Chapter 4 Trigonometric Functions

Transcription

1 Cater Trigonometric Functions Cater Trigonometric Functions Section. Angles and Teir Measures Eloration. r. radians ( lengts of tread). No, not quite, since te distance r would require a iece of tread times as long, and >.. radians Quick Review.. C= #.= in.. C= #.=9. m.. r = # 8 = ft. (a) s=. ft (b) s=9. km. (a) v=. m/sec (b) v=8.0 ft/sec r = # = m 0 mi r # 80 ft mi # mi r # 80 ft mi # 8.8 ft sec # ft sec # mi 80 ft mi 80 ft Section. Eercises r sec r sec # 00 sec r. '= a + =. 0 b. '= a + =. 0 b = 88 ft>sec = ft>sec # 00 sec r = m = 90 m. 8 '"= a 8 + = b. 8 0'"= a = b.. = (0 # 0.)'= '. 9. =9 (0 # 0.)'=9 '. 8. =8 (0 # 0.)'=8 9.' =8 9'(0 # 0.)"=8 9'" =99 (0 # 0.)'=99.' =99 '(0 # 0.)"=99 '" For #9, use te formula s=r, and te equivalent forms r=s/ and =s/r # rad 80 = # rad 80 =. 0 # rad 80 =. 0 # rad 80 =.. #.8 rad # 0.0 rad 80. '= a + =. #.0 rad 0 b 80. 0'= a + 0 =. #. rad 0 b # 80 = 0 # 80 = 0 # 80 = 8 # 80 = 08 9 # 80 = 0 0 # 80 = # 80 L.9.. # 80 L.8. s=0 in.. s=0 cm. r=/ ft 8. r=./ cm 9. = radians 0. = radians 0. r= cm

2 Section. Angles and Teir Measures. s=( ft)(8 ) a ft 0 b =. = s >r = 9 rad and s = r =. = s >r =. rad and r = s > = km. Te angle is 0 #, so te curved side 80 = 8 rad measures Te two straigt sides measures in. 8 in. eac, so te erimeter is ++ L inces. 8. Te angle is 00 #, so 80 = 9 rad Ten. Five ieces of track form a semicircle, so eac arc as a central angle of / radians. Te inside arc lengt is r i > and te outside arc lengt is r o >. Since r o > - r i > =. inces, we conclude tat r o - r i =.> L. inces. 8. Let te diameter of te inner (red) circle be d. Te inner circle s erimeter is. inces, wic equals d. Ten te net-largest (yellow) circle as a erimeter of d + + = d + =. + L. inces. 9. (a) NE is. (b) NNE is.. (c) WSW is.. 0. (a) SSW is 0.. (b) WNW is 9.. (c) NNW is... ESE is closest at... SW is closest at.. Te angle between tem is =9 '= radians, so te distance is about s=r =()(0.9). statute miles.. Since C = d, a tire travels a distance d wit eac revolution. (a) Eac tire travels at a seed of 800 d in. er minute, or a 800d in. min ba0 min r = 9 r. r = L cm. Veicle d Seed.8d Taurus.. m Carger m Mariner m mi ba b L.8d mi>r.,0 in. d in. (b) a so eac mile rev ba mi,0 in. b = d,0 mi>rev,,0 requires L 0,8 revolutions. d d 0,8 Taurus: revolutions. L 0.9 0,8 Mariner: revolutions 8.9 L 9. Te Taurus must make just over more revolutions. (c) In eac revolution, te tire would cover a distance of d new rater tan d old, so tat te car would travel d new >d old = d new >d old = 8>. L.0 miles for every mile te car s instruments would sow. Bot te odometer and seedometer readings would be low.. v= ft/sec and r= in., so =v/r= a ft # 0 sec sec min b a in. # ft # rad in. rev b 8.8 rm. S. (a) mm. W = R 00 S = WR 00. mm= in., so S = WR # in. 00. = WR 0 WR (b) D+S=D+ a WR =D+ in. 0 b 0 (c) Taurus: D = + # 0 L. in. 0 Carger: D = 8 + # 0 L 8. in. 0 Mariner: D = + # 0 L 8.9 in. 0 Ridgeline: D = + # L 9. in. 0. =000 rm and r= in., so v=r = a in. # teet b # in. a 000 rev # rad # min,. teet er min rev 0 sec b second. 8. c. a mi 0 mi # 8 b. 9 stat mi 0. naut mi 9 statute miles 0,800 naut mi 0, stat mi # naut mi 8 nautical miles 9 stat mi. (a) Lane as inside radius m, wile te inside radius of lane is 8 m, so over te wole semicircle, te difference is 8-=. m. (Tis would be te answer for any two adjacent lanes.) (b) 8-=.08 m.

3 Cater Trigonometric Functions. (a) s=r =()()= 0. in., or.89 ft. (b) r =.8 ft.. s=r =() a = ft 80 b rev min. (a) =0 rad # # = rad/sec min rev 0 sec (b) v= R =( cm) a rad =8 cm/sec sec b (c) =v/r= a 8 cm ( cm)= rad/sec sec b rev. (a) = # min rad # =. rad/sec min rev 0 sec (b) v= r =(. m) a. rad =. m/sec sec b (c) Te radius to tis alfway oint is r*= r=0. m, so v=r* =(0. m) a. rad =. m/sec. sec b. True. In te amount of time it takes for te merry-goround to comlete one revolution, orse B travels a distance of r, were r is B s distance from te center. In te same time, orse A travels a distance of (r)=(r) twice as far as B. 8. False. If all tree radian measures were integers, teir sum would be an integer. But te sum must equal, wic is not an integer. 9. = a rad Te answer is C. 80 b = If te erimeter is times te radius, te arc is two radii long, wic imlies an angle of radians. Te answer is A.. Let n be te number of revolutions er minute. in. rev min mi a ban ba0 ba rev min r,0 in. b L 0.0 n m. Solving 0.0 n=0 yields n L 9. Te answer is B.. Te size of te circle does not affect te size of te angle. Te radius and te subtended arc lengt bot double, so tat teir ratio stays te same. Te answer is C. In #, we need to borrow and cange it to 0' in order to comlete te subtraction.. '-8 '=8 0'. 09'- 0'= 09'. 9 '-8 9'=9 '-8 9'= '. 0'-80 '= 08' In # 0, find te difference in te latitude. Convert tis difference to minutes; tis is te distance in nautical miles. Te Eart s diameter is not needed.. Te difference in latitude is 0'- '= 0' =80 minutes of arc, wic is 80 naut mi. 8. Te difference in latitude is '- '=9 9' =89 minutes of arc, wic is 89 naut mi. 9. Te difference in latitude is 9'-9 '= 0' =90 minutes of arc, wic is 90 naut mi. 0. Te difference in latitude is 0'- '=8 ' = minutes of arc, wic is naut mi.. Te wole circle s area is r ; te sector wit central angle makes u / of tat area, or # r = r.. (a) A= (.9) a.8=0.9 ft. b (b) A= (.) (.)=. km.. B 0 0 mi A. Bike weels: = v >r= ft>sec # in.>ft ( in.). rad/sec. Te weel srocket must ave te same angular velocity: =. rad/sec. For te edal srocket, we first need te velocity of te cain, using te weel srocket: v L in.. rad>sec 8.8 in./sec. Ten te edal srocket s angular velocity is = 8.8 in.>sec (. in.) 8.9 rad/sec. Section. Trigonometric Functions of Acute Angles Eloration. sin and csc, cos and sec, and tan and cot.. tan. sec.. sin and cos Eloration. Let =0. Ten sin = 0.8 csc =. cos = sec = tan =. cot = 0.. Te values are te same, but for different functions. For eamle, sin 0 is te same as cos 0, cot 0 is te same as tan 0, etc.. Te value of a trig function at is te same as te value of its co-function at 90 -.

