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Chter Trigonometric Functions Chter Trigonometric Functions Section. Angles nd Their Mesures Exlortion. r. rdins ( lengths of ted). No, not quite, since the distnce r would require iece of ted times s long, nd >.. rdins Quick Review.. C= #.= in.. C= #.=. m.. r = # = ft. () s=. ft (b) s=. km. () v=. m/sec (b) v=.0 ft/sec... 0. r = # = m 0 mi # 0 ft mi # mi # 0 ft mi #. ft sec # ft sec # mi 0 ft mi 0 ft Section. Exercises 00 00 sec sec # 00 sec. '= + =. 0 b. '= + =. 0 b = ft>sec = ft>sec # 00 sec = mh = 0 mh. '"= + =. 0 + 00 b. 0'"= + 0 =. 0 + 00 b.. = (0 # 0.)'= '.. = (0 # 0.)'= '.. = (0 # 0.)'=.' = '(0 # 0.)"= '".. = (0 # 0.)'=.' = '(0 # 0.)"= '" For #, use the formul s=r, nd the equivlent forms r=s/ nd =s/r.. 0 # rd 0 = 0. 0 # rd 0 =. 0 # rd 0 =. 0 # rd 0 =.. #. rd 0.. # 0.0 rd 0. '= + =. #.0 rd 0 b 0. 0'= + 0 =. #. rd 0 b 0... 0.... # 0 = 0 # 0 = 0 # 0 = # 0 = 0 # 0 = 0 0 # 0 = # 0 L... # 0 L.. s=0 in.. s=0 cm. r=/ ft. r=./ cm. = rdins 0. = rdins 0. r= cm

Section. Angles nd Their Mesures. s=( ft)( ) ft 0 b =. = s >r = rd nd s = r =. = s >r =. rd nd r = s > = km. The ngle is 0 #, so the curved side 0 = rd mesures The two stright sides mesures in. in. ech, so the erimeter is ++ L inches.. The ngle is 00 #, so 0 = rd Then. Five ieces of trck form semicircle, so ech rc hs centrl ngle of / rdins. The inside rc length is r i > nd the outside rc length is r o >. Since r o > - r i > =. inches, we conclude tht r o - r i =.> L. inches.. Let the dimeter of the inner (red) circle be d. The inner circle s erimeter is. inches, which equls d. Then the next-lrgest (yellow) circle hs erimeter of d + + = d + =. + L. inches.. () NE is. (b) NNE is.. (c) WSW is.. 0. () SSW is 0.. (b) WNW is.. (c) NNW is... ESE is closest t... SW is closest t.. The ngle between them is = '=. 0. rdins, so the distnce is bout s=r =()(0.). sttute miles.. Since C = d, tire trvels distnce d with ech revolution. () Ech tire trvels t seed of 00 d in. er minute, or 00d in. min b0 min = r. r = L cm. Vehicle d Seed.d Turus.. mh Cger.. mh Mriner..0 mh mi b b L.d mi>.,0 in. d in. (b) so ech mile rev b mi,0 in. b = d,0 mi>rev,,0 requires L 0, revolutions. d d 0, Turus: revolutions. L 0. 0, Mriner: revolutions. L. The Turus must mke just over more revolutions. (c) In ech revolution, the tire would cover distnce of d new rther thn d old, so tht the cr would trvel d new >d old = d new >d old = >. L.0 miles for every mile the cr s instruments would show. Both the odometer nd seedometer redings would be low.. v= ft/sec nd r= in., so =v/r= ft # 0 sec sec min b in. # ft # rd in. rev b. rm. S. () mm. W = R 00 S = WR 00. mm= in., so S = WR # in. 00. = WR 0 WR (b) D+S=D+ WR =D+ in. 0 b 0 (c) Turus: D = + # 0 L. in. 0 Cger: D = + # 0 L. in. 0 Mriner: D = + # 0 L. in. 0 Ridgeline: D = + # L. in. 0. =000 rm nd r= in., so v=r = in. # teeth b # in. 000 rev # rd # min,. teeth er min rev 0 second.. c... mi 0 mi # b. stt mi 0. nut mi sttute miles 0,00 nut mi 0,00. stt mi # nut mi nuticl miles stt mi. () Lne hs inside rdius m, while the inside rdius of lne is m, so over the whole semicircle, the difference is -=. m. (This would be the nswer for ny two djcent lnes.) (b) -=.0 m.

