\\ Chapter 4 Trigonometric Functions

Transcription

1 9 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.. r = # = m. (a) s=7. ft (b) s=9.77 km Section. Eercises. '= a + =. 0 b. 8 '"= a8 + = b.. = (0 # 0.)'= ' =8 (0 # 0.)'=8 9.' =8 9'(0 # 0.)"=8 9'" For #9, use te formula s=r, and te equivalent forms r=s/ and =s/r # rad 80 =. 0 # rad 80 =. 7.7 #.8 rad 80. '= a +.07 rad 0 b =. # mi r # 80 ft mi # 8.8 ft sec # # 80 = 0 0 # 80 = # 80 = 0 00 r = 88 ft/sec sec mi # 00 sec 80 ft r = m. # 80 L.9. s=0 in. 7. r=/ ft 9. = radians 0. r= cm. = s /r = 9 rad and s = r u=. Te angle is 0 #, so te curved side 80 = 8 rad measures Te two straigt sides measure in 8 in. eac, so te erimeter is ++ L in 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. 9. (a) NE is. (b) NNE is.. (c) WSW is 7... ESE is closest at... Te angle between tem is =9 '= radians, so te distance is about s=r =()(0.9). statute miles.. v= ft/sec and r= in., so =v/r= a ft # 0 sec sec min b, a in. # ft # rad in rev b 87.8 rm. 7. =000 rm and r= in., so v=r = a in# teet in b # a000 rev # rad # min,.7 teet er min rev 0 sec b second mi mi. 89 stat mi # 0,800 naut mi 778 nautical miles 9 stat mi. (a) s=r =()( )= 0. in, or.89 ft Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

2 Section. Trigonometric Functions of Acute Angles 9 (b) r =.8 ft Eloration rev. (a) =0 # rad # min v = rad/sec. Let =0. Ten min rev 0 sec (b) v= =(7 cm) a rad sin = 0.8 csc =. Rv =8 cm/sec sec b cos = sec = (c) =v/r= a8 cm v ( cm)=7 rad/sec sec b tan =.7 cot = True. In te amount of time it takes for te merry-goround to comlete one revolution, orse B travels a. Te values are te same, but for different functions. For distance of r, were r is B s distance from te center. eamle, sin 0 is te same as cos 0, cot 0 is te same In te same time, orse A travels a distance of as tan 0, etc. (r)=( r) twice as far as B.. Te value of a trig function at is te same as te value 9. = a rad Te answer is C. 80 b = of its co-function at Quick Review.. Let n be te number of revolutions er minute. in. rev min a ban ba0 rev min r ba mi. = + = 0=,0 in. b. = 0-8 = L n m. Solving n=0 yields n L ft # in = 00.8 in ft Te answer is B. 7. a=(0.88)(0.)=7.9 km In #, we need to borrow and cange it to 0' in order to comlete te subtraction. 9. Å=. #..00 (no units).. '-8 '=8 0'. 9 '-87 9'=9 7'-87 9'= 7' Section. Eercises In #7 70, find te difference in te latitude. Convert tis difference to minutes; tis is te distance in nautical miles. Te. sin =, cos =, tan =, csc =, sec =, Eart s diameter is not needed. 7. Te difference in latitude is 0'- '= 0' cot =. =80 minutes of arc, wic is 80 naut mi. 9. Te difference in latitude is 9'-9 7'= 0'. sin =, cos =, tan = ; csc =, =90 minutes of arc, wic is 90 naut mi. 7. Te wole circle s area is r ; te sector wit central sec =, cot =. u angle makes u / of tat area, or # r = ur.. Te yotenuse lengt is 7 + = 70, so sin =, cos =, tan = ; csc =, sec =, cot = Te oosite side lengt is - 8 = 7, so B 0 mi A sin =, cos =, tan = ; csc =, sec =, cot =. Section. Trigonometric Functions of Acute Angles Eloration. sin and csc, cos and sec, and tan and cot.. tan. sec.. sin and cos Using a rigt triangle wit yotenuse 7 and legs (oosite) and 7 - = 0 = 0 (adjacent), 0 we ave sin =, cos =, tan = ; csc =, sec =, cot =. 0. Using a rigt triangle wit yotenuse and legs (adjacent) and - = 9 = (oosite), we ave sin =, cos =, tan = ; csc =, sec =, cot =. Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

