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# Trigonometry

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### Trigonometry

1. 1. Chapter 13 Trigonometric Functions of Angles Tirigonometry is usually considered as a study of the relationships between the sides and angles of a triangle. This chapter will discuss properties of trigonometric functions as applied to problems about triangles. We will Assume some familiarity with material from high school geometry, such as the Pythagorean theorem And the properties of similar triangles. The concept of trigonometric functions of angles will be extended to trigonometric notions of real numbers in the next chapter. 3.1 Distance in the Plane Let P1(: c;, m) and Pg(I3,ﬂ2) be two points in the zy-plane, where the coordinates n the axes reprment distance in the some units of measurement (see Figure 13.1.1). The intance between P1 and P2, denoted by d or | P.P2|, is the length of the line segment that ins P, and P2. The line segment that joins P, and P; will be denoted by P1P2. Y Y2 Figure 13.1.1. The distance d between P, and P2. 308 IV-igorwntetv-ic ﬂsnclinm of Angle: [ Chop. 1.9] 12. The perpendicular bioeclor of a line segment AD in the line that wutains all the p0ll| Ll that are equidimim to A and B. Show that point P is on the perpondinilu biptflur of AB in each of the following: a. P(3.3]. A(2.0). B(0,2l l2. l’('4.—3). .‘l(3.‘)). B(—l. -‘ll 13.2 Anglos nnd Degree Measure Let U and A be two points on a line L. We rlxzliiic my Uri to be the paint (7 and all the points on L beyond 0 uni containing point .4. Point 0 is called the‘ in‘ ial point of my 011, An angle A08 in the union of mu rays 0.4 and 0b‘. The common i itial point 0 in called the vertex of angle AOB. Figure lJ5.2.l illiistrates these deﬁnitions. Figure 13.2.1 We rerall that my linr divides the plant‘ into hall-planes. Ciwn All angle A03, L, izuntaiuing ray OA determines n hnlf~plnno Irhirh auntai is poiul. B (sue Figure 1 Similarly, the line L, containing ray 05 determines a h. all»pl. iu-iv which llailll A 13 .2(b)). The intersection of their half-pluws is rolled the interior of 410*’ (Figure l3.2.2(r: ]). (5) L3 dnerminci a N“-DIHK @ immam cl (3) L, determines . hall-plan: ooﬂtlinirig Point A hlﬂvolanﬂ oumeuan. poem 3 Figure 13.2.2. The interior of angle AOB
2. 2. [ Sec. 1.9.2] Angle: and Deyne Mooaum 309 Let the circumfemicc of a circle of radius 1 (also calla-<1 a unit circle) be divided into 360 equal parts. The meanun: of one such pun is deﬁned as one dagree. wriucn l°. Eath dense in divided into 60 minutes. written 60', and each minute is djvidn. -d into 60 semuila. wriwen 60'. The muuuve in degree: of an angle A03 in the nmauxe in dogmas of the are P0 inleroeptad by the angle on a unit circle with center at (). The arc PO is undinxtted, and angle A08 is usually referred to as an undincted angle or umigncd angle. Example 13.2.]. A right ungle has meurun: 90°. and A ulnnght angle (formed when rays OA and 03 are collinear) has measure 180". 4 Q . . 3 no” ){ 130" Q ‘5 «T - , .. Example 13.2.2 An Angie with uieaaurc 30" is ouothird of .1 90’-angle. .- 45"-angle is onehnll‘ of a 90°-angle. AuangieAOBinsa. idwbes'n ltondovd pon'tionifl. heAng1e‘ndnwninl. he: z-9-plane so that vertex 0 coincides with the origin and my 0.4 coincides with the positive 2-uis (Figure 13.23). Riny 0A is usually refermd to u the initial my or initial lid: of the anglc. nndray OB ismllodthe terminal myor terminal aide. 312 1hgo' nometﬁc Function; a] Angk, [ CM’. 1’ "°““ P “°’°“"°“ "Y W W" Wrrlinntu (r.0') Tnnv um I» dax-ribod by (r 9° 2, ago- '~hmkinany‘te . ll: hp 1 - ' ~ » ' . (mm mﬂin zer no at e no In mnrdnunno of; pom m n. .- pim an: um. umq - P(r.9) or (v.0"+360") Polar axis ﬁgure 13.3.1. The pom coovdinatas of a point at M . .., ... ,.. E"""P" 13-3-1 1-mm the pain: p whose polar coordinates are (3 —-339°) rand ﬂmhff 1"“ °‘ WW °°°rdin-was (r. 9) than ducribel P me}. am the absolute value: or}; is mix '. ... .m. Solution. r=3and9=-330' andpoimPiuhmm' thrﬁ hh ' . _ ‘, mulbobedﬂcribedwchaou)‘ In gunn : - w Vs» nu that I’ P(3.—3I7°) or (3.30'’) Exercile 13.3 "WM" W6 ‘Mk _I'I= point with the given Polar mmwmm. rm. ﬁnd Ill lam tum gum pam 0/ Mar aundmalu to dw. -rib: the paint. " W130”) 2- (5.-120") 3. (a,315°) & 0.405”) 6. (7.-559°) -, -> (g‘m5.. , 4. 9,270") 8. (ﬂ. is’) 1:]
3. 3. ) .41 13.4 Sine and Cosine of Angles Sine and Coach: of Angle: 313 Let in draw the polar courdinue system and the 11; mordjnazc axes so that the pok- and the origin coincide and the polar axis and the positive :1:-axis We observe that if paint 1’ 1m mctangirlrsr mordimm (.17, y) and polar coordinates (no). men r = /17+ R (see Figure 13.4.1). If P in nm the origin. we define the u‘n¢ of9 (wriw. -n 55119) and the amine M0 (written 5080) ms fnlloivaz sin6= E r and Flynn 13.4.1. Polar and rectangular coordinans We will show that in and me are [um-. tinne of the direclnd angle 0 rind an independent of the dioioe ol‘ (ha point I’ on the terminal my of 0. For suppose that I" were uinlher point an the Ienninnl ray ol'0 (refer to Figure 13.4.1). Lc1(r‘,0'). |nd(: r', y')b9 polar and recungulnr mordirnuao. l'l! l|)eCllVl'ly, of P’. Using pmpatics of similar triangle. in» have - -1 I -2-“ 1un0— r H and «:69 r — H.
4. 4. ' [ 55:. 13.4 ] Sine and Coaine of Angie: - 315 Y . .., _< F(0.1) ——>X Y Y J P(l. I) Pu/5.1) 2 . A ]. .x _. I —»x 1 /3 Flguru 13.4.2. Sine and cosine of some special angics. Ramark 1. It In choose point P(. z. y) on the terminal side of 9 so that r - 1, than «m0 = 2 nnd sin9 A :1. Remark 2. Sim‘: r is the hypotenuse of the right trinuglc with sides of length lzf and [34]. we have 1:] 5 r nnd M 5 r. Thus. -lsuinﬂgl and 'l<, mn9Sl- The anglu oonsidemd in Example 13.4.2 are angle: which lie in the first qtuldrnnt. Construr- y tion uf appxoprina congruent triangle: will yield 1- thesine nnduninnolnhucdanglesinmhnrquut rants. This ‘u illmtntcd in the maxi. exaxnplz. 70,1‘/5/3) Example 13.4.2 Cmnpuu: the nine and uzsine of 9 _v 2411!. V5” Solution. Th? cu-mm: side 01' 9 — 240" makes ~ , ,, 3 an angle of 60' with the negative z-axis (see fig- ure below). We cotnuilct the 60‘ angle in the ﬁnd. quadrant Ind choose the point P(: .y) = (1/2./5-! /2]. when 1' 7 l. Sinus the triangle in Q aw niganmmmc nmcumu a1 Angle: 1 amp. isj the lint quadnnt is congruent to the trinnglu in the third qiwdxanl. the mordiuau-is ul‘ point Q on the terminal ride of 0 = 2-10° can be determined based on tlu: ﬁmhquadrautal angle. Than. the rcctnngulnr rxxrrdinatm uf Q area: = ~§ andy = -1?. Sim-L-r = 1. we haw and nin240“ : y-—£. ou24D‘= 2 Rsmnrk 3. The signs of sin9 and C099 depend on the sign: of the . r-monlinnu» nnd the y-coordinate of the chooen point P on tlu: Icnninnl side of 0. Thin, the xigna depend on the quadrant Irlun the terminal side of0 lien and an shown in the liuﬂowing table. Quadrant II Quadrant III Quadvan: IV The 1-ectangulu mmdizims (I. y) of n point ran bv dctcrmixwsl from its point coordi- nuel (r.9) thmugli the lurnnilnio 2:"-ou0 and u= rnin0. The polar umdinuu (r.0) can be obtained Emu: the nscumgulu mnrdinnmi (z, y) by r = V/ :3 + y’ and 9 xnuzh that sin! ) -= 1% and cow - Example 13.4.3 Find the rectangular rxxmlinau: of the point whme pulnr ruuniiumcs an r = B and 0 = 240". Solutinn. Using the Inuit: of Example 13.4.2 for uin240" and eos2¢0". we hl'l' z-rooao-amu24o° = s(—%) --4 y = rsin9= 8sin240" = B(~ = —4x/3. Enmph 13.4.4 Find two pain af polar coordinates to: the point P(: :,y) — (-2‘/ :3, —2).
