Section 4.7 Inverse Trigonometric Functions

Transcription

1 Section 7 Inverse Trigonometric Functions 89 9 Domin: 0, q Rnge: -q, q Zeros t n, n nonnegtive integer 9 Domin: -q, 0 0, q Rnge: -q, q Zeros t, n non-zero integer Note: te gr lso suggests n te end-bevior smtote = [ 05, ] b [, ] 9 Domin: [, ] Rnge: [0, 9] (roimtel) Zeros t nd [, ] b [, ] Section 7 Inverse Trigonometric Functions [ 5, 5] b [ 05, 5] 9 Domin: -q, 0 0, q Rnge: roimtel [ 0, ) Horizontl smtote: =0 Zeros t n, n non-zero integer [ 5, 5 ] b [ 05, ] 9 Domin: -q, 0 0, q Rnge: -q, q Horizontl smtote: =0 Verticl smtote: =0 Zeros t n, n non-zero integer [, ] b [ 05, 05] 95 Domin: -q, 0 0, q Rnge: roimtel [ 0, ) Horizontl smtote: = Zeros t, n n non-zero integer [, ] b [ 0, ] Elortion tn = = tn =tn = b + (b te Ptgoren teorem) sin(tn ())=sin( )= + 5 sec(tn ())=sec( )= + Te otenuse is ositive in eiter qudrnt Te rtios in te si bsic trig functions re te sme in ever qudrnt, so te functions re still vlid regrdless of te sign of (Also, te sign of te nswer in () is negtive, s it sould be, nd te sign of te nswer in (5) is negtive, s it sould be) Quick Review 7 sin : ositive; cos : ositive; tn : ositive sin : ositive; cos : negtive; tn : negtive sin : negtive; cos : negtive; tn : ositive sin : negtive; cos : ositive; tn : negtive 5 sin = tn = 7 cos = 8 sin = sin = 0 cos = Section 7 Eercises For #, kee in mind tt te inverse sine nd inverse tngent functions return vlues in c -, nd te inverse, d cosine function gives vlues in [0, ] A clcultor m lso be useful to suggest te ect nswer (A useful trick is to comute, eg, sin ( /) nd observe tt tis is 0, suggesting te nswer /) sin = sin - = b b tn (0)=0 cos ()=0

2 90 Cter Trigonometric Functions 5 cos = tn ()= b 7 tn ( )= 8 cos - b = 5 9 sin - 0 tn - = - b = - cos (0)= sin ()= ro ro 07 5 ro 85 ro 00 7 ro 7 8 ro 57 9 ro ro 59 =tn ( ) is equivlent to tn =, /<< / For to get ver lrge, s to roc / So lim nd Sq tn- = > lim S-q tn- = > =(tn ) is equivlent to = tn ;, 0 < / For to get ver lrge, in te ositive or negtive direction, s to roc /So lim tn - = > nd lim tn - = > Sq S-q cos sin =cos = b sin(tn )=sin = 5 sin cos =sin = b b cos cos 7 cos b = b = 7 cos sin =cos # b = b 8 sin[tn ( )] =sin = b 9 rcsin cos =rcsin = b 0 rccos tn =rccos =0 b costn = cos b = tn (cos )=tn ( )=- Domin: [, ] Rnge: [ /, /] Incresing Smmetric wit resect to te origin (odd) Absolute mimum of /, bsolute minimum of / No smtotes No end bevior (bounded domin) Domin: [, ] Rnge: [0, ] Decresing Neiter odd nor even (but smmetric wit resect to te oint (0, /)) Absolute mimum of, bsolute minimum of 0 No smtotes No end bevior (bounded domin) 5 Domin: ( q, q) Rnge: ( /, /) Incresing Smmetric wit resect to te origin (odd) No locl etrem Horizontl smtotes: = / nd = / End bevior: lim nd Sq tn- = > lim S-q tn- = - > Domin: ( q, q) Rnge: (0, ) Decresing Neiter odd nor even (but smmetric wit resect to te oint (0, /)) No locl etrem Horizontl smtotes: = nd =0 End bevior: lim nd lim S-q cot- = Sq cot- = 0 7 Domin: c - Rnge: c - Strting from,, d d =sin, orizontll srink b 8 Domin: c - Rnge: [0, ] Strting from, d =cos, orizontll srink b stretc b (eiter order) nd verticll 9 Domin: ( q, q) Rnge: - 5 Strting from, 5 b = tn, orizontll stretc b nd verticll stretc b 5 (eiter order) 0 Domin: [, ] Rnge: [0, ] Strting from =rccos =cos, orizontll stretc b nd verticll stretc b (eiter order) First set =sin nd solve sin =, ielding = for integers n Since =sin must be in + n c -, we ve sin =, so =, d First set =cos nd solve cos =, ielding =cos Ten solve cos =cos, wic gives =+n or = +n, for ll integers n Divide bot sides of te eqution b, leving sin =, so =sin L 079