3 9 Cater Trigonometric Functions. 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 + = 0, we ave sin =, cos =, tan = ; csc =, sec =, cot =. 7. Using a rigt triangle wit yotenuse and legs 9 (oosite) and - 9 = 8 = 87 (adjacent), we ave sin =, cos =, tan = ; csc =, sec =, cot = =. sec = /cos L.. Squaring tis result yields.0000, so sec =. 7. csc (/) = /sin (/) L.7. Squaring tis result yields. or essentially /, so csc (/) = / = / = /. For #9 0, te answers marked wit an asterisk (*) sould be found in DEGREE mode; te rest sould be found in RADI- AN mode. Since most calculators do not ave te secant, cosecant, and cotangent functions built in, te recirocal versions of tese functions are sown *. 0.9* cos 9 L.* 7. tan 0.89 L tan (/8) L.. =0 =. =0 =. =0 = 7. =0 = 9. = sin L.8. y = tan 7 L y = /sin 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 a a b = c - a =c cos Å=c sin ı=a tan ı= tan b c = a + b a = cos a = a sin b = b sin a = b cos b If one angle is given, subtract from 90 to find te oter angle. a. b = tan b =. tan 0 L.79, a c = sin b =. L.9, a = 90 - b = 70 sin 0 7. b=a tan ı=.8 tan., c = As gets smaller and smaller, te side oosite gets smaller and smaller, so its ratio to te yotenuse aroaces 0 as a limit.. = tan 7 0. ft. A = # L 7. ft sin. AC=00 tan ft 7. False. Tis is only true if is an acute angle in a rigt triangle. (Ten it is true by definition.) 9. sec 90 = is undefined. Te answer is E. cos 90 = 0 7. If te unknown sloe is m, ten m sin =,so m = - = -csc u. Te answer is D. sin u 7. 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. 7. 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. 77. One ossible roof: (sin u) + (cos u) = a a c b + a b c b = a c + b c = a + b c = c c = a cos b =.8 cos L 7., b = 90 - a = d B cos a = d A - cos 0 = - cos 0 (Pytagorean teorem: a +b =c.) Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

4 Section. Trigonometry Etended: Te Circular Functions 9 Section. Trigonometry Etended: Te Circular Functions Eloration. Te side oosite in te triangle as lengt y and te yotenuse as lengt r. Terefore sin u = o. y = y r. cos u = adj y = r. tan u = o adj = y 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+ ). 7. 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 = = 7. csc = 9. Using a rigt triangle wit yotenuse and legs (oosite) and - = (adjacent), we ave sin =, cos =, tan = ; csc =, sec =, cot =. Section. Eercises. Te 0 angle lies on te ositive y ais (0-0 =90 ), wile te oters are all coterminal in Quadrant II. In #, recall tat te distance from te origin is r= + y.. sin =, cos =, tan = ; csc =, sec =, cot =.. sin =, cos =, tan =; csc =, sec =, cot =. 7. sin =, cos =, tan = ; csc =, sec =, cot =. 9. sin =, cos =0, tan undefined; csc =, sec undefined, cot =0.. 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: lf 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 7 0). (b)+(i.e., cos t 7 0). (c)+(i.e., tan t 7 0).. Tese angles are in Quadrant III. (a). (b). (c) +. For #7 0, use strategies similar to tose for te revious roblem set. 7. is in Quadrant II, so cos is negative rad is in Quadrant II, so cos is negative A (, ); tan = y = Q y =. 7. A (, ); is in Quadrant III, so and y are bot negative. tan 7 =. 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 Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

5 9 Cater Trigonometric Functions 7. 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 = =. 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. 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 = =.. cos =cos =. sin =sin = 7. 0 is coterminal wit 70, on te negative y-ais. (a) (b) 0 (c) Undefined 9. 7 radians is coterminal wit radians, on te negative -ais. (a) 0 (b) (c) 0-7. radians is coterminal wit radians, on te ositive y-ais. (a) (b) 0 (c) Undefined. Since cot 7 0, sin and cos ave te same sign, so sin u sin = + - cos u =, and tan =. cos u = sin u. cos = + - sin u=, so tan = cos u = and sec = =. cos u 7. Since cos 0 and cot 0, sin must be ositive. Wit =, y=, and r= + =, we ave sec = and csc =. 9. sin a =sin a b = + 9,000b. cos a,, b = cos a b = 0. 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 8. Â=.9 sin (b) Wen t=, d=0.e 0. cos 0.8 in 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.. 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.. If sin =0., ten sin ( )+csc = sin + sin u = 0.+ =.. Te answer is E. 0.. (sin t) +(cos t) = for all t. Te answer is A. 7. Since sin 7 0 and tan 0, te terminal side must be in Quadrant II, so =. 9. Since tan 0 and sin 0, te terminal side must be 7 in Quadrant IV, so =. 7. 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 /. 7. 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. 7. 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 t (, 0) 7. (a) Wen t=0, d=0. in Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

6 ` 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 ƒ sin u -u ƒ of, or by graing ƒ sin u ƒ ` sin sin - sin - sin u Tis Taylor olynomial is generally a very good aroimation for sin in fact, te relative error (see Eercise 77) is less tan % for œ œ (aro.). It is better for close to 0; it is sligtly larger tan sin wen 0 and sligtly smaller wen 7 0. sin - sin -a - b 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: +,-,+,-. -0 # 80 = - 7. Starting wit te gra of y, vertically stretc by to obtain te gra of y. 9. Starting wit te gra of y, vertically srink by 0. 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 stretc by a factor of, reflection across te -ais.. Amlitude 0.7; vertical srink by a factor of 0.7. In #7, 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, a as te same effects as in #. 7. Period /; orizontal srink by a factor of /. 9. Period /7; orizontal srink by a factor of /7, reflection across te y-ais.. Period /= ; orizontal srink by a factor of /. Also a vertical stretc 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 /( ). [, ] by [, ] Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