5. 5. — [ 521:. 1.9.4 1 Sine and Colin: of Angle: 317 in 1hnonom¢tﬁ'c NMW ‘'1 -"'1'" I c'”"" "1 Exurulu 18.4 InPvnNem0 1«12,aI= ¢«cIaMea»gIesav~dﬁn-1W-*'I<¢'l4°°""‘ "°"“’* 1 150- 2. 120- 3. 135° 4. 210». 5‘ m. 5. mo" 7. -as‘ 8- -180" , _ 43,. . m .459» 11. 955- 12. -1510’ 1,. Pmblgym 1.9-:3, ﬁnd lhe ruumgular wonﬁnatu (x. u), given an polar manlinntn (r.0). ”_ “J5” 1¢_ (5'1zo-) 15. (4.-135") m_ (/1-2700) 17. (1o, —-ms’) 13~ (8.457-‘>‘l In Pmblemr mu. [End (In pair: ofpalur movdiuutu (r. 9) yivcn nu mm-gular w-vniivwm (my). w_ (DJ) 39, (_4,_4/5) 21. (-31.20) 22. Ni. -1) 2:. (-3.0) M. (f2.—~/5) Find the solution: oflhz aquarium in Prvblaru A5-28. 25.. -inc-:9 2s. duo--4 27. cou0=—l/2 2s. nun0=—l In rmblenu 2940, find the distance between no mm min alwmb» 29. (Asa-1-nd(a, m') so. (s. lao') and (4.-140‘) Solution. We have r— ‘/ :’+g7= <2»/3)? Q (-211 =4. isin0=g= ¥=—é and m. e=§= T = The ﬁrs! -quadranl. a.l angle a such that zinc: : Q, n. ndoosu= =§isu= :ur'_11.epoimPi: snn. « fourth quadrant. so we slu-'u: .h n mnym-nc angle in the fourth quadrant such that the terminal sklu forms as 30" angle with the x-axis. Two angles that describe tlleptnilitrnoflhcunminalsideueei = —3U°nnd 0; 2 330". Thus. two pairs ofpular morrlimun-s for Ibis given point are (4. -30’) and (-1.330‘). Example 13.4.5 Find all angles 0 such that siu9 — 1/2. Sohninn. A first quadmnzal angle that iumxﬁes the equation is 0 _ :10". Note that the angle may also be in the newnd quadrant since the‘ pains in Alan positive for second—quadrn. ntal anglr. -. We construct a cougmrnt trial-glr in the second quzulranl. as . -lhuwn in Pigune 13.4.3. We haw Hm fulluuring solutions: 9 3 am + 360% . 150" + 350"): when I: is any integer. Y 18.5 Other llrignnomecric Phnctiolu wnecalleuzeheumulamnmotadmcud-nsleﬂmdeﬁn-xib7"***m"" auo= !and tn0=£» 1‘ T , ,0 the ruclanguhr and palm: coordinatn. nIl>t<‘-iV'1)¥- 8 P05“ 0'1 : ,l: ::mmn1(z‘v) nine and the amine funmian are deﬁned fur all Ilsud I085“- We nlsomgntiuned innction 13.4 that Figure 13.4.3. Some solutions of lsinf} : _1 < ma (1 ‘M T I SW05 L ; ..u. -um with the linennd cosine ofan angle 0 an the following = rix°n°Iw'**'- ‘“°°“°""
6. 6. [ Sec. 1.9.: ] Other Iﬁgonovnetric Phnctiona 319 1. The tangent of 9 (writoen umﬁ). which Is deliucd hy the formula xinﬂ coco, when @6950. Lane: '2. The olnangenl oft? (written col 9)‘ which is deﬁnal by the furmula cot9= L — —'—°‘9 “mg sing-. when sinﬁyéo. 3. The Juan! of? (wriuzvn 92419). which E dcﬁnul by the formula at-c9= it cogs. wbrre nno0#0. 4. The column! uf0 (written cscﬁ). which is deﬁned by the formula where siuﬂ 7‘ 0. l 9- —. ac s‘m9 Example 13.5.1 Compute the values of the six Lrigunomclrir functions or 120°‘ Solution. The u-nninnl side of 9 : 120° nmma n so-' angle with the ncgmw . r—a. xis. “'1' coustrun A 50‘ anglc in the ﬁns: quadrant (see Figurc 13.5.! ) and choose I’ on the mrmmnl aids such that 7‘ 2 1. Thus‘ the horizzontnl side is z = C0560” * 1/2 and tlw vertical side in y = sin 60-’ = s/3/2. By r. homing my point Q on :1» terminal side or 0 . — 120" and: mu r 2 L and by thp ronyuonoc of triangles. we have s1n12o°— Q and oc. s12u"— » 2 Thus. _Vain127’_/ .7/2_ , ,_ 1 _ 1 ‘““‘2°'E‘_1/2 ‘/5 “"”“"u. ..12o~ ‘a 1 1 2 '°°‘l2°°= §i20-> ‘-2 °"’l2°a ‘. an12o~‘ﬁ ___¢ 321 320 Iﬁganometric Ihmzﬁovu 0/ Angle: Y tmna 111» of (hr 1 Fﬁun 13.5.1 1' Tluorom 13.5.1 nm 1 i. The tangent and Joan: functions are duﬁned for all angles «mm: 90'' 1» 360°! ) or « —9o' +360”! -1, n e z. 1 . 11. Thcoolangentandzxzaecantfunctiansuvdcﬁaedfatnllmglcscxmpc 1so°n, ne z. Pmo1o1(I). s1mun9= 35 andnec9=- ; ‘,. bo111un; unandoooecmuann1«1ennea whmzvernoo0=0. 'l'hiaocnurawl: cnevcrth¢ecr1ninnliidcofBl'1annthey-udz. Tluu-. V wI0=0i[| ndonlyif0=90" +. ’.l60‘n1.1rH= —90°+360"n, wheren inlay integer, I vmtov II . (E ' 1 l ( ) urcle. U Ur ‘ Theorem 15.6.2 awn 1. Fbxevu-yv| lueaf9lbrwhi1:hnec3Andtan0amdeﬁned, sec’9=l+tan‘9. “K 1 11. Far evu-y value on M vlzidi 13:0 and com um deﬁned‘ cséo _ 1 + mﬂa. . Prooi or (1). By Theorem 13.-1.1. oos= o+. in’a - 1. By dividing 11.1.. nquazion by “““ on-’9(oos9 54 0). In: 11:: 1 sin'9 1 1 nine 2 . ».= a=”. ».= av °‘ (m)= ”(; .ﬁ)~ ‘w .51 Hence soc'o-1+un7a. I ma of (Ii). (1-zxemne. )
7. 7. : Sec. 1.9.6] Right Ihﬁxngks 323 Ind c. ,.5-9‘*i_-°&~'-3-2 hypmumue c‘ nu. .. sinA = cos B. The proufof (ii) and (iii) will he lcﬂ u an-cine. I Example 13.5.1 S01v9. for the unknown parts of triangle ABC. given that nudge C — 90". I a 10 and b = 25. Sﬁnion. We cumider Criauglr. ABC in Figure | ;.s. :s with the pi! !! labeled aaourdjng m the given " publan. Then, _ V_ _ __ c 3210 = = / ;n7+b3=/10'-‘+25’ = fr25_—5/29 oppocivesid-. _2_1g_§ A C “'4 uijaoentusde b‘ *‘s‘ ‘'= ?5 Fqun 13.6.3 zﬁing I calculmor. we get A = 21.8‘ and B = 90" A = 90’ — 21.8 ‘ 68.2‘. 1.3.6.2 A xegular In-Jugon ‘us inscribed in n circle of radius Ill centimeters. Find an enclosed by the hcxagnn. jinn. Fhxm Figure l3.6.-1. we see that the rugu]. u' hmmgon 4.-an be divided into six 9. iuclcehi triangle» with two sides oflcngth 10 and that third Ride of length 5. [At he the altitude of A component triangle. Tlwn. - _‘/ _2_: “"30" 10 “'20 ‘ . =2om3o'=2nG)-1ocm. m. m~= g n—1oooa3o"=1o(€ =5/ firm.
8. 8. 324 Iﬁgonometric function: a] Angler [ Chap. 1.! ] The area A1 of a component triangle is A, = , :;(uam)(m-sgzrr) — ; uo)(s/5) ; 25»/5 sq. mi. ma. hence, the area A nf the 1.. -argon is A = 5,4, —. ms‘/5) 2 150»/5 sq. rm. Smne pmhlerm requirr the measurement of the height or depth of objects abuw or below a. horixonml plane. For 1-xunple, if llw elevation of point B in Figure l3.6.F: were nzqrrinvd, than it would be nemuuy to detamine angle BALI. 11’ we wen: station: -d al. poim 3. angle ABD would dojus: as well angle» BAG md .4130 III equal. An angle of eleoutirrn or depvenion D - - - - - - - - - - ' B ist. henn; lebetweu: z|tlu: Iinr. -ufniglrtafnnnlxnernmrnnda hotixomnl line drawn through (lac point ufolxurrvltitm. when Llwanglainabmelhzhariwnkd plane, ititcalledananplc ofdevatiomandwlxrnthpnngkisbclnwiliudbdm nngk of depvluion. ln Figllfe 13.6.5. Angle BAG in ur * angle of elevuian, Ind angle ABD is In angle of depression. pg. " “£5 Example 13.6.3 ﬁom the top of a cliﬂ 1%) inner: high the nude of depnmion ol A boat uuunrnirzltl‘. Findthedistanoe(w1.heneucstumu) unbr. -boa: fa-rnnthehueofthe cliﬂ. Solution. We draw a diagram fur the pmhlam (nee Figure l3.6.6). The distance 3 ix given hr . 1: tanli0'= r6. z=120t: n60'. sinoerursw = Iin60‘/ cna60'— L, “/; —’ vu. -abuin: I:= lﬁ)/ _-2(l8I'n: lers. 5:. .. 13.6.6 Example 13.6.4 A surveyor sunning a: point A ﬁnds the M1514: ur 1-lozvntinn M the cap of a tree to be 30'. Vl'a. Ilring 20 means towards the tree to point B, be nous the urglv nf elevltion of the top of the (rec to be 60". What in the height at the tree it we assume um. point: .4 ma 3 are on um: pound? (Refer to Figure 116.7.) ‘T . ._. ,_ [ Sac. 1.9.6] Right Thianplza 325 Solution. We Inlve for tbs height /1 of the tree using the furmuh acdjaccntaide ‘ f r-ozangem. or-n acute angle = upped“ “M”. E I I . We have — A 20 III 3 d h ' Figun 13.6.7 I. Thus, oot30"= ¥ +cot60". Since ool.30" ; ./ Ema vul,60° = 1/r/3. vuvhxve 20 . 2 2:27? ~: h= l05mct. errr. . /5-29; -‘-_ h+3_o/5‘-ah 3 Exercise 1.1.6 In Prublenu [-4. wine for the rmlmmam parts of III: riglil triangle with angle: A. B and C = 90‘ and anvuponding opposite ride» a. b and c. l. n=9.c=1U 2.u: m0,b=5o0 3.A-30'. b-Io 4.zr_34J~. b=u In Pmblems 5 8, ﬁnd the per-imzfer and arm of the myular polygon. 5. A regular hexagon iimcrihed in a circle of radius 5. . A regular octagon inacrihed in a circle of radius 10. . A regular hexagon circumscribed about in circle of radius 10. A nxular octagon cirrurmaibed about at circle uf radius 5. ll‘ a lighthouse is I00 meters alnm! nan level and the Angle of elevation of tho liglrtliuunr Ml observed from in ship is 15‘, how far in the ship from the lighthouse? 10. A lower stands on mp at’ or building 20 malrnl high. Fhim a point uu the ground 40 meters ﬂour the building, the Angle of elevation M the top of the tuner is 35? Huw high is ll}: tlvwcr? H. A boy standing on la-val y’0und measures the angle of elevation tn the my oh; ﬂagpole whom". Hemcuuuruvthcdistarwr-Inthel'ootul't. beﬂ. n4a>oleanLlﬁnrlnix. v;ob«:25frv¢1. Hawhighiatheflagpulaiflheboyisﬁfratuill? ens: -e»