4 Section. Trigonometric Functions of Acute Angles Quick Review.. = + = 0=. = 8 + = 08=. = 0-8 =. = - = =. 8. ft # in = 00.8 in. ft. 90 ft # mi = L 0.80 mi 80 ft. a=(0.88)(0.)=.9 km 8. b =.9 L.89 ft. 9. Å=. #..00 (no units). 0. ı=.9 #..899 (no units) 8. Section. Eercises. sin =, cos =, tan =, csc =, sec =, cot = sin =, cos =, tan = ; csc =, 8 sec =, cot =. 8. sin =, cos =, tan = ; csc =, sec =, cot = sin =, cos =, tan = ; csc =, 8 sec =, cot =. 8. Te yotenuse lengt is + = 0, so 0 sin =, cos =, tan = ; csc =, sec =, cot =.. Te adjacent side lengt is 8 - = 8 =, so sin =, cos =, tan = ; csc =, sec =, cot =.. Te oosite side lengt is - 8 =, so 8 sin =, cos =, tan = ; csc =, 8 8 sec =, cot = Te adjacent side lengt is - 9 = 88 =, 9 9 so sin =, cos =, tan = ; csc =, 9 sec =, cot = Using a rigt triangle wit yotenuse and legs (oosite) and - = 0 = 0 (adjacent), 0 we ave sin =, cos =, tan = ; 0 0 csc =, sec =, cot = Using a rigt triangle wit yotenuse and legs (oosite) and - = (adjacent), we ave sin =, cos =, tan = ; csc =, sec =, cot =.. Using a rigt triangle wit yotenuse and legs (adjacent) and - = 9 = (oosite), we ave sin =, cos =, tan = ; csc =, sec =, cot =.. Using a rigt triangle wit yotenuse 8 and legs (adjacent) and 8 - = 9 (oosite), we ave sin =, cos =, tan = ; csc =, sec =, cot =. 9. Using a rigt triangle wit legs (oosite) and 9 (adjacent) and yotenuse + 9 = 0, we ave 9 0 sin =, cos =, tan = ; csc =, sec =, cot =. 9. Using a rigt triangle wit legs (oosite) and (adjacent) and yotenuse + =, we ave sin =, cos =, tan = ; csc =, sec =, cot =.. Using a rigt triangle wit legs (oosite) and (adjacent) and yotenuse + = 0, we ave sin =, cos =, tan = ; csc =, sec =, cot =.

5 8 Cater Trigonometric Functions. Using a rigt triangle wit yotenuse and legs (oosite) and - = 9 (adjacent), 9 we ave sin =, cos =, tan = ; 9 9 csc =, sec =, cot =. 9. Using a rigt triangle wit yotenuse and legs 9 (oosite) and - 9 = 8 = 8 (adjacent), we ave sin =, cos =, tan = ; 8 8 csc =, sec =, cot = Using a rigt triangle wit yotenuse and legs (adjacent) and - = = (oosite), we ave sin =, cos =, tan = ; csc =, sec =, cot = = =. sec = >cos L.. Squaring tis result yields.0000, so sec =.. sin 0 L Squaring tis result yields 0.00=/, so sin 0 = > = >.. csc > = >sin > L.. Squaring tis result yields. or essentially /, so csc > = > = > = >. 8. tan > L.0. Squaring tis result yields.0000, so tan > =. For #9 0, te answers marked wit an asterisk (*) sould be found in DEGREE mode; te rest sould be found in RADIAN mode. Since most calculators do not ave te secant, cosecant, and cotangent functions built in, te recirocal versions of tese functions are sown * 0. 0.*. 0.9*. 0.9* cos 9 L.*.. tan 0.89 L tan>8 L. 0. sin 9 L.0* cos. L.09 sin>0. =0 =. =0 =. =0 =. = = L.. =0 =. = =. =0 = 8. =0 = 9. = 0. z = sin L.8 cos 9 L 9.0. y =. = sin L 9. tan L 0.8. y = >sin L 0.. = 0 cos L 0. For # 8, coose wicever of te following formulas is aroriate: b a = c - b =c sin Å=c cos ı=b tan Å= tan ı a b = c - a =c cos Å=c sin ı=a tan ı= tan Å a c = a + b = cos ı = a sin Å = b sin ı = b cos Å If one angle is given, subtract from 90 to find te oter angle.. b = a tan Å =. tan 0 L.9, c = a sin Å =. L.9, ı = 90 - Å = 0 sin 0. a=c sin Å=0 sin., b=c cos Å=0 cos., ı=90 -Å=9. b=a tan ı=.8 tan., c = a cos b =.8 L., Å = 90 - ı = cos 8. b=a tan ı= tan 9 8., c = a Å=90 -ı= cos ı = cos 9 L 9., As gets smaller and smaller, te side oosite gets smaller and smaller, so its ratio to te yotenuse aroaces 0 as a limit. 0.. As gets smaller and smaller, te side adjacent to aroaces te yotenuse in lengt, so its ratio to te yotenuse aroaces as a limit.. = tan 0. ft. =+0 tan 8.8 ft. A = # L. ft sin. =0 tan ft. AC=00 tan 8.80 ft. Connect te tree oints on te arc to te center of te circle, forming tree triangles, eac wit yotenuse 0 ft. Te orizontal legs of te tree triangles ave lengts 0 cos..8, 0 cos.0, and 0 cos Te widts of te four stris are terefore,.8-0=.8 (stri A).0-.8=. (stri B) =.8 (stri C) 0-9.9=0. (stri D) Allen needs to correct is data for stris B and C.