Chter Trigonometric Functions. () s=r =()()= 0. in., or. ft. (b) r =. ft.. s=r =() = 0.0 ft 0 b rev min. () =0 rd # # = rd/sec min rev 0 sec (b) v= R =( cm) rd = cm/sec (c) =v/r= cm ( cm)= rd/sec rev. () = # min rd # =. rd/sec min rev 0 sec (b) v= r =(. m). rd =. m/sec (c) The rdius to this hlfwy oint is r*= r=0. m, so v=r* =(0. m). rd =. m/sec.. True. In the mount of time it tkes for the merry-goround to comlete one revolution, horse B trvels distnce of r, where r is B s distnce from the center. In the sme time, horse A trvels distnce of (r)=(r) twice s fr s B.. Flse. If ll tee rdin mesures were integers, their sum would be n integer. But the sum must equl, which is not n integer.. x = x rd The nswer is C. 0 b = x 0. 0. If the erimeter is times the rdius, the rc is two rdii long, which imlies n ngle of rdins. The nswer is A.. Let n be the number of revolutions er minute. in. rev min mi bn b0 b rev min,0 in. b L 0.0 n mh. Solving 0.0 n=0 yields n L. The nswer is B.. The size of the circle does not ffect the size of the ngle. The rdius nd the subtended rc length both double, so tht their rtio stys the sme. The nswer is C. In #, we need to borrow nd chnge it to 0' in order to comlete the subtrction.. '- '= 0'. 0'- 0'= 0'. '- '= '- '= '. 0'-0 '= 0' In # 0, find the difference in the ltitude. Convert this difference to minutes; this is the distnce in nuticl miles. The Erth s dimeter is not needed.. The difference in ltitude is 0'- '= 0' =0 minutes of rc, which is 0 nut mi.. The difference in ltitude is '- '= ' = minutes of rc, which is nut mi.. The difference in ltitude is '- '= 0' =0 minutes of rc, which is 0 nut mi. 0. The difference in ltitude is 0'- '= ' = minutes of rc, which is nut mi.. The whole circle s re is r ; the sector with centrl ngle mkes u / of tht re, or # r = r.. () A= (.).=0. ft. b (b) A= (.) (.)=. km.. B 0 0 mi A. Bike wheels: = v >r= ft>sec # in.>ft ( in.). rd/sec. The wheel srocket must hve the sme ngulr velocity: =. rd/sec. For the edl srocket, we first need the velocity of the chin, using the wheel srocket: v L in.. rd>sec. in./sec. Then the edl srocket s ngulr velocity is =. in.>sec (. in.). rd/sec. Section. Trigonometric Functions of Acute Angles Exlortion. sin nd csc, cos nd sec, nd tn nd cot.. tn. sec.. sin nd cos Exlortion. Let =0. Then sin = 0. csc =. cos = sec = tn =. cot = 0.. The vlues re the sme, but for different functions. For exmle, sin 0 is the sme s cos 0, cot 0 is the sme s tn 0, etc.. The vlue of trig function t is the sme s the vlue of its co-function t 0 -.