9. 9. J26 Iﬁgomnnztv-ic Flmctiovu of Angle: [ Chap. 13] 12. A lighzhoum 40 for: high stands on a small island. The lighthouse keeper M me my mm: the angle of depression of 3 nhip din-rtly east of (luv lightlmnno an be ll? " and Lhr angle 0|’ depression of a ship dil-only was: of the lighthouse in be 20“ How fax npan an: the ships? 13. A stream run: between poiulls A and B. nu that zhv dixlunu» .43 cannot ho vnenruxretl din-ctly (mun l3.6.8). Point C is dcwrluinod amch um AC is pcx1)cndicula: lu AB. 1r pain: C L»: 121) mevars away from point A and BC lnakcs an angle ol‘50° with AC. how fu apart In A Ind B’! Figure 13.6.! Fqure 13.6.9 14. A surveyor standing . : . point. A mensunts me anglr orolmmu of an» mp of It mwu m be GO”. vmuug so feel uphill wwnrds the lower a» palm. H, be limls um aw z-Inge tuft-lrvation of me mp of um mm is 75° ma ills‘ nngl! nfdvprrss-0|) of point .4 is 30' U-"igurv 116.9). '}un is the height of we um. -r (rum llrwl Bi’ 15. Two adjoining sidvﬁ Of A Pwsllelograuu am 3 and 14 ucnliuwu-rls long. Find the net of the pamlklogxmu if the anglv botvm-n these Sldes is 70". [ Sec. 1.7.7] The Sin: and Cosine Lawn 3'27 13.7 The Sine and Cosine Laws W1‘ discused the anlutitm 0! right lriulglea in uectiam l3.6. VVO will not consider techniques fol’ solving any triangle. including right triangles. using the Law uf Sines and the Law of Cosinea. We recall that two angina A nml B of a triangle are I-uppkvnentary angle: vlhnu-'vx . -l + 5 = 130". For trxunple, 30° uxd l50“ are oupplemmtary angles. The following l. hLv. ;n*| u givrs ll relationship between the ldnz and uuuiuc of oupplrmrnuxy nngha. Lemm. 1:. -1.1 7m and 8 . ;re nlpplemuntary angles. than 1' sun! = ‘ml? ii. cos A = — cos B Prool. In 0 5 A 5 90° and .4 + 3 = 1304.’ Aufgl. -. B is a s«voudvqu: ulI: Lnlal nnglr -muse terminal side forms an angle -ml. Inmnuuxre A with the negum : —m. . (Figun: l3,T. l). By mngmcnoe of triangles. we have xiuB = shut and cos}? = A 11:14.4. Y Fgurt 13.7.1. Sin: and cosine of supplcmcntary angles. Example 13.7.1 Since IN)‘ and 1% are Kupplnnzntary angles. then uinlM. 'r°-| in30"= % ma unl50"= ~oou30’= —%§. l’I‘hgm-urn 13.1.2 um. of Sinea) :4»: the interior mm of . "u'. '.. {gic be 4.3 and C. and lot the oonrsponding opposite shit: be mb and r. mspertivwly. Then sinA _ 9:13 _ §inC _ b R r a
10. 10. i 323 Tﬁgnnamelrir: Functimu oy Augie: [ Chap. 1.9] «hr, .~qm'n1I-nﬁ,7, " ‘ ’ a b r six? /1 siI| B zinc" Pronl. mm Figure 13.7.29), w. hAveainA - wt» and mu - Ix; /u. Thus, 1., — om; a u isiu 17. no um Figure 13.7.2 mm Figure l3.T. '£(h). sin I; = Ii, /c and ainusm —C) = A, /b. n, -Imnu1.1.7.1, niu(12l0" C’) V. uinC. Tm. A, ; muzz _ bzaiuL' so am siuH_xsiu(' b ’ o‘ - Tharetorc. sun A sin I! .1in(‘ I u b . - Thur Law nfsines is mod in nnlving triangles when tlw fulluwing infvirnlmilm arr km (x) ‘W0 nae. and an uppmito anglt-. or (2) Own M1311: and 0114- mid». I/ ha-n two sides and an opposite Angle are given, .u| _v sides a and b and angle A. we mu» “"5 = “"4 . nr. sqnivalrmlv. smn - ""“" . b a v a Nole than Rh’: gives rise to thrve possibilities: (‘mac l. sinl? ;~ 1. Sim: n- 1 5 ml? 5 l. 17 in um . mu'1.1.«. Tm-« vzuw hm ; ... m1.. um. . Case 2. aiuB — 1, Then. B -= 90’ and line Lriauugle in a right lti.1n; :l<'. I Sen 13-7 l The Sine and cman. Law 329 Case 3. sinB < I. By benlxnn 117.1, siu(l80" » 5) _— sin 8. Thus, that! will be [via possible valu_n: an acute angle 0 and in supp]: -men: H’ = 150’ — 13. The mute angle nu always a solution. Thu suppliexumt B‘ is a second wlutiun whcnzvar A + H’ < 180“. Example 13.1.2 Is it pulllible to construrt triangle ABC with . . = 1o_ 1, 4 mg 3 = 45¢. Solution. L-"sing welnve aainu _ ll)uin45' b ’ 4 ’ Since sin A > 1. there is up solution for A. The triangle that, satixﬁqﬁ ‘hp. gum, ., .,, ,d, ’g; .,, ,_. g unpumxible to oonstnxcs. sinA»- Exampk 13.7.3 Use the Law of Sims to solve for angle A of the right triangle ABC will: n=3.c——5undC=90'. Solution. By the Law of Sins. sn. A_5'. nc n ‘ I: ' Thur, mo‘ = - _ -M_ _ r o 5 Using a calculator or a Irigonnmetric table. we have A _ 36.9” and A’ 180" — ,4 : I80" — : w.9° = 143.1°. Since G + A’ - 90° +1411" > 130°, A’ it mu . solution. The unique suluticm is A = 36.9’. sinA ; Example 13.1.4 Find males 8 and (7 of triangle ABC, given um 41 ; 7, b = 3 3.. .; A = 52°. Solution. I-Yam
11. 11. run THgonomdn'c nmcumu of Angle: [ Chap. zsj Thus. B = 6-1.2“ and H’ = ISO'’— 1? = 180’ — 61.2" = 115.8". Sinrs A 4 B’ A 5? 6 11:13" <: 130“. H’ is a second solution. Whrn B = 64.2”. F =1ao° — (A + m=1w“ 152" ~ 5-1.2“) : 53.3" when B’ ; 115.3”. ("=1s0° -1.1 5» 12') = 1210" 152" +1153’) : 12.1: Both triangles ABC um AH’(, ”’ satixfy 11:» giwn conditions, The Law of Sin: does not apply when (hr: -o sidnx of I triangle am giwn. nr when hm 811194 and the included angle am uivvu. VVP will umv d'L-M1131-1 thv Lzuv of ('01Iin¢-s which ia useful in than cues. Thoonsm 13.7.3 (Law of Cotinu) in 51.11» 11.1 And angle at .1: trinnglw ABC 1,. "- givrn. T11.-11. r’= u’+b’ —-2a. hms('. Proof. Lat ma.1.1gl1- .400 be drawn um the 1:1;-planv no that angle (7 is a posnivv a11gle- in standard position (Figur! l3.7.3). The pnlu (‘00l’(ﬁIl1ItPﬂ of vvrtiuu A and H are . -1(b.0°jI and Bin. (‘L The corn-xsponding recuulgulnr roorciilxnlm law A(h. 0) and B(ams(‘. asin ('1 Th! dismuw ht"? -VPQTI lhc paints A and B, vrhirh is z‘. i: ulvtninrd using the tlixtanre fommln I04 follmrs. ‘ Vﬁiigﬁrmi - K— .1510 If — 1? A'2abvusC +a’coa''(: . r1’.1n’(' . 1»? — 2uboo1x(' 4 a'(m‘C + 5-. ..= o, ByThm1-um 13.4.1. cos7C + . -1n'c = 1. Thus. 8 -a‘ob‘—2abma<7. I . .<QI‘ l3¢C- 1-7-7} The Sin: and Coaine Lain 331 Y Figure 13.7.3. 13,- relabcllixxg vemx c of 1-‘agum 13.7.3. we may also write the Law uf Canines as follows: ‘ a’= b2+¢*"-‘I11-uouA b’ - a‘ 4 c’ — zcmcoas The Law of Cannes may be usrd to solve a triangle when the {nllowiug information an: H’) clue: sides. or (2) two sides and the included angle. Example 13.7.5 Find the unknown parts or triangle ABC, given A : am, 1» = 7. n = 1n. Sol-lien. The given pan: are way aida and ch: included angle (Figure 13. _. )_ 5,. up Law of Casiucs. C a'= b*+. ?-2bcoos.4 . — 7’ + 10’ — 2(1)(1o) c0380" = 49 + 100 — 14o(o.17355) r 124138926 In x I V a a —_ ¢ : l1.l66 By the Law M Sines, ainB oinA A 5 ' . , «=10 Finn! 13.7.4
12. 12. we ‘ ’ Aug‘ | ' Soc. 1 Chapter 1.9 Review Pnobema 33.‘! 1hgo' name Phnchoru o u 8. A vuln-Ann in the midrib rrf A plain has .3 conical shape with xidrs having 40 dc-gn-e Hlupn. bsinA 7.«inB0" _ 7(U.98«l8l)_ On at site .5 kilometers away from thv fool of the volcano. the angle of elevation of the . l.. 13 = = T1 1? _ H M . . top of the volcano is 15'. Find the hcixlll cl u. .- vulcmw and (hf mlnn «pf in. lm-. ‘‘ ' ’ 4 I my B, 9. nmm nn elevated palm loo rneters away {mm a tower. the angle oi dcpnslsiun or the Tlms. B — 38-1° and H’ -' I80’ - 334" -1“-9‘ Sm" " * 5 ‘ 5°” * “”'9° > ‘ base of the tower is 45'. AndI1w nnglr or elmulnn of the cup in 30°. Find the height of is not Iuollltioll. When 8 = 38.1“. ghg “, ,,e, _ 10. A plane ﬂifli It an lﬂtilude of 'lll. (Il(J fort. Al. some Instant the pilot S003 hilllsulf between lwo ships: mu: lship luving an angle nf dc-vpn§lK'Inn 21)". and tho othvr has angle at‘ deprvmion 10'. Find the distance between tlac ships. C: ma"—(A4-3)-1w—uso~+3a. l')= o1.s°. Exercise 1.1.7 ll. Au u. u.l. iu: n.su. I TV station whose mm-l. lorntion is unknown makes an inwerluiluem hmndcaat of banned materials. The police has him c| u'(runir d('Vi(‘(‘1£. can of whirl: In Pmblema I 3. mac tr-‘angk ABC yiveﬂ Me iv-diwlr-1 wnvﬁﬁwl can dc-tact the direction of the TV signal, but no: zlm location or the scum» One such 1. .4 = 5o°, u = 7.b= s 2. B = 20". a = lo. b= 7 device is placed at location A, mtul tlxr nun-r. at Ioratiun I1, 10 kilornetom away from 3. A = sow. = 8.b =12 4. c = l1l0". b = 5,c — 2 A. The unknown location of the source of the uunuumrlmcl T’ bruiulcaal. is lnlx-lk-d (1 5- A-um-3~c=4 6- ”= '"°"~ = “~€= “ S17.‘5£\$? ';£1.“§. “‘. .;. i;"§f. .'Z. ?;1.‘L¥'“; .o-”'? vEI‘; “;‘l. ”Z‘13;”).3'3.‘i£’; .§’3;; .'f‘‘H; L‘’3;2:321‘; 7. 4- = .b= 4.c=5 3- *' '‘= ‘’‘ 5'‘= ‘’ -15’ at location B, with thae angles dnxuu to be interior axlglt-as of AABC. Give un- 9. The diagonal: or a pmllelogrnn an: 10 and 14 acnlillleterl lmlg. Find an r--rinu-wt 0' . ,., .,, .n. ..m or c. the pnnllclogrun, ll‘ tlw diagonals form n 45-’ Magic. 12. l. .: A. H and C be interior angles on: triangle -ml. oppnnn~ nan: a, b and 1', mp. -mn»ly. 10. A nulio tower stands between twn points A and B on the ymmd. The angle of elevation 21. [fa = 7, I: = 4 and H - IN’. ﬁnd b. oftbe. topuflhn: lowrr fmmpuint A is -15‘. nnd son B 30'. How high 1- the M-M. lawn b. ll’ A = 30', c = 45- and c — 4. ﬁnd a. elm A ma am so nelmap-rt? (Wriw your--=2-v-r to the n-amt u-ever-l c. In _— 60', 5 = 20 ml r. - 15. find 1:. l3. Two adjacent side! ol' :5 trinnglc lnm-. lengths an rm and 50 cm. ml they form an . a.. ,;lo of 30'. Find tho urn of the triangle. Chapter 13 Review Problems I. Find the distulc: between palm (—l,2) and (4,5) _ 2. Given an point: .4(o. o) and Bu. 1. ), ldenlfy “" ‘M Wm of two -q---1-stml txinnglas with onmrnnn tide AB‘ 3. Find the mrtnngulax coordinates of the point with (3. -135-')» -1. Givc a pair of polar coordinates to Mar puiul with rrrtangulzlr coordinates ( —. ';,5). 5. Find the vnlnas or six trlgnnolnrtric functions of 9 : 740'. 5. How fall away l. up hnriwn as seen by one vl-In) is looking luvmrds the npn mm from the lop ofa rlill which ‘n 400 meters above the an Incl’! Give you mm In | uJD"'°'m~ [Usr thr fact Lllat the such has an averagv ntlilu of \$370.5 | d|(m1rIm1.] 7. A: pound level so nu. 'I. cI's away fmm n man. the ulglr of elrvuiou of thv 0017 UT 1111‘ We is 60'. What is me heigln nfzlw txu-7
