive if nd hve opposite signs. This occurs in qudrnts II nd IV Common Error: cos ; clcultor must e in rdin mode (clcultor in degree mode) (clcultor in rdin mode) 7. sin( 7'") = sin 0 00 = Flse. The domin of the wrpping function is the set of ll rel numers. 5. Flse. The wrpping function is not one-to-one. W(0) = W(π) ut 0 π. 5. Flse. The reciprocl of sin x is csc x, unless sin x = Flse. The secnt function is not one-to-one. sec 0 = sec π, ut 0 π. 57. Flse. The domin of sin x is ll rel numers; the domin of csc x excludes ll integer multiples of π. 59.

6 5 CHAPTER TRIGONOMETRIC FUNCTIONS Since there is no secnt utton, we enter y = sec x s y = /cos x. There re no zeros nd turning points. They occur when the grph is t height or. Since sec x =, sec x = ± when cos x = ±. This cos x occurs for x = π, π, nd π, so the turning points re (π, ), (π, ), nd (π, ).. There re 5 zeros. Since tn x = sin x, the zeros occur when sin x = 0. The x vlues re 0, π, π, π, nd π. cos x There re no turning points. For Prolems - 7, the following sketches re useful. Sketch : counterclockwise wrpping Sketch : clockwise wrpping. Using sketch nd.57 < <., we see tht W() is in the second qudrnt. Hence, if W() = (, ), then < 0 nd > 0, tht is, is negtive nd is positive. 5. Using sketch nd.57 < <., we see tht W() is in the second qudrnt. Hence, if W() = (, ), then < 0 nd > 0, tht is, is negtive nd is positive. 7. Using sketch nd.7 < 5 <.8, we see tht W(5) is in the fourth qudrnt. Hence, if W(5) = (, ), then > 0 nd < 0, tht is, is positive nd is negtive. 9. Using sketch nd. <.5 <.57, we see tht W(.5) is in the third qudrnt. Hence, if W(.5) = (, ), then < 0 nd < 0, tht is, nd re negtive. 7. Using sketch nd.8 <. <.7, we see tht W(.) is in the first qudrnt. Hence, if W(.) = (, ), then > 0 nd > 0, tht is, nd re positive.

7 SECTION v 0 (, 0) u From the sketch, if W(x) = (, 0), 0 x < π, then x = 0. Since W(x) = W(x + kπ), k ny integer, if there re no restrictions on x, then x = 0 + kπ or x = kπ, k ny integer. 75. From the sketch, if W(x) =,, 0 x < π, then x =. Since W(x) = W(x + kπ), k ny integer, if there re no restrictions on x, then x = + kπ, k ny integer. 77. W(x) is the coordintes of point on unit circle tht is x units from (, 0), in counterclockwise direction if x is positive nd in clockwise direction if x is negtive. W(x + π) hs the sme coordintes s W(x), since we return to the sme point every time we go round the unit circle ny integer multiple of π units (the circumference of the circle) in either direction. 79. sin x < 0 in qudrnts III nd IV; cot x < 0 in qudrnts II nd IV; therefore, oth re true in qudrnt IV. 8. cos x < 0 in qudrnts II nd III; sec x > 0 in qudrnts I nd IV; therefore, it is not possile to hve oth true for the sme vlue of x. 8. cos x = is lwys defined. There re no vlues for which it is undefined. 85. tn x = is undefined if nd only if = 0. This occurs t points on the verticl xis. The only vlues of x etween 0 nd π for which W(x) is on the verticl xis re nd. 87. sec x = is undefined if nd only if = 0. This occurs t points on the verticl xis. The only vlues of x etween 0 nd π for which W(x) is on the verticl xis re nd. 89. Two points re given; to find the eqution of the line we first find the slope, then use point-slope form. 0 m = y 0 = (x 0) 0 y = x y = x x y = 0

8 5 CHAPTER TRIGONOMETRIC FUNCTIONS 9. Given n =, r = 5, we use the given formul A = nr sin n to otin: A = ()(5) sin = 50 sin = 50 = 75 squre meters 9. Given n =, r =, we use the given formul A = nr sin n to otin: A = ()() sin = sin = = 0.78 squre inches 95. = 0.5 = + cos = cos 0.5 =.7758 = + cos = cos.7758 =.5959 = + cos = cos.5959 = = + cos = cos = = to six deciml plces. Section. Given two sides, or one cute ngle nd side, solving for the remining three quntities is clled solving the right tringle.. Yes, digonl of the rectngle prtitions it into two right tringles. 5. Any numer of similr tringles cn hve given set of ngles, s long s the sides hve equl rtios, thus, the length of the sides is not determined. 7. sin θ = Opp 7 9. csc θ = Hyp Hyp = sec θ 7. Adj Hyp 5 Opp 7. tn θ = Opp 7. Adj Adj = cos θ 5 Hyp Adj = cot θ Opp Solve for α: α = = 7. Solve for : sin β = c sin 7.8 =.5 =.5 sin 7.8 =.05 Solve for : cos β = c cos 7.8 =.5 =.5 cos 7.8 =.8 7. c 0' Solve for α: α = 90 0' = 0' Solve for : tn β = tn 0' = = tn 0' = Solve for c: sec β = c c sec 0' = c = sec 0' = 9 9. c 5 Solve for β: β = 90 0' = 7 0' Solve for : tn β = tn 7 0' = 5 = 5 tn 7 0' = 7 Solve for c: csc α = c c csc 0' = 5 c = 5 csc 0' = 8

9 SECTION Solve for β: β = =.79 Solve for : tn α = Solve for c: sec α = c tn 5. =.8 =.8 tn 5. =.85 c sec 5. =.8 c =.8 sec 5. = Solve for β: tn β = tn β = β = tn 8. = 5.7 or 5 0'.00 Solve for β: sin β = c sin β = 0.0. β = sin 0.0 = 5.5 or 5 0'. Solve for α: α = ' = 5 0' Solve for c: csc β = c c csc 5 0' = 8. c = 8. csc 5 0' = 0. Solve for α: α = ' = 7 0' Solve for : cot β = cot 5 0' = 0.0 = 0.0 cot 5 0' = Flse. Knowing only the three ngles cn't tell us nything out the lengths of the sides. The two tringles t the right hve the exct sme ngles ut totlly different side lengths. 9. True.. Flse. c 5 According to the digrm, sin α = Opp Hyp = Adj nd cos β = c Hyp = c. According to the digrm, sec α = Hyp = 5 Adj nd cos β = Adj Hyp = 5.