6 Section. Trigonometry Etended: Te Circular Functions 9. False. Tis is only true if is an acute angle in a rigt triangle. (Ten it is true by definition.) 8. False. Te larger te angle of a triangle, te smaller its cosine. 9. sec 90 = is undefined. Te answer is E. cos 90 = 0 o 0. sin = Te answer is A. y =.. If te unknown sloe is m, ten m sin =,so m = - = -csc u. Te answer is D. sin u. For all, cos. Te answer is B.. For angles in te first quadrant, sine values will be increasing, cosine values will be decreasing and only tangent values can be greater tan. Terefore, te first column is tangent, te second column is sine, and te tird column is cosine.. For angles in te first quadrant, secant values will be increasing, and cosecant and cotangent values will be decreasing. We recognize tat csc (0 )=. Terefore, te first column is secant, te second column is cotangent, and te tird column is cosecant.. Te distance d A from A to te mirror is cos 0 ; te distance from B to te mirror is d B =d A -. Ten PB = = - L.9 m.. Let P be te oint at wic we sould aim; let Å and ı be te angles as labeled in #. Since Å=ı, tan Å=tan ı. P sould be inces to te rigt of C, were is cosen 0 - so tat tan Å= tan ı=. Ten = 0 0=(0-), so =0, wic gives =8. Aim 8 in. to te rigt of C (or in. to te left of D).. One ossible roof: sin + cos = a a c b + a b c b = a c = a + b c = c c d B cos ı = d A - cos 0 = - cos 0 + b c (Pytagorean teorem: a +b =c.) = 8. Let be te lengt of te altitude to base b and denote te area of te triangle by A. Ten a = sin =a sin Since A=, we can substitute =a sin to get b A = ab sin. Section. Trigonometry Etended: Te Circular Functions Eloration. Te side oosite in te triangle as lengt y and te yotenuse as lengt r. Terefore sin = o. y = y r. cos = adj y = r. tan = o = y adj r r. cot = ; sec = ; csc = y y Eloration. Te -coordinates on te unit circle lie between and, and cos t is always an -coordinate on te unit circle.. Te y-coordinates on te unit circle lie between and, and sin t is always a y-coordinate on te unit circle.. Te oints corresonding to t and t on te number line are wraed to oints above and below te -ais wit te same -coordinates. Terefore cos t and cos ( t) are equal.. Te oints corresonding to t and t on te number line are wraed to oints above and below te -ais wit eactly oosite y-coordinates. Terefore sin t and sin ( t) are oosites.. Since is te distance around te unit circle, bot t and t+ get wraed to te same oint.. Te oints corresonding to t and t+ get wraed to oints on eiter end of a diameter on te unit circle. Tese oints are symmetric wit resect to te origin and terefore ave coordinates (, y) and (, y). Terefore sin t and sin (t+) are oosites, as are cos t and cos (t+).. By te observation in (), tan t and tan(t+) are ratios y -y of te form and, wic are eiter equal to eac - oter or bot undefined. 8. Te sum is always of te form + y for some (, y) on te unit circle. Since te equation of te unit circle is + y =, te sum is always. 9. Answers will vary. For eamle, tere are similar statements tat can be made about te functions cot, sec, and csc. Quick Review tan = =. cot =

7 0 Cater Trigonometric Functions. csc = 8. sec = 9. Using a rigt triangle wit yotenuse and legs (oosite) and - = (adjacent), we ave sin =, cos =, tan = ; csc =, sec =, cot =. 0. Using a rigt triangle wit yotenuse and legs (adjacent) and - = 8 (oosite), we ave 8 8 sin =, cos =, tan = ; csc =, 8 sec =, cot =. 8 Section. Eercises. Te 0 angle lies on te ositive y ais (0-0 =90 ), wile te oters are all coterminal in Quadrant II.. Te - angle lies in Quadrant I -, + = wile te oters are all coterminal in Quadrant IV. In #, recall tat te distance from te origin is r= + y.. sin =, cos =, tan = ; csc =, sec =, cot =.. sin =, cos =, tan = ; csc =, sec =, cot =.. sin =, cos =, tan =; csc =, sec =, cot =.. sin =, cos =, tan = ; csc =, sec =, cot =.. sin =, cos =, tan = ; csc =, sec =, cot =. 8. sin =, cos =, tan = ; csc =, sec =, cot =. 9. sin =, cos =0, tan undefined; csc =, sec undefined, cot =0.. sin =, cos =, tan = ; csc =, sec =, cot =.. sin =, cos =, tan = ; csc =, sec =, cot =. For #, determine te quadrant(s) of angles wit te given measures, and ten use te fact tat sin t is ositive wen te terminal side of te angle is above te -ais (in Quadrants I and II) and cos t is ositive wen te terminal side of te angle is to te rigt of te y-ais (in quadrants I and IV). Note tat since tan t= sin t>cos t, te sign of tan t can be determined from te signs of sin t and cos t: if sin t and cos t ave te same sign, te answer to (c) will be + ; oterwise it will be. Tus tan t is ositive in Quadrants I and III.. Tese angles are in Quadrant I. (a)+(i.e., sin t 0). (b)+(i.e., cos t 0). (c)+(i.e., tan t 0).. Tese angles are in Quadrant II. (a) +. (b). (c).. Tese angles are in Quadrant III. (a). (b). (c) +.. Tese angles are in Quadrant IV. (a). (b) +. (c). For # 0, use strategies similar to tose for te revious roblem set.. is in Quadrant II, so cos is negative is in Quadrant III, so tan 9 is ositive. 9. rad is in Quadrant II, so cos is negative rad is in Quadrant II, so tan is negative.. A (, ); tan = y = y =.. B (, ); tan is in Quadrant II, = y = -. so is negative.. C (, ); is in Quadrant III, so and y are bot negative. tan =.. D (, ); 0º is in Quadrant IV, so is ositive wile y is negative. tan -0 = -. For #, recall tat te reference angle is te acute angle formed by te terminal side of te angle in standard osition and te -ais.. Te reference angle is 0. A rigt triangle wit a 0 angle at te origin as te oint P(, ) as one verte, wit yotenuse lengt r=, so cos 0 = =. r. Te reference angle is 0. A rigt triangle wit a 0 angle at te origin as te oint P(, ) as one verte, y so tan 00 = =. 0. sin =0, cos =, tan =0; csc undefined, sec =, cot undefined.

8 Section. Trigonometry Etended: Te Circular Functions. Te reference angle is te given angle,. A rigt triangle wit a radian angle at te origin as te oint P(, ) as one verte, wit yotenuse lengt r=, so r sec = =. 8. Te reference angle is. A rigt triangle wit a radian angle at te origin as te oint P(, ) as one verte, r wit yotenuse lengt r=, so csc = =. y 9. Te reference angle is (in fact, te given angle is coterminal wit ). A rigt triangle wit a radian angle at te origin as te oint P(,) as one verte, y wit yotenuse lengt r=, so sin = =. r 0. Te reference angle is (in fact, te given angle is coterminal wit ). A rigt triangle wit a radian angle at te origin as te oint P(, ) as one verte, wit yotenuse lengt r=, so cos = =. r. Te reference angle is (in fact, te given angle is coterminal wit ). A rigt triangle wit a radian angle at te origin as te oint P(, ) as one verte, - y so tan = =.. Te reference angle is. A rigt triangle wit a radian angle at te origin as te oint P(, ) as one verte, so cot = =. y. cos =cos =. cos =cos =. sin =sin = 9. cot =cot =. 0 is coterminal wit 0, on te negative y-ais. (a) (b) 0 (c) Undefined 8. 0 is coterminal wit 90, on te ositive y-ais. (a) (b) 0 (c) Undefined 9. radians is coterminal wit radians, on te negative -ais. (a) 0 (b) (c) 0 0. radians is coterminal wit radians, on te negative y-ais. (a) (b) 0 (c) Undefined -. radians is coterminal wit radians, on te ositive y-ais. (a) (b) 0 (c) Undefined. radians is coterminal wit 0 radians, on te ositive -ais. (a) 0 (b) (c) 0. Since cot 0, sin and cos ave te same sign, so sin sin = + - cos =, and tan =. cos =. Since tan 0, sin and cos ave oosite signs, so cos = - sin =, and cos u cot = =. sin u sin. cos = + - sin =, so tan = cos = and sec = =. cos. sec as te same sign as cos, and since cot 0, sin must also be negative. Wit =, y=, and r= + = 8, we ave sin = and 8 cos =. 8. Since cos 0 and cot 0, sin must be ositive. Wit =, y=, and r= + =, we ave sec = and csc =. 8. Since sin 0 and tan 0, cos must be negative. Wit =, y=, and r= + =, we ave csc = and cot =. 9. sin a =sin a b = + 9,000 b 0. tan (,,)-tan (,,) =tan ()-tan ()=0. cos a,, b = cos a b = 0-0,000. tan a b =tan a =undefined. b. Te calculator s value of te irrational number is necessarily an aroimation. Wen multilied by a very large number, te sligt error of te original aroimation is magnified sufficiently to trow te trigonometric functions off.. sin t is te y-coordinate of te oint on te unit circle after measuring counterclockwise t units from (, 0). Tis will reeat every units (and not before), since te distance around te circle is. sin 8. Â=.9 sin sin. sin = 0... (a) Wen t=0, d=0. in. (b) Wen t=, d=0.e 0. cos 0.8 in.