14. 14. Ryan 14.1.3. beg“. .. Mi“ in u was and arch» . but P(z. v) be an nrbitnry pom an mm-cmd. z'+y’-1.0m-idmheuc «flex-mh mom n(1,o) In P(z, y) (9, pi‘. “_“‘ l41‘-5)‘ We deﬁne trigaomnetrir: func- numofrul lumbwtuﬁ-ﬂloug: m'nt= v actzl/ y mat = 2' sec! =1/2 um! —. y/1: cm; = , /y qanaidcxing the an that AOAP is . riyu umnglr or -5:11 hypomauae 1. n have sinl —. sint"’ clcl = c1-)ctl') ms! : mu"! .9: «mt -. um cw = ,.. ,,r>: amt = cute") Tllme eqlunlities imply that trigonomatric function: of real numbers uld trigonomet- ric functions af Angle: have iduttical pmp anion. {Ch-=9» I4.’ Hun 14.1.! Spc<; n'aI_: n:s u-. a um! arde. —T— : Sec. 14.1 Radian Manure and Han! Numbcrl 337 Tahlv 1-LL2 gives 5111!. um: and am: for some : <p(<'iz1l v. :1.1:1m of the real uu1:1ku-r L 1! . “" is nut :1 xpg. -cia. l angle we may use :1 calculator w ﬁnd KppYO'H'lla19 values ufLIigJ11()ml‘1rI(' (cl. {ﬁll} functions of r (see Example I-1.1. Example 14.1.2 Ln'Isin[—§]w= si11(—3D“} _ -5, m. n225° _: la. sin -1 = .=| n —z "‘ — ' rm mm . w42o° —m-so“ 2 6 6 I 3 mwsinr 1251:: .-0.9490 {cl . -men: 11.95.24 Exercise 14.1 In Pmblrms I .7, xkatcll Mr angle and / ind the mwsunr m dr. yrrr. ! n] (hr anglr whose mdum mensurrs are given. 1 X49171‘ ‘«5Tv)z« 1.. Pmblcvrxs 1 a: find lhr . -um-m m mdums Hf U1: unylex wllusr dz-gwr mrruvmrs an‘ , ,.. .r. . -1. 210°. 210°, 270° .5 -13.5‘. 495°. 215' IS. 13". 72". law“ In rmumu 7 13 find mrh mmmmx 1.-nlur 1/ .4 exists. 7.s11n'~”‘) S rux(5:] ‘J Lani? ! 11. n~. c(Jg 12 1:«'l "; '1 14. 111.197") 15. rm( 7 "fl 15. siu[“T') 17. not 20.: 1» m{“?7'. ~ in Pmb-‘em: 19’21 ﬁnal uu uppmzimutizm in / our drrxmal plarrt for writ mdxmtrd mlur uszng a ralrulatar. I9. sH‘lU 20. siu4 2| 1'ml.71r '32 mu’! 1.‘ 21!. lax-190 21 mLl—~1.T911l . __, ,,. ,1,
15. 15. 338 1h'gmImn¢¢n'4: Ihncléoiu af Raul Numbers [ Chap. It : 14.2 Variations of I)-iganometric Functions We «fer to Figure 14.1.5. As we inrmmv tlw nizv of I from 0 to ‘hr lry muiting the line segment 0)’ muuwmlockwiiw. 14 goes {mm 0 lu 1. then goes bad: to D, dawn tn -1. thvn badi again no 0. making is minplcw cycle. Th'n4 ryrlv is rvpoazzd awry time I is in(‘r(‘. I!fKl by annthrr 21:. Since lint = y. (hm nint varim the Way y duet. Sixiiilarly. cost vanes the way I dues. Thus. mac mitts with value 1, than it gm: M) D. in -1. hark to 0. then In I again 1111 t gnes (rum 0 to 2:1. The varim. inn1u of other Lrigouometnr fumiiuus may be deu-nninvd using Ibo variation uf 1: nr y. Take, lb: example. the run at taut. We start with! _: 0, thus. (ant = W‘: ~ 051 = (1. A15 t uiovea taimrda «f2, 5; gets closer In 1 and .5 in :1 positive IPJLI number that tends In 0. Thus, y/ ':: is positive and mrrws uwranh 4-0:. In the nomnd qumlrzuil, as r gt-‘is 1- nwr to rr, "2. :1 moves in I and z in :1 negative rral numb: -1 whirl: gels rlrx-M m 0. Thu: yfx ll n ncgmirc xval number that mud: in —: o. . -. s L inows tmmrds 1:. y winds to U hut nriuaius POSIUVP and 2' gets rlnuur to —l. lpencc y]: is nrgatiw and moves mmmlx 0. Siinilar explanations amount to: the behnvim at am e in me other two qliadrautx. We su1nm1uim- the variations M the mix uigonomctrir flliwtinxns in ’l':1|>l- H.2.L [Quwwi F’ 11 111 IV ‘ 5-41? I V4921 3;-ozn 0—»1 1—po W —1—+n 110 04-1 11»o 0-»! - 0—«+rx: —rxc—ol) o—o+oo 1 <: i.—>0 I +on_. n 'o—»—cn +m. .o '04-. » l1—+lc: o —oo—y-1 17-v—zx; l—b-+0.1: ]«ac—»1 l—r+aa 7 -31-14-1 | —1)—-. xJ Table 14.2.1. Variations af trigonometric functions. l 5=€- 1&3 ] Gmplu of 1!-igonometric Function: 339 14.3 Graph: of 'IHgono1netric Functions DOMAINS AND IMAGES The functions sin 1: and cos .1‘ 111-0 dcﬁuu! for any n-:11 uumln-r .7 In-nw mu dmnaiu of hut in flulchullts in R. The s-nluaa of bath funrtiinns vary from -1 to 1. hence their iiuagc IS 7-1.1]. on n. .- ullwr lmnd. tan . — _ sin :1" nos 1-. lwnce [an 1715 g + Zn: deﬁned for all real uumhun cxurpl whlrn «us: 0, / — -x that is. wlwn z = :; +2nsr. n e z. "rim, 11.. » / domaiuorunzin . / . . +. ‘ ‘ i1=00St#Ul= {J-’= -l= #t§+2nn. v1EI} = {:I: ::: #g+n¢r. n€l) V (2n+l]1r . __ _/ _{: r.1# 2 . neZ} ~31’ +2mr Tnbla: 14.2.1 «haw: aim the image onuu i1IR. '|"he drnmnins anti images ofolher Irigonomctric functions ran lw .1.-1.. -rnnnui silliilarly. Example 14.3.1 Find 1.lu' domains 0! me funrtimut 1 and ,7 giw-n hy (al I[: ) = 3rin-It (bl g(J= ] = uu1(: r/2). Solution. (A) Not? that ﬂill4Z is deﬁned for All 1. Hence, d] = R (11) u1n(: /2) in daﬁnud if and only ifrnn(1/2) ,4 n in ; + 2.”: —§ + 2n1r no-Imr —: l+4un ‘ "(L z/2ae{ '1' herefore. . "Ll. In z#{ d, =(: r:: #2n+-Imr. nel}. Rumnrk. The domain of ain(z/2) oos(r/2) ylw) ‘ ‘MI! /2) ‘
16. 16. 340 1|-inonometv-ic nmctioru 0] Real Number: [ Chap. u 1 can also he dumrrmim-d by (mating 9(1) as n qumient uf the sine and the cosine fun<-ti-m. <. Sim? ﬂu‘ domain: of the sine and the minim: functions we both R, (hr domain of Lauﬁr/ '2) u d, ;R {x: :cus(. r_. ’2)= U) = R—(r: z - z= r+-Inn. n€ Z}. GRAPHS In slwtrhing the gxapb of any lrignnmnru-it function. it is important Lu ('unsid('r thr rrptrliti - pattern of variation. Fur «ample. sin: varies in vxnrtly lhc same manner in L111‘ ink-rval 30.243] an it does in [-23.0] or I211. 411']. Thus. Ln graph the function y—sin: . —2!r5z-_:2‘. r. wv slu-la-h the curve in [(l,21r] ﬂu-n skrlrh its replica in {- 2w.0]. We us? uh» cahlv of -ullms bx-law. The graph it! nlnntu in Figum 1-1. . . 5 _ I vl°| ’:-‘I Ftgure 1:31. Graph 0' y - sin: Exampbe 14.3.2 Sketch the grnph of y - mum. on thr Inlorrvnl ‘__ Solulion. Since um: = l'In. r[ro»z, um graph has vv-rticzal asyxupmxes whenowr nus: = 0. Thus. the folkvwing linrs an vonical nxyrnpmmar 1: - —¥—. 1! = Z — and x : The vnrinliou uflauz in the inu'r1Ll( ; .;) is mpenll-d m (— ’; . ; ) and m (53751 w. - mz. -h tbc curve on (-5 ; ) using the um. » n! values h. -nu. -. rm graph is givvu in Figure 14.3.2. [ 5°C‘ ‘"31 GNIWH U] Iﬁnonovnclric ﬁmctiovu 341 0 i f o 43]. ?" ‘ N! !! . z_‘ -3 -3 —% —% v ~ —/5 -1 _: § § s/5 ""“‘»1n Fgure 14.3.2. Graph of V ; gun- Since cac: V. 1/sin: nnd |5inz| < 1, gh, >, M” . ”,7"'P'f"' " “=7 = “M -in 0-“mauzn ~11: €1.39‘. .. . .'. .:.7a: ..'. ?.. ’.. l’ :3: n“h'p"°"‘n°°. "f‘h‘ST\$P1|¢>fy-coczfxxnn¢hA&ofy——: dnz. Pg“). “,1; lhoﬂﬂlheﬂlphlofysgnggndgstxczunlhﬁsilnﬂplnne. ¢'--—<—_ n ~--«-«- to =1 Figure 14.3.3. Graphs of 1, = mg, ".4 1, _. uh, ‘
17. 17. 342 Iﬁpmwnutﬁc ﬁmctiona of Rank Number: [ Chap. 14 ] Simﬂuobacrvaxiomrnnln-mnde[urthervciproolfunn:1.ion. sy= rm1And; /== P<'1. nndI0ry= I&nxandy= cut. z PERIOD . - nonmnuut function f is periodic if there exists in puuilivl: ma! munlu-. r k such that [(3 + k) = [(2) V: C 41/ _l-1.31] The smallest. I: Ihnt calhﬁs :14.3.l] in called the of I. The Irigonmnclric function given by y . - xinz is periodic. for siu(: +21) = xinx for every rml number at. All ache: Lrigonometrir ﬁmrtiuna are periodic. 14.8.1 SIIIJPQC f is a function with period 12. Then b; any 1' E 41;. V ﬂzinp) = /(2). n =0,1.2. PIBOI‘. Rnu=0,wehnvennidcutitysoweneetlmgmyvcthetherwvmuuly[orn 3 l. Vt‘ will do it hy mntlncnulical induction. Fhr u-1. we luv! - / (I +7) = NI): /(= = -)7) = I((r-9) +9) = )‘(r + (-p+p) = /(1). Thus, nu») ! (z)< [M313 Suppose um Rn -umc nbitruy nonnqcaziw: inxcger I: . fitikr) = l{r)- ‘IN-3-3] Then / I: t (Ir 4' up) = I((x iv) i hr) = f(: : <2 p). by (14.3.3) = [(1). by 114.32]. I Thoorcnal 2 Tbeﬁxncfn. uy= :inz. Indy= rn-zhnveperiod2:r. V Pnoc. For any ml number; mm + 2»); .inz. wxV. ei p be the period aflhc function 1, = sin 4:. Thu 17 5 2!. But ifp < 2:. an in have uuclly oneoﬂhe following pouibilitieu: %+p= Ir. %+p= !}. ;+p=2r. or%+pliuinl| nyof(hc4q1|Id1'anLs. ln| nyLmn. I Sec. 14.3] Graph: of ID-igonometn'c nmcuom 343 sins; + p) 3‘ 1; am is. sin(; + p) 2* xiu(; ). This. [I = 2". A similar argument will pmw am the perind nfthe function 1, = mix is also 2x. I Corollary 14.3.3 The ﬁlnctlnn y _ mun: has prriod 3:. V __: 7 I Proof. Now that Thus, if Inn: has period 1:, then I: f. it. But if I: < : r. then either Ic = 5, 1: lies in the [int quadrant. or I: lie: in the second qllndfnnt. Ifk - 3, than la. u(0 + k) — tAn(; ) does not wxist. If I: lira in i-izher the first quadrant or the somnd qiwlrnnz, than zuim » 1:) ¢ 12ml). } Thus. k = x. I 1 The corullnry lluu. fulluwx is an immediate couxequexmr of Theorem 14.3.2 nnd Corollary 14.3.3. Corollary 14.34 The period oruzértunl-non; y -_ so” and y - mu is 2»; my in a or we funrrinn y = rm z is If. Pmoc. (Exercise) . Example 14.3.3 Find the ptzriod of / (:3 A siu~I: . ‘ Snlntinu. Using the rm am y _ sin): has period 2.. we haw siudz — siu(-11 + 2”). Thus, f[: ) - sinéz _ sin (4: + 21:) = siu4(r + —_ , '(z» Thrrrﬁm-_ the period of I is Argumem similar In an: umd 1.. the preceding ! xu11pll* will wmI.1i. i. the folltvwmg . ~.m.11.. .~, ». Camila} 14.3.5 Let B and C be mu munb . r.{ run The functions y = sin(Bz 4 C]. y ~ ros(b‘z « CL :1 ~ PK-(‘(31 ~ (‘I and V = :‘sr(Rz + (7) haw pwiou Zr/ B: rm Fha fnnctiom: y = tn. n(Bu: + C‘) and y : roI. (I31 A ('1 hair pi-mu u; 'H.