7 98 Cater Trigonometric Functions. For y= (/) sin, te amlitude is /, te eriod is /=, and te frequency is /. 7. Period 8, amlitude=, frequency=/(8 ) y [, ] by [, ] Note: te frequency for eac gra in #7 is /( ). 7. Period, amlitude= y 9. Period, amlitude= y. Period, amlitude=0. y Period ; amlitude.; [, ] by [, ].. Period ; amlitude ; [, ] by [, ].. Period ; amlitude ; [, ] by [, ].. Maimum: aat and ; b minimum: aat and. b Zeros: 0,,. 7. Maimum: (at 0,, ); minimum: 7 aat and. Zeros:,,,. b 9. Maimum: aat, ; minimum: (at 0,, ). b 7 Zeros:,,,.. y=sin as to be translated left or rigt by an odd 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. Period, amlitude=, frequency=/.. Period /, amlitude=0., frequency=/ y y. [, ] by [ 0.7, 0.7] For #. Starting from y=cos, orizontally stretc by, vertically srink by, reflect across -ais. Te eriod is. Possible window: [, ] by [, ] [, ] by [, ] For # Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

8 Section. Gras of Sine and Cosine: Sinusoids Starting from y=cos, orizontally srink by and vertically stretc by. Te eriod is. Possible window: [, ] by [.,.]. (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 7: A.M.). 77. (a) Te maimum d is aroimately.. Te amlitude is (.-7.)/=7.. Scatterlot: [, ] by [.,.] For Eercise 7 9. Starting wit y, vertically stretc by.. Starting wit y, orizontally srink by. For #, gra te functions or use facts about sine and cosine learned to tis oint.. (a) and (b). (a) and (b) bot functions equal cos In #7 0, for y=a sin (b(-)), te amlitude is a,te eriod is / b, and te ase sift is. 7. One ossibility is y= sin. 9. One ossibility is y=. sin (-).. Amlitude, eriod, ase sift, vertical translation unit u.. Rewrite as y= cos c a - bd Amlitude, eriod, ase sift, vertical 8 translation units u.. Amlitude, eriod, ase sift 0, vertical translation unit u Amlitude, eriod, ase sift -, vertical translation unit down. 9. y= sin (a=, b=, =0, k=0). 7. (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. 7. Te eigt of te rider is modeled by =0- cos a, were t=0 corresonds 0 tb to te time wen te rider is at te low oint. =0 - wen =cos a. Ten, so t 0 tb 0 t L sec. 7. (a) A model of te det of te tide is d= cos c (t - 7.) d + 9, were t is ours since. midnigt. Te first low tide is at :00 A.M. (t=). [0,.] by [7, ] (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)= 7. cos ( t/0.8)+.. (d) [0,.] by [7, ] 79. One ossible answer is T =. cos a ( - 7)b Start wit te general form sinusoidal function y = a cos (b( - )) + k, and find te variables a, b,, and k as follows: 79 - Te amlitude is ƒ a ƒ = =.. We can arbitrarily coose to use te ositive value, so a=.. Te eriod is monts. = Q ƒ b ƒ =. ƒ b ƒ = Again, we can arbitrarily coose to use te ositive value, so b =. Te maimum is at mont 7, so te ase sift = Te vertical sift k = = 7.. [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. Te minimum and maimum values differ by twice te amlitude. Te answer is D. Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

9 00 Cater Trigonometric Functions 8. For f()=a sin (b+c), te eriod is / b, wic ere equals /0= /0. Te answer is C. 87. (a) [, ] by [.,.] (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. 89. (a) = sec = (b) f = ( cycles er sec ), or Hertz (Hz). sec (c) [0, 0.0] by [, ] 9. (a) a-b must equal. (b) a-b must equal. (c) a-b must equal k. 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., 0b 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, 0b 97. (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 cba( - ) + # b bd + k = a sin cba - a - as te same b bbd + k gra as y = a cos [b( - )] + 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 sin [b( - ) + ] + k = -a sin [b( - )] + k. We terefore coose H = -. b (e) Part (b) sows ow to convert y = a cos [b( - )] + k to y = a sin [b( - H)] + k, and arts (a) and (d) sow ow to ensure tat a and b are ositive. Section. Gras of Tangent, Cotangent, Secant, and Cosecant 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 7 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 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: = 7. Zero:. Asymtotes: = 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: ( ) += + 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= csc must be vertically stretced by and orizontally srunk by comared to y=csc,so y = csc and y =csc.. Te gra of y=tan results from srinking te gra of y = tan orizontally by a factor of. Tere are Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

10 Section. Gras of Tangent, Cotangent, Secant, and Cosecant 0 vertical asymtotes at = -.,,, [, ] by [, ] 7. 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 [, ] 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,, 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. : q : -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. : q : -q. Starting wit y=tan, vertically stretc by.. Starting wit y=csc, vertically stretc by.. Starting wit y=cot, orizontally stretc by, vertically stretc by, and reflect across -ais. 7. Starting wit y=tan, orizontally srink by and reflect across -ais and sift u by units. 9. sec = cos = [, ] by [, ]. Te gra of y=csc a results from orizontally b stretcing te gra of y=csc by a factor of. Tere are vertical asymtotes at = -, -, 0,,... = cot = - tan = - = csc = sin = = [, ] by [, ]. Gra (a); Xmin= and Xma=. Gra (c); Xmin= and Xma= 7. 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). tan =. L cot = 0. tan = Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