10 5 CHAPTER TRIGONOMETRIC FUNCTIONS. See the figure. A line of positive slope m will form right tringle ABC s shown. tn θ = m. Here tn θ = θ = tn =. m 5. See the figure. A line of positive slope m will form right tringle ABC s shown. tn θ =. Here tn θ = 5 θ = tn 5 = See the figure. A line of negtive slope m will form right tringle ABC s shown. tn θ = Here tn θ = = θ = tn =. 9. See the figure. An ngle θ cn e formed y either of two lines, one with positive slope m nd the other with negtive slope m. Thus m = ±tn θ Here m = ±tn 0 = ±0. 5. See the figure. An ngle θ cn e formed y either of two lines, one with positive slope m nd the other with negtive slope m. Thus m = ±tn θ Here m = ±tn 80 = ± See the figure. An ngle θ cn e formed y either of two lines, one with positive slope m nd the other with negtive slope m. Thus m = ±tn θ Here m = ±tn 0 = ± (A) In tringle OAD, cos θ = Adj Hyp = OA = OA (B) In tringle OED, ngle EOD = 90 θ, ngle OED = 90 (90 θ) = θ. Thus cot OED = Adj Opp = DE = DE = cot θ (C) In tringle ODC, sec θ = Hyp Adj = OC = OC 57. (A) As θ pproches 90, OA = cos θ pproches 0. (B) As θ pproches 90, DE = cot θ pproches 0. (C) As θ pproches 90, OC = sec θ increses without ound. m.

11 SECTION (A) As θ pproches 0, AD = sin θ pproches 0. (B) As θ pproches 0, CD = tn θ pproches 0. (C) As θ pproches 0, OE = csc θ increses without ound.. Lel s shown t the left. In right tringle ADC, cot β = x h In right tringle ABC, cot α = d x h Hence x = h cot β nd d + x = h cot α d = h cot α x d = h cot α h cot β d = h(cot α cot β) d h = cot cot. Sketch figure: 5. Sketch figure: feet Let h = height of tree From the figure, it should e cler tht h tn 5. = 05 h = 05 tn 5. = 8 feet Let h = how fr trin clims. h sin ' = 580 h = 580 sin ' = 7.5 feet 7. Sketch figure: Note α = (') = ' dimeter = d = r tn α = r D r tn ' = 9,000 r = 9,000 tn ' d = (9,000)tn ' = 5 miles R D r Alterntively, we cn write sin α = r R r sin ' = 9,000 r = 9,000 sin ' d = (9,000)sin ' = 5 miles Although it is not cler whether 9,000 miles is to e interpreted s D, R, or D r, t this ccurcy it does not mtter. 9. Sketch figure: We will find θ nd doule it. sin θ =.5 v 7. We use g = t sin with v =., t =.0, θ = 8.0. g =.0sin8.0

12 58 CHAPTER TRIGONOMETRIC FUNCTIONS.5 θ = sin θ = sin.5 = g = 9.8 meters/second 7. A (A) We note tht the cle consists of wter section IB, nd shore section CB = 0 mi AB. Let y = IB nd d = CB = 0 x In right tringle ABI y x sec θ = nd tn θ = mi mi y = sec θ mi x = tn θ mi Cost of Wter Numer of Cost of Shore Numer of Thus the cost of the cle = + Section Per Mile Wter miles = y Section Per MileShore miles = d C(θ) = (5,000 dollrs )( sec θ mi) + (5,000 dollrs )(0 tn θ mi) mi mi C(θ) = 75,000 sec θ + 00,000 5,000 tn θ (B) θ C(θ) $8, $,5 $0, $0, $, Sketch figure: t 5 x C B r r E.0 D In tringle ABC, we hve r.0 = sin 5 In tringle ADE, we hve x r x Eliminting x, we see tht r x = = cot 5 (reciprocl identity).0 tn5 x = cot 5 r Sustituting, we hve r = sin 5 cot5r r = sin 5 (cot 5 r) r = sin 5 cot 5 sin 5 r r( + sin 5 ) = sin 5 cot 5 sin5cot5 r = sin5 r = 0.77 meters Section = tn 5

13 SECTION An eqution is n identity if it holds true for ll replcements of the vrile or vriles y rel numers for which oth sides re defined.. To evlute sin x nd (sinx), find sin nd squre the result. To evlute sinx, find nd then sine of the result. Clcultor in rdin mode. 5. Since one cn only see portion of the grph of function, it is not possile to tell from grph whether or not the function is periodic. 7. sine: π; cotngent: π; cosecnt: π 9. (A) Since the rnge of the cosine function is [, ], the lrgest nd smllest y vlues on its grph re, respectively, nd. The lrgest devition of the function from the x xis is therefore unit. (B) Since the rnge of the tngent function is ll rel numers, the grph devites indefinitely fr from the x xis. (C) Since the rnge of the cosecnt function is ll rel numers y or y, the grph devites indefinitely fr from the x xis.. (A) π, π, 0, π, π (B),,, (C) No x intercepts. (A) Defined for ll rel x (B),,, (C) π, π, 0, π, π 5. (A) There re no verticl symptotes (B),,, (C) π, π, 0, π, π 7. (A) A shift of π/ to the left will trnsform the cosecnt grph into the secnt grph. [The nswer is not unique see prt (B).] (B) The grph of y = csc(x π/) is π/ shift to the right nd reflection in the x xis of the grph of y = csc x. The result is the grph of y = sec x. The grph of y = csc x is π/ shift to the left nd reflection in the x xis of the grph of y = csc x. The result is not the grph of y = sec x. 9. Let f(x) = tn x x f( x) = tn( x ) ( x) tn x = x = tn x x = f(x) y = tn x is even. x. Let f(x) = csc x x f( x) = csc( x) x = sin( x) x sin x = x csc x = x = csc x x = f(x) y = csc x is even. x. Let f(x) = sin x cos x f( x) = sin( x)cos( x) = sin x cos x = f(x) y = sin x cos x is odd. 5. Let f(x) = sin x cos x f( x) = sin( x) cos( x) = sin x cos x Since f(x) f( x) nd f(x) f( x), 7. Let f(x) = x + sin x f( x) = ( x) + sin ( x) = x + ( sin x) = x + sin x 9. Let f(x) = x sin x f( x) = ( x) sin( x) = x ( sin x) = x sin x