9 Cater Trigonometric Functions 8. Wen t=0, =0. (rad). Wen t=., =0. cos rad. 9. Te difference in te elevations is 00 ft, so d=00/sin. Ten: (a) d= ft. (b) d=00 ft. (c) d 9. ft. 0. January (t=):.+. sin =0.. Aril (t=):.+. sin.8. June (t=):.+. sin =.. October (t=0):.+. sin 8.9. December (t=):.+. sin =.. June and December are te same; eras by June most eole ave suits for te summer, and by December tey are beginning to urcase tem for net summer (or as Cristmas resents, or for mid-winter vacations).. True. Any angle in a triangle measures between 0 and 80. Acute angles (<90 ) determine reference triangles in Quadrant I, were te cosine is ositive, wile obtuse angles (>90 ) determine reference triangles in Quadrant II, were te cosine is negative.. True. Te oint determines a reference triangle in Quadrant IV, wit r = = 0. Tus sin =y/= /0 = 0.. If sin =0., ten sin ( )+csc = sin + sin u = 0.+ =.. Te answer is E. 0.. If cos =0., ten cos ( +)= cos = 0.. Te answer is B.. (sin t) +(cos t) = for all t. Te answer is A.. sin = - - cos, because tan =(sin )/(cos )>0. So sin = - - = -. B 9 Te answer is A.. Since sin 0 and tan 0, te terminal side must be in Quadrant II, so =. 8. Since cos 0 and sin 0, te terminal side must be in Quadrant IV, so =. 9. Since tan 0 and sin 0, te terminal side must be in Quadrant IV, so =. 0. Since sin 0 and tan 0, te terminal side must be in Quadrant III, so =.. Te two triangles are congruent: bot ave yotenuse, and te corresonding angles are congruent te smaller acute angle as measure t in bot triangles, and te two acute angles in a rigt triangle add u to >.. Tese coordinates give te lengts of te legs of te triangles from #, and tese triangles are congruent. For eamle, te lengt of te orizontal leg of te triangle wit verte P is given by te (absolute value of te) - coordinate of P; tis must be te same as te (absolute value of te) y-coordinate of Q. Q( b, a). One ossible answer: Starting from te oint (a, b) on te unit circle at an angle of t, so tat cos t=a ten measuring a quarter of te way around te circle (wic corresonds to adding > to te angle), we end at ( b, a), so tat sin t + > = a. For (a, b) in Quadrant I, tis is sown in te figure above; similar illustrations can be drawn for te oter quadrants.. One ossible answer: Starting from te oint (a, b) on te unit circle at an angle of t, so tat sin t=b ten measuring a quarter of te way around te circle (wic corresonds to adding > to te angle), we end at ( b, a),so tat cos t + >= b= sin t. For (a, b) in Quadrant I, tis is sown in te figure above; similar illustrations can be drawn for te oter quadrants.. Starting from te oint (a, b) on te unit circle at an angle of t, so tat cos t=a ten measuring a quarter of te way around te circle (wic corresonds to adding > to te angle), we end at ( b, a), so tat sin t + >= a. Tis olds true wen (a, b) is in Quadrant II, just as it did for Quadrant I. P(a, b) Q( b, a) t + π y y t + π P(a, b) t t t (, 0) (, 0). (a) Bot triangles are rigt triangles wit yotenuse, and te angles at te origin are bot t (for te triangle on te left, te angle is te sulement of -t). Terefore te vertical legs are also congruent; teir lengts corresond to te sines of t and -t. (b) Te oints P and Q are reflections of eac oter across te y-ais, so tey are te same distance (but oosite directions) from te y-ais. Alternatively, use te congruent triangles argument from art (a).

10 ` ` Section. Gras of Sine and Cosine: Sinusoids. Seven decimal laces are sown so tat te sligt differences can be seen. Te magnitude of te relative error is less tan % wen œ œ 0. (aroimately). Tis can be seen by etending te table to larger values sin - of, or by graing sin 8. Let (, y) be te coordinates of te oint tat corresonds to t under te wraing. Ten + y +(tan t) =+ a y = = =(sec t). b (Note tat +y = because (, y) is on te unit circle.) 9. Tis Taylor olynomial is generally a very good aroimation for sin in fact, te relative error (see #) is less tan % for 9. œ œ (aro.). It is better for close to 0; it is sligtly larger tan sin wen 0 and sligtly smaller wen 0.. sin sin sin - sin sin - sin Tis Taylor olynomial is generally a very good aroimation for cos in fact, te relative error (see #) is less tan % for œ œ. (aro.). It is better for close to 0; it is sligtly larger tan cos wen Z 0. cos - + cos -(- + ) Section. Gras of Sine and Cosine: Sinusoids Eloration. / (at te oint (0, )). / (at te oint (0, )). Bot gras cross te -ais wen te y-coordinate on te unit circle is 0.. (Calculator eloration). Te sine function tracks te y-coordinate of te oint as it moves around te unit circle. After te oint as gone comletely around te unit circle (a distance of ), te same attern of y-coordinates starts over again.. Leave all te settings as tey are sown at te start of te Eloration, ecet cange Y T to cos(t). Quick Review.. In order: +,+,-,-. In order: +,-,-,+. In order: +,-,+,-. # 80 =. -0 # 80 = -. 0 # 80 =. Starting wit te gra of y, vertically stretc by to obtain te gra of y. 8. Starting wit te gra of y, reflect across y-ais to obtain te gra of y. 9. Starting wit te gra of y, vertically srink by 0. to obtain te gra of y.

