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# Trigonometry MAT 106

CSU Pomona

GPA 3.53

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This 97 page Class Notes was uploaded by Dr. Elody Harris on Saturday October 3, 2015. The Class Notes belongs to MAT 106 at California State Polytechnic University taught by Charles Hale in Fall. Since its upload, it has received 138 views. For similar materials see /class/218196/mat-106-california-state-polytechnic-university in Mathematics (M) at California State Polytechnic University.

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Date Created: 10/03/15

CHAPTER 7 Section 71 Solutions 1 5e06 1 20506 Z 0056 7 51n6 3 3 0506i 4 5e06iiZ 51n6 06 3 0056 08 4 1 1 1 1 6 00t6 2 500t6tt955 t605 7 Slnce 5050 weseethat 8 Sin0e se06 1 we seethat 2 51n6 2 0056 2 2J3 2 NE 51n6 0056 5 5 JH 11 10 tan6 0t6 12 0506 51116 14 00t6cf Sg 51n6 2 39 16 tan6 Sm6 0396 5 2 0056 8 4 18 Using Exercise 15 observe that cott91155 H n9281n96 z tang T T17 T 0056 2 6 5 5 T 6 20 tan9sm63 b 0 0056 b 22 0053 600563 013 0001 24 tan2 6 tan 62 52 Chapter 7 25 45f 26 4 27 Solving the equation sin2 9 cos2 9 1 for cos 9 yields cos 9 iVl sin2 9 Since the terminal side of 9 is in QIII we use cos 9 1 sin2 9 Hence cos6 m 28 Solving the equation sin2 t9cos2 9 1 for cos 9 yields cost9 iVI sinz t9 Since the terminal side of 9 is in QIII we use cos 9 1 sin2 9 Hence 29 Solving the equation sin2 9 cos2 9 1 for sint9 yields sin 9 ixl cos2 9 Since the terminal side of 9 is in QIV we use sint9 1 cos2 9 Hence sint9 quotl Z g 30 Solving the equation sin2 9 cos2 9 1 for sin 9 yields sin 9 ixll cos2 9 Since the terminal side of 9 is in QIV we use sint9 1 cos2 9 Hence sint9 ll 2 Jjg 31 Solving the equation ltan2 t9 sec2 9 for sect9 yields sect9 ixlltan2 t9 Since the terminal side of 9 is in QIII cost9 lt 0 and so sect9 lt 0 So we use sect9 ltanzt9 Hence sect9 xltan2 quotl42 32 Solving the equation ltan2 t9 sec2 9 for sect9 yields sect9 ixltan2 t9 Since the terminal side of 9 is in QII cost9 lt 0 and so sect9 lt 0 So we use sect9 ltanzt9 Hence sect9 Iltan2 quotl 52 33 Solving the equation lcot2 t9 csc2 9 for csct9 yields csct9 ixllcot2 t9 Since the terminal side of 9 is in QIII sint9 lt 0 and so csct9 lt 0 So we use csct9 lcotzt9 Hence csct9 lcot2 quot122 34 Solving the equation lcot2 t9 csc2 9 for csct9 yields csct9 ixllcot2 t9 Since the terminal side of 9 is in QII sint9 gt 0 and so csct9 gt 0 So we use csct9xl1cot2 9 Hence csct91cot2 quotl 3 2 796 Section 71 35 Solving the equation sin2 t9cos2 9 1 for cos 9 yields cost9 i 1 sin2 19 Since the terminal side of 19 is in QII we use cos 9 1Sin2 9 Hence mew 42 i Thus tant9smg 15 8 8 161 cost9 161 4161 161 15 36 Solving the equation sin2 t9cos2 9 1 for cos 9 yields cost9 i 1 sin2 19 Since the terminal side of 19 is in QIII we use cos 9 1 sin2 19 Hence 2 7 176J176 cost9 4 1 15 m 15 7 Thustant9smgi cost9 4176 4176 45 44 15 37 Solving the equation sin2 19 cos2 9 1 for sin 19 yields sin 19 ixll cos2 9 Since the terminal side of 19 is in QIII we use sin 19 l cosz t9 Hence sin6 ll Z ms scg15i 71 my 176 4J 44 38 Solving the equation sin2 19 cos2 9 1 for sin 9 yields sin 19 ixl cos2 9 Since the terminal side of 19 is in QII we use sine Ik cos2 6 Hence sint9quot1 Z 11150309 1 i 15W s1n9 161 161 39 Solving the equation sin2 t9cos2 9 1 for sint9 yields sint9 ixll cos2 t9 Since the terminal side of 19 is in QIV we use sin 9 1 cos2 9 From the given 3 1 sec i i hallEl information observe that cos 9 Hence we obtain 797 Chapter 7 40 Solving the equation sin2 t9cos2 19 l for sint9 yields sinB r 17 cos2 I9 Since the terminal side of 19 is in QII we use sin 6 17 cos2 6 From the given information 1 Hence we obtain secH7a Sm9 1ZJ observe that cos 9 41 Since the terminal side of 9 lies in QIV we know that cost9 gt 0 Since sin2 t9cos2 t9 1 we see that l COSZ t9 1 so that cost9 So we conclude that 42 Since the terminal side of 9 lies in QIV we know that sint9 lt 0 Since sin2 t9cos2 t9 l we see that 0152 sin2 t9 1 so that sin 43 Since the terminal side of 9 lies in Q1 and tanB weknowthat cost9gt0 Consider the following diagram Hence sec 19 44 Since the terminal side of 9 lies in 1 QIV and cot t9 75 we know that Hence sin 19 if sin 19 lt 0 Consider the following diagram 26 798 Section 71 45 Since the terminal side of 6 lies in J5 QIH and csc 6 4J5 we know that Hencev 5 g 7m cos 6 lt 0 Consider the following diagram 46 Since the terminal side of 6 lies in Q1 J5 and csc6 weknow that cos gt0 Hencev 565 Consider the following diagram LL JG 47 Consider the followrng diagram Since an 6 ii and me terminal side of 6 lies in QH we conclude that 4 25 Sln I 9 3 773 cos67 7 JR 5 Chapta7 4839 C md me f ll wmg 5 since tan a 7 and the terminal side of 5 lies in Q11 we concludethat sin c050 71 5 5 Since lane 2 and the terminal side of 5 hat 49 Consider the following diagram lies in QlJI we concludet 50 Consider the following diagram since tan a 5 and the tenninal side of 5 lies in QlJI we conclude that sin 9 L J27 26 c050 7 7E J27 26 52mm 1 51 Smce tanS06 5 ande termmal sxde of S has m QHI we conclude that 3 N34 5 m 52 Smce tanS08 andLhetermmal 75 sxde of a has m QHL we conclude that 4 51119 741 41 5 5J41 cosaii J41 41 chaptei 7 53 Consider the following diagram yew Since c0167 and the