14 0 CHAPTER TRIGONOMETRIC FUNCTIONS y = sin x cos x is neither even nor odd.. (, ) = (, 8) r = = sin θ = r = 8 0 = 5 cos θ = r = 0 = 5 8 = 00 = 0 csc θ = r = 0 8 = 5 sec θ = r = 0 = 5 = f(x) y = x + sin x is even. r (, 8) = f(x) y = x sin x is odd. tn θ = = 8 = cot θ = = 8 =. (, ) = (, ) r = = sin θ = r = cos θ = r = = tn θ = = ( ) ( ) = = csc θ = r = sec θ = r = = cot θ = = = = α = 0 00 = 0 α = 7 π = α = 5 + π =. cos θ = r = ecuse r > 0. (-, ) is negtive in the II nd III qudrnts. The smllest positive θ is ssocited with 0 reference tringle in the II qudrnt s shown in 0 the figure. - θ = 0 or π rdins. sin θ = r = ecuse r > 0. is negtive in the III nd IV qudrnts. The smllest positive θ is ssocited with 0 - reference tringle in the III qudrnt s drwn. θ = 0 or 7-0 rdins (-, -)

15 SECTION - 5. csc θ = r = ecuse r > 0. is negtive in the III nd IV qudrnts. The smllest positive θ is ssocited with 0 reference tringle in the III qudrnt s drwn. θ = 0 or rdins (-, - ) 7. Since sin θ = 5 > 0 nd cos θ < 0, θ is II qudrnt ngle. We sketch reference tringle. Since sin θ = r =, we know tht = nd r = 5. Use the Pythgoren theorem to find. 5 + = 5 = = ( must e negtive ecuse θ is II qudrnt ngle) Using (, ) = (, ) nd r = 5, we hve cos θ = r = = tn θ = 5 5 = = sec θ = r = 5 = 5 csc θ = r = 5 cot θ = = = 9. Since cos θ = 5 < 0 nd cot θ > 0, θ is III qudrnt ngle. We sketch reference tringle. Since cos θ = r = 5 we know tht = 5 nd r =. Use the Pythgoren theorem to find. ( 5 ) + = = = ( must e negtive ecuse θ is III qudrnt ngle) Using (, ) = ( 5, ) nd r =, we hve sin θ = r = sec θ = r = 5 tn θ = = = cot θ = 5 5 = 5 5 = csc θ = r = = 5. In these situtions P(, ) is restricted so tht = 0. In this cse, functions for which is in the denomintor re not defined. These functions re tngent nd secnt. 5. cos θ = r = = Thus (, ) = (, ) or (, ) θ is ssocited with 0 0 reference tringle in the II qudrnt or the III qudrnt s drwn: - 0 θ = 50 or 0

16 CHAPTER TRIGONOMETRIC FUNCTIONS 55. tn θ = = = Thus (, ) = (, ) or (, ). θ is ssocited with reference tringle in the I qudrnt or the III qudrnt s drwn: 5 θ = or 57. Flse. f(x) = x + is neither even nor odd. 59. Flse. sin x nd cos x hve period π; however, sin x = tn x hs period π. cos x. True. If f( x) = f(x) nd g( x) = g(x), then (fg)( x) = f( x)g( x) = [ f(x)][ g(x)] = f(x)g(x) = (fg)(x) for ll x in the domin of fg(x).. If f(x + p) = f(x) for ll x, then (x + p) + = x + p + = = 0 Hence f(x) =, ny rel numer. 5. If f( x) = f(x) for ll x, then ( x) + = x + 0 = x Hence must e 0, f(x) =, ny rel numer 7. (A) (B) The x intercepts do not chnge. (C) The devition of y = cos x from the x xis is unit; the devition of y = cos x from the x xis is units; the devition of y = cos x from the x xis is units. (D) The devition of the grph from the x xis is chnged y chnging A. The devition ppers to e A. 9. (A) (B) period of y = sin x ppers. periods of y = sin x pper. periods of y = sin x pper. (C) n periods of y = sin nx would pper. 7. (A) (B) The grph of y = cos x is shifted C units to the right if C < 0 nd C units to the left if C > For ech cse, the numer is not in the domin of the function nd n error messge of some type will pper.

17 SECTION Here re grphs of f(x) = sin x nd g(x) = x, x. (A) The grphs ecome more indistinguishle the closer x is to the origin. (B) x sin x (A) Since θ R = s r nd r = rdius of circle =, we hve θ R = 7 or.75 rdins (B) Since sin θ = r nd cos θ =, we cn write r = r cos θ = cos 7 = 0.7 = r sin θ = sin 7 =.9 (, ) = ( 0.7,.9) 79. We know tht s = r θ. (, ) = (, ). From the Pythgoren theorem, r = = ( ) = Since tn θ = = =, θ =. Hence s = r θ = = π units. 8. From the figure, the following reltions re cler: + = r r = = cos θ = sin θ Using the Pythgoren theorem in the right tringle whose hypotenuse is the rod connecting the piston to the wheel, we hve (y ) + = (y ) = y = y = + Since = cos θ nd = sin θ nd θ = πt, we hve y = sin πt + (cos t) 8. I = k cos θ For θ = 0, I = k cos 0 = k For θ = 0, I = k cos 0 = 0.8k For θ = 0, I = k cos 0 = 0.5k 85. A = n tn 80 n (A) For n = 8, A = 8 tn 80 =.7 8 For n = 00, A = 00 tn 80 =. 00 For n = 000, A = 000 tn 80 = For n = 0,000, A = 0,000 tn =.59 0,000 (B) s n, A seems to pproch π(=.59 ), the re of the circle. 87. (A) Using the formul given: For θ = 88.7, m = tn θ = tn 88.7 =.07 For θ =., m = tn θ = tn. = 0. (B) Using the formul for inclintion, the slope m is given y m = tn 7 = 0.9. We now use the