18. 18. 3-14 1H9onome¢n'c Function: of Raul Number: I Chap. 14 ', When a function f has a period p, f(z] varies in surh ll way that a complete pzalcru ix. allalual in an inn-rvnl [a. r: +11}. This pattern is n‘pL‘zI1.ed throughout ch» surtvsslvt‘ mu»rv: d,< ja + p. a + 2p]. [0 + 2p, ll 0 3p}, cu». Thus. to sketch the graph of we function wlimr pvriud is lmmrn. wv sketch 3 colllplclv pzllwnl in an interval and duplicate this pattern an uc'dEd in utlurr intervals. Exavnple 14.3.4 skmli mo graphs y — sin-Lr. u g r 5 2w. (‘ompm this function wuh V = 9m: by smelling al-an tho ynph or y = .13 in um same plmlv Solution. The period of of llll‘ fuulcliuu 1/ ; sink is 1,; = 1} -— g {see Examplv H 3.31 "c plot the palms and skl-tr): the graph on [0, Ilzdug llw fulluwillg llllvlv of V‘. ‘ll1’ﬁ fur S! = EiD~1Z- The values of 2: an- mmpuwd Fmm values assigned to 4: which rangv nvor om- pvriod of the sin! functioll from 0 to 2m 41‘ o ; r O ; y — mu 4; 0 1 We rvpelll me pawem in we iuu~rvu. l:s 1; . ii]. [IL 5, Figum 14.3.4). in , 7 ,5 Figurn 14.3.4. Graphs of y ; wind: and y = SIIIJ Isec. 14.3 1 Gvupha of 1»-iponomelrac nmcciom us AMPLITUDE For any real number n, (hr value of sinu and cusu vary from -1 to 1 Thus. given real numbers A, B. and C, and for any real llumbvr 1'. the snlilc of A l. lu(Bz « (‘V ': I.ry from 4.4,‘ ta l.4|. The graph or the function y : Asia (51 - C: is a W: Vv: ‘ (hm gm-N up to tho unujlnulll IA! and down to the minimum of —. ’.-1| at rv-gular ll'I('l'V'v| l7~. 1 in Mull! ’ olxiéwvntinn holds for tllv graph of y ’ :1 (‘OS [B1 ' (W. Ve (‘all L4‘ Lhv uinpliludc of Hm fluu1im. l y : A sin (B! + Cl and y = A ms [[31 A (""1 '11.» amplihlde L41 uf the functions 3/ — .4:<ln(H: + F: and y . lm. »;u. » - (aw i. .- the lnrgcst. displzlccnwnz of the yaphn of then» flun-Lions [mm the : —:u<i “'0 . ~;-ly that the laugoilx. cotangent, BPCn. |'ll and ousmiu: fuucliuns have no my-pliziulo gnuls or iii. - graphs of thaw functions lid. ‘ beyond any dismnrv al-mvv nr bvluw Lllv : —.xxL'< ' 14.3.5 The fundion y _ (‘UH 1 Lugs : u.upliLud(‘ 1 and y ; Licnexx has amplitudc 3 Fr sham A rompaxilson of lllcir graphs Y 5 ii/ =3cus: y/ =co«: r 0 " 4!. . _ / Jrj: / Figun 10.3.5. Graph or y ; Lzcos: and , , 7 rns: PHASE SHIFT Given a function 1 and II will uuullwx Ii. let 5] be a fullcunn given by y(. r\$ = /(I ~ h} for all z E d, . Fmm the doliuiliou ul‘ y. we obtain f(: l 1 f((. r+h§—hl =57l1'+h!
19. 19. 3-I0 Tﬁgonmnetric Hlncliom of Real Number: [ Chap. 1‘ ] vi-lliczh wt‘ may intcxprcz grnphiclllly . u shmrm in Figure 14.3.5. Y — l/ */I-Fl . . . . . .,, =_qm —1(. :—h) Figure 14.3.6. Graphs of y - fur) and y - f(. r — fl). is > U Sim .2 is an uxbitraxy elﬁmrm 4,. we may . .l. uuu the graph 0; y = [(1 up hy Ihiﬂing nu graph ufy — , r(. -) m ll (lislikunf h. Tl. » . ,u-mu involws Lrztrlxlaliuu adullg «ii. » I ll. .. hy A uniln. The slim is toward the rlglll ill: is p¢»: iti~l- and umuil this left ifh is negative. x'. « ml] Pl the plume lhift M . , —_- [lie — in). Tim. the fllnnionx y = sin A[r —M. 5. _ ms Au — ii), _. , = Lau Au — M. l‘KL'. . have phrllw slim h. Example 14.3.5 skmdi the graph of y a tan [1 — 51. Solution. wo l'I‘fl"l’ to Figure 14.3.2. Tlw ruucmn y _ mun: A g) has plmw Shift ; ,m the graph emu: ruucliuu can he uma; i.ul by shifting u. the right by . . auum of 3 nu graph ofy = tlulr. The grnph of y : tan (3 — 1) is given in Figure 14.3.. -. Now that W9 my nlsu may »l1.«. .rlm. ;l. nu gflph using nu» folkywillg uni. » n! values fur y = Lal'| (1T _ ; » wlwr» 4» - ; vaxk-s (WHY mlr pa-rind of nu l. iu| gl’ul rumiuu. Since lhv period umie tangrnt fllurliun u vr. wr dlousc In have ; < I —; «. ;. -l-1.» vnluni of . » are miupumi from the values of I — §. o ¢ . u - up: six 2' [ Sac. 14.3] Gvuyha of TY-igauornelric Ihncﬁavu Figure 11.3.7. Graph of y = tau(z — fl. Example 14.3.6 Detrnnine the period. the amplitude and the phnw shjﬁ of y _- 4siu(gz 4 3), and use mu inlormmon m smell one oompldc cycle ofthc graph oflhr equation. Solutinn. The equation my be rewrimsu u y = mu [; (z — (—‘; ))]. Thmi. um | 'um'. tion has lunplitude 4. period 21r/ g s 4:, and phase shift — The graph of the nivﬁn equation may be obtained by mining to the Hi In a dinnnoe cl; the graph uf y = «Iu'ln(; z] on in, 41:] (see Figure 1«l. .'l.8).