tenninal side of 0 lies in QIV we conclude that 2J2 sin6r 54 Consider the following diagram since cot 0 s 01 and the tenninal side of 0 lies in QIH we conclude that Sinai 1 77 101 Jl01 101 01 01Jl01 cos 7 7 Jl01 101 55 Considerthe following diagiam since cot and the tenninal side of 0 lies in Q1 we conclude that 4 4E Sm 17 cos 1 s l J 17 Section71 56 Consider the following diagram Since cot 722 and the terminal side of 6 lies in QII we conclude that 5 g 1584 584 1 m 22 224584 cos 6 7 7 11 584 584 1 57 sec cot6 pas sin 1 58 csc tan M cos sin2 6717 icosz 6 7 2 59 tanz isecz i 2 2 2 cos 6 cos 6 cos 6 cos 6 Equivalently you can compute as follows tan2 67 ltan2 6M717 71 osz 1 icosz ilii inz 60 cotz icsc26c 27 2 7 2 sin 6 sin 6 sin 6 in 6 Equivalently you can compute as follows cot2 671 cot2 6 M717 6 71 lisin2 6 61 csc isin isin sin 6 sin 6 2 62 sec icos 1 7cos6 cos 6 cos 1 csc a m 64 cot 6 cos 6 sin cos 62 sin2 62sin6cos6cos2 6 sin2 6cos2 6 2sin6cos6 1 2sin6cos6 Chapter 7 66 sint9 cost92 sin2 t9 2sint9cost9cos2 t9 sin2 t9cos2 t9 2sint9cost9 1 2sint9cost9 67 1tant921tan2t92tant9sec262tant9 12 Mw cos 9 cost9 cos 9 68 1cott921cot262cott9csc2t92cott9 12 27086 w sm 9 smt9 sm 9 sint9cost9 sint9cost9 0 tant9cott9 5 quot f 69 cott9tant9 2 353 cos2 t9sin2 9 W sect9csct9 c0159 W 70 sin 9 cos 9 cos2 9 sin2 9 sin 9 cos 9 39 Z 2 sm 6cos t9 s1n2 t9cos2 9 cos2 9 sin2 9 c050 71 Observe that r 8sin t9cos2 t9 8sint91 sin2 t9 8sint9 8sin3 t9 9 r 30 r8sin30 8sin330 8 834 13 60 r 8sin60 8sin3 60 8 83 45 35 J3 90 r8sin90 8sin390 81 8138 80 72 Observe that r 8sin t9cos2 t9 8sint91 sin2 t9 8sint9 8sin3 t9 9 r 30 r 8sin 30 8si113 30 8 8 3 41 3 60 r 8sin 6o 8sin3 60 8 8 3 4J 3J J 90 r 8sin 90 8sin3 90 8 1 8 13 88 0 73 Cosine is negative in QIII so that cos 9 2 1 w 804 Section 71 74 Must use the Pythagorean identity here not the ratio identities Further note that 4 is not a possible value for cos 19 since the only tenable values are in the interval 11 In this case the correct values would be 1J 4 4J sin19 cos19 J 17 J 17 75 False Since csc19 16 they have the same sign for a given value of 19 sin 76 False Since sec19 1 they have the same sign for a given value of 19 c os19 1 sin 19 for which eithersin 19 1 or sin 19 1 The only value of 19 such that 0 lt 19 S 180 that satis es either one of these equations is 19 90 m The equation sin 19 1 has no solution when 19 is restricted to this interval 77 Observe that sin 19 csc 19 6 so that sin2 19 1 Hence we seek those values of 1 78 Observe that cos 19 sec 19 6 so that cos2 19 1 Hence we seek those values cos of 19 for which either cos 19 1 or cos 19 1 The only value of 19 such that 0 lt 19 S 180 that satis es either one ofthese equations is 19 180 Note Even though cos0 1 0 is not included in the interval and so the equation cos 19 1 has no solution when 19 is restricted to this interval cos19 iVl sin2 19 79 cot 19 where the sign is chosen appropriately depending on s1n 19 s1n 19 the angle 19 1 1 80 csc 19 where the si is chosen a ro riatel de endin on me g W P y P g the angle 19 81 J64 8sin192 641 sin219 J64cos119 8cos19 o E 5 amp 82 J36 6cos192 J361 cos2 19 J36sin2 19 39 s3 J255tan192 J251tan26 25sec219 5sec19 cos2 19 1 csc2 19 cos2 19 cot2 19 4 cot 19 cos2 19 1 tan2 19 sec2 19 sin2 9 tan2 19 1tan2 19 402 a 805 Chapter 7 85 True Observe that sec2 9 12 6 is the smallest when cos2 9 is largest but the cos largest cos2 9 can be is 1 Hence sec2 9 Z l 86 False Observe that l csc2 t9 cot2 9 S 0 for any value of 9 87 a sin22 m 03746 b cos22 m 09272 c m 04040 cos22 d tan 22 m 04040 Yes the values in parts c and d are the same 88 a sin160 E 03420 b cos16 m 09613 c M m 03558 cosl6 d tan 10 m 01763 No the values in parts c and d are not the same 89 a sin35 m05736 b l cos35 m01808 c Jl cos235 m05736 The values in parts a and c are the same 90 a tan68 m24751 b ltan68 z34751 c Jltan268 m26695 Part c does equal sec68 but none ofthe other parts Section 72 n 39 quot 1 sinxcscx m 2 tanxcotxf Equot sec x cot x 1 cos x l W l cos x sin x J sin x sinx m sin x 4 tan xcos x M sm x 9244 5 csc xs1n x sinx s1n x cot xtanx cotxtanx 1261M 0 l secxcos xtan2xMWtan2xltan2x x gt 9 1 sec x tan x cos x tan 94934 91364 806 Section 72 9 sin2 xcot2 x1 sin2 xcsc2 x irf ini j 10 cos2 xtan2 x1 cos2 xsec2 xW w gj 11 sinx cos xsinxcos x sin2 x cos2 x 12 sinxcosx2sinzx2inxcosxcos2xsinzxcoszx23inxcosx 1 1 cscx 39 1 secx 1 13 W 5 14 W cotx 005x cosx tanx Slnx smx x x l cot x 1cotx 16 1539 2 2 2 Z 1cotx 1cotx sec x tan xsec x tan x sec2 x tan x2 sec2 x tan2 x 1tan2 x tan2 x 1 0054x 1 0052 9 2 2 quot x 1Sin4x l sinzx 1 39 x 18 1 sin2x 1sin2x 1cot2x W 19 lco x x 20 1 tan x 1 tan x4 1 1an4x 1tan2xW l tangx 1 tan2x 1 tan2x M 1tan2x 21 sinzx licoszx 1 x 1cosx 1 1 1 1 1 171 licosx licosx W cosx cosx 807 Chapter 7 22 cos2 x l sin2 x W1Slnx 1 1 1 1 1 smx ls1nx ls1nx W l lsinx 23 39 2 2 sum cosx s1n x cos x tanx cotx W 2c0s2x c9sx Smx 2c0s2x 2cos2x tanxc0tx smx005x s1n xc0s x cosx sinx W sin2 x coszx 2cos2xsinzx c0s2x2c0s2x s1n xcos x 1 sin2 xc0s2 x 24 39 2 2 sum cosx s1n x cos x tanx coth 2 cosx Sj xL 2 s osx 2 ws 2 2 vua x tanxcotx s1nx cosx s1n xc0s x cosx sinx 539 osx sinzx coszx 2 2 2 2 cos xs1n x cos xcos x s1n xcos x 1 25 2 2 s1nxcosx s1nx cosx 2 2 2 2 s1n xWcos xs1n x Wcos x 2 What remains after cancellation 1 What remains a er