18 CHAPTER TRIGONOMETRIC FUNCTIONS point-slope form of the eqution of line. y y 0 = m(x x 0 ) y 5 = 0.9[x ( )] y 5 = 0.9x.7 y = 0.9x +.8 Section 5. Motion with the sme frequency nd mplitude indefinitely is simple hrmonic motion.. The grph of y = A cos(bx + C) is the grph of y = A cos Bx shifted horizontlly y n mount C/B. 5. Amplitude = A = = Period = = = π B 7. Amplitude = A = = Period = = = π B 9. Amplitude is not defined for the cotngent function. Period = = B. Amplitude is not defined for the tngent function. Period = = B 8 = 8. Amplitude is not defined for the cosecnt function. Period = = = π B 5. Amplitude = A = =. Period = = B = The sic sine function sin t hs zeros when t = kπ, k n integer. We exmine πx = kπ, x nd find x = k flls in this intervl when x =,, 0,,. 7. Amplitude is not defined for the cotngent function. Period = = π. The sic cotngent function cot t hs zeros when t = + kπ, k n integer. We exmine x = + kπ, 0 < x < π, nd find x = π + kπ flls in this intervl when x = π nd x = π. 9. Amplitude = A = =. Period = = = π B The sic cosine function cos t hs turning points when t = kπ, k n integer. k We exmine x = kπ, π x π, nd find tht x = flls in this intervl when x = π,, 0,, π. The turning points re therefore ( π, ),,, (0, ),,, (π, ).. Amplitude is not defined for the secnt function. Period = = B =. The sic secnt function sec t hs turning points when t = kπ, k n integer. We exmine πx = kπ, x, nd find tht x = k flls in this intervl when x =, 0,,,. The turning points re therefore (, ), (0, ), (, ), (, ), (, ).. A = P = =. Hence B =, y = sin x, x. 5. A = 0 P = =. Hence B = π, A = 0, since the grph hs the form of the stndrd sine curve turned upside down. y = 0 sin πx x.

19 SECTION A = 5 P = 8π = π = π. Hence B =, y = 5 cos x π x 8π. 9. A = 0.5, P = 8 = π = π. Hence B =, A = 0.5, since the grph hs the form of the stndrd cosine curve turned upside x down. y = 0.5 cos x 8. y = cos x Amplitude = A = Period = π Phse Shift = 0. y = sin x completes one cycle s x + vries from x + = 0 to x + = π, tht is s x vries from to + π. Amplitude: A = Period: π Phse shift: Divide the intervl, into four equl prts nd sketch one cycle of y = sin x. Then extend the grph to cover [ π, π], deleting the smll portion eyond x = π. 5. y = cot x completes one period s x vries from x = 0 to x = π, tht is s x vries from to + π. Period: π Phse shift: Sketch one cycle of y = cot x, the grph of y = cot x shifted units to the right, on the intervl,. Then extend the grph to cover [ π, π], deleting the portion eyond π. 7. y = tn x completes one period s x vries from x = to x = tht is, s x vries from to. Period: Phse shift: 0 Sketch one cycle of y = tn x, the grph of y = tn x stretched verticlly y fctor of nd shrunk horizontlly y fctor of, on the intervl,. Then extend the grph to cover [0, π].

20 CHAPTER TRIGONOMETRIC FUNCTIONS x 9. y = π sin x completes one cycle s x vries from x = 0 to tht is, s x vries from 0 to. Amplitude = A = π Period: Phse shift: 0 Divide the intervl [0, ] into four equl prts nd sketch one cycle of x y = π sin, then extend the grph to cover [0, ]. = π. y = sin x completes one cycle s π x π x = 0 to π x = π vries from x + = 0 x + = x = x = Amplitude = A = = Period: = Phse shift: Divide the intervl, into four equl prts nd sketch one cycle of y = sin x -- n upside-down sine curve. Then extend the grph to [, ].. y = sec(x + π) hs period π nd completes one period s x + π vries from x + π = 0 to x + π = π, tht is, s x vries from π to π. Phse shift: π The required grph is this one period, the grph of y = sec x shifted π units to the left. 5. y = 0 csc πx completes one period s πx vries from πx = 0 to πx = π, tht is, s x vries from 0 to. Period: Phse shift: 0 Sketch one period of y = 0 csc πx, the grph of y = csc x stretched verticlly y fctor of 0 nd shrunk horizontlly y fctor of, on this intervl. Then extend the grph to [0, ]. 7. True. If x = 0, y = A sin B(0) = 0. Thus the point (0, 0) is on the grph. 9. Flse. The function y = sin x is neither even nor odd. 5. True. Every function of form A sin(bx + C) or A cos(bx + C) hs period. B

21 SECTION Here is grph of y = cos x sin x, π x π. The grph hs mplitude nd period π. It ppers to e the grph of y = A cos Bx with A =, nd B = π P = π π =, tht is, y = cos x. 55. Here is grph of y = sin x, π x π. The grph hs mplitude 0 = nd period π. It ppers to e the grph of y = cos x turned upside down nd shifted up one unit, tht is, y = cos x + or y = cos x. 57. Here is grph of y = cot x tn x drwn y grphing clcultor. The grph ppers to hve the form y = A cot Bx. Since the period is, set = B to otin B =. The grph of y = A cot x shown ppers to pss through, 8, thus = A cot 8 = A cot = A The eqution of the grph cn e written y = cot x. 59. Here is grph of y = csc x + cot x drwn y grphing clcultor. The grph ppers to hve the form y = A cot Bx. Since the period is π, set π = to otin B = B. The grph of y = A cot x shown ppers to pss through,, thus = A cot = A cot = A The eqution of the grph cn e written y = cot x.. Here is grph of y = sin x + cos x cot x drwn y grphing clcultor. sin x. Here is grph of y = cosx drwn y grphing clcultor. The grph ppers to hve the form y = A csc Bx. Since the period is, set = to otin B =. The grph of y = A csc x shown ppers to B pss through,, thus = A csc = A csc = A The eqution of the grph cn e written y = csc x. The grph ppers to hve the form y = A tn Bx. Since the period is, set = to otin B =. The grph of y = A tn x shown ppers to pss B through, 8, thus = A tn 8 = A tn = A The eqution of the grph cn e written y = tn x.