11 0 Cater Trigonometric Functions 9. csc = sin = 0. 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 sin (t - ) -b (c) tan(t- )= =. -a = b cos (t - ) 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 #).. 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. 0. (a) d=0 sec = ft cos (b) d,8 ft For #7 0, te equations can be rewritten (as sown), but generally are easiest to solve graically. 7. sin =cos ; ; cos = ; ;.07 or ;.0. False. f()=tan is increasing only over intervals on wic it is defined, tat is, intervals bounded by consecutive asymtotes.. 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.. y=k/sin and te range of sin is [, ]. Te answer is D. 7. On te interval [, ], f 7 g on about ( 0., 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)= 7 f(b)=.. csc =sec a - (or csc =sec a - a b + nbb for any integer n) Tis is a translation to te rigt of or a units. + nb 0. d=0 sec = cos N kg (. m) a00 a9.8 m m = m b sec b kg (.7* 0 - m)sec Ï 0.07 sec Ï, so sec sec Ï.990, and Ï 0.89 radians.9. Section. Gras of Comosite Trigonometric Functions Eloration [, ] by [ 0, 0] [ 0., 0. ] by [0, 00] y = sin + cos [, ] by [, ] Sinusoid y = sin - cos y = sin - cos [, ] by [, ] Sinusoid y = sin ( + ) - cos [, ] by [ 0, 0] [, ] by [, ] Not a Sinusoid [, ] by [, ] Sinusoid Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

12 Section. Gras of Comosite Trigonometric Functions 0 y = cos a 7 - b + sin a 7 b y = cos + sin 7 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 [, ] Sinusoid Not a Sinusoid Quick Review.. Domain: ( q, q); range: [, ]. Domain: [, q); range: [0, q). Domain: ( q, q); range: [, q) 7. As : - q, f() : q; as : q, f() : f g()= () -=-, domain: [0, q). g f()= -, domain: ( q, ] [, q). Section. Eercises. Periodic. [, ] by [, ] [, ] 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 [, ]. Domain: ( q, q). Range: [0, ]. [, ] by [.,.]. Not eriodic. [, ] by [ 0.,.]. Domain: all Z n, n an integer. Range: [0, q). [, ] by [, 0]. Not eriodic. [, ] by [ 0., ] 7. Domain: all Z an integer. Range: ( q, 0]. + n, n 7. Periodic. [, ] by [, ] [, ] by [ 0, 0.] [, ] by [ 0, 0] Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

13 0 Cater Trigonometric Functions In #9, te linear equations are found by setting te cosine term equal to ; y + [ 0, 0] by [ 0, 0] 9. (a). (c). Te daming factor is e, wic goes to zero as gets large. So daming occurs as : q.. Te amlitude,, is constant. So tere is no daming. 7. Te daming factor is, wic goes to zero as goes to zero. So daming occurs as : f oscillates u and down between. and.. As : q, f() : y - 0. [ 0, 0] by [, 8] 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 ) 7. 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)].. A., b=, and 0., so f(). sin[ (-0.)].. A., b=, and., so f(). sin(+.).. Te eriod is. [0, ] by [, ]. f oscillates u and down between and -. As : q, f() : 0. [0, ] by [.,.]. Period : sin[(+ )]+cos[(+ )]= sin(+ )+ cos(+ )=sin + cos. Te gra sows tat no < could be te eriod. [, ] by [.,.8]. Period : sin[(+ )+]-cos[(+ )-] = sin(++ )-cos(-+0 ) = sin(+)-cos(-). Te gra sows tat no < could be te eriod. [, ] by [.,.] 7. Te eriod is. [, ] by [, ] [, ] by [, ] Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

14 Section. Gras of Comosite Trigonometric Functions 0 7. Period : ` sin c 7. ( + ) d ` += ` sin a += ` -sin. ` + = ` sin + bd ` + Te gra sows tat no < could be te eriod. [, ] by [, ] 9. Not eriodic. [, ] by [, ]. Not eriodic. (a) tis is obtained by adding to all arts of te inequality - sin. In te second, after subtracting from bot sides, we are left wit -sin sin, wic is false wen sin is negative. 77. Gra (d), sown on [, ] by [, ] 79. Gra (b), sown on [, ] by [, ] 8. False. Te beavior near zero, wit a relative minimum of 0 at =0, is not reeated anywere else. 8. Te negative ortions of te gra of y=sin are reflected in te -ais for y = ƒ sin ƒ. Tis alves te eriod. Te answer is B. 8. f( )= +sin ( )= -sin = f(). Te answer is D. 87. (a) Answers will vary for eamle, on a TI-8: = 0.0 Á L 0.07; 7. on a TI-8: = 0.08 Á L 0.07; 7 on a TI-8: = Á L 0.0; on a TI-9: = 0.0 Á L (b) Period: = /=0.0». For any of te TI graers, tere are from to cycles between eac air of iels; te gras roduced are terefore inaccurate, since so muc detail is lost. 89. Domain: (- q, q). Range: [, ]. Horizontal asymtote: y=. Zeros at ln a, n a non negative integer. + nb [, ] by [ 7, 7] For # 70, 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). 7. sin must be nonnegative, so te domain is Á [, ] [0, ] [, ] Á ; tat is, all wit n (n + ), n an integer. Range: [0, ]. 9. Tere are no restrictions on te value of, since sin Ú 0, so te domain is (- q, q). Range: [0, ]. 7. (a) [, ] by [.,.] 9. Domain: [0, q). Range: (- q, q). Zeros at n, n a nonnegative integer. [ 0., ] by [, ] [0, ] by [ 0., 0.] (b) For t>0. (aroimately). 7. No. Tis is suggested by a gra of y=sin ; tere is no oter section of te gra tat looks like te section between and. In articular, tere is only one zero of te function in tat interval (at =0); nowere else can we find an interval tis long wit only one zero. 9. Domain: (- q, 0) (0, q). Range: aroimately [ 0., ). Horizontal asymtote: y=0. Zeros at n, n a non zero integer. [, ] by [ 0.,.] Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