cancellation 1 as claimed 26 l sin xl sin x 1 sin2 x cos2 x as claimed 27 cscxlcscx l csc2 x l 1c0t2 x l cot2 x as claimed 28 secxlsecx l sec2 x l 1tan2 x l tan2 x as claimed 29 sinx cosx sin2 xc0s2 x 1 tan x cot x cosx s1nx s1nxcosx s1nxcosx as claimed 1 sinx l COS x cscxsecx Section 72 30 cscx sinx l l sin2 x cos2 x cosx smx cosxcotxcosxas x sm x sm x sin x claimed 39 2 1 1 sin2 x 2 2 2 sm x lcos x 1 cos x 31 secxcosx cosx cosx cosx cosx cosx 2 2 cos2 x 11 COS 95 lsmZ x 1 sm2 x 32 cscxsmx smx smx smx sm x smx 33 cos x 1 1 cos x cos x 1 cos x l cos2 x l 1 I 34 tan xcotx tanxcotx W m l cosx 35 sec xcotx cos x sinx sin x cosx sinx cosx csc x 1 cos x sinx cosx sinx sin x 36 cotz x 0502 x1 Mcscxl cscxl cscx lcsc x las csc xl cscxl c 1 claimed 1 l l 2 2 37 2 2 s1n xcos xlas cla1med csc x sec x 1 sin2 x cos2 x 38 l 1 tan2 x cot2 x 2 2 sin2 x cos2 x 2 2 tan 99 0 95 2 2 cot x tan x tan xcot x cos x sm x 1 1 r 4 4 2 2 2 2 gm xcos x sm x cos x sm xcos x sin2 xcos2 x sin2 xcos2 x M 993 1 1 2 2 2 2 sec x csc x Wcosz x sin2 XW 005 x 5111 x as claimed 1 1 lsinxl sinx 2 l sinxlsinx lsinxl sinx l sinzx coszx 2sec2 x as claimed Chapter 7 1 1 lc0sx l cosx 2 40 2 T2csc2xas l cosx lc0sx lc0sxl c0sx l cos x sm x claimed sinzx l coszx 1 x 1cosx 41 lc0sxascla1med l cosx l cosx W 2 2 cos x l sm x 1 x 1Slnx 42 lsmxasclalmed l smx l smx W 43 secxtan x secx tan x sec2 x tan2 x secxtanx secx tanx secx tanx lW W 1 secx tanx secx tanx as claimed 44 cscxc0t xcscx cot x csc2 x cot2 x cscxc0tx cscx cotx cscx cotx lM M 1 cscx cotx cscx c0tx as claimed 1 Sinx cosx sinzx 45 cscx tanxsinxcosx W c0sx sin2x 2 2 as claimed secxc0tx 1 00536 smxc0s x smxc0s x cosx sinx W l sinx lsinx 46 secxtanxcosx cosx cosx W smx smxtanx as 39 cscxl 1 1 1Slnx cosx W cosx sinx sinx claimed 47 coszx1sinx lisinzxlsinx 7sinzx7sinx72 2 cos x3 lis1n2x3 7sm2x74 Slnx1M sinxl lsinx 7Msinx2 sinx2 2sinx as claimed 810 Section 72 48 s1nxl cos2x Slnx l39l l Sln 5 Sin2 xs1nx an coszx l sinzx l sinzx l sinxl 39 x sinx l s1nx as claimed 1 r l sinx cosx l sin2xcos2x l l secxtanxcotx 49 cosx cosx s1nx cosx cosxs1nx cosx cosxs1nx l l l cscx 2 2 WP 2 cos x s1nx cos x cos x as claimed sinx l l sinzx coszx 50 tanxcscx s1nx s1nx cosx as cosx s1nx cosx M cosx claimed 51 Observe that the leftside of the equation simpli es as follows cos2 xtan x sec xtan x sec x cos2 xtan2 x sec2 x cos2 xtan2 x 1tan2 x cos2 x Since cos2 x is in fact never equal to 1 being a nonpositive quantity the given equation is a 39 39 52 Observe that cos2 xtanx secxtanxsec x cos2 xtan2 x sec2 x cos2 xtan2 x ltan2 x cos2 x l sin2 x sin2 x l So the given equation is an identity 53 Observe that cotx cscxcotx sinx l l l l cosx l 3 1 cotx 1 cotx cosx cotx cot x secxtanx smx smx tanx smx tanx itanx tanx c a cosx cosx cm cotx So the given equation is an identity 54 The equation sin xcos x 0 is a conditional since for instance it does not hold if x since the leftside reduces to in such case Chapter 7 55 The equation sin x cos x J5 is a conditional since for instance it does not hold if x 0 since the leftside reduces to 1 56 The equation sin2 x cos2 x l is a known identity 57 Observe that tan2 x sec2 x tan2 x ltan2 x 1 1 So the given equation is a quot 39 which 39 39J quotJ is never true 58 Observe that sec2 x tan2 x 1tan2 x tan2 x 1 So the given equation is an identity 59 The equation sin x 11 cos2 x is a conditional To see this observe that the right side reduces as follows Vl cos2 x Isin2 x lsin xl As such the given equation is not true for x 3T for instance Note The equation IS an identity if one restricts attention to only those xvalues for which sinx gt 0 60 The equation csc x 41 cot2 x is a conditional To see this observe that the right side reduces as follows 41 cot2 Icch x lcsc xl As such the given equation is not true for x 3T for instance Note The equation IS an identity if one restricts attention to only those xvalues for which sin x gt 0 and hence cscx gt 0 61 Observe that lsin2 x cos2 x J 1 So the given equation is an identity 62 Observe that sin2 xcos2 x J 1 So the equation W sin x cos x is an conditional Indeed take for instance x Then the leftside 1 while the rightside inE Note A common algebraic error is to say W a b which is false 63 Conditional Note that it is false when x 0 64 Conditional This holds only when x 2 7239 2 65 Let x asint9 where S6S Then 2 la2 x2 la2 asin t92 la21 sin2 t9 la2 cos2 9 lallcos 9 Note that lcos 6 cos 9 for all S 9 S Hence we conclude that la2 x2 acost9 812 Section 72 66 Let x atant9 where lt6lt Then Val x2 laz atant92 lczzltan2 t9 Iaz sec2 9 lallsec l Note that lsec 6 sec 9 for all lt 9 lt Hence we conclude that Val x2 asect9 67 Simpli ed the two fractions in Step 2 incorrectly The correct computations are cosx cosx cos2 x sinx cos x sinx cosx sinx cos x cos x sin x sin x sin2 x 1 cosx sinx cosxsinx cosx sin x sinx 68 There are two errors as follows l l secx not cosx smx l 2 sinx 1 1 1 not a l smx s1nxl s1n x smx sm2 x l sm2 x 69 Need to observe that tan2 x l is a conditional which does not hold for all values of x Verifying the truth of the equation for a particular value of x here x is not enough to prove that it is an identity 70 Verifying the truth of the equation for particular values of x here odd multiples of is not enough to prove that it is an identity Indeed