11 Cater Trigonometric Functions 0. Starting wit te gra of y, translate down units to obtain te gra of y. Section. Eercises In #, for y=a sin, te amlitude is a. If a >, tere is a vertical stretc by a factor of a, and if a <, tere is a vertical srink by a factor of a. Wen a<0, tere is also a reflection across te -ais.. Amlitude ; vertical stretc by a factor of.. Amlitude /; vertical srink by a factor of /.. Amlitude ; vertical stretc by a factor of, reflection across te -ais.. Amlitude /; vertical stretc by a factor of /, reflection across te -ais.. Amlitude 0.; vertical srink by a factor of 0... Amlitude.; vertical stretc by a factor of., reflection across te -ais. In #, for y=cos b, te eriod is / b. If b >, tere is a orizontal srink by a factor of / b, and if b <, tere is a orizontal stretc by a factor of / b. Wen b<0, tere is also a reflection across te y-ais. For y=a cos b, aas te same effects as in #.. Period /; orizontal srink by a factor of /. 8. Period /(/)=0; orizontal stretc by a factor of /(/)=. 9. Period /; orizontal srink by a factor of /, reflection across te y-ais. 0. Period /0.=; orizontal stretc by a factor of /0.=., reflection across te y-ais.. Period /=; orizontal srink by a factor of /. Also a vertical stretc by a factor of.. Period /(/)=; orizontal stretc by a factor of /(/)=/. Also a vertical srink by a factor of /. In #, te amlitudes of te gras for y=a sin b and y=a cos b are governed by a, wile te eriod is governed by b, just as in #. Te frequency is /eriod.. For y= sin (/), te amlitude is, te eriod is /(/)=, and te frequency is /().. For y= (/) sin, te amlitude is /, te eriod is /=, and te frequency is /.. For y= sin (/), te amlitude is, te eriod is /(/)=, and te frequency is /(). Note: te frequency for eac gra in # is /().. Period, amlitude= 8. Period, amlitude=. 9. Period, amlitude= 0. Period, amlitude= [, ] by [, ] [, ] by [, ] y y.. y y. Period, amlitude=0.. Period, amlitude= y y 0. [, ] by [, ]. For y= cos (/), te amlitude is, te eriod is /(/)=, and te frequency is /(). 0. [, ] by [, ]

12 Section. Gras of Sine and Cosine: Sinusoids. Period, amlitude=,. Period, frequency=/ amlitude=, frequency = /().. Period /,. Period /, amlitude=0., amlitude=0, frequency=/ frequency = / y y. Period 8, 8. Period /, amlitude=, amlitude=8, frequency=/(8) frequency = /() y. 0. y 0 0 y y Maimum: at, ; minimum: (at 0,, ). Zeros:,,,. 0. Maimum: at -, ; minimum: at,. Zeros: 0,,.. y=sin as to be translated left or rigt by an odd multile of. One ossibility is y=sin ( + ).. y=sin as to be translated rigt by lus an even multile of. One ossibility is y=sin (-/).. Starting from y=sin, orizontally srink by and vertically srink by 0.. Te eriod is /. Possible window: c - by c -.,, d d. Starting from y=cos, orizontally srink by and vertically stretc by.. Te eriod is /. Possible window: c - by [, ]., d Period ; amlitude.; [, ] by [, ] 0. Period /; amlitude ; c - by [, ], d. Period ; amlitude ; [, ] by [, ]. Period ; amlitude ; [, ] by [ 0, 0]. Period ; amlitude ; [, ] by [, ]. Period ; amlitude ; [, ] by [, ]. Maimum: at and ; minimum: at and. Zeros: 0,,.. Maimum: (at 0); minimum: (at ). Zeros:.. Maimum: (at 0,, ); minimum: at and. Zeros:,,,. 8. Maimum: at and ; minimum: - at - and. Zeros: 0,,. [, ] by [ 0., 0.] [, ] by [ 0., 0.] For # For #. Starting from y=cos, orizontally stretc by, vertically srink by, reflect across -ais. Te eriod is. Possible window: [, ] by [, ].. Starting from y=sin, orizontally stretc by and vertically srink by. Te eriod is 0. Possible window: [ 0, 0] by [, ]. [, ] by [, ] [ 0, 0] by [, ] For # For #. Starting from y=cos, orizontally srink by and vertically stretc by. Te eriod is. Possible window: [, ] by [.,.].

13 Cater Trigonometric Functions 8. Starting from y=sin, orizontally stretc by, vertically stretc by, and reflect across -ais. Te eriod is 8. Possible window: [ 8, 8] by [, ]. [, ] by [.,.] [ 8, 8] by [, ] For # For #8 9. Starting wit y, vertically stretc by. 0. Starting wit y, translate rigt units and vertically srink by.. Starting wit y, orizontally srink by.. Starting wit y, orizontally stretc by and vertically srink by. For #, gra te functions or use facts about sine and cosine learned to tis oint.. (a) and (b). (a) and (b). (a) and (b) bot functions equal cos. (a) and (c) sin a + b =sin ca - =cos a - b + d b In # 0, for y=a sin (b(-)), te amlitude is a, te eriod is / b, and te ase sift is.. One ossibility is y= sin. 8. One ossibility is y= sin (/). 9. One ossibility is y=. sin (-). 0. One ossibility is y=. sin (-). Amlitude, eriod, ase sift, vertical translation unit u.. Rewrite as y=. sin c a -. bd - Amlitude., eriod, ase sift, vertical translation unit down.. Rewrite as y= cos c a - bd Amlitude, eriod, ase sift, vertical 8 translation units u.. Amlitude, eriod, ase sift, vertical translation units down.. Amlitude, eriod, ase sift 0, vertical translation unit u.. Amlitude, eriod, ase sift 0, vertical translation units down.. Amlitude, eriod, ase sift -, vertical translation unit down. 8. Amlitude, eriod 8, ase sift, vertical translation unit u. 9. y= sin (a=, b=, =0, k=0). 0. y= sin[(+0.)] (a=, b=, =0., k=0).. (a) Tere are two oints of intersection in tat interval. (b) Te coordinates are (0, ) and (,. ) (.8, 0.9). In general, two functions intersect were cos =, i.e., =n, n an integer.. a= and b =.. =. Te eigt of te rider is modeled by =0- cos a, were t=0 corresonds 0 t b to te time wen te rider is at te low oint. =0 - wen =cos a. Ten, so t 0 t b 0 t L sec.. Te lengt L must be te distance traveled in 0 min by an object traveling at 0 ft/sec: L=800 sec # ft 0 9,000 ft, or about 8 miles. sec =. (a) A model of te det of te tide is d= cos c t -. d + 9, were t is ours since. midnigt. Te first low tide is at :00 A.M. (t=). (b) At :00 A.M. (t=): about 8.90 ft. At 9:00 P.M. (t=): about 0. ft. (c) :0 A.M. (t=. alfway between :00 A.M. and : A.M.).. (a) second. (b) Eac eak corresonds to a eartbeat tere are 0 er minute. (c) [0, 0] by [80, 0]