20. 20. sm Iﬁgofwrncﬁric nuncuom of Real Numberl j Chap. 11 j 5 y-4.«au(§: ) Y Y F , y=4sin(gx+ 3) T ; .—. ,' »«1x Figun 14.3.8. Graph 01 y = 4.-1in(§I + z). Example 14.3.7 Sina‘ ms: - sin[: r + g). um nMaix| the graph of the Pquamm y : 41151’. by shifting to Ihv left by A dxsunre of Ihv gtrapll of y : sm fr ADDITION OF V—COORD| NAYES The graph)! uf sums of trignnomo-trir {um-Lium umy be ukmrhed by a rm-Lhud known M addition of y-—caanIx'nat¢: . The addmon rsferrvd to in dnnr Ivy gmmcmr means mﬂu-r than by m-tual c-ompuumou, We mll ulhmnm- this method in the m-X1. Pxmnplr. Example 14.3.3 Sh-trh lbv graph u{ y = ' I11 + sin 21, a Solution. Now that y — ’- sin: has nmplimdr and petiod '2n. and y - sin 2: has ampliludv 1 and period 1r, ‘9 skvtrh the graph of both 4-qualions nu thv saxur plmw. To pm the point. ‘ of y _ ; siuz +sin2:r. we chomp A nuﬂicicnt number of walnut nfx. For rad: valuauf 2:. we combine the the y—comdinM. no urmo yaphs nfy = gain: and y : .vin2t w u| .vI. n.iu mat of y = %5in1' + n1'n2z (we l"igum 14.3.9). Graph: of Iﬁgonametric hmctiovu Figure 14.3.9. Graph of , , = gm, . , ,.; ,, 2,‘ km-cise 14.8 In Problem: I 5, ﬁnd tilt domain of the g-ixrgn funding. ‘ l. I(1l=4sin2:: 2, / (,), _2 . gm” 3‘ , (I)_2w| (.uA1’ ‘< /1‘? — = °¢(*-4:) 5~/ (1)-l+aoc(dx 1r) 6. f(a: ) = 3¢, c(, r,, , ’ 7. Show that y = /1005 (B: + (7) has period 8. Show that y = saute: 4-C) luau period 5-. In Prvblzm-I 9 I7‘ ﬁnd the period. the amplitude (when applimbk) and my ylmsr xhsﬂ. Sketch a complete period of ﬂu graph. 9~ II “N21 l04v= :-as-:1: 11 y=3rmI'21' 12»u=3sin(2z-rr) ]3.y= cuu(z+" 14.y—| ,;n(3_-, +1} 15. y_~2co: (2z—; ) 16- y= %sec(3z'>r) :7, y= :m. :(-u+"2x; In Problem: 15 20, akzlrll the ymplu wing (hr mzthod of addslium 0/ y m. mm. .m, l8.y—'4sinz—muz l9.y= =z+2siut 20.y=2+L. m3;
21. 21. 3543 Iﬁganornctvic nmcaom of mu Number: 1 Chap. 14 1 14.4 Addmon Fhmzulas Let ,4. B. C, and n be point»: an llu: unit cam. .. shown in mum 14.4.1. Not: that 1,151 =1cn1, Iwuu-, 1.4312 = 1cp1=. Bul 1.431’ = (cow — ms)’ + [saw — me)’ = 'Z—2(nm1arm9+ninwxin9) And ICDI’ = [m9(10 ‘ 9)~1}’ + >dn’(w - 9) = 2 — 2ons[: p — 9). Thus. 2 - 2(noawcu1s0 + uinwsinﬂ) = Z — 2cos(g: — 9). bl) than 1'. m(p -9) = msgacnu9+aingasin9. 34.4.1] X nun) ’ F’-gun u.4.l. 1.131 = |(. ‘D| Now, we mnuidut the on man cos[Az') : ms: and Iin(—r) — ~sm: r for em-1. mu number 2. With [14.»1.1], we nhuiin: wI(¢ + 9) = = mIi<P - (-3)] = cos<; :eo1I(—9) + sinwain( 0) = I>1uwnn1i9~uinwaiu0 11.1.4.2] / Sec. 13.; ] Addition Fbnmdaa 351 Slim: tin: - oos(§ —z). and max = siu(§ -2) £0: allz, wcqaninobtain [mm [14.4.1;: 3'i“((P"9) =0¢l'[§ ~ (W-9)1=°0H{(§ ~ *9) +9] = caa(§ — w)c¢n9 - siu(; — 1p]ain8 = - sinwcmﬂ — ooogasin0 [14,43] Thus. Idn(1p + 9) = u1'n[(p — (—a)] = uin: pcus( 0) cunq>xin(—8) = xi| wu: un0+ muypuinﬂ {X-1.4.4] We combine [i4.4.1) and {x-1.4.2] to term the following equuaom which we call up aunim fvnnulaa for coainu: cos(w: tﬂ) = oosgamnI0\$sinq2sin9 We do the same for (14.4.3) and (14.4.4) to obtain the addition [ow-nwlaa fvr nincn sin(y: t9)= ainwcas0immpuin€ Using the addition fonnulas for both the sine and the cosine. we get sin(¢i:0) _ sinwooootcongasinﬂ cuu(wi9) " cuupaJu9=Fuiuw2tin0 = [uingsoosaﬁwnwxinﬂ I/ (oocwcocﬂ) = tanzpttanﬂ 00lwooa9\$sin4p. -I'In9 I/ (mapcms) 1:: ¢m«puu0' ti-lI(<Pt9) = This givnc the addition formulas for tangents: tamps-_h. n9 '-*"‘1°*°>= i Exampie 15.4.! Use an addition fm-rnuln to find con 75°. Snlunbn. cns'I‘. 's' ; cus(45° 1 30’) - oua45'wu30' — sin45'sin2ﬂ)' «*2 1/5 . /i 1 _ /6-‘/2 2 2 2 2 4
22. 22. 352 1HgononIeh'ic ihndiona of Real Number: [ Chap. 14] Example 14.4.2 Use an addition furmula Mr ﬁnd um(f, ). Solution. , _ , ._, _¢In“>tI. n-'5_/5-1‘ m(ﬁ)—tm(§-. )- —1+ﬁ‘1 Exercise 14.4 In Pmblnna I-I2, find and: mu: by naming an addition formula. 1. s1nﬁ—; 2. mg :1.1u. §—; 4. wt 5,3,1 5. sec ﬁ 6. etc ‘-3-’ 7. ms L}; s. are 345° 9. sin 105‘ 10. nu-195' 11. 1.11115“ 12. cosl95' In Problnnx 13-15. simplity each expmxion using ui additimn somuia. 1:1. mu; + 17) 14. cot(9+2x) 15. 1:11.11 — 3) 10. . ..-. (o _ g) 17. cuu(0+ 5) 1s. csc(0 + 5) 14.5 Other Formula: The addition fonnulu yield sin29 = sin(0 4 0) = Iin0u. Is0+0u3!i. n0 = ‘lsinﬂmsﬂ ooi29= cns(9+9) - mamas — uinﬂsinﬂ — was -i1n‘9. [mm which we obtain the doubt: -angle ﬁn-mular. IiIiZ9 -2sin9oou0 oouzo - oos’9—ain'0 =1—'zsin’9 T2cu’0—l I’-inn the double-mxle ﬁoqniuh for casings we obtain mp0: 1 -2.121.719) ma 1:050: 2ooa’1;9)—1. [5¢9- 1‘-5] Other nvrmnlao Thcle nlulll yiald lb: luau-augk/ op-m. |¢¢ l—amB Z I-I-411139 7'? ain§o= t august Theﬁotmulu -in(1p + 0) I Iimpaoo0+ couprina 3i'| (IP—5) = IiIIwoos9- congusinﬁ may be addcd toga: -i(w+9) +I'in(1p—9) = 2tin¢: cou9. Hence. -iww-9- %I'in(v+0) +Iin(w-0)]- Similarly. -eaddand-ubmctcheibxmiuu °°'(i0+0) = mogpoou0—: iugo-inﬂ 0000-9) = G-‘lwoou9+Iingpn’n0 nudge: m5(V+9)+0U! (1p-0)=2cuI1pcou0 w-(1a+0)—oo-(«: —0) - —2lin4pIi. n0. Thu. we obtain the bﬂawin‘ human wwco-0= Moo-(v+0)+ooa(w—0)l -inw-i-h 910°-(v—0)-on-(w+9)1 11.. tmnulu [14,51]. [M51]. and [14.5.: :] In culled the pvodu. -1 fol-Inuku. Enmplu 14.5.1 Find sin(-15') min; ; m| t.. ng¢ mrmuh Snllniwl. Siam -15‘ in in the third quldnnt. then nin(—l5') is ngtive. 1'hug, 153 114.51] mm} 114.544]
23. 23. 35., c Ihnctiouu of am: Number: I Chain 14] sample 14.5.2 Find sin 75' 011115‘. Solution. U-in; I woduﬁ formnh. '9 '18" 1 . . - . . Iin 75'unul5' = §[sin(75‘ I 15‘) + uin(75' —15')] = 5 ['“‘(’° V “"0” ll 1 ‘/3’ 1 2 + J5] 2 + ~/ ti = 5 [1 + T] - [ 2 , ,__. 4 Exercbe 14.5 1.Findud1ofthafoﬂv-ingbymirkgllllll-IIIGl€5°Tm“l* ‘_ w.15' I). half)’ .1. sin; 0» = m’z‘ 2. Una pmduct Scn'InIIlAn¢oonmputetlIefoll«1I‘ID£‘ll'|5~ ' . _ w.1,', 'gin15' b. con75'oou45' C. IilI('%l""‘ #3" 3. Daiveaformuhﬁurusachufth following. _ , _ may b. -mo en s}n:9 d. Ian30 5- “'3 L ""3 4. Shoi that L (, .m, _m, )2=1_; ,i, ,g. ,., .o b. ne. c’0;l+an'9 ¢. .,¢’0-1+eoL’9 d- “"g= “i"9/(1"'“"9) e. = (1 —ooI9)/5in0 14.0 '11-igonometrlc Equation: -, o,, ,,m. ¢.-u_~' . ¢u¢¢um' mﬁern in an equation that lnvulus one or morv trigono- mgu-ic To solve an aquuion involving trigonomeu-ic functions, ll ls_neoeuII-'11 I0 , I In “C, H“, mm mama m: For Iuppuuc that n pnrindu: funacuon I 1.. . aperInd' p. I{. :., inanolutimulanequlion [(1) = °v “*9 fl“) _“' “' By 14_; _1, “,0 ¢, .,) - [(19) an. n = _ o, 1, 1, 3,. .. Thus. {(211 t 7-17) '— & Th“ "19"" ‘ 2:9 1 up is A nolution ofthe eqllllnm N! ) = 0- Exampu 14.6.1 Solve Lbeequuion sin: = 1;. [ Sec. 146 ] ﬁigonometv-in Equation: \$ socuuon. since sin: ix positive, the solutions lie truly in the link and in the second quad: -:. nLs. In theﬁntqundnun wenaveuang = :4, audinthc second qundrlnt we have sin’; — 'I'IJus’§' and 3; are Iolutiom in the am and second q — rams, Since -an: 1.. . period 2». _{ §+2mr — . nel ’, "+2mv nrr the solutions of the Examph 14.6.2 Solve the equatinn unz = -1. Sohniun. Sim: can: is In. -gnlivv. the solu- lions lie only in the 1-uamd and in Iln: fuunh quadrants. We identify i‘; tau be u solution in the neamrl quadnnt Ind -% to be A solution in the fourth quadrant. Slum tum: has period x. zhenz= ¥+: ur. uezucan1uziomm‘ the equ. -nun. Note mu. —} +1 = -31, um. —§ + rm amtrilmles no addition-II solution. Tlumlare. the solutions of the equaliun an z=3Tﬂ+nn, n€l. Example 14.6.3 Solve the equation cuu(§x) = -1. Soiltion 1. WannuIh. a4cnu(§z)-—l when §z_n. orz =21r. T| ms, h’iIlIc-lulion of the nqnation. Sinu: r. uu(§z). has period 5'; - 41, then . t=21r+4n1r, nEl are the mlutions nl the equation.
24. 24. 336 Iﬁgmnmnetvic Fhnctimu 0] Real Number: [ Chap. 11 I Solution 2. coa(; z]= —1<: .: =vr4-‘zmr. nez as 1:21!-+-Imt. n('l. . ;._ Example 14.6.4 Sulwt the x'qu.1l. iou 2sin' z + ﬁsinr — 3 = 0. Soluﬂon. Nun: that the equation is in quadralit‘. fonn. We (umr the lvit-Inzunl xidr and solve fnr sin : . (sin: + 3](2ninr -1) = o The lirsl. [actor gives the eqnuution sin: a» 3 = U. or ninz . . 3. No value of: satisﬁes this equation fur uinz waxing only [mu -1 lu 1. Tina. the tint Tutor contribuirs no solution. The second Lvmr gives 25in 2: — 1 = 0 sin: _ g Sinrt: win: in ]. Kn¢iO. i'l: , thou I linzzoeilhcr in the tin! or in the second quadrant. Sims : .1'n. r 2 when : e g or r = '67". bod: ofzhese values are solutions. Thus. tIu- wlnlimnn um. » vqualion u. gm by { ; mm. HE z [First Quadrant] z = ﬁg +2mr, n e Z. [Socoud Quadrant] Exercise 14.6 \$010! rarh 47] file / ollouring Iquatitnu. / I. sin: -I 2. Ian: ’ '_/3 /2.-on-2z= _ﬁ .1. ninI| z_ 1/2 rm-"2. 2col. .t; .-0 6. zoos’: -¢cou= s—san*: 7.-| ma’x—3=0 s. .x’x—2m. z—o 9.: au’z= sec. :—1 xu. .mu2mn’2; 2-0 14. 7 Identitiea we rvnsll am an idr ray is an cquiui/ an whose solution set L: Pqmd to in; rr1:lan'nn'vI| . set. The equations tun: = 1 and mu _ x-in1/ ms: haw the xulw mplwcmout sci. R—(r: ¢.‘osa: =D}= {;r: I#%+7x! r.7:61,. I 3% 14-7) ldcntitia 35-, T"“’99l-lnanleutsctiauihemlntionn. -tftau = ~. - . . identity. Bu: tlm solution sec (1: an = 50 » nnzu 'f'I": ,T', ',, :'": ‘,'. '.’, ',', ,', ",'(, '} ﬁr revlamum not, Thus, an: = 1 a. not an identity. "‘ ‘R: pmvc An identity we usually do eilha of the bllalving: I. Show um ﬂu: expression on om: side or we equll-ion (3--vfmbly, the our whidn a. ‘Wmm‘m: ?Ilnnmcuthccxpreuianantbemhnuidefurwuyvaluwxn 2. Show that the . um‘ ‘ ‘. .]A ~ . idmmy‘ on! “'3 em to some equannn that in known to be an En"-Me 14.7.1 Pmvu the identity =1—| inz. [M111 Proof 1. Note that the atpauninn on the leh-land side ' . m . . = _ , _ _ Izun¢; leﬁus>dwhcnl+2unz=0. mf‘, _:_'. n;'; °"u§"; fm eq‘_; ‘3;"'; 'i‘b'I’P'n° If and only -rz = -5 + Zmr. n e 2. thus, the U‘R""‘i""""U = {"5R11?‘1§"\$2n1r. utl). Usinxtheidm. |ciLysin’z+: na3z;1_w¢h“E a'2z=1_sinz: mum 00-’: _ t_(1—sinz)(l+Iin1:) _ l+ainz rum ‘ “"'“”- Pmolz. WithxEU. I+‘ o. I. ~ - . w°bm“h”qumhmq= “:; :_ I Ill, 'e0anmnlnplybo¢hs| duol'| l4.7.1]by 1+. ..” 1 — linz)(1 + sin 2) A l — sin’: = out’: This sbmvu that [1411] is equjvuc-nz u; .1. olwim. W_. ,,m_y_ Ex 5.. - - _'-WI’ mph! 12 Pmvcthc-Idalmy 2oo. 'z_1_hﬁ_ nod. Thenumionunzudasnedomy-hen: ¢1=,1Jz, nél. Forulymxchx. l"“”"" "55", !/'06‘: ma’z—sin': n z/ eou3:r_ n: )|’z+ain7; -cu‘z—sin7x—ouu7z—(l—cus‘1:) —.2cou’x—1.