the given equation is actually a conditional To see this take x 0 Then Isin 0 cos 0 l at l 71 False The equation sin2 x l is not an identity since even though it is true for x rm 7 is an integer it is false for all other values ofx for which sin x is defined 72 False The equation sin x 1 has infinitely many solutions but it is not always true 73 The equation cos 9 Vl sin2 9 is true whenever cos 9 gt 0 since 1 sin2 9 Icos2 t9 lcos 6 This occurs for those angles 9 whose terminal side is in Q1 or QIV 74 The equation cos 9 11 sin2 9 is true whenever cos 9 lt 0 since ll sin2 t9 Icos2 t9 lcost9 This occurs for those angles 9 whose terminal side is in QII or QIII 813 Chapter 7 75 csc t9 Icsc2 9 occurs for those angles 9 whose terminal side is in QIII or QIV 76 sect9 ltan2 9 occurs for those angles 9 whose terminal side is in Q1 or QIV 77 2 2 as1nxbcosx bs1nx acos x a2 sin2 x2absirrxmb2 cos2 xb2 sin2 x galasirrx ma2 cos2 x a2 b2 sin2 xa2 b2cos2 x a2 b2sin2 xcos2 x 1 78 lcot3 x lcot3 xcot xlcot x cotx lcotx lcotx lcot3 xcotxcot2 x lcotx 1cot xcot3 xcot2 x lcotx lcotxcot2 xcotxl lcotx 1 eo tfjlcot2 x x csc2 x 79 The equation sinA B sinA sinB is not true in general For instance let A 30 and B 60 Then sin 30 60 sin 90 1 whereas sin30 sin60 I M3 1 2 l l 80 The equation cos Ecos A is not true in general For instance let A Then cos cos 1 whereas 1cos 1 0 2 2 4 2 2 2 81 No For instance take A 5 Note that tan2A is not de ned whereas 2tan 2 814 Section 72 82 No they are not the same as is apparent from the difference in their graphs shown below 83 The correct identity is cosA B cos AcosB El sinA sin B Consider the graphs of the following functions Note that the graphs of y1 BOLD and y3 are the same The gmph of y2 is dotted Foosba yz 005xcos jsinx 51515 4 84 The correct identity is cosA7 B cosA cos B sinA sin B Consider the graphs of the following functions Note that the graphs of y1 BOLD and y3 are the same The gmph of y2 is dotted y Cos x7 yz Mm Elsimml yi ms 0005 sin gm 4 Chapta7 as The correct 1dmt1ty IS smA 5 smAcosB Consxderthe gaphs othe followmg funchons Note LhaLLhe gaphs of y BOLD and y2 are the same The gaph of y 15 mm 1 y sm x 4 2 smx cos j cos y smxcos r cos S6 The correct ldmtlty IS smA7 5 smAcosB E cosAsmB Consxdethe gaphs othe followmg functlons None LhaLLhe gaphs of y BOLD and y are the same The gaph of yzxsdotz y e 4 Secnun7 z 37 The correct tdenttty ts cos2 A Constder the gaphs ofthe followmg fuhettons Note that the gmphs of y BOLD and y are the same The gmph of y2 1511011211 yt Z 1 7 C05 2X y 7 y cos2 x as The carted Identity stn2 AEc Constder the gaphs ofthe followmg fuheuons Note that the graphs of y2 BOLD and y are the same The graph of y 15 mm yti Z 7 lrcost y 7 2 y3sm2x chapter7 tan x tan 17 Lanx 0mg Consrder the gaphs ofthe followmg functrons Note that the graphs of y BOLD and y are the same The graph of y2 1511011211 yr law 9 39 The correct rdentrty rs tanxg tanxr tan y 1tarrtane y31rtanxtang tan x7 tan g 1Lan Lan Consrder the gaphs ofthe followmg functrons Note that the graphs of y BOLD and y2 are the same The graph of y rstzomz 90 The correct rdentrty rs LanOCi x ytarrx 39 7 tan xrtang y 1tanxtarrg 7 an tarr y3 17tanxtarrg Section 73 Section 73 n quot 1 Using the addition formula sinA BsinAcosB cosAsinB yields l l 7 l l l E S1I1 SlIl SlIl COS COS S1I1 12 3 4 3 4 3 4 E l 2 4M 7 2 2 2 2 7 2 2 2 7 4 2 Using the addition formula cosA BcosAcosBsinAsinB yields 7T 7T 7T 7T 7T 7T 7T COS COS COS COS SIl S1I1 b 4 lld lld 1 i 2 1 2M 7 2 2 2 2 7 2 2 2 7 4 3 Using the even identity for cosine along with the addition formula cosAB cosAcosB sinAsinB yields cola awecommomraj singling J J JJJ J 4 Using the odd identity for sine along with the addition formula sinA B sin A cos B cos A sin B yields moosecaramel l lgll lglllgl l 819 Chapter 7 5 First we need to compute both sin and cos Then we use the de nition of tangent to compute tan To this end we have the following Step 1 Using the odd identity for sine and Exercise 1 yields mearme2l ga Step 2 Using the even identity for cosine and the addition formula cosA B cos A cosB sinAsinB yields 7239 7239 7239 7239 7239 7239 7239 7239 cos cos cos cos cos s1n s1n l 12 a a 4 a 4 a 4 i 1 M 2 2 2 2 2 2 2 4 Step 3 Using the de nition of tangent and the computations in Steps 1 and 2 yields 12 2223 rm 6J JE JE 62J JE 2 84J UEh kE JE 6 2 4 6 Using the periodicity and odd identities of tangent followed by Exercise 5 yields tAmMAarhAlH 223 7 Using the addition formula sinA B sin A cosB cos A sinB yields sin105 sin60 45 sin60 cos 45 cos 60 sin45 l lel l 2 2 2 2 4 820 Section 73 8 First applying the addition formula cosA B cos A cos B sin A sin B yields cos195 cos225 30 cos225 cos30 sin225 sin 30 1 Next observe that the reference angle for 225 is 45 Using this observation and affixing appropriate signs we further simplify 1 as follows cos 45 cos 30 sin 45 sin 30 Jl ll tewe 9 First we need to compute both sin 105 and cos 105 Then we use the de nition of tangent to compute tan 105 To this end we have the following Step 1 Using the odd identity for sine and Exercise 7 yields o 0 J5 J3 s1n 105 s1n105 T m2 Using the even identity for cosine and the addition formula cosA B cos A cos B sinA sinB yields cos 105 cos 105 cos 60 45 cos60 cos45 sin60 sin45 GJ J J J Jg 2 2 2 2 4 