3 Section 7 Inverse Trigonometric Functions 9 If tn =, ten =tn( ) L For n in [0, ], cos(cos ),= Hence, = For n in, sin c - (sin )= Since is in, d 0 c -, =, d 0 7 Drw rigt tringle wit orizontl leg, verticl leg (if 7 0, drw te verticl leg u ; if 0, drw it down), nd otenuse + Te cute ngle djcent to te leg of lengt s mesure =tn (tke 0 if 0), so sin =sin(tn )= + > 0 < Use te sme tringles s in #7: drw tringle wit orizontl leg, verticl leg (u or down s 7 0 or 0), nd otenuse + Te cute ngle djcent to te leg of lengt s mesure =tn (tke 0 if 0), so cos =cos(tn )= + 9 Drw rigt tringle wit orizontl leg -, verticl leg (if 7 0, drw te verticl leg u ; if 0, drw it down), nd otenuse Te cute ngle djcent to te orizontl leg s mesure =rcsin (tke 0 if 0),so tn =tn(rcsin )= - > 0 < 0 50 Drw rigt tringle wit orizontl leg (if 7 0, drw te orizontl leg rigt; if 0, drw it left), verticl leg -, nd otenuse If 7 0, let be te cute ngle djcent to te orizontl leg; if 0, let be te sulement of tis ngle Ten =rccos, so cot =cot(rccos )= - > 0 < 0 5 Drw rigt tringle wit orizontl leg, verticl leg (u or down s 7 0 or 0), nd otenuse + Te cute ngle djcent to te leg of lengt s mesure =rctn (tke 0 if 0), so cos =cos(rctn )= + > Drw rigt tringle wit orizontl leg (if 7 0, drw te orizontl leg rigt; if 0, drw it left), verticl leg - 9, nd otenuse If 7 0, let be te cute ngle djcent to te orizontl leg; if 0, let be te sulement of tis ngle Ten =rccos, so sin =sin(rccos )= - 9 > 0 9 < 0 5 () Cll te smller (unlbeled) ngle in te lower left Å ; ten tn Å=, or Å=tn (since Å is cute) Also, +Å is te mesure of one cute ngle in te rigt tringle formed b line rllel to te floor nd te wll; for tis tringle tn( +Å)= Ten +Å=tn (since +Å is cute), so =tn -Å=tn -tn (b) Gr is sown Te ctul mimum occurs t L 59 ft, were L 859 [0, 5] b [0, 55] (c) Eiter L 8 or L 5 tese round to ft or 5 ft 5 () is one cute ngle in te rigt tringle wit leg lengts (oosite) nd (djcent); tus tn =, nd =tn (since is cute) (b) Gr is sown (using DEGREE mode) Negtive vlues of corresond to te oint Q being usore from P ( into te icture) insted of downsore (s sown in te illustrtion) Positive ngles re ngles tt oint downsore; negtive ngles oint usore [ 0, 0] [ 90, 90] (c) =tn < 0

4 9 Cter Trigonometric Functions 55 () =tn s 500 (b) As s cnges from 0 to 0 ft, cnges from bout 58 to 90 it lmost ectl doubles ( 999% increse) As s cnges from 00 to 0 ft, cnges from bout 80 to 78 n increse of less tn, nd ver smll reltive cnge (onl bout 5%) (c) Te -is reresents te eigt nd te -is reresents te ngle: te ngle cnnot grow st 90 (in fct, it roces but never ectl equls 90 ) 5 () Sin() eists for ll, but sin () is restricted to [, ] Te domin of f() is [, ] Te rnge is [, ] (b) Since te domins of sin () nd cos () re [, ], te domin of g() is [, ] Te rnge is e f (c) Since sin() for ll, () eists for ll nd its domin is ( q, q ) Te rnge is c -, d (d) Sin() eists for ll, but cos () is restricted to [, ] Te domin of k() is [, ] Te rnge is [0, ] (e) Since 0sin()0 for ll, q() eists for ll nd its domin is ( q, q) Te rnge is [0, ] 57 Flse Tis is onl true for, te domin of te sin function For < nd for >, sin (sin ) is undefined 58 True Te end bevior of =rctn determines two orizontl smtotes, since lim rctn = - > nd S-q rctn = > lim Sq 59 cos (5 /)=- >, so cos - - > = 5 > Te nswer is E 0 sin (sin )=sin 0=0 Te nswer is C sec (tn ) = + tn tn - = + Te nswer is C Te rnge of f()=rcsin =sin is, b definition, [ /, /] Te nswer is E Te cotngent function restricted to te intervl (0, ) is one-to-one nd s n inverse Te unique ngle between 0 nd (non-inclusive) suc tt cot = is clled te inverse cotngent (or rccotngent) of, denoted cot or rccot Te domin of =cot is ( q, q) nd te rnge is (0, ) In te tringle below, A=sin nd B=cos Since A nd B re comlementr ngles, A+B= / Te left-nd side of te eqution is onl defined for 5 () Domin ll rels, rnge [ /, /], eriod [, ] b [ 05, 05 ] (b) Domin ll rels, rnge [0, ], eriod [, ] b [ 0, ] (c) Domin ll rels ecet /+n (n n integer), rnge ( /, /), eriod Discontinuit is not removble [, ] b [, ] () Let =sin () Ten te djcent side of te rigt tringle is - = - - cos( )= = - (b) Let =tn () Ten te otenuse is + sec ( )= + + b =+ (c) Let =cos Ten te oosite side of te rigt tringle is - = - - sin( )= = - A B