22 8 CHAPTER TRIGONOMETRIC FUNCTIONS 5. A =. Hence A = or. The grph completes one full cycle s x vries over the intervl [, ]. Since C B is required etween 0 nd, we cnnot simply set C =. We must (mentlly) extend the B curve so tht the phse shift is positive. Then the (extended) grph is tht of n upside down sine curve tht completes one full cycle s x vries over the intervl [, 5]. Hence A = C B = C B + = 5 B C = B = B = = B The eqution is then y = A sin(bx + C) y = sin x C = 7. A =. Hence A = or. The grph completes one full cycle of the cosine function s x vries over the (mentlly extended) intervls [ π, 7π] or [π, π]. Since the phse shift is required etween 0 nd π, we must set C = π. Then the (extended) grph hs the form of stndrd cosine curve. Hence B A = C B + = π B C B = π = 8π B = B 8 C = πb = C = πb The eqution is then y = A cos(bx + C) y = cos x 9. y =.5 sin ( 0.5) t A =.5 Solve (t + 0.5) = 0 (t + 0.5) = π t = 0 t = t = 0.5 t = =.5 Phse shift Period P = The grph completes one full cycle s t vries over the intervl [ 0.5,.5]. 7. y = 50 cos[π(t 0.5)] A = 50 Solve π(t 0.5) = 0 π(t 0.5) = π t 0.5 = 0 t 0.5 = t = 0.5 t = =.5 Phse shift Period P = The grph completes one full cycle s t vries over the intervl [0.5,.5].

23 SECTION Here is grph of y = sin x + cos x, π x π. It ppers tht this is sine curve shifted to the left, with A = nd, since P = nd P ppers to e π, B = = B P =. The x intercept closest to the origin, to three deciml plces, is To find C, sustitute B = nd x = into the phse-shift formul x = C B nd solve for C: x = C B = C C = The eqution is thus y = sin(x ). 75. Here is grph of y = sin x cos x, π x π. It ppers tht this is sine curve shifted to the right, with A = nd, since P = nd P ppers to e π, B = = B P =. The x intercept closest to the origin, to three deciml plces, is 0.5. To find C, sustitute B = nd x = 0.5 into the phse-shift formul x = C B nd solve for C: x = C B 0.5 = C C = 0.5 The eqution is thus y = sin(x 0.5). 77. Here is grph of y =.8 sin x. cos x, π x π. It ppers tht this is sine curve shifted to the right, with A = 5 nd, since P = nd P ppers to e π, B = = B P =. The x intercept closest to the origin, to three deciml plces, is 0.. To find C, sustitute B = nd x = 0. into the phse-shift formul x = C B nd solve for C: x = C B 0. = C C = 0.8 The eqution is thus y = 5 sin(x 0.8). 79. The mplitude is decresing with time. This is often referred to s dmped sine wve. Exmples re the verticl motion of cr fter going over ump (which is dmped y the suspension system) nd the slowing down of pendulum tht is relesed wy from the verticl line of suspension (ir resistnce nd friction). 8. The mplitude is incresing with time. In physicl nd electricl systems this is referred to s resonnce. Some exmples re the swinging of ridge during high winds nd the movement of tll uildings during n erthquke. Some ridges nd uildings re destroyed when the resonnce reches the elstic limits of the structure.

24 70 CHAPTER TRIGONOMETRIC FUNCTIONS A = P = = When t = 0 y = 8. Hence 8 = A cos B(0), tht is, A = 8. Since the period is 0.5 seconds, = 0.5, B B = = π. Hence the eqution is y = 8 cos πt. 0.5 n 89. The grph is the sme s the grph of y = cos, shifted.5 units up. A = P = π = 5 The grph shows the sesonl chnges of sulfur dioxide pollutnt in the tmosphere; more is produced during winter months ecuse of incresed heting. 9. I = 5 cos 0 t A = 5 Solve 0πt + = 0 0πt = 0πt + = π 0πt = + π The grph completes one full cycle s t vries over the intervl, 0 0. t = 0π t = 0π t = 0 Phse Shift = 0 t = = 0 or 80 Period P = 0 9. If the disk rottes through n ngle θ in t seconds, we see

25 SECTION - 7 θ = revolutions π rdins t seconds = πt rdins second revolution y Then = sin θ = sin πt R y = R sin πt Since the disk hs rdius, y = sin πt. This function hs A =, P = =. 95. (A) In tringle MNP, we cn write (B) The grph hs n symptote t cos θ = 0 t =. Sketch the portion of c t y = sec on [0, ). c cos θ = 0 0 c = cos c = 0 sec θ t Since θ =, the eqution for c in (C) The length of the light em strts t 0 ft nd increses slowly t first, then increses rpidly without end. t terms of t is c = 0 sec. The eqution is vlid for 0 t <, since sec t is undefined when t =. 97. (A) (C) 0 (B) From the tle, Mx y = 9:5 nd Min y = :5. Hence A = Mx y Min y = =.7 B = = = Period C =.75 from the grph k = Min y + A = = 8. 0 x Thus, y = sin(.75) (D) Compute the regression eqution. Section 5 Grph the dt nd the regression eqution.. On this intervl, the function is not one-to-one since, for exmple sin 5 = sin ut 5.. Yes ecuse the rnge of tn x is ll rel numers. 5. The grphs of f nd f re reflections of one nother in the line y = x. 7. y = cos 0 is equivlent to cos y = 0 0 y π y =

26 7 CHAPTER TRIGONOMETRIC FUNCTIONS 9. y = rcsin is equivlent to sin y = y = y. y = rctn is equivlent to tn y = < y < y = (see sketch for prolem 9). y = sin is equivlent to sin y = y = y 5. y = rccos is equivlent to cos y = 0 y π y = 0 7. y = sin is equivlent to sin y = y y = rccos x is not defined if x >. 5. y = rctn( ) is equivlent to tn y = < y < y = 7. y = cos ( ) is equivlent to cos y = 0 y y = 9. rcsin( ) is not defined. is not in the restricted domin of the sine function.. cot [ cos ( 0.700) ] = = tn[cos ( 0.700)] 5. tn(tn 5 ) = 5 y the tngent-inverse tngent identity. 7. sin sin y the sine-inverse sine identity. 9. cos cos is not defined. is not in the restricted domin of the cosine function.