Step 3 Using the definition of tangent and the computations in Steps 1 and 2 yields J5 J3 sin 105 4 tan 105 J6 5 cos 105 J2 xB J2 xB 4 43 5 JEJ3 6 2J J3 2 8 4J 2 J35J3 2 6 4 Chapter 7 10 Using the addition formula tanA B w yields 1 tan AtanB tanl65 tanl20 45 w 1 l tanl20 tan 45 Now observe thattan 45 1 and since the reference angle of 120 is 60 we have affixing the appropriate signs cos 120 cos 60 sin120 sin 60 J tan120 NlH So continuing the computation in 1 yields J 1 31 J 1 1 J3 2J 31 165m 1J m H 11 Using the odd identity of tangent followed by Exercise 5 yields 1 l l Whigmmz tmt azW 1 1 26 2J 26 2 x3 2 J3 2J 4 3 822 Section 73 12 First we need to compute both sin and cos Then we use the definition of cotangent to compute cot To this end we have the following Step 1 Using the odd identity for sine and Exercise 3 yields 57239 xg J2 cos 12 4 Step 2 Using the odd identity for sine along with the addition formula sinA B sin A cos B cos A sin B yields sint i zt smc zmeat mucosaWigwam l ll J J Step 3 Using the de nition of tangent and the computations in Steps 1 and 2 yields 12 4 JE JE JE JE6 2 E28 4 JM 6 4 823 Chapter 7 13 First we need to compute cos 111 2 Then we use the de nition of secant to compute sec To this end we have the following Step 1 Using the even identity for cosine and the addition formula cosA B cos A cos B sin A sinB yields 1 117239 117239 7239 cos7239cos s1n7239s1n cos 11 12 To 12 12 Using the addition formula cosA B cos A cosB sin A sin B again now yields 7239 7239 7239 7239 7239 7239 7239 cos cos cos cos s1n s1n EJJ J J JE J Hence using this in 1 yields cos M 12 4 Step 2 Using the definition of secant and the computation in Step 1 now yields 4 117239 1 1 3 cos M m 12 4 sec 4 Ji Ja4 JE 39m W39T 824 Section 73 14 First we need to compute cos 113 2 Then we use the de nition of secant to compute sec To this end we have the following Step 1 Using the even identity for cosine and the addition formula cosA B cos A cos B sin A sinB yields 117239 117239 7239 cos7rcos s1n7rs1n cos 11 12 To 12 12 Using the addition formula cosA B cos A cosB sin A sin B again now yields oedemagmomgranola Jl ff fjw 1 Hence using this in 1 yields cos Step 2 Using the definition of secant and the computation in Step 1 now yields sec 13 4 COS13 JihE Eh3 12 4 4 54g 4J JE 1 1 sin 45 300 sin45 cos 300 cos 45 sin 300 1 GJ f M JE1J JE JE 15 csc 255 Chapter 7 sin45 60 sin45 cos 60 cos45 sin60 16 csc 15 1 lel awaa mgt 17 Using the addition formula cosA B cosA cosB sinAsinB with A 2x and B 3x followed by the even identity for cosine at the very end yields sin 2xsin 3x cos 2xcos 3x cos2x 3x cos x 18 Using the addition formula cosA B cosAcosB sinAsinB with A x and B 2x yields sinxsin 2x cos xcos 2x cos xcos 2x sin xsin 2x cosx 2x 19 Using the addition formula sinA B sinAcosB cosAsinB with A x and B 2x followed by the odd identity for sine at the very end yields sinxcos 2x cos xsin 2x sinx 2x sin x 20 Using the addition formula sinA B sinA cosB cosAsinB with A 2x and B 3x yields sin 2xcos 3x cos 2xsin 3x sin2x 3x 21 Using the addition formula sinA B sin A cos B cos A sin B with A 7239 x and B x yields sin xcos7r xcos xsin7r x sinx n39 x sin 7239 22 Using the addition formula sinA B sin A cosB cos A sin B with A x and B x yields sinfxcos x cos xsin x sin x x sinx 23 Observe that sinA sinB2 cosA cosB2 2 sin2 A ZsinAsinBsin2 Bcos2 A ZcosAcosBcos2 3 2 sin2AcfA sin2B iB 2sinAsinBcosAcosB Z 1 1 2 sin AsinB cosAcos B 2cosAcos B sinAsin B 2cosA B 826 Section 73 24 Observe that sinA sinB2 cosA cos B2 2 sinZA2sinAsinBsinZBcos2A2cosAcosBcos2B2 sin2Agp5A sinzBJrisjB 2sinAsinBcosAcosBZ 1 1 2sinAsinBcosAcosB 2cosAcosBsinAsinB 2cosAB 25 Observe that 2 sinAcosB2 cosAsinB2 2 sin2 A 2sinAcosBcos2 B cos2 A 2cosAsinB sin2 B 2 sin2Ac A sin2Bchi 2sinAcosBcosAsinB 1 1 2 sin A cosB cos A sin B 2sinA B 2 9 Observe that 2 sinA cosB2 cosAsinB2 sin2A 2sinAcosBcos2 B cos2 A 200sAsinBsin2 B sin2A9osZA sin2BgpsB 2sinAcosB cosAsinB 1 1 2sinAcosB cosAsinB 2sinA B 2 2 27 Using the addition formula tanA B m with A 49 and ltan AtanB o tan 49 tan 23 0 o o B 23 1e1ds tan 49 23 tan 26 y 1 tan 49 tan 23 tan A tanB o 28 Us1ng the add1t10n formula tanAB w1th A 49 and 1 tan AtanB o tan 49 tan 23 0 o o B 23 1e1ds tan 49 23 tan 72 y 1m490m230 lt gt l 29 Observe that cosa cosacos sinasin J sinasm We 3 need to compute both sina and sin We proceed as follows 827 Chapter 7 Since the terminal side of a is in QIH we Since the terminal side of is in Q11 we know that sina lt As such since We know that sin p gt 0 As such since we 1 are given that cosa 73 the diagram is am given that mp 7 me ding is From the Pythagorean Theorem we have From the Pythagorean Theorem we have 2271232sothatz7 2271242sothatz15 2 Thus sinar 2 3 5 Thus sin p T Hence weWeathermanMi l 12 12 Section 73 1 30 Observe that cosar cosacos slnrzslnp 7 z slnrzslnp We neeolto compute both sin and slnp We proceed as follows Since the terminal side of II is in QIV we know that slnrz lt 0 As such since we 1 are grven that cosrz the dragram ls as 3 follows From the Pythagorean Theorem we have 2212 32sothat 272 2 2 2 Thus sinarTJ Hence owGlamourallwllf ifsTo 12 12 Since the terminal side of p is in Q1 we know that sin p gt 0 As such since we are given that cosp 7 the diagram is as follows From the Pythagorean Theorem