15 0 Cater Trigonometric Functions 9. Domain: (- q, 0) (0, q). Range: aroimately 7. tan ( )= [ 0., ). Horizontal asymtote: y=. Zeros at, n n 9. sin a - a non zero integer. b = -. cos (0)=. aro... aro aro aro [, ] by [ 0.,.]. y=tan ( ) is equivalent to tan y=, /<y< /. For to get very large, y as to aroac /. So and Section.7 Inverse Trigonometric Functions Eloration. tan = =. tan =tan a = b lim tan - ( ) = /. : -q lim : q tan -() = /. cos asin - =cos = b. sin acos =sin a = b b. + (by te Pytagorean teorem). sin(tan ())=sin( )= +. sec(tan ())=sec( )= +. Te yotenuse is ositive in eiter quadrant. Te ratios in te si basic trig functions are te same in every quadrant, so te functions are still valid regardless of te sign of. (Also, te sign of te answer in () is negative, as it sould be, and te sign of te answer in () is negative, as it sould be.) Quick Review.7. sin : ositive; cos : ositive; tan : ositive. sin : negative; cos : negative; tan : ositive. sin = 7. cos = - 9. sin = Section.7 Eercises For #, kee in mind tat te inverse sine and inverse tangent functions return values in c -, and te inverse, d cosine function gives values in [0, ]. A calculator may also be useful to suggest te eact answer. (A useful trick is to comute, e.g., sin ( /) and observe tat tis is 0., suggesting te answer /.). sin a = b. tan (0)=0 7. cos a sin =cos a # b = b 9. arcsin acos =arcsin = b. cos (tan ) = cos a b =. Domain: [, ] Range: [ /, /] Continuous Increasing Symmetric wit resect to te origin (odd) Bounded Absolute maimum of /, absolute minimum of / No asymtotes No end beavior (bounded domain). Domain: ( q, q) Range: ( /, /) Continuous Increasing Symmetric wit resect to te origin (odd) Bounded No local etrema Horizontal asymtotes: y= / and y= / End beavior: lim tan - = / and : q lim tan - = -/ : -q 7. Domain: c -. Range: c -. Starting from,, d d y=sin, orizontally srink by. 9. Domain: ( q, q). Range: a -. Starting from, b y= tan, orizontally stretc by and vertically stretc by (eiter order).. cos a = b Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

16 Section.7 Inverse Trigonometric Functions 07. First set =sin and solve sin =, yielding = for integers n. Since =sin must be in + n, we ave sin c - =, so =., d. Divide bot sides of te equation by, leaving sin =, so = and for all integers n. + n + n,. For any in [0, ], cos(cos ),=. Hence, =. 7. Draw a rigt triangle wit orizontal leg, vertical leg (if 7 0, draw te vertical leg u ; if 0, draw it down), and yotenuse +. Te acute angle adjacent to te leg of lengt as measure =tan (take 0 if 0), so sin =sin(tan )=. + > 0 < 0 + θ θ + 9. Draw a rigt triangle wit orizontal leg -, vertical leg (if 7 0, draw te vertical leg u ; if 0, draw it down), and yotenuse. Te acute angle adjacent to te orizontal leg as measure =arcsin (take 0 if 0),so tan =tan(arcsin )=. - > 0 < 0 θ θ. Draw a rigt triangle wit orizontal leg, vertical leg (u or down as 7 0 or 0), and yotenuse +. Te acute angle adjacent to te leg of lengt as measure =arctan (take 0 if 0), so cos =cos(arctan )=. + =tan -Å=tan -tan. (b) Gra is sown. Te actual maimum occurs at L.9 ft, were L 8.9. [0, ] by [0, ] (c) Eiter L.8 or L. tese round to ft or ft.. (a) =tan s. 00 (b) As s canges from 0 to 0 ft, canges from about.8 to.90 it almost eactly doubles (a 99.9% increase). As s canges from 00 to 0 ft, canges from about.80 to.78 an increase of less tan, and a very small relative cange (only about.%). (c) Te -ais reresents te eigt and te y-ais reresents te angle: Te angle cannot grow ast 90 (in fact, it aroaces but never eactly equals 90 ). 7. False. Tis is only true for, te domain of te sin function. For < and for >, sin (sin ) is undefined. 9. cos ( /)=- /, so cos - (- /) = /. Te answer is E.. sec (tan ) = + tan (tan - ) = +. Te answer is C.. Te cotangent function restricted to te interval (0, ) is one-to-one and as an inverse. Te unique angle y between 0 and (non inclusive) suc tat cot y= is called te inverse cotangent (or arccotangent) of, denoted cot or arccot. Te domain of y=cot is ( q, q) and te range is (0, ).. (a) Domain all reals, range [ /, /], eriod. > 0 < 0 + θ θ +. (a) Call te smaller (unla- [, ] by [ 0., 0. ] beled) angle in te lower left Å ; ten tan Å=, or Å=tan (since Å is acute). Also, +Å is te measure of one acute angle in te rigt triangle formed by a line arallel to te floor and te wall; for tis triangle tan( +Å)=. Ten +Å=tan (since +Å is acute), so (b) Domain all reals, range [0, ], eriod. [, ] by [ 0, ] Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