27 SECTION - 7. sin (sin.5) =.5 y the sine-inverse sine identity.. cos [cos ( π)] = cos ( ) = π 5. tn tn is not defined. is not in the domin of the tngent function. 7. y = sin ( ) is equivlent to sin y = 90 y 90 y = 0 9. y = tn ( ) is equivlent to tn y = 90 < y < 90 y = 5 5. y = rccos( ) is equivlent to cos y = 0 y 80 y = 50 Clcultor in degree mode for prolems sin (sin ) =.. For the identity sin (sin x) = x to hold, x must e in the restricted domin of the sine function; tht is, x. The numer is not in the restricted domin. 59. True. A periodic function cnnot e one-to-one, since f(x + p) = f(x) ut x + p x.. Flse. None of them re periodic.. True. sin ( x) = sin (x) for ll x, x

28 7 CHAPTER TRIGONOMETRIC FUNCTIONS (A) (B) The domin of cos is restricted to x ; hence no grph will pper for other x Let y = sin x. Then x = sin y, y. If sin y = r = x = x, then let = x, r =. + = r + x = = x = x ( must e positive since y is I or IV qudrnt ngle) cos y = cos(sin x) = r = x = x 77. Let y = rctn x. Then x = tn y, < y <. If tn y = = x = x, then let = x, =. + = r + x = r r = x (r is lwys tken positive) cos y = cos(tn x) = r = x 79. f(x) = + cos(x ). x + π. With this restriction, f is one-to-one (proof omitted).

29 SECTION - 75 Solve y = f(x) for x: y = + cos(x ) x + π y = cos(x ) 0 x π y = cos(x ) cos(x ) cos y = x cos(x ) x = + cos y = f (y) + cos(x ) rnge of f: y Interchnge x nd y f (x) = + cos x + cos(y ). Thus rnge of f = domin of f : x. 8. (A) (B) The domin for cos x is (, ) nd the rnge is [, ], which is the domin for cos x. Thus, y = cos (cos x) hs grph over the intervl (, ), ut cos (cos x) = x only on the restricted domin of cos x, [0, π]. 8. For 8mm lens, x = 8, thus θ = tn. 8 =.57 rdins. In deciml degrees, θ = 80 D (.57 rd) = For 00 mm lens, x = 00, thus θ = tn. 00 = 0. rdins. In deciml degrees, θ = 80 D (0. rd) = (A) 50 (B) 59. mm From the ove figure, the following should e cler: Finlly, Length of elt

30 7 CHAPTER TRIGONOMETRIC FUNCTIONS Length of elt = [rc A B + B B + rc B A ] rc A B = d (θ) rc B A = D (π θ) To find B B we note: C C hs length C B E is constructed prllel to C C. EC is prllel to B C. Hence EB C C is prllelogrm. EB hs length C. EB B is right tringle. Thus D d BE () cos θ = EB = D d = C C BB () sin θ = EB, so B B = EB sin θ = C sin θ L = [rc A B + B B + rc B A ] d D = C sin ( ) = dθ + C sin θ + D(π θ) L = πd + (d D) θ + C sin θ nd (from () ove) θ = cos D d C Sustituting the given vlues, we hve D =, d =, C = (clcultor in rdin mode) θ = cos = cos L = π + ( )cos + sin cos.59 inches 89. (A) 5 (B) 7. inches (A) Following the hint, we drw AC. Then, since the centrl ngle in circle sutended y n rc is twice ny inscried ngle sutended y the sme rc, ngle ACB hs mesure θ. Thus, d = r θ = r θ In tringle ECP, tn θ = x r, hence θ = tn x r d = r tn x r (B) Sustituting the given vlues, we hve r = 00 x = 0 d = 00 tn 0 0 = 00 tn = 00 tn (0.) (clcultor in rdin mode) = 7.0 feet D A Light E r x P C r Shdow d B F CHAPTER REVIEW. θ = s 5 centimeters = r centimeters = 5 =.5 rdins ( ). s = r θ = ( centimeters)(.5 rdins) = 7.5 centimeters ( ). Solve for α: α = = 5.8 Solve for : Solve for : cos β = c cos 5. = 0. sin β = c sin 5. = 0.

31 CHAPTER REVIEW 77 = 0. cos 5. =.5 ft = 0. sin 5. =. ft ( ). (A) (B) = -0 α = α = 80 + ( 0 ) = 0 (C) (D) = 0 α = = = 0 80 = 0 ( ) ( ) 5. (A) = sin θ < 0 if < 0. This occurs in qudrnts III, IV. r (B) = cos θ < 0 if < 0. This occurs in qudrnts II, III. r (C) = tn θ < 0 if nd hve opposite signs. This occurs in qudrnts II, IV.. + = r + ( ) = r 5 = r r = 5 sin θ = r = = 5 5 sec θ = r = 5 P(, ) = (, -) ( ) cot θ = = = 7. rd sin cos tn csc sec cot ND* ND ND ND ND ND 70 0 ND ND ND ND *ND = not defined (, )

32 78 CHAPTER TRIGONOMETRIC FUNCTIONS 8. (A) π (B) π (C) π ( ) 9. (A) Domin = ll rel numers, Rnge = [, ] (B) Domin is set of ll rel numers except x = k π, k n integer, Rnge = ll rel numers ( ) 0.. ( ) ( ). The centrl ngle in circle sutended y n rc of hlf the length of the rdius. ( ). If the grph of y = sin x is shifted units to the left, the result will e the grph of y = cos x. ( ). θ D = Solve for β: θ R = 80 (.7) = ( ) Solve for c: tn β = tn β =. 5.7 β = tn. 5.7 = 0. Solve for α: α = = 9.7 sec β = c c sec 0. = 5.7 c = 5.7 sec 0. = 0. cm ( ). (A) Since 70 < 0 < 80, (C) Since. <. <.7, this is II qudrnt ngle. this is III qudrnt ngle. (B) Since 5 is coterminl with, this is qudrntl ngle. ( ) 7. (A) Since = 0, this is coterminl with 0. (B) Since π =, which is equivlent to 50, this is not coterminl with 0. (C) Since 80 (0 ) = 0, this is coterminl with 0. ( )