we have 22 712 42 sothat z 5 Thus sin T Chapter 7 31 obsewethat sinar sinacos rcosasin gjcos rcosaE We neeoto compute both cosa and cosp We proceed as follows Since the terminal side of a is in QIH we Since the temnna1 side of p is in Q1 we know that cosalt 0 As such since we know that cos 5 gt0 As such since we 3 are gwen that Sm a 2 the diagram 15 am given that mp 13 me diagmm is as 5 follows as follows F m Pym Th h From the Pythagorean Theorem we have mm e agorean eorem we ave 2 2 2 2 1 5 sothat zJ 4 223252sothatz74 2 5 Thus we Thus cask 5 Hence sewszjeowwtzhHWHQGJifs Section 73 3 1 32 Observe that sinzz6 SanZ cos coszzsin cos COSZZ We need to compute both coszz and cos3 We proceedasfollows Since the terminal side of z is in QHI we Since the terminal side of is in QII we As such since we know that cos3 lt 0 As such since we 3 are given that sin 12 r the diagram 15 are given that sin 6 the diagmm is as as follows follows From the Pythagorean Theorem we have From the Pythagorean Theorem we have 22e3257750 ha12e4 zz1252sothatzr Thus coszz JEENE 2 6 Thus cos iT Hence SWtacosmemetateitataee Chapter 7 33 tan a tan p y A 1 r anatan p cosrz and sin 5 39 39 values of sin and cos p in order to compute tan and tan 5 We proceed as follows since the terminal side of a is in QIH we since the terminal side of p is in Q11 we know that cosa lt 0 As such since we know that sin p gt 0 As such since we are given that sin r1 7 the diagram is are given that mg l the diagram is a From the Pythagorean Theorem we have 224712 42 sothat zJ1 J1 4 As such cosar and so As such sin r and so 4 5 39n391 y3 mpg5 From the Pythagorean Theorem we have 2273257 sothat 274 J1 tana t cosa mp cos 7 Hence 374455 4 4 law Li 51215 1927255 5 7119 3745 473J1 5 12716J1 579 43J1 473all 5 16791 Section 73 34 Using the values of tana and tan obtained in Exercise 29 we obtain J 3JE tana tan ltanatan 1JE 4 34xl 5 tana 34J15 43JE1216J159JE1215 19225J 4 3JE 43JE 16 905 I 119 35 Observe that sinABsinA B sinAcosBWsinAcosB W 2 sin A cosB So the given equation is an identity 36 Observe that cosABcosA B cosAcosB WcosAcosBW 2 cos A cos B So the given equation is an identity 37 The equation sinx cos x is a conditional since it is false for x g for instance 38 The equation sin x j cos x is a conditional since it is false for x g for instance 39 Observe that sin x sin xcos sin cos x sin x cos x So the given equation is an identity 40 The equatlon 1s a condltlonal s1nce 1t 1s false for x 3 for 1nstance 41 Observe that 1 cos 2x l cosxcosx sinxsinx1 30S2 x l COSZ 95 2 2 2 2 200s2 x T So the given equation is an idem l cos2xsin2x 833 Chapter 7 42 Observe that 1cos 2x 1cosxcosx sinxsinx11Sin2 xSin2 x 2 2 2 2 2 sin2 x T So the given equation is an identity l sin2 xcos2 x 43 Observe that sin 2x sinx x sin xcosxsin xcosx 2sinxcosx So the given equation is an identity 44 Observe that cos 2x cosxx cosxcosx sinxsinx cos2 x sin2 x So the given equation is an identity 45 The equation sinA B sinA sin B is a conditional since it is false for A B g for instance 46 The equation cosAB cosA cosB is a conditional since it is false for A B g for instance 47 Observe that tan tan 0tan l tan tan l 0tan So the given equation is an identity tan tan 48 Observe that tanA tanA tann39 tanA O tanA ltanAtan7r ltanA0 So the given equation is an identity 49 Observe that l cot37rx tan37rx tanx So the given equation is an identity 50 The given equation is never true because 1 1 csc 2x 7secxcscx s1n 2x 2s1nxcosx So it is a conditional in a big way 51 The given equation is a conditional since it is false when x for instance 834 Section 73 52 Observethat l 7 l 7 litanxi litanx 12nx1 tanxtzn itanxliltanxi litanxtzn Q So the given emla on i anidentity 53 Using the addition formula sinA B sinA cos B cosAsin B With A x and 31 yields y cos i sinx sin 5 cosx sin x 3 3 3 3 So the graph is the graph of y sinx shi ed to the le unis as seen below 54 Using the addition formula sinA 7 B sinAcos B7 cosA sin B With A x and 3 yields y c05 j sin xisin cosx sin x7 So the graph is the graph of y sinx shi ed to the right 1 unis as seen below 3 Chapter 7 55 Using the addition formula cosArB cosAcosB sinAsinB WithA x and 3 5 yields 4 It It It 7 Ir ysinxsin cosxcos cosxcos smxsin cos xi 4 4 4 4 4 So the graph is the graph of y cosx shi ed to the right units as seen below 56 Using the addition forrnula cosA B cosA cosB isinAsinB With A x and 39 B 1d 4 yie s I I I I I ysinxsin rcosxcos 7 cosmos isinxsm rcos x 4 4 l 4 4 E 4 So the graph is the graph of y cosx shi ed to the left E units and then re ected over the x axis as seen below Section73 57 Using the addition formula sinA 13 sinAcosB cosAsinB with A x and B 3x yields 7 7 r W7 7 A 7 7 L l7 If So the general shape ofthe graph is that of y sinx but with period is g and then re ected over the xaxis as seen below 58 Using the addition formula cosA7 B cosAcosB sinAsinB with A x and 7 0 y sin x sin 39 39 7 So the general shape ofthe graph is that of y cosx but the perio is 2 2 It as seen below chapter 7 1 t 59 Observe that J auxtanxif so Observe that aw 14am 1J3tanx So the graph is as follows so the graph is as follows a a s 39 I I I t 3 I I I litan x 62 Observe that tar 7x 1tanx So the graph is as follows to I I I I I a 3 I i I 63 Observe that SmX h e slrl x 7 slrl