17 08 Cater Trigonometric Functions (c) Domain all reals ecet /+n (n an integer), range ( /, /), eriod. Discontinuity is not removable. y =. (b) Te two orizontal asymtotes of te gra on te rigt are y = and y =. [, ] by [, ] 7. y= -tan. (Note tat y=tan a does not ave te correct range b for negative values of.) 9. In order to transform te arctangent function to a function tat as orizontal asymtotes at y= and y=, we need to find a and d tat will satisfy te equation y=a tan +d. In oter words, we are sifting te orizontal asymtotes of y=tan from y = - and y = to te new asymtotes y= and y=. Solving y=a tan +d and y= for tan in terms of a and d yields =a tan +d; so, - d = tan -. We know tat y= is te lower a orizontal asymtote and tus it corresonds to y = -. - d - d So, = tan - = - Q = -. Solving a a tis for d in terms of a yields d = + a ba. Solving y=a tan +d and y= for tan in terms of a and d yields =a tan +d; so, - d = tan -. We know tat y= is te uer a orizontal asymtote and tus it corresonds to y =. - d - d So, = tan - = Q = Solving tis a a. for d in terms of a yields d = - a ba. If d = + a and d = - a ten ba ba, + a = - a So, 8= a, and a = 8. ba ba. Substitute tis value for a into eiter of te two equations for d to get: d = + a =+9= or ba8 b d = - a =-9=. ba8 b Te arctangent function wit orizontal asymtotes at y= and y= will be y = 8 tan (a) Te orizontal asymtote of te gra on te left is (c) Te gra of y = cos - a will look like te gra on b te left. (d) Te gra on te left is increasing on bot connected intervals. y y (, π) (, π) (, 0) (, 0) Section.8 Solving Problems wit Trigonometry Eloration. Te arametrization sould roduce te unit circle.. Te graer is actually graing te unit circle, but te y-window is so large tat te oint never seems to get above or below te -ais. It is flattened vertically.. Since te graer is lotting oints along te unit circle, it covers te circle at a constant seed. Toward te etremes its motion is mostly vertical, so not muc orizontal rogress (wic is all tat we see) occurs. Toward te middle, te motion is mostly orizontal, so it moves faster.. Te directed distance of te oint from te origin at any T is eactly cos T, and d=cos t models simle armonic motion. Quick Review.8. b= cot.9, c= csc 9.. b=8 cot 8-8 cot., c=8 csc 8 9., a=8 csc comlement: 8, sulement: Amlitude: ; eriod: Section.8 Eercises All triangles in te sulied figures are rigt triangles.. tan 0 =, so =00 tan 0 =00 00 ft 9. ft.. Let d be te lengt of te orizontal leg. Ten tan 0 0 ft 0 =, so d= =0 cot ft. d tan 0 Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

18 Section.8 Solving Problems wit Trigonometry 09. Let / be te wire lengt (te yotenuse); ten ft cos 80 =, so /= = sec ft. O cos 80 Let be te tower eigt (te vertical leg); ten tan 80 =, so = tan ft. ft 7. tan 80 =, so =8 tan 80 8 ft 0 ft. 9. tan 8 =, so =00 tan 8 89 ft. 00 ft. tan =, so =0 tan. m. 0 m 8 ft 00 ft LP. tan =, so LP=. tan.98 mi.. mi. Let be te elevation of te bottom of te deck, and be te eigt of te deck. Ten tan 0 = and 00 ft + tan 0 =, so =00 tan 0 ft and += 00 ft 00 tan 0 ft. Terefore =00(tan 0 -tan 0 ). ft. 7. Te two legs of te rigt triangle are te same lengt (0 knotsá r=0 naut mi), so bot acute angles are. Te lengt of te yotenuse is te distance: 0 L 8.8 naut mi. Te bearing is 9 + =0. 9. Te difference in elevations is 097 ft. If te widt of te 097 ft canyon is w, ten tan 9 =, so w w=097 cot 9 8 ft. 097 ft 80 8 º 0 m w 9º. Te acute angle in te triangle as measure / 80-7 =, so tan =. Ten ft O= tan 8 ft.. If is te eigt of te vertical san, tan =,so. ft =. tan 9.8 ft.. Let d be te distance from te boat to te sore, and let be te sort leg of te smaller triangle. For te two triangles, te larger acute angles are 70 and 80. Ten d d tan 80 = and tan 70 =, or =d cot and +0=d cot 70. Terefore 0 d= 9 ft. cot 70 - cot 80 v 7. (a) Frequency: =8 cycles/sec. = (b) d= cos t inces. (c) Wen t=.8, d.8; tis is about. in left of te starting osition (wen t=0, d=). 9. Te frequency is cycles/sec, so v = # = radians/sec. Assuming te initial osition is d= cm: d= cos t.. (a) Te amlitude is a= ft, te radius of te weel. (b) k= ft, te eigt of te center of te weel. v (c) rotations/sec, so = /0 radians/sec. = 0. (a) Given a eriod of, we ave =. ƒbƒ ƒbƒ = so ƒbƒ =. We select te ositive = value, so b =. (b) Using te ig temerature of 8 and a low temerature of 8, we find ƒaƒ = = so ƒaƒ = and we will select te ositive value k = = (c) is alfway between te times of te minimum and maimum. Using te maimum at time t=7 and te 7 - minimum at time t=, we ave =. So, = + =. (d) Te fit is very good for y = 7 sin a (t - )b +. [0, ] by [, 88] (e) Tere are several ways to find wen te mean temerature will be 70. Graical solution: Gra te line t=70 wit te curve sown above, and find te Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