Section. Trigonometric Functions of Acute Angles Quick Review.. x = + = 0=. x = + = 0=. x = 0 - =. x = - = =.. ft # in = 00. in. ft. 0 ft # mi = L 0.0 mi 0 ft. =(0.)(0.)=. km. b =. L. ft.. Å=. #..00 (no units). 0. ı=. #.. (no units). Section. Exercises. sin =, cos =, tn =, csc =, sec =, cot =.. sin =, cos =, tn = ; csc =,. sin =, cos =, tn = ; csc =,. sin =, cos =, tn = ; csc =,. The hyotenuse length is + = 0, so 0 sin =, cos =, tn = ; csc =, 0 0 0. The djcent side length is - = =, so sin =, cos =, tn = ; csc =,. The oosite side length is - =, so sin =, cos =, tn = ; csc =,. The djcent side length is - = =, so sin =, cos =, tn = ; csc =,. Using right tringle with hyotenuse nd legs (oosite) nd - = 0 = 0 (djcent), 0 we hve sin =, cos =, tn = ; 0 0 csc =, 0 0. Using right tringle with hyotenuse nd legs (oosite) nd - = (djcent), we hve sin =, cos =, tn = ; csc =,. Using right tringle with hyotenuse nd legs (djcent) nd - = = (oosite), we hve sin =, cos =, tn = ; csc =,. Using right tringle with hyotenuse nd legs (djcent) nd - = (oosite), we hve sin =, cos =, tn = ; csc =,. Using right tringle with legs (oosite) nd (djcent) nd hyotenuse + = 0, we hve 0 sin =, cos =, tn = ; csc =, 0 0 0. Using right tringle with legs (oosite) nd (djcent) nd hyotenuse + =, we hve sin =, cos =, tn = ; csc =,. Using right tringle with legs (oosite) nd (djcent) nd hyotenuse + = 0, we hve sin =, cos =, tn = ; 0 0 0 0 csc =,

Chter Trigonometric Functions. Using right tringle with hyotenuse nd legs (oosite) nd - = (djcent), we hve sin =, cos =, tn = ; csc =,. Using right tringle with hyotenuse nd legs (oosite) nd - = = (djcent), we hve sin =, cos =, tn = ; csc =,. Using right tringle with hyotenuse nd legs (djcent) nd - = = (oosite), we hve sin =, cos =, tn = ; csc =,. 0..... = =. sec = >cos L.. Squring this result yields.0000, so sec =.. sin 0 L 0.0. Squring this result yields 0.00=/, so sin 0 = > = >.. csc > = >sin > L.. Squring this result yields. or essentilly /, so csc > = > = > = >.. tn > L.0. Squring this result yields.0000, so tn > =. For # 0, the nswers mrked with n sterisk (*) should be found in DEGREE mode; the rest should be found in RADIAN mode. Since most clcultors do not hve the secnt, cosecnt, nd cotngent functions built in, the recirocl versions of these functions re shown.. 0.* 0. 0.*. 0.*. 0.*. 0.. 0.0. cos L.*.. tn 0. L 0.0.. tn> L. 0. sin L.0* cos. L.0 sin>0. =0 =. =0 =. =0 =. = = L.. =0 =. = =. =0 =. =0 =. x = 0. z = sin L. cos L.0. y =. x = sin L. tn L 0.. y = >sin L 0.. x = 0 cos L 0. For #, choose whichever of the following formuls is rorite: b = c - b =c sin Å=c cos ı=b tn Å= tn ı b = c - =c cos Å=c sin ı= tn ı= tn Å + b = cos ı = sin Å = b sin ı = b cos Å If one ngle is given, subtrct from 0 to find the other ngle.. b = tn Å =. tn 0 L., sin Å =. L., ı = 0 - Å = 0 sin 0. =c sin Å=0 sin., b=c cos Å=0 cos., ı=0 -Å=. b= tn ı=. tn., cos b =. L., Å = 0 - ı = cos. b= tn ı= tn., Å=0 -ı= cos ı = cos L.,. 0. As gets smller nd smller, the side oosite gets smller nd smller, so its rtio to the hyotenuse roches 0 s limit. 0.. As gets smller nd smller, the side djcent to roches the hyotenuse in length, so its rtio to the hyotenuse roches s limit.. h= tn 0. ft. h=+0 tn. ft. A = # L. ft sin. h=0 tn. 0.0 ft. AC=00 tn.0 ft. Connect the tee oints on the rc to the center of the circle, forming tee tringles, ech with hyotenuse 0 ft. The horizontl legs of the tee tringles hve lengths 0 cos.., 0 cos.0, nd 0 cos... The widths of the four stris re therefore,.-0=. (stri A).0-.=. (stri B).-.0=. (stri C) 0-.=0. (stri D) Allen needs to correct his dt for stris B nd C.

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