25. 25. .158 Iﬁgonomehic Ilmctiona a] Real Number: . _ 1 . Enmpb 14.1.3 pm 11.. . udenmy —: _—__”§% - l f": :". Solution. Note that: ungandsccz andeﬁned 4» nuuz#0 :9 zyéﬂ, -"-‘1r. n€Z unzyén es sinz;6O ¢-9 zgﬁmnnel | —c1.1nzgHJ 0» 1:00:94! as 2:#2mr. 1162 '1'1uu.1.be1-eplmexnentn-eti. IU-{z: z#m1. z#“5‘-‘I. 7|5l)-F°'| "Y1€U- I+wc. : (1031 111115: 1-eon (1 +vec1)(1 —oosz)=1u:1.an’: : 1-ooIz+u: c:1:—secxcunz-coaznn'z 1 l~——l omz= oous(ﬁnzz -m"+u111z ens: cos’: 1 sin‘: —oo8.'e+ — - — 2 (nu: -cos 1+1 5 1in‘_x was we: n11"z= sin7z Exercise 14.7 In Pmbleml I-U. prove each :1] file given identity. cot: 1. ainxo1)tzI: cz=1 2- ; z=t= c= sen: 3., E,, ‘;f“_’. 4. tanz= —— an, ace: ninz coo: '°C1+°°“’ "".3.? .*§= ‘ ’-es’: --we-*2 71—cooz= tun’: 8. C3¢3‘_*"____, :_‘ ’ eon: 1+-Inc: 0'31 H70‘? InPmNem11 u—12. llIa111:l. AaVteaclI¢1IlM! """'°‘l"‘“““""-'”“ an-’dm1-'19» 94.. ... -‘/1-.11.‘: 111. sin‘I1=2si11z 11. si11(: + . ,)= s1.. z-may 12- In-nz1=1-um [Chap-Ill ~=1-nczoot: [ Soc. 14.: ] 14.8 Invaru ﬁigonometric Functions tvwene Iﬁgonometric Funcﬁovu 359 Sin: an-ry trigonomcuic function I in periodic, it is not onmtoone. Thus. /" k rlnk a function. However. we may choose A g 4, so am the nxxriclion 1|. of J is nnumronu. -. hence fl‘ " in I function (lee Sectim 10.5). b¢v. y=Sinzbe¢herest1'ictionofy—1duzto[—g-, §]. Thus. Sin 2 -sin: V: E [—-§. 1}]. The graph ofy I Sin: is shown in Figunz 14.8.1. Note zlul y = Sins ilaonnmoone function f1'nm[—; -,ﬂ In [-1.1]. T11cinIuuz= Sinvianfum,1.ion. Wcdenotcby y= uuinzthzinvu-: efunn:1.iouofv= Sinz. Thun. y- grain: to 1.-= Siny. yE[—-3,; -]. The domain of y = nrrninz 1. [-1.1] and 11.: image is [-1.1]- (a) y - Sin 1‘ (I7) I = 3111]] Fhum 14.3.1. Graph of y a Sin 1 and the invus: relation z = Sin y. V_11'e can similarly nstrict ﬂu: other 11-igumxnnctric [un<*l. ion.1 and dcliur I. |1cir curm- spondlng invmue lunctiona accordingly. We represent them by Cos .2. 11111 1. Cat 2. Set 1 and Car 2. and denote tbs immr fnnctians by arrtcnsz, nrcunz. unxzvt. -1:, nruuucz and amcscx, nspectivcly. "I‘n. b!a: 14.8.1 givu the dcﬁnitiom of the six invvnc triguuumavtrit functions. together with their domains and images.
26. 26. 360 ﬂiponmnelric Ehnztiom 0] Hal Number: I Chap. 11 j | ;l| ctioI‘f Deﬁnition Dun. -min Image | y= main: z= Siny [-1.1] {—§, §| | y- amoosz z= Cosy [-1.1] V [Om] y= an: nnz z= 'nmyV R y= arenas: zrcol y R _(n, ) yrs u-caucz : =Secy (—ou. —l]Uil. +oc) [—1r. —§)U[0.; ) ly= Inner V: I.'—'Cst: y (-00.-l]U[| ,+oa) (—ar, —§]LJ(0.; ] Taﬂl 14.9.1 Imus: mgonomeuic functions. “'1-ran sketch the graph of lrigrmulurtrht function: in the usual way: that In, by plouuxg of points. Huwuw. -r, we may instead use our funiliarity with the graphs of trigonomotrir funrtimns and comic! !! the (am. that nu: graphs nf invent: fInr(ionx4 are mlrmr images of each other with raped to the [inn y = .r. Exampb 14.8.1 Sh-uh the graph of y = arcainz. Solution 1. W9 nkztch the ynph uf y = ‘in 1. when: —; g x 5 5. The mu. -«am. with respect to the line y = 4: in the graph 0! y = auxin: (as: Figure H.812 (n)]. Solution 2. The graph or y - Armin: my h- uhtaiued by pinning the graph of un- aquivnlenz equntinn z = Sin y, — 3 _< y 5 ; . Thr fuﬂmling table will help lame important points in the graph. The 9-mph in ahmrn in Figure H.3.2 (I1). v -&—: —;lso: l%s; ‘ “sin: -1 -4 —= ?|—; 0 H4 91, (sec. 14.1] 361 (a) obuaned by reﬂecting y - Sin 1. (5) 0M= 'M1 by plotting : = Sin y, Flynn 103.2. Graph of y = uuinz. ExnnIplul4.l.2 Sheu: htheyaphofy- uchn. -L sulunhin. wukadnheynphofy= '1hn: ,where -5525; nndlahiuxdeclion Ivitluupecttnth: limy= z. 'l‘hcrdectionuuharninFigure!4.8.. '5isIhegrnpho[ y= uuAnz. (WemqAhoIhu: hthep-| pho(y= ucun; b, p|guin‘u. .5.-. P|, .-, {;; , equivulenteqmtimz-'Ihny, —}5yg; -.) Y Fqut 143.3. GEM! of y = Irctuz.
27. 27. 362 Tﬁgonomclv-ic Functiovu of Raul Number: Exampie 14.8.3 Find (a) An1sn(—l), (h) m-ms(ma §). 4:) We willdloose y E {—-g. ;] such um Tm y : —1.Clea. I'1y. y = —% thus ar<*tan(-1) = —§. (h) amm«(wa§] = g s1mecoa§= ca ‘, '. and Cut: and uwmx m. inwme fInu‘— Iioms. (cl uvoo6[°0e(-%)] = m’mI(§) ‘ 5'- Thc prlecvding znmple shown that although Co: I and atoms: in iuvvrpc funmiunx, um-cog[cos(—§)] 74 —§. This in because —§ in nm. in the restricted domain of y = Cos I which is 10.21:]. Eunlple 14.8.4 (a) a. rcsiu4 is undefined airman: 4 in not in me domain [-1. 1] of arcsln 1. (M Using a calculawr. we get ar(: uI. u 10 w 1.471]. It in lint pmnihlr to ﬁnd the nut vllue ohzcuui 10 since there in no special ange W’ with "DH: 11 = N1 Exampie 14.8.5 Solve the equation axuin[2z — 1) V. ;. sohniou. The equation is equivaicnt no _ 1 J5 2r—l= S|n(§)—T. /5 2+s/3 2:_1+7_ 2 2+f3 Z= 4 . The nolution an is (3-94). Example 14.8.6 Find the invent function I" of [(1') = 20013:. 0 5 :1: S % , ¢hcn 0S3zS1,t. hul, um3z= C<n3x. WBIEC Solution. Note that since 0 3 z s an [ 5°€- 1&5] Invert: 1Hyonomz¢n'c ﬂmctiovu 36.’! II = 200331 and solve for 2. '21: until! = cm ll: Intros = arc; -naf(‘A-nw 3:} -3: . z= §nrrros(%) The inwrrserclation is y_- §arc: :us(§). Hz-um, 1-u, -) ; §m. .,. ;(§), Example 14.8.7 Shaw um. uin(u-ccmz) - /1 — :2, -1 3' 2 g 1. Solution]. L9! ¢- nrroosz. Sinoclgzgluht-1105 art-t: us25x;1.haL: .s. 0 S t 5 sr. Thus 1 = cm: = cost. mu: um nu’: = l—ca:7t —1—ﬂ.11.~m-» sin! = 1»/1~: ’. But since 0 5 t 5 1r. then sin! is nomu-gazivv, Thma, Sill! _ 1— 1’, Cousrqnemly, sin(axtms. r] = Solution 2. Cumlider the triangle xshnwn in Figure 14.8.4. Now that ain¢= uint(" = i/7-7. Aha ' eoue= con"”-z. Sincetisbe2weenDu. ndx. wat= co. g:, he. w.; g= arccunz. Tha'¢£me. ain(atcoouz)= / | -1-‘. Exercise 14.8 In rmum. 1 12, find the mama mam, L arcain(-1) 2. axmmsé 4- am= nr(—/ E) 5. unset‘) 7. a: ccus(Sin §) 3. armin(Cos 3)