xcosh slow cosxr slnx slnxcoshr slnxslnhcosx h 7 slnxcoshr1slnhcosx f 7 slnxcosh1 smhcosx h h s hgt 1eosrs hgt h h sWWmem h h Section 73 64 Observe that cosx h 7 cos x 7 cos xcosh 7 sin xsinh 7 cos x 7 COS XCOSh7 COS x7 Sin xSinU l h 7 h 7 h cos xcosh7l7sinxsinh cosh71 sjnh f cos x Tj7sm 7008 x 17 coshjisinx sinh isillx sinh7COS x l7 cosh h h h h tant92 tant91 mz m1 65 Observe that tant92 91 1tan 62 tan 61 1 m2m1 tant91 tant92 ml m2 66 Observe that tant91 62 1tan6tan6 1 mlm 1 Z Z 67 E Acoskz ct Acoskzcosctsinkzsinct If i n an integer then 2 n1 so that 2 knl Zmz39 Hence sinkz 0 So the formula forE simpli es toE Acos kz ct Acos k2cosct 68 E Acoskz ct Acoskzcosctsinkzsinct So in particular E0 Acos 69 tanA B i tan A tanB Should have used tanA B tan A tan B 1 tan A tan B 70 Tangent is an odd function not an even function Hence the very rst step should have been tan 7 tan The remaining calculations are correct 71 False Observe that cos15 cos45 730 cos45 cos30 sin45 sin30 Q E 2 1 JEN 7 2 2 2 2 7 4 while cos45 cos30 2 2 72 False Since sin 1 sin l g and sin and it is clear that 3 61 2 l at the given equality is not true Chapter 7 73 False Observe that m tanx l tan x l ltanx tan x l tanx39 74 True Observe that cot x 75 Approaches may vary slightly here but they will result in equivalent answers as long as the identities are applied correctly The nal form of the simpli cation depends on how you initially choose to group the input quantities For instance the following is a legitimate simpli cation sinABCsinABC sinABcosCsinCcosAB sinAcosBsinBcosAcosCsinCcosAcosB sinAsinB sinAcosBcosCsinBcosAcosCcosAcosBsinC sinAsinBsinC 76 Approaches may vary slightly here but they will result in equivalent answers as long as the identities are applied correctly The nal form of the simpli cation depends on how you initially choose to group the input quantities For instance the following is a legitimate simpli cation cosABCcosABCcosABcosC sinCsinAB cosAcosB sinBsinAcosC sinCsinAcosBsinBcosA cos AcosBcosC sinBsinAcosC sinAcosBsinC sinBcosAsinC 77 Observe that sinA B sinAcosB sin BcosA We seek values ofA and B such that the rightside equals sinA sinB Certainly if we choose values of A and B such that cosB cos A 1 then this occurs And this occurs precisely whenA and B are integer multiples of 27239 and they need not be the same multiple of 27239 So the solutions are IB 2m7239 A 27272 where n and m are integersl 78 Observe that sinA B sin Acos B sin B cos A We seek values ofA and B such that the rightside equals sinA sinB Certainly if we choose values of A and B such that cosB cos A 1 then this occurs And this occurs precisely whenA and B are integer multiples of 27239 and they need not be the same multiple of 27239 So the solutions are IB 2m7239 A 27272 where n and m are integersl 840 Section 73 3979 a The following is the graph of s1n1 licosl y cosx TjisinxTj along with the graph of y cosx dotted 11 c The following is the graph of ice s1n001 sh licos001 y 001 001 39 along with the graph of y cos x dotted gmph b The following is the gmph of sinO1 licos01 isinx 0 01 along with the graph of y cos x dotted gmph y2 cosx Refer back to Exercise 63 for a development of this difference quotient fo ula Looking at the sequence of graphs in parts a 7 c it is clear that this sequence of gmphs better approximates the graph of 841 Chapter 7 80 a The following is the graph of b The following is the graph of sinl 17cos1 sin01 17cos01 y17s1nx 7cosx y27s1nx 7cosx l l 01 01 along with the graph of y 7sinx along with the graph of y 7sinx dotted dotted graph graph c The following is the graph of sinOOl 1415001 y37slnx 7cosx 001 001 along with the graph of y 7 sin x dotted graph Refer back to Exercise 64 for a development of this difference quotient formula Looking at the sequence of gmphs in arts 21 7 c it is clear that this sequence of graphs better approximates the graph of y i 7 sin x as 0 842 Section73 81 Observe that sm 2xh7sm 2x 7 sm 2xcos2hsm2hcos 2pm 2x f sm 2xcos2h7lsm2h 05 2x f sm 2xcos2hl sm2h052x h h ms 2xsm2hi 5m 2X licos2h h h a The following is the graph of b The following is the graph of y cos 2x Sln2rsin 2x y cos 2xs39quot0392r sin 2x 1 01 along with the graph of y 2905 u along with the graph of y 2cos 2x dotted domed graph graph 4 4 c The following is the graph of y cos ZX Sm0 0Zgtjr sm Zx 0 0 along with the graph of y 2cos Zr Looking at the sequence ofgraphs in parts a domed graph 7 c it is clear that this sequence ofgraphs better approximates the graph of y2cos2x as hgt0 4 Chapter 7 82 Observe that cos2xh7cos 2x 7 cos 2xcos2h7sin 2xsin2h7 cos2x h 7 cos2xcos2h717sin 2xsin2h h coszxcos2hhiljisin 2X sm22h 7 cos 2 17 M isin 2x s quoth a The following is the graph of b The following is the graph of 7 39 17 02 39 02 smog along with the graph of y 2905 2x along with the graph of y 2cos 2x dotted dotted graph graph c The following is the graph of y 7 6052417 cos0 02 gm smlt0 02 0 01 0 01 along with the graph of y 2cos 2x Looking at the sequence ofgraphs in parts a dotted graph 7 c it is clear that this sequence ofgraphs better approximates the graph 0 y725in2x as h Section 74 Section A q 39 1 1 1 Since sinx and cosx lt0 the 2 Since sin and cosx lt0 the J3 J3 terminal side ofx must be in Q1 terminal side of must be in QH The diagram