19 0 Cater Trigonometric Functions intersection of te two curves. Te two intersections are at t.7 and t 9... Model te tide level as a sinusoidal function of time, t. r, min=7 min is a alf-eriod, and te amlitude is alf of -9=. So use te model f(t)= cos ( t/7)- wit t=0 at 8: P.M. Tis takes on a value of 0 at t=. Te answer is D.. (a) [0, ] by [, 88] Algebraic solution: Solve for t. 7 sin a (t - )b + = 70 7 sin a (t - )b + = 70 sin a (t - )b = 7 (t - ) = sin - a 7 b (t - ) L 0.99,.8 Note: sin =sin ( - ) t.7, 9. Using eiter metod to find t, find te day of te year.7 9. as follows: # L 9 and # L 87. Tese reresent May 9 and October.. (a) Solve tis graically by finding te zero of te function P = t - 7 sin a t. Te zero occurs at aroimately b.. Te function is ositive to te rigt of te zero. So, te so began to make a rofit in Marc. (b) Te first is te best. Tis can be confirmed by graing all tree equations. (c) About oscillations/sec. = L 9. Te 7-gon can be slit into congruent rigt triangles wit a common verte at te center. Te legs of tese triangles measure a and.. Te angle at te center is, so a=. cot cm. = 7 7 L. 7. Coosing oint E in te center of te rombus, we ave AEB wit rigt angle at E, and mjeab=. Ten AE=8 cos in, BE=8 sin in., so tat AC=AE. in and BD= BE.9 in 9. =tan (a) [0, 0.00] by [ 0., ] 7 mi u (0.0)(7) = 0. mi [0, ] by [ 0, 0] (b) Solve tis graically by finding te maimum of te function P = t - 7 sin a t. Te maimum occurs b at aroimately 0.7, so te so enjoyed its greatest rofit in November. [0, ] by [ 0, 0] 7. True. Te frequency and te eriod are recirocals: f=/t. So te iger te frequency, te sorter te eriod. 9. If te building eigt in feet is, ten tan 8 =/0. So =0 tan 8 L 80. Te answer is D. [0, 0.009] by [.,.] (b) One retty good matc is y=.97 sin[7(t-0.000)] (tat is, a=.97, b=7, =0.000). Answers will vary but sould be close to tese values. A good estimate of a can be found by noting te igest and lowest values of Pressure from te data. For te value of b, note te time between maima (aro =0.007 sec); tis is te eriod, so b L. Finally, since L is te location of te first eak after t=0, coose so tat 7( ). Tis gives L (c) Frequency: about 9 Hz. L L It aears to be a G. (d) Eercise ad b, so te frequency is again about 9 Hz; it also aears to be a G. Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

20 Cater Review Cater Review. On te ositive y-ais (between Quadrants I and II); # 80 =0.. Quadrant III; # =. 80. Quadrant I; 78 # = Quadrant I; # = or radians For #, it may be useful to lot te given oints and draw te terminal side to determine te angle. Be sure to make your sketc on a square viewing window.. =tan a =0 = radians b. =0 = radians. =0 +tan ( ) 9..7 radians 7. sin 0 = 9. tan( )=. sin =. sec a - = b. csc 70 = 7. cot( 90 )=0 9. Reference angle: 0 ; use a 0 0 rigt triangle wit = side lengts, ( ), and (yotenuse). sin a - =, cos a - =, tan a - = ; b b b csc a - =, sec a - = ; cot a - =. b b b. Reference angle: ; use a rigt triangle wit side lengts ( ), ( ), and (yotenuse). sin( )=, cos( )=, tan( )=; csc( )=, sec( )=, cot( )=.. Te yotenuse lengt is cm, so sin Å=, cos Å=, tan Å=, csc Å=, sec Å=, cot Å=. For #, since we are using a rigt triangle, we assume tat is acute.. Draw a rigt triangle wit legs 8 (adjacent) and, and yotenuse 8 + = 89= sin =, cos =, tan = ; csc =, sec =, cot = radians For #9, coose wicever of te following formulas is aroriate: b a= c - b =c sin Å=c cos ı=b tan Å= tan b a b= c - a =c cos Å=c sin ı=a tan ı= tan a a a b b c= a + b = = = = cos b sin a sin b cos a If one angle is given, subtract from 90 to find te oter angle. If neiter Å nor ı is given, find te value of one of te trigonometric functions, ten use a calculator to aroimate te value of one angle, ten subtract from 90 to find te oter. 9. a=c sin Å= sin 8.0, b=c cos Å = cos.87, ı=90 -Å= a. b=a tan ı=7 tan , c= cos b 7 = 0., Å=90 -ı= cos 8. a= c - b = 7 - = =.90. For te angles, we know cos Å= ; using a calculator, we find 7 Å., so tat ı=90 -Å.8.. sin 0 and cos 0: Quadrant III 7. sin Ú 0 and cos 0: Quadrant II 9. Te distance OP=, so sin =, cos =, tan = ; csc =, sec =, cot =.. OP=, so sin =, cos =, tan = ; csc =, sec =, cot =.. Starting from y=sin, translate left units. Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