14 Section. Gras of Sine and Cosine: Sinusoids. (a) Te maimum d is aroimately.. Te amlitude is (.-.)/=.. Scatterlot: [0, ] by [0, 80] [0,.] by [, ] (b) Te eriod aears to be sligtly greater tan 0.8, say 0.8. (c) Since te function as a minimum at t=0, we use an inverted cosine model: d(t)=. cos (t/0.8)+.. (d) [0,.] by [, ] 8. (a) Te amlitude is., alf te diameter of te turntable. (b) Te eriod is.8, as can be seen by measuring from minimum to minimum. (c) Since te function as a minimum at t=0, we use an inverted cosine model: d(t)=. cos (t/.8)+.. (d) [0,.] by [9, 8] 9. One ossible answer is T =. cos a - b +.. Start wit te general form sinusoidal function y = a cosb - + k, and find te variables a, b,, and k as follows: 9 - Te amlitude is 0 a 0 = =.. We can arbitrarily coose to use te ositive value, so a=.. Te eriod is monts. = 0 b 0 =. 0 b 0 = Again, we can arbitrarily coose to use te ositive value, so b =. Te maimum is at mont, so te ase sift =. 9 + Te vertical sift k = = One ossible answer is y = cos a - b +. Start wit te general form sinusoidal function y = a cosb - + k, and find te variables a, b,, and k as follows: 8-0 Te amlitude is 0 a 0 = =. We can arbitrarily coose to use te ositive value, so a=. Te eriod is monts. = 0 b 0 =. 0 b 0 = Again, we can arbitrarily coose to use te ositive value, so b =. Te maimum is at mont, so te ase sift = Te vertical sift k = =. [0, ] by [0, 80] 8. False. Since y=sin is a orizontal stretc of y=sin by a factor of, y=sin as twice te eriod, not alf. Remember, te eriod of y=sin b is / b. 8. True. Any cosine curve can be converted to a sine curve of te same amlitude and frequency by a ase sift, wic can be accomlised by an aroriate coice of C (a multile of /). 8. Te minimum and maimum values differ by twice te amlitude. Te answer is D. 8. Because te gra asses troug (, 0), f()=0. And lus eactly two eriods equals 9, so f(9)=0 also. But f(0) deends on ase and amlitude, wic are unknown. Te answer is D. 8. For f()=a sin (b+c), te eriod is / b, wic ere equals /0=/0. Te answer is C. 8. Tere are solutions er cycle, and 000 cycles in te interval. Te answer is C. 8. (a) [, ] by [.,.]

15 8 Cater Trigonometric Functions (b) cos Te coefficients given as 0 ere may sow u as very small numbers (e.g.,.*0 ) on some calculators. Note tat cos is an even function, and only te even owers of ave nonzero (or a least non-small ) coefficients. (c) Te Taylor olynomial is - ; te + = coefficients are fairly similar. 88. (a) For #9 9, note tat A and C are one eriod aart. Meanwile, B is located one-fourt of a eriod to te rigt of A, and te y-coordinate of B is te amlitude of te sinusoid. 9. Te eriod of tis function is and te amlitude is. B and C are located (resectively) units and units to te rigt of A. Terefore, B=(0, ) and C= a., 0 b 9. Te eriod of tis function is and te amlitude is.. B and C are located (resectively) units and units to te rigt of A. Terefore, B= a and,. b C = a 9, 0 b [, ] by [.,.] (b) sin Te coefficients given as 0 ere may sow u as very small numbers (e.g.,.*0 ) on some calculators. Note tat sin is an odd function, and only te odd owers of ave nonzero (or a least non-small ) coefficients. (c) Te Taylor olynomial is - ; = - 0. te coefficients are somewat similar. 89. (a) = sec = (b) f = ( cycles er sec ), or Hertz (Hz). sec (c) [0, 0.0] by [, ] 90. Since te cursor moves at a constant rate, its distance from te center must be made u of linear ieces as sown (te sloe of te line is te rate of motion). A gra of a sinusoid is included for comarison. [0, ] by [.,.] 9. (a) a-b must equal. (b) a-b must equal. (c) a-b must equal k. 9. (a) a-b must equal. (b) a-b must equal. (c) a-b must equal k. 9. Te eriod of tis function is and te amlitude is. B and C are located resectively units and units to te rigt of A. Terefore, B= a and C= a, b, 0 b 9. Te first coordinate of A is te smallest ositive suc n + tat -=n, n and integer, so = must equal. Te eriod of tis function is and te amlitude is. B and C are located (resectively) units and units to te rigt of A. Terefore, A= a,, 0 b B = a., b and C = a, 0 b 9. (a) Since sin ( )= sin (because sine is an odd function) a sin [ B(-)]+k= a sin[b(-)]+k. Ten any eression wit a negative value of b can be rewritten as an eression of te same general form but wit a ositive coefficient in lace of b. (b) A sine gra can be translated a quarter of a eriod to te left to become a cosine gra of te same sinusoid. Tus y = a sin c b a - + # b bd + k = a sin c b a - a - as te same b bbd + k gra as y = a cosb - + k. We terefore coose H = -. b (c) Te angles + and determine diametrically oosite oints on te unit circle, so tey ave oint symmetry wit resect to te origin. Te y-coordinates are terefore oosites, so sin( +)= sin. (d) By te identity in (c). y = a sinb k = -a sinb - + k. We terefore coose H = -. b (e) Part (b) sows ow to convert y = a cosb - + k to y = a sinb - H + k, and arts (a) and (d) sow ow to ensure tat a and b are ositive.

16 Section. Gras of Tangent, Cotangent, Secant, and Cosecant Section. Gras of Tangent, Cotangent, Secant, and Cosecant 9 Eloration. Te gras do not seem to intersect.. Set te eressions equal and solve for : k cos =sec k cos =/cos k(cos ) = (cos ) = /k Since k 0, tis requires tat te square of cos be negative, wic is imossible. Tis roves tat tere is no value of for wic te two functions are equal, so te gras do not intersect. Quick Review.. Period. Period. Period. Period For # 8, recall tat zeros of rational functions are zeros of te numerator, and vertical asymtotes are found at zeros of te denominator (rovided te numerator and denominator ave no common zeros).. Zero:. Asymtote: =. Zero:. Asymtote: =. Zero:. Asymtotes: = and = 8. Zero:. Asymtotes: =0 and = For #9 0, eamine gras to suggest te answer. Confirm by cecking f( )=f() for even functions and f( )= f() for odd functions. 9. Even: ( ) += + 0. Odd: = Section. Eercises. Te gra of y= csc must be vertically stretced by comared to y=csc, so y = csc and y =csc.. Te gra of y= tan must be vertically stretced by 0 comared to y=0. tan, so y = tan and y =0. tan.. Te gra of y= csc must be vertically stretced by and orizontally srunk by comared to y=csc,so y = csc and y =csc.. Te gra of y=cot(-0.)+ must be translated units u and 0. units rigt comared to y=cot,so y =cot(-0.)+ and y =cot.. Te gra of y=tan results from srinking te gra of y = tan orizontally by a factor of. Tere are vertical asymtotes at =..., -....,,, [, ] by [, ]. Te gra of y= cot results from srinking te gra of y=cot orizontally by a factor of and reflecting it across te -ais. Tere are vertical asymtotes at =....,-...., -, 0,, by [, ],. Te gra of y=sec results from srinking te gra of y=sec orizontally by a factor of. Tere are vertical asymtotes at odd multiles of., by [, ] 8. Te gra of y=csc results from srinking te gra of y=csc orizontally by a factor of. Tere are vertical asymtotes at =....,-, -...., 0,, [, ] by [, ]