33 CHAPTER REVIEW (B) nd (C), since rdins is equivlent to the rel numer, nd cosine is periodic with period π. (A) is not the sme s cos, since is equivlent to, not. 80 ( ) 9. (A) = tn x is not defined if = 0. This occurs if θ =,. (B) = cot x is not defined if = 0. This occurs if θ = 0, π. r (C) = csc x is not defined if = 0. This occurs if θ = 0, π. ( ) 0. Since the coordintes of point on unit circle re given y P(, ) = P(cos x, sin x), we evlute P(cos( 8.05), sin( 8.05))--using clcultor set in rdin mode--to otin P( 0., 0.900). Note tht x = 8.05, since P is moving clockwise. The qudrnt in which P(, ) lies cn e determined y the signs of nd. In this cse P is in the third qudrnt, since is negtive nd is negtive. (, ). (, ) = (, 0) r = tn 0 = = 0 = 0 (, 0). (, ) = (0, ) r = sec 90 = r = 0 (0, ) Not defined ( ). y = cos is equivlent to cos y = 0 y π y = 0. ( ) ( ) cos = r = = = ( ) 5. y = sin is equivlent to sin y = y sin y = r = = r = + ( ) = = = (positive since y is qudrnt I or IV ngle) y = y (, ) = (, ) ( )

34 80 CHAPTER TRIGONOMETRIC FUNCTIONS (, ) = (, - ) r = csc 00 = r = = or ( ) 7. y = rctn is equivlent to tn y = < y < tn y = = = = r = + ( ) r = r = y = y ( ) 8. sin 570 = r = ( ) (, ) = (-, -) r = 9. y = tn ( ) is equivlent to tn y = < y < tn y = = = = r = ( ) + r = r = y = ( ) y - 0. cot = = = or ( ). y = rcsin sin y = sin y = r = y = is equivlent to y = r = = y - ( )

35 CHAPTER REVIEW 8. y = cos is equivlent to cos y = cos y = r = y = 5 0 y π = r = = ( ). cos(cos 0.) = 0. y the cosine-inverse cosine identity. ( ). Let y = tn ( ), then tn y =, < y <. Using the drwing in prolem 9, we hve (, ) = (, ), r =. csc y = csc[tn ( )] = r = = ( ) 5. Let y = rccos, then cos y =, 0 y π.. Let y = sin, then sin y =, y. 5 5 Drw the reference tringle ssocited with y, Drw the reference tringle ssocited with y, then sin y = sin rccos cn e determined then tn y = tn sin 5 cn e determined directly from the tringle. directly from the tringle. cos y = r = = = r = = sin y = r = = r = 5 = 5 sin y = r = tn y = = y - 5 y - ( ) ( ) ( ) 8. tn = 5.7 ( ) 9. sec(.07) = =.077 ( ) ( ) cos(.07). rccos x is not defined if x < ( )..557 ( )..095 ( ). Since tn.5 =.5 >, sin (tn.5) is not defined. ( ) 5. (A) θ = rcsin is equivlent to sin θ = θ θ =

36 8 CHAPTER TRIGONOMETRIC FUNCTIONS Using the reltion θ D = 80 θ R, we hve rd sin = 80 = 0 (B) θ = rccos cos θ = is equivlent to 0 θ π θ = Using the reltion θ D = 80 rd θ R, we hve ( ) rccos = 80 = 0.. Clcultor in degree mode: (A) Θ = 5.0 (B) Θ = 8.8 ( ) 7. cos [cos( )] = For the identity cos (cos x) = x to hold, x must e in the restricted domin of the cosine function; tht is, 0 x π. The numer is not in the restricted domin. ( ) 8. Amplitude = = Period = = x 9. y = + sin x For the grph of y = sin, we note: A =, P = π = π, phse shift = 0. x We grph y = sin, then verticlly trnslte the grph down units. ( 5) ( 5) 50. A = P = π = π. 5. A = 0.5 P = = π π. Hence B = π A = 0.5, since the grph hs the form of the stndrd sine curve Hence B = y = cos x; x π ( 5) turned upside down. y = 0.5 sin πx; x ( 5) 5. If the grph of y = tn x is shifted units to the right nd reflected in the x xis, the result will e the grph of y = cot x. 5. (A) sin( x) cot( x) = sin( x) cos( x ) sin( x) Quotient Identity = cos( x) Alger = cos x Identities for negtives ( )

37 CHAPTER REVIEW 8 (B) sin sin x x sin x = cos x Pythgoren Identity sin x = cos x Alger = tn x Quotient Identity ( ) x 5. y = sin A = x x Solve + = 0 + = π x x = = + π x = π x = π + π Phse shift Period The grph completes one full cycle s x vries over the intervl [ π, π]. 55. y = cos x mplitude = A = = 5. Domin = [, ] Rnge = [0, π] ( 5) Solve x = 0 x = π x = x = x = x = + π x = x = + Phse shift Period ( 5) ( ) 57. Here is the grph of y = tn x clcultor. in grphing The grph hs mplitude 0 = nd period π. It ppers to e the grph of y = cos x shifted up unit, tht is, y = cos x +. ( 5) 58. (A) Here is the grph of y = clcultor. sin x sin x in grphing The grph ppers to hve the form y = A tn Bx. Since the period is π, B =. The grph of y = A tn x shown ppers to pss through,, thus = A tn = A

38 8 CHAPTER TRIGONOMETRIC FUNCTIONS The eqution of the grph cn e written y = tn x. 59. (B) Here is the grph of y = clcultor. cos x sin x in grphing The grph ppers to hve the form y = A cot Bx. Since the period is π, B =. The grph of y = A cot x shown ppers to pss through,, thus = A cot = A The eqution of the grph cn e written y = cot x. ( 5) 59. (A) f( x) = tn ( x) = ( tn x) The grph is symmetric out the y-xis so f(x) is even. = = f(x) tn x f(x) is even. (B) g( x) = = tn( x) tnx This is not equl to g(x) nor is it equl to g(x) which would e. tnx So g(x) is neither even nor odd. 0. Flse. 5 The grph is not symmetric out the y-xis nor is it symmetric out the origin. ( ) According to the digrm, sin α = Opp Hyp = Hyp while csc β = 5 Opp = 5. ( ). True. The two cute ngles in right tringle dd up to 90 so if they're equl, oth re 5. sin 5 = 0.7 cos 5 = 0.7 tn 5 = cot 5 = sec 5 =. csc 5 =. ( ). (A) Since θ R = s r nd r = rdius of circle = distnce of A from center = 8, we hve θ R = 0 8 =.5 rdins. (B) Since sin θ = r nd cos θ =, we cn write = r cos θ = 8 cos.5 =. (clcultor in rdin mode) r = r sin θ = 8 sin.5 =.79 (, ) = (.,.79) (, )