21 ƒ ƒ ƒ ƒ ƒ ƒ Cater Trigonometric Functions [, ] by [.,.]. Starting from y=cos, translate left units, reflect across -ais, and translate u units. [, ] by [, ] 7. Starting from y=tan, orizontally srink by. [ 0., 0. ] by [, ] 9. Starting from y=sec, orizontally stretc by, vertically stretc by, and reflect across -ais (in any order). [, ] by [ 8, 8] For #, recall tat for y=a sin[b(-)] or y=a cos[b(-)], te amlitude is a, te eriod is, ƒbƒ and te ase sift is. Te domain is always ( q, q ), and te range is [ a, a].. f()= sin. Amlitude: ; eriod: ; ase sift: 0; domain: ( q, q); range: [, ].. f()=. sin ca -. Amlitude:.; eriod: 8 bd ; ase sift: ; domain: ( q, q); range: [.,.]. 8. y= cos ca -. Amlitude: ; eriod: ; bd ase sift: ; domain: ( q, q); range: [, ]. For #7 8, 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. 7. a.7, b=, and., so f().7 sin(-.). 9. L 9.99 L 0.87 radians 7. = radians 7. Starting from y=sin, orizontally srink by. Domain: c -. Range: c -.,, d d 7. Starting from y=sin, translate rigt unit, orizontally srink by, translate u units. Domain: c0,. d Range: c -., + d 77. = 79. = 8. sin 8. As ƒƒ : q, 0. : 8. tan(tan )=tan = 87. tan, were is an angle in c -, asin- b = sin u cos u d wit sin =. Ten cos = - sin u= 0. =0.80 and tan = Periodic; eriod. Domain Z, n an integer. + n Range: [, q). 9. Not eriodic. Domain: Z, n an integer. + n Range: [ q, q). 9. s=r =() a b = 9. tan 78 =, so =00 tan m. 00 m 00 m 78 Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

22 Cater Project 0 ft 0 ft 97. tan = and tan 8 =, so d + =0 cot and d+=0 cot 8. Ten d=0 cot 8-0 cot 9 ft. Cater Project Solutions are based on te samle data sown in te table.. 0 ft 8 d 99. See figure below. 0. tan 7 '=, so = tan 7 ' 9. ft ft ft nort tower 8 7 sout tower 0. s=r =( in) a # 80 b = L. in. 0. Solve algebraically: Set T()= and solve for. 7. sin c ( - ) d + = sin c ( - ) d = 7. ( - ) = sin - a 7. b ( - ) L 0.,.98 Note: sin =sin( - ) L, 87 Solve graically: Gra T()= and T() = 7. sin c ( - ) d + on te same set of aes, and ten determine te intersections. [ 0.,.] by [0, ]. Te eak value seems to occur between =0. and =0., so let =0.. Te difference of te two etreme values is =0., so let a L 0./ L 0.. Te average of te two etreme values is ( )/=0.7, so let k=0.7. Te time interval from =0. to =., wic equals 0.8, is rigt around a alf-eriod, so let b= /0.8 L.9. Ten te equation is y L 0. cos (.9(-0.)) Te constant a reresents alf te distance te endulum bob swings as it moves from its igest oint to its lowest oint. And k reresents te distance from te detector to te endulum bob wen it is in mid-swing.. Since te sine and cosine functions differ only by a ase sift, only would cange.. Te regression yields y L 0. sin (.87-0.)+0.7. Most calculator/comuter regression models are eressed in te form y=a sin (b+f)+k, were f/b= in te equation y=a sin (b(-))+k. Here, te regression equation can be rewritten as y L 0. sin (.87 (-0.0))+0.7. Te difference in te two values of for te cosine and sine models is 0., wic is rigt around a quarter-eriod, as it sould be. [0, ] by [ 0, 00] [0, ] by [ 0, 00] Using eiter metod, we would eect te average temerature to be F on day (May ) and day 87 (October ). Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.

23 Coyrigt 0 Pearson Education, Inc. Publising as Addison-Wesley.