28. 28. an 1HgaIwmutrI'c Function: olﬂeal Ncmbzro {Cr-op. 14] no. c-n<-mm) 11.-ens: -1-an-(=31 12 -«us-[sen (-31 In Pmblemn 1.! -I5, men aulculatorta/ indou apprnivnctian énlmxrdaimalplacu/ urtllc 13. u-ooo¢(-g) 14. u-min; 15. u'ctan7 In Pmblaru 16-21, Jlradl the graph ofeach / unction given. 16.11-nnxecz l7.y= uoootz 19- ll=1+‘-“*3“2‘ 19. 1,: amount 20. v = an-inb-int) 21- v= Irv-=0-(00-2) In Problem: 2247. wlvlelluzyivunaquatiavns/ vrz. 22. 3uuec(3z+2)= n 23.araxr(4-3¢)=1f- 24. ﬁarcung-=1! 25.2uuin(u-1)= —g 25. ucm¢Z2=’; 27- IK°°¢: ="T In Pmblenu za-:1, [ind an apmuion [vr I"(z), 23. / (z)= oin§: . -yszsli 29‘ ! (=l= °°'1=»°S¢S§ so. /(z)-nu-. z,o<z<; or«<z<i;1 31‘ ! (z)=2mz~-§<= <§ 32. Show that uin(u-cunt) = :13. sm. .u. .n. ..(. m.z). =£f, :'? —. —1szs1.z¢o. 14.9 Complex Numbers Inptevimndupu: n.vIucxptundugcnuII>l¢l nnmbazintheﬁn-mz-x+iywhaez. vER Anywmplexnunrbernouprunediauidmbein mctnng-ular-[m-m. G|-q7himI]y, xinm1naanledby Adimctaodlinelmnenlfxomtheorigintoihepoint (. '., y)| .l5hmvnin Fignn 14.9.1. Wuwillnowexprazintcrmlofrmda. Note cm 7.14 = ./ :1 + y’. 2 = re-0. _ 1! ysrﬂl9. Ind nm0.. l. H’-‘"33 Thus, x—rcm9+| '(r: in0) -r(coI0+iIin0) IN-94] [Sec 14.41] Cmuplea Nurnbtra 355 Aoomplu numbetzint]| gt'aI'mgiven| :y[l4.9.l] ncuniduobehnpolarfonu Formu- venience. we an A shone: symbol cisﬂ lo denote oos0+x'ninﬂ. Using this uymhol. [1-1.9.1] may be in-men u z - rcisﬂ. Examp| al0.9.1 WH1zz=3—3iinpn| arform. - Solution. Wewi1lﬁndrand0mehatIvec: nwn'ta: =rcia8. Int}-iscasc. z; ma 1; = 3. The mum. of: in = / §1+(—3)5= 3»/ i. Noteum the point (3.- is in the fourth qundrnnt and Iinm my - y/ z — -3/3 _ -1, mm 9 = ’T"" (or 315-) ThIll. IIIecanvn'ilezinpoluIurmnnz=3/ icix7;l. VM-mayn. l:ochooIr0=—3Inthat z =3/ §d: (—}). In; encmI.9=-§+2u1r, ne z. ’l11ux, z , » 3/ iris —§ 4 2n1r). n€Z. Example 14.9.2 Write : = 4ciq15u° in rccta. ugul-r farm. Sohlim. I — 463150’ 2! {(013 150' + x'xin15U') OVERATIONS ON COMPLEX NUMBERS IN POLAR FORM Given 2. = nchsa, md 12 = qciasﬂa, addition 3 pcdorrmsd by exprtxing both mm plgx numbe. -nt in rezclangulu {arm and adding their real conupuncnu and their imaginary uan. I'ponenta. Subtnclfnm is deﬁned similarly. 'l‘I. nn, 2; +13 - (ncola, 6iT, lin9;)+(T: ﬁ>I01OI! '>; lin9;) = (r, 013:9; + rgwsﬂg) + i(n shad; + rgninﬂg) :1 — 1/1 = (r; 0038. - 1'gctﬁ91)+i(f‘1!i1I0; — r, uin9,) Mllltiplimtipn and division are conveniently done in polar farm. Using the zuldiliun formula: fm nine and social: we derive lhc follvwing formnln for the product of z. and 21. l1'lq = [r; (ms9; + inin0,)] - [r; (ma02 + isiu0g)] - r. r,[(ma, mm, — sinﬂ. mm) + . '(m. o,. ane, + sane. sh-61)] I Tx"2[°°‘(9I +92)+535l| (91 4‘ 02"‘ = 7'| ":d! (9x +92)
29. 29. 366 Iﬁgvnanccvic Ihnctioru of Raul Nsmkn [ Chap- 14 X lfzz; i0,vn: d:tivetbequotien¢ofx, Andhuhlbn: :. r; (eoI0g +(uinD, ) = r, (ooo0; + in‘n0,) _ 0:209; —Vilin92 9, = r; (urI0z+I'tin9z) rz(uus02+I'Iin92) 00632—ilin92 ] 3 r. (m. a.m. a,_+ nnmmog) - -‘(: in_o, I119: — m-9xsi_-_x§3)_ r; (onII’9; +: ini0;] -'_I__: ___l1 “""*"’*’+"""“"“"’ = :_: [.. .(o. —e1>+. ~.a-ta, vol rg-1 = 5‘-a'. (9.—Oz) 7‘: Eumpie M33 (1) (8cil40')(2ciI30’) xx s(z)ciI(40' +30‘) = 166-70‘ (5) (3.349-)/ (253.30-) = gcis(40' — 30') = 445.310’ DE MOIVRES THEOREM Theorem 14.9.: (D: Moivrn'n i-"nan. -am) cilia -ncr mmwh; numb! ’-' rm? ’ "4 PW . ‘. . 'm"mwn' (r4:i|9)“-r'cilnl. Pvuol, Wawillp: wethethuxanbymnhanuimlindIwt. iuu. I1:1 z -rciao. Whazns 1. , I-, =ns. a=r'ci. (1-0). 'I'hus, thetl': aatuni| truelotn= l. Supponv. hnthtnoInein*4%'= ‘’‘21- ; * =1-"duke. ,-+-= ,~. ,=[»ca. u:g. [rca. a1 = ,lHc; .(gg+9)= r"“c'-(k+l)9. I awn-uh 14.9.4 compme(-) («ti-1:)’ (Ia) (—2—2=‘)‘u-insD= M°i"'*’”"'°°"'“- Sduﬁl-. (a) (waif)? =4'u'a[2(¥)l - Wait‘ I Sec- 14.0 : Complex Number: :51 (b) We ﬁm cxpmus ~2 ~ 2:‘ polar fonn. In um cast‘. 1: _ -2 and y = -2. Nnu- man me = 5 = :g = 1. The paint (-2, -2) is in we third quaalmnl, um. » 42 - g. Tlw modulus of the camplzx nun. lb('r is r = ‘/ (—2')i + ((2)5 — / ni 03 = 2»/ é Thus. -2 — 2: = zﬁcis %. By Dc Mam‘; Theorem. ' I , (-2- 2")‘ ~ (2/ ids ; (2/21%; . (4 - = o4us5« =64(<.1>a5:r c uin5x) = m(A1+- 0) = 4:4 is; -ollu-y 14.9.2 Let x = uﬂiund _ 9 2:: u. =r! /~c. s( 4*" "), I=_o.1.2.. ... n—1. Then (u. )" = :. Front Using the De Muivn": Thmn-m, for I: = 0. 1,2. . ... n — I. Rmnnrk. Sin: -u. the canine And the sine have period 21!. U0. u1.. ... u,, _1 an- all <li. xLin<1. End: In is called an nth not ufx. Example 14.9.5 Find um um. » cube men of 3, and plot the room in tho complex plun- salmon. —24 = s(—1+o-. ')= S(cusw + iiiuw) — 811511. Using C-nrollaxy 14.9.2 we obtain the following cube roots: . I: Zcilc "ff 7. A-7 0.1.2
30. 30. 368 ﬁndwnco]ﬂﬂNmtn We compute these cube mot: Izpanlely. §+x's; ..’§') —2(; +-§)=1+eﬁ “ u. - 2.5.? =2(en11r+t'ainx) -21-1+s~o)= —2 '2 w. =2ca4-"'31-2(oo-"%+1.an"‘; ') =26-i£)=1-1»/5 uo=2c'a; =2(eo| 2 Figure 14.9.2 xho-vs the cube root! of -B on H! I "31 C M of _8. the wmplu phne. u '00“ nu-an-k. Enn| p|e14.9.5 inequinlcnt to solvingtheequubn :3 = -3. In generd, me prohlemofﬁndingthcnlhmmofakequivﬂcnltolﬂvinglheequlinn z"—a=0wheren iaapunitive integer. Thnntlz xootuamupond ﬁnch: verliouofangulu 11-901: inau-ibed inadrrhwilhoeulnstthenﬁginmdrndim c/ F1. Enmp| :I4.9.6 Solve 2‘-z‘+1oz'-16-0. Snllllinn. We(u: marLhe| eR-hu1dISdeof¢heu1uaIion. z‘(a. -* — 1)+1o(: ‘ - 1) - o 1:‘ +1s)(x’ -1): 0 Thu. z‘+1e=0 or 1'-1=l), snthaz? .heno1uLi0naal'thcequAtim.1uel, —landtl4¢ ﬁ)un. hmouo| ‘—l6. -11- 141-1 +11.) _ 16(cnut+ mu) - 15:1.» By Comlhty 14.9.2. the fourth not: of -16 me via. (' 2"’), k = o.1.2,: I 5"‘ 1"’ 1 Cmﬂrlw Number: '; ¢;g For 1- - U. yr-21;-1 ‘mr. a»( )- 2.. -.(; ) 2 2[. ».. .(§) +. ... .(; )]- 21? . .%fJ -/2+1ﬁ= .FZ(l+tI . Fuxlc 1. V1’o’c1s (" 32"’) : 2.-1s(¥) = ~ +: .<in(”_; ')]»_ 2L - 13 ME] For k —. '1. =2[«~(”I) ~ m-«(€111 — ——/ ‘E—nF = / §(—l -11 For k —:1. “w~<‘*. .’’‘">W12‘)~»1~»(%>+m»<¥>1=21¥ 4?} -/2-11/2:/ i(l-z; The solution so! is {L 4- f2u+n. »/ it-1-131. / i1—1 —. -.1/211-1)}. Exercise 14.9 In Pmblrms l—A, urnle mdx mmplez numbrr 1n nvrfrmgultrr / arm. 1. 5111.-4% 2‘ L. h,.1rJ-. ~ 1 41-11330" 4. '2cis1B0" P‘ ""’“"'*f' 5 "’~ W’-" mh «ample: number m polar Iovm. Use 8 cnlrululnr .1 my uvnglr .5 not mean}. ‘ 5- 1 ‘ 1 6. -.1 + 2: 7. 1 - 21 5_ 1 ‘/3;
31. 31. [Sam j Chayter 14 11mm Pmbcnu 371 .1. Fmd «he dolnain. 11113119. period. I1n1pl1nu'lp {_i(appI11'al>1vf. and the pm. » nun. 11.. -1. sketch the graph :1 y=4sin1 1». y—3tan{*. r . » y: - . ~<'1‘.1 11. 5- -3:-osJ. r 1» y= up 1. 5, , 2m. .1; - 3-1112: 5. Fund 1111' domain and 1111- i111ag1-nl'L2w.4ivo11 l'1m1‘tiun xhvu skctrh 1111- graph 21. y : ‘4(L>1u2.r 11 1,1 : 2‘ A axccusx r 1; 13111-12111: 6. G11-r an 1-xprc-ssiuu (ox / “Lzt for the giwm fun 2 h. f1j. r1 -«u»<‘2J. 11-. .~ . . _/ '(z)=3si111:, -; 5 1» 1- f|1)=21n'l. 'u1(2:w-3) .1 1111-; —a| ’1‘: <i11.r 7 s«-1»-1» :1.» gm; (-quatinn 11. L'11s? .1*r~lu1«z-3:0 1..1-soc; -am“ . ~ k'L>§I5~l1A3.(’U 11. . »m’z-M2; =1 g. arClAu(-11+ 1) = § 15. Prove the given idemitv ru'r—I1r. I (D11 * m” 2: _ g. ~——, —‘__m X —2LAn. r 1. “H - v. us»: .-_. ..i w k ‘ ’ : t1m‘1 pm. 1 9. Pvrfonu 111.» 111.111-1.1.11 optwauuus u>1ug 1). » .11nm. »"s T11.»1.m1. 11.11 rzilnpllfy 3. (;1c1s§‘. ’1‘zcix§1 b. g~l1*i. ~4;H31‘1S: ')3'I . .-.5!‘ 1-. 121-1» 55) j-1 11 (2«»i~. '»"-‘: m.x‘°'; "[m: 111' 1] 11>. Pmd the fuliuw11.11.: ' 1.11 .1“. “,1.. . r. ,.,1,a . .r 9 .31.; (ha 1'n11nh r1>4rt<11f5'251 Ir] the squme roots or 1 - 1 11111 the mm mm. » of l H Salvo H14‘ givuu 1-qu: t1m1. . a.: °n1=u 15.15’ 1 . ;*~. —"—2.r‘-. .1;3.x. .— {: ‘.;11V'3:1—«