17 80 Cater Trigonometric Functions 9. Te gra of y= cot results from srinking te gra of y=cot orizontally by a factor of and stretcing it vertically by a factor of. Tere are vertical asymtotes at =....,-, -...., 0,, [, ] by [, ] 0. Te gra of y= tan results from stretcing te gra of y=tan orizontally by a factor of and stretcing it vertically by a factor of. Tere are vertical asymtotes at =....,-,,,.... [, ] by [, ]. Te gra of y=csc results from orizontally stretcing te gra of y=csc by a factor of. Tere are vertical asymtotes at =....,-, -, 0,,.... [, ] by [, ]. Te gra of y= sec results from orizontally srinking te gra of y=sec by a factor of and stretcing it vertically by a factor of. Tere are vertical asymtotes at odd multiles of. 8 [, ] by [, ]. Gra (a); Xmin= and Xma=. Gra (d); Xmin= and Xma=. Gra (c); Xmin= and Xma=. Gra (b); Xmin= and Xma=. Domain: All reals ecet integer multiles of Range: ( q, q) Continuous on its domain Decreasing on eac interval in its domain Symmetric wit resect to te origin (odd) Not bounded above or below No local etrema No orizontal asymtotes Vertical asymtotes = k for all integers k End beavior: lim cot and lim cot do not eist. Sq S-q 8. Domain: All reals ecet odd multiles of Range: ( q, ] [, q) Continuous on its domain On eac interval centered at an even multile of : decreasing on te left alf of te interval and increasing on te rigt alf On eac interval centered at an odd multile of : increasing on te left alf of te interval and decreasing on te rigt alf Symmetric wit resect to te y-ais (even) Not bounded above or below Local minimum at eac even multile of, local maimum at eac odd multile of No orizontal asymtotes Vertical asymtotes = k/ for all odd integers k End beavior: lim sec and lim sec do not eist. Sq S-q 9. Domain: All reals ecet integer multiles of Range: ( q, ] [, q) Continuous on its domain On eac interval centered at = (k an integer): + k decreasing on te left alf of te interval and increasing on te rigt alf On eac interval centered at : increasing on te + k left alf of te interval and decreasing on te rigt alf Symmetric wit resect to te origin (odd) Not bounded above or below Local minimum at eac =, local maimum + k at eac =, were k is an even integer in + k bot cases No orizontal asymtotes Vertical asymtotes: = k for all integers k End beavior: lim csc and lim csc do not eist. Sq S-q 0. Domain: All reals ecet odd multiles of Range: ( q, q) Continuous on its domain Increasing on eac interval in its domain Symmetric wit resect to te origin (odd) Not bounded above or below No local etrema No orizontal asymtotes Vertical asymtotes =k for all odd integers k End beavior: lim tan> and lim tan> Sq S-q do not eist.

18 Section. Gras of Tangent, Cotangent, Secant, and Cosecant 8. Starting wit y=tan, vertically stretc by.. Starting wit y=tan, reflect across -ais.. Starting wit y=csc, vertically stretc by.. Starting wit y=tan, vertically stretc by.. Starting wit y=cot, orizontally stretc by, vertically stretc by, and reflect across -ais.. Starting wit y=sec, orizontally stretc by, vertically stretc by, and reflect across -ais.. Starting wit y=tan, orizontally srink by and reflect across -ais and sift u by units. 8. Starting wit y=tan, orizontally srink by and vertically stretc by and sift down by units. 9. sec = 0. csc =.. cos = = sin = = cot = - tan = - = sec = - cos = - =. csc = sin = =. cot = tan = = -. tan =. L 0.9. sec =. cos =. L.. cot = 0. tan = csc =. sin = -. -( 0.).8 9. csc = sin = 0. or tan = or (a) One elanation: If O is te origin, te rigt triangles wit yotenuses OP and OP, and one leg (eac) on te -ais, are congruent, so te legs ave te same lengts. Tese lengts give te magnitudes of te coordinates of P and P ; terefore, tese coordinates differ only in sign. Anoter elanation: Te reflection of oint (a, b) across te origin is ( a, b). sin t (b) tan t=. cos t = b a sint - -b (c) tan(t-)= =. -a = b cost - a = tan t (d) Since oints on oosite sides of te unit circle determine te same tangent ratio, tan(t_)=tan t for all numbers t in te domain. Oter oints on te unit circle yield triangles wit different tangent ratios, so no smaller eriod is ossible. (e) Te tangent function reeats every units; terefore, so does its recirocal, te cotangent (see also #).. Te terminal side asses troug (0, 0) and (cos, sin ); sin - 0 sin te sloe is terefore m= = =tan. cos - 0 cos. For any, a (+)= = = a (). f b f + f f b Tis is not true for any smaller value of, since tis is te smallest value tat works for f.. (a), (b) Te angles t and t+ determine oints (cos t, sin t) and (cos(t+), sin(t+)), resectively. Tese oints are on oosite sides of te unit circle, so tey are reflections of eac oter about te origin. Te reflection of any oint (a, b) about te origin is ( a, b), so cos(t+)= cos t and sin(t+)= sin t. sint + -sin t sin t (c) tan(t+)= = = =tan t. cost + -cos t cos t In order to determine tat te eriod of tan t is,we would need to sow tat no satisfies tan(t+)= tan t for all t. 0. (a) d=0 sec = ft cos (b) d,8 ft

19 8 Cater Trigonometric Functions 800. (a) =800 cot y= ft tan (b),0 ft (c) # 80 0 = 9 For # 0, te equations can be rewritten (as sown), but generally are easiest to solve graically.. sin =cos ; ; cos =sin ; 0. or. 9. cos = ; ;.0 or ;.0 0. cos =sin ;.08 or.00. False. f()=tan is increasing only over intervals on wic it is defined, tat is, intervals bounded by consecutive asymtotes.. True. Asymtotes of te secant function, sec =/cos, occur at all odd multiles of > (were cos =0), and tese are eactly te zeros of te cotangent function, cot =cos /sin.. Te cotangent curves are saed like te tangent curves, but tey are mirror images. Te reflection of tan in te -ais is tan. Te answer is A.. sec just barely intersects its inverse, cos, and wen cos is sifted to roduce sin, tat curve and te curve of sec do not intersect at all. Te answer is E.. y=k/sin and te range of sin is [, ]. Te answer is D.. y=csc =/sin as te same asymtotes as y=cos /sin =cot. Te answer is C.. On te interval [, ], f g on about ( 0., 0) (0., ). 0. Tey look similar on tis window, but tey are noticeably different at te edges (near 0 and ). Also, if f were equal to g, ten it would follow tat = cos = =on tis interval, wic we know to be f g false.. csc =sec a - (or csc =sec a - a + n bb b for any integer n) Tis is a translation to te rigt of [, ] by [ 0, 0] [0, ] by [ 0, 0] a or units. + n b. cot = tan a - (or cot = b tan - a for any integer n). + n b Tis is a translation to te rigt of a or units, + n b and a reflection in te -ais, in eiter order. 0. d=0 sec = cos [, ] by [ 0, 0] 8. On te interval [, ], f g on about (,.) a - a., 0 b,. b [, ] by [ 0, 0] 9. cot is not defined at 0; te definition of increasing on (a, b) requires tat te function be defined everywere in (a, b). Also, coosing a= > and b=>, we ave a b but f(a)= f(b)=. [ 0., 0.] by [0, 00]. (a) For any acute angle, cos =sin a te sine - b of te comlement of. Tis can be seen from te rigt-triangle definition of sine and cosine: if one of te acute angles is, ten te oter acute angle is, since all tree angles in a triangle must add - to. Te side oosite te angle is te side adjacent to te oter acute angle. (b) (cos t, sin t)