39 CHAPTER REVIEW 85. (A) cos x = r = =. is negtive in the II nd III qudrnts. The lest positive x is ssocited with reference tringle in the II qudrnt s drwn: x = (B) csc x = = = r, is negtive in the III nd IV qudrnts. The lest positive x is ssocited with reference tringle in the III qudrnt s drwn: x = 5. The dshed curve is the grph of y = cos x. The solid curve is the required grph of y = sec x, < x < 5. ( ). y = 5 tn x Solve πx + = 0 πx = x = Phse Shift = πx + = π Period P = πx = + π x = + ( ) ( 5) 8. From the figure, it should e cler tht cos( x) = cos x (P nd P hve the sme x-coordinte) sin( x) = sin x (P nd P hve opposite y-coordintes) Therefore tn( x) = sin( x ) sin x = = tn x cos( x) cos x It follows tht the grph of (A) sine hs origin symmetry (B) cosine hs y xis symmetry (C) tngent hs origin symmetry Domin = ll rel numers Rnge =, ( ) x 7. One cycle of y = csc is completed s x vries from 0 to π. Solve ech eqution for x: x x = 0 = π x x = = + π x = x = + π (-, 0) Phse shift = Period = π ( 5) (0, ) (0, -) R = x rd 0 cos x P(cos x, sin x) sin x x units (, 0) x units P (cos(-x), sin(-x)) ( )

40 8 CHAPTER TRIGONOMETRIC FUNCTIONS 9. Let y = sin x. Then x = sin y, y. If sin y = r = x, then let = x, r = + x = = x = x We choose the positive sign ecuse y is I or IV qudrnt ngle. sec y = sec(sin x) = r = ( ) x 70. For ech cse, the numer is not in the domin of the function nd n error messge of some type will pper. (, ) 7. A =. Hence A = or. The grph completes one full cycle s x vries over the intervl 5,. Since C B is required etween nd 0, we cnnot simply set C B = 5. We must (mentlly) extend the curve so tht the (extended) grph is tht of stndrd sine curve tht completes one full cycle s x vries 7 over the intervl,. Hence A = C B = C B + 7 The eqution is then y = A sin(bx + C) = B y = sin B x ( 5) C = = B = = π C = B 7. Here is the grph of y =. sin x +. cos x in grphing clcultor. It ppers tht this is sine curve shifted to the left, with A = nd, since P = nd P ppers to e π, B = = B P =. From the grphing clcultor, we find tht the x intercept closest to the origin, to three deciml plces, is 0.. To find C, sustitute B = nd x = 0. into the phse-shift formul x = C B nd solve for C. x = C B 0. = C C = 0.98 The eqution is thus y = sin(x ). 7. (A) (B) ( 5)

41 CHAPTER REVIEW In one yer the line sweeps out one full revolution, or π rdins in 5 dys. In 7 dys the line sweeps out 7 7 of one full revolution, or 5 5 π = rdins. ( ) Sketch figure. From geometry we know tht θ = (0 ) = 5 8 s = r sin 5 = r s Hence P = s = 8 = 8r = 8(5.00) = 8. cm ( ) 7. d = 0 feet, so c = πd = 0π feet In 80 revolutions, the wheel turns totl of 80 π rdins = 0π rdins. 0 rd min rd = 8.8 min 0sec sec In 80 revolutions, the liner distnce trveled is 80 0π feet = 00π feet. 00 feet min ft = 7.55 min 0sec sec 77. When t = 0, I = 0. Hence 0 = A cos B(0), tht is, A = 0. Since the period is 0 second, = B 0, B = π(0) = 0π. Hence the eqution is I = 0 cos 0πt. ( ) ( 5) 78. A 0 L C B E 5 D (A) In the figure s leled, L = AC + CE. In tringle ABC, since sin θ = 0 AC, AC sin θ = 0, AC = 0 = 0 csc θ sin In tringle CDE, since cos θ = 5 CE, CE cos θ = 5, CE = 5 = 5 sec θ cos Then L = AC + CE = 0 csc θ + 5 sec θ 0 < θ < (B) θ rdins L feet From the tle, the shortest distnce L to the nerest foot, is 5 feet. This is the length of the longest log tht cn mke the corner. (C) Here is grph of L = 0 csc θ + 5 sec θ from grphing clcultor. The minimum vlue of L is shown s 5. feet, to one deciml plce. (D) As θ 0, L = 0 csc θ + 5 sec θ pproches n symptote of csc θ; L increses without ound. As θ, L = 0 csc θ + 5 sec θ pproches n symptote of sec θ; L increses without ound. (, )

42 88 CHAPTER TRIGONOMETRIC FUNCTIONS Rmx Rmin 79. (A) A = = 7 =. Hence A = or. The grph ppers to e shifted up from grph of n upside down cosine curve tht completes one full cycle s t vries 8 over the intervl [0, ]. Thus A =. Then P = = π = π. Hence B =. 0 Rmx Rmin k = = 7 =. Thus R(t) = cos t. (B) The grph shows the sesonl chnges in soft drink consumption. Most is consumed in August nd the lest in Ferury. ( 5) (A) 90 0 (B) ymx ymin A = = = 8.5. Hence A = 8.5 or 8.5. ymx ymin k = = =.5. P = = π = π. Hence B =. The x intercept closest to the origin is estimted from the grph s.5. C To find C, sustitute B = nd x =.5 into the phse shift formul x = nd solve for C. B x = C B.5 = C (C) C =.5 =. With this vlue of C, the grph is seen to e shifted up from the grph of stndrd sine curve, thus A = 8.5. The eqution is thus y = sin. x. 90 (D) 90 0 Compute the regression eqution. 0 Grph the dt nd the regression eqution. ( 5)