5 Section 7 Inverse Trigonometric Functions 9 (d) Let =cot () Ten te otenuse is csc + + = + ( )= b = (+ )= - + (e) Let =sec ( ) Ten te oosite leg of te rigt tringle is - = - - tn( )= = - 7 = -tn (Note tt =tn does not ve te correct b rnge for negtive vlues of ) 8 () cos(sin ) or sin(cos ) (b) sin(tn ) or cos(cot ) + (c) tn(sin ) or cot(cos ) 9 In order to trnsform te rctngent function to function tt s orizontl smtotes t = nd =, we need to find nd d tt will stisf te eqution = tn +d In oter words, we re sifting te orizontl smtotes of =tn from = - nd = to te new smtotes = nd = Solving = tn +d nd = for tn in terms of nd d ields = tn +d; so, - d = tn - We know tt = is te lower orizontl smtote nd tus it corresonds to = - - d - d So, = tn - = - = - Solving tis for d in terms of ields d = + b - d = tn - We know tt = is te uer orizontl smtote nd tus it corresonds to = - d - d So, = tn - = = Solving tis for d in terms of ields d = - b If d = + nd d = - ten b b, + = - So, 8=, nd = 8 b b Substitute tis vlue for into eiter of te two equtions for d to get: d = + 8 =+9= or b b d = - 8 =-9= b b Te rctngent function wit orizontl smtotes t = nd = will be = 8 tn () As in Emle 5, cn onl be in Qudrnt I or Qudrnt IV, so te orizontl side of te tringle cn onl be ositive (b) tn sin - bb = s = - (c) sin cos - bb = s = - (See figure below) 7 () Te orizontl smtote of te gr on te left is = (b) Te two orizontl smtotes of te gr on te rigt re = nd = (c) Te gr of = cos - will look like te gr on b te left (d) Te gr on te left is incresing on bot connected intervls (, π) s (, π) (, 0) (, 0) Solving = tn +d nd = for tn in terms of nd d ields = tn +d; so,

6 9 Cter Trigonometric Functions 7 () Te orizontl smtote of te gr on te left is = 0 (b) Te two orizontl smtotes of te gr on te rigt re =0 nd = (c) Te gr of = sin - will look like te gr b on te left (d) Te gr on te left is decresing on bot connected intervls (, π/) (, π/) Section 8 Solving Problems wit Trigonometr Elortion Te rmetriztion sould roduce te unit circle Te grer is ctull gring te unit circle, but te -window is so lrge tt te oint never seems to get bove or below te -is It is flttened verticll Since te grer is lotting oints long te unit circle, it covers te circle t constnt seed Towrd te etremes its motion is mostl verticl, so not muc orizontl rogress (wic is ll tt we see) occurs Towrd te middle, te motion is mostl orizontl, so it moves fster Te directed distnce of te oint from te origin t n T is ectl cos T, nd d=cos t models simle rmonic motion Quick Review 8 b=5 cot 9, c=5 csc 9 =5 cos 8 95, b=5 sin 8 80 b=8 cot 8-8 cot 5, c=8 csc 8 59, =8 csc 008 b= cot - cot 8 0, c= csc 077, = csc comlement: 58, sulement: 8 comlement: 7, sulement: Amlitude: ; eriod: 0 Amlitude: ; eriod: / Section 8 Eercises All tringles in te sulied figures re rigt tringles tn 0 =, so =00 tn 0 =00 00 ft 59 ft (, π/) (, π/) tn =, so =00 tn 00 ft ft Let d be te lengt of te orizontl leg Ten tn 0 0 ft 0 =, so d= =0 cot ft d tn 0 90 ft 90 tn =, so d= =90 cot ft d tn 90 ft d 5 Let / be te wire lengt (te otenuse); ten 5 ft 5 cos 80 =, so /= =5 sec ft O cos 80 Let be te tower eigt (te verticl leg); ten tn 80 =, so =5 tn 80 8 ft 5 ft ft cos =, so /= = sec 08 ft O cos tn =, so = tn 009 ft ft l ft 7 tn 80 =, so =85 tn ft 05 ft ft 8 Let be te eigt of te smokestck; ten tn 8 =, so =580 tn 8 ft 580 ft 9 tn 8 =, so =00 tn 8 89 ft 00 ft 8 00 ft