20 Section. Gras of Comosite Trigonometric Functions 8 (c) Using ^ODA ~ ^OCB (recall ~ means similar DA BC to ), =tan t=, so BC=tan t. OC = BC OD OD OC (d) Using ^ODA ~ ^OCB, =cot t= OB = OA OB, so OB= =sec t. cos t (e) BC is a tangent segment (art of te tangent line); OB is a secant segment (art of a secant line, wic crosses te circle at two oints). Te names cotangent and cosecant arise in te same way as cosine tey are te tangent and secant (resectively) of te comlement. Tat is, just as BC and OB go wit jboc (wic as measure t), tey also go wit jobc (te comlement of j BOC, wit measure ). - t N kg = (. m) a 00 a 9.8 m m m b sec b kg (.* 0 m)sec Ï 0.0 sec Ï, so sec sec Ï.990, and Ï 0.89 radians.9.. (a) = = # = cos(b)= y a secb a secb a sin(b+/) a (b) y=0. sin a + b (c) a=/0.= and b=/ (d) y= sec a. Te scatter lot is sown below, and b te fit is very good so good tat you sould realize tat we made te data u! [ 0., 8.] by [., ] y = sin - cos [, ] by [, ] Not a Sinusoid y = cos a - b + sin a b [, ] by [, ] Sinusoid Sinusoid Not a Sinusoid Quick Review.. Domain: ( q, q); range: [, ]. Domain: ( q, q), range: [, ]. Domain: [, q); range: [0, q). Domain: [0, q); range: [0, q). Domain: ( q, q); range: [, q). Domain: ( q, q); range: [, q). As S-q, f Sq; as Sq, f S As S-q, f S-q; as Sq, f S føg()= -=-, domain: [0, q). gøf()= -, domain: ( q, ] [, q). 0. føg()=(cos ) =cos, domain: ( q, q). gøf()=cos( ), domain: ( q, q). Section. Eercises. Periodic. y = sin + - cos [, ] by [, ] y = cos + sin [, ] by [, ] Section. Gras of Comosite Trigonometric Functions Eloration y = sin + cos y = sin - cos. Periodic. [, ] by [.,.] [, ] by [, ] Sinusoid [, ] by [, ] Sinusoid [, ] by [.,.]

21 8 Cater Trigonometric Functions. Not eriodic. 9. Since te eriod of cos is, we ave cos (+)=(cos(+)) =(cos ) =cos. Te eriod is terefore an eact divisor of, and we see graically tat it is. A gra for is sown: [, ] by [, 0]. Not eriodic. [, ] by [, 0]. Not eriodic. [, ] by [, ] 0. Since te eriod of cos is, we ave cos (+)=(cos(+)) =(cos ) =cos. Te eriod is terefore an eact divisor of, and we see graically tat it is. A gra for is sown: [, ] by [, ]. Not eriodic. [, ] by [.,.]. Since te eriod of cos is, we ave cos + = cos + = cos = cos. Te eriod is terefore an eact divisor of, and we see graically tat it is. A gra for is sown:. Periodic. [, ] by [, ] 8. Periodic. [, ] by [ 0, 0] [, ] by [, ]. Since te eriod of cos is, we ave cos (+) = (cos(+)) = (cos ) = cos. Te eriod is terefore an eact divisor of, and we see graically tat it is. A gra for is sown: [, ] by [ 0, 0] [, ] by [, ]. Domain: ( q, q). Range: [0, ]. [, ] by [ 0.,.]

22 Section. Gras of Comosite Trigonometric Functions 8. Domain: ( q, q). Range: [0, ] y - 0. [, ] by [ 0.,.]. Domain: all Z n, n an integer. Range: [0, q). [ 0, 0] by [ 0, 0] y - 0. [, ] by [ 0., ]. Domain: ( q, q). Range: [, ]. [ 0, 0] by [, 8]. y + [, ] by [.,.]. Domain: all Z an integer. Range: ( q, 0]. + n, n [, ] by [ 0, 0.] 8. Domain: ( q, q). Range: [, 0]. [, ] by [.,.] In #9, te linear equations are found by setting te cosine term equal to ; y + [ 0, 0] by [ 0, 0] For # 8, te function y + y is a sinusoid if bot y and y are sine or cosine functions wit te same eriod.. Yes (eriod ). Yes (eriod ). Yes (eriod ). No. No 8. No For #9, gra te function. Estimate a as te amlitude of te gra (i.e., te eigt of te maimum). Notice tat te value of b is always te coefficient of in te original functions. Finally, note tat a sin[b(-)]=0 wen =, so estimate using a zero of f() were f()canges from negative to ositive. 9. A., b=, and 0.9, so f(). sin[(-0.9)]. 0. A., b=, and 0., so f(). sin[(+0.)].. A., b=, and 0., so f(). sin[(-0.)].. A., b=, and 0.0, so f(). sin[( +0.0)].. A., b=, and., so f(). sin(+.).. A., b=, and 0., so f(). sin[(-0.)]. [ 0, 0] by [ 0, 0]

23 8 Cater Trigonometric Functions. Te eriod is. [, ] by [.,.]. Te eriod is. [0, ] by [, ] 0. f oscillates u and down between and. As Sq, f S 0. [, ] by [, ]. Te eriod is. [0, ] by [, ]. f oscillates u and down between and -. As Sq, f S 0. [, ] by [, ] 8. Te eriod is. [0, ] by [.,.]. f oscillates u and down between e and e. As Sq, f S 0. [, ] by [, ] 9. (a) 0. (d). (c). (b). Te daming factor is e, wic goes to zero as gets large. So daming occurs as Sq.. Te daming factor is, wic goes to zero as goes to zero (obviously). So daming occurs as S 0.. Te amlitude,, is constant. So tere is no daming.. Te amlitude,, is constant. So tere is no daming.. Te daming factor is, wic goes to zero as goes to zero. So daming occurs as S Te daming factor is (/), wic goes to zero as gets large. So daming occurs as Sq. 9. f oscillates u and down between. and.. As Sq, f S 0. [0,.] by [, ]. Period : sin[(+)]+cos[(+)]= sin(+)+ cos(+)=sin + cos. Te gra, sows tat no < could be te eriod. [, ] by [.,.8]

24 Section. Gras of Comosite Trigonometric Functions 8. Period : cos[(+)]- cos[(+)-] = cos(+)- cos(-+) = cos - cos(-). Te gra, sows tat no < could be te eriod. 8. Not eriodic [, ] by [ 0, 0] 9. Not eriodic [, ] by [, ]. Period : sin[(+)+]-cos[(+)-] = sin(++)-cos(-+0) = sin(+)-cos(-). Te gra, sows tat no < could be te eriod. [, ] by [, ] 0. Not eriodic [, ] by [, ]. Period : cos[(+)-]- sin[(+)-] = cos(-+)- sin(-+) = cos(-)- sin(-). Te gra, sows tat no < could be te eriod. [, ] by [, ]. Not eriodic [, ] by [, ] [, ] by [ 8, ]. Not eriodic. Period : sinb += + R sin a += -sin. + = sin + + b Te gra, sows tat no < could be te eriod. [, ] by [ 0, ] [, ] by [, ] For # 0, gras may be useful to suggest te domain and range.. Tere are no restrictions on te value of, so te domain is (-q, q). Range: (-q, q).. Tere are no restrictions on te value of, so te domain is (-q, q). Range: (-q, q).. Tere are no restrictions on te value of, so te domain is (-q, q). Range: [, q).. Tere are no restrictions on te value of, so te domain is (-q, q). Range: (-q, q).