is as follows The diagram is as follows From the Pythagorean Theorem we have 2 From the Pythagorean Theorem we have 2 2212J5 sothatz72 22125 sothatz72 As such cosx 772 Therefore As such cosx Therefore 5 sin 2 Zsinxcosx 2 chapter7 3 since cosx and sinxlt0the terminal side of must be in le The diagram 15 as follows From the Pythagorean Theorem we have 2252 132 sothat z 712 As such sinx 7 so that tan cos Therefore 4 since cosxr and sinxlt0the terminal side ofx must be in an The diagram is as follows From the Pythagorean Theorem we have 22 is2 132 so that z 712 As such sin 7E so that 13 tanx 5quot 13 2 cos 7 13 Therefore tan2x 2 Section 74 5 Since tanx 12 5 5 and n ltx lt37 the diagram is asfollows From the Pythagorean Theorem we have 752 e122 z2 so that z 13 12 5 As such smxr and cosxr 13 13 Therefore 12 5 sm2x25mxcosx2 7 7 E 131 o From the Pythagorean Theorem we have 752 7122 22 sot 12 5 As such smx 7 and cos e 13 13 Therefore 2 2 5 12 cos2cos2rsm2x 7 r r 13 13 chapter 7 7 Since secx J5quot it follows that 1 J 1 5 andso cosx cos J3 also sin x gt 0 the terminal side of must be in Q1 The diagram is as follows Since From the Pythagorean Theorem we have 2 2212 JE sothat 22 As such sin x i so that J5quot Therefore tan 2 1 3 Since sec J5 it follows that JE and so cos Since 3 also sinx lt 0 the terminal side of must be in QIV The diagram is as follows cos From the Pythagorean Theorem we have 2 2212 sE sothat 275 As such sin x mg so that J3 45 sin tanV as c L J3quot Therefore taan Ztanx lrtmzx 1 Sectlon 7 4 9 since cscr 4J5quot it follows that 1 72 5 andso 5 5 2J3 10 51 Since also cosx lt0 the terminal side ofx 39 The diagram is as must be in QI follows From the Pythagorean Theorem we have 227J 2 102 so that FriE r As such cosx 7 Therefore 7 10 sinZXZsinxcosr2 100 10 Since cscrr 3it follows that Since sinr E alsocosr gt 0 the terminal side ofx must be in le The diagram is as follows From the Pythagorean Theorem we have 22 712 so that 2 IE J1 2 As such cosx Therefore 5 sianZsinxCOSX2 71E7 ll chapter7 11 Since cscr lt 0 it follows that sinx lt 0 Since alsocosr 7 the ermlnal 394 The diagram ls as follows From the Pythagorean Theorem we have 7122 132 so that z 75 As such sinx i so that 13 sinx 5 tan cosr 12 Therefore 2 com e i Ztanx tmzx 7 Ztanx inquot 12 Since cotr lt o and sin it follows that cosr lt 0 and so the terminal follows From the Pythagorean Theorem we have 22122 132 so that 275 5 As such cosx 7 13 1 1 csclx 51 2 251m cost 7 1 2 2 i 1 13 13 169 Therefore Section 74 2tanA 13 Using tan 2A 2 with l tan A A 15 yields 2tan15 1 tan2 15 1 sin 30 5 tan2l5 tan30 cos30 J5 3 2 14 Using tan 2A with A 1 l tan 8 yields 2tan l 8 WW4 L i 15 Using sin 2A 2 sin A cos A with 7239 A ields 8 y mmamp l 2 Q 16 Using sin 2A 2sinAcos A with A 15 yields sin 15 cos 1 5 2sin15 cos 15 imp15 sin3o 17 Using cos 2A cos2 A sin2 A A 2x yields with cos2 2x sin2 2x cos2 2x 18 Using cos 2A cos2 A sin2 A with A x 2 yields cos2 x 2 sin2 x 2 cos2 x 2 cos2x 4 22 2sin2 5T l cos25T cos57 i 2 851 Chapter 7 23 24 2cos2 1 0 1 cos 271 1 2cos2 l cos 2l cos 7T cos7T cosl 25 Observe that csc2Al lcscAsecA sin2A 2sinAcosA 2 sinA cosA 2 26 Observe that 2 2 cot2A 1 1 1 tan Al 1 tan A lcotA tanA tan2A man1 2tanA 2 tanA tanA 2 1 tan2 A 27 Observe that sin x cos xcos xsin x sin2 x cos2 x cos2 x sin2 x cos 2x 28 Observe that 2 s1nxcos x s1n2 x2s1nxcosxcos2 xs1n2 xcos2 x2s1nxcosx sin 2 1 1sin2x 29 Note that starting with the rightside is easier this time Indeed observe that 1 cos 2x 1 cos2 x sin2 x T f 1 sin2 x cos2 x f cos2 xcos2 x f 2cos2x 2 cos x 30 Note that starting with the rightside is easier this time Indeed observe that 1 cos 2x 1 cos2 x sin2 x T f 1 cos2 x sin2 x 2 sin2 xsin2 x T 2sin2 x sin2 x 852 Section 74 3 1quot Observe that cos4 x sin4 x cos2 x sin2 xcosZ xsin2 x cos 2x 1 3 Equot Note that starting with the rightside is easier this time Indeed observe that 1 1 1 s1n22x 1 2s1nxcos x2 2 2 1 2sin2 xcos2 x 2 sin2 xcos2 x 2sin2 xcos2 x sin4 x3sinZx 2cos4 x W sin4 x cos4 x 33 Observe that 8sin2 xcosZ x 24sin2 xcos2 x 2 2sin xcos x2 2 sin 2x2 2 1 cos2 2x 2 2 cos2 2x 11 2cos2 2x cos2 2x sin2 2x1 2 cos2 2x 1 cos2 2xsin2 2x 1 cos2 2x sin2 2x 1 cos22x 1 cos 4x 34 Observe that cos 2x sin 2xsin 2x cos 2x cos2 2x sin2 2x cos2 2x cos 4x 35 Note that starting with the rightside is easier this time Indeed observe that 2 sin2 x 2 sin2 x 2M 1 1 2s1n2 xcsc2 2x sec2 x 2 2 2 Sln 2x 2s1nxcosx 4 xcoszx 2005 x 2 853 Chapter 7 36 Note that starting with the rightside is easier this time Indeed observe that sec xcsc x 1 cos 2x cos xsin xcos 2x 1 1 E 2 cos xs1n x cos 2x 1 1 s1n 2xcos 2x 1 ll2sin2xcos2x 2 2 1 4csc4x Sm4x s1n4x 37 Note that starting with the rightside is easier this time Indeed observe that sin x4cos2 x 1 sin x3 cos2 x cos2 x 1 sin x3 cos2 x 1 cos2 x sin x3 cos2 x sin2 x 3sin xcos2 x sin3 x sin xcos2 x sin2 x 2 sin xcos2 x sin xcos 2x 2 sin xcos x cos x sin xcos 2x sin 2xcos x sinx 2x sin 3x 854 Section 74 38 Observe that tan3xtanx2xm 1 tanxtan2x 2tanx l tanx 2tanzx l tan x tanx1 tan2 x 2tanx R x l tanzx 2tan2x 3tan x tan3 x 1 3tan2 x tan x3 tan2 x W 39 Note that starting with the rightside is easier this time Indeed observe that 2sinxcosx 4sin3 xcosx 2sinxcosx1 2sin2 x EH cos2 I 2sinxcosx1 sin2 x sin2 x sin 2xcos2 x sin2 x sin 2xcos 2x 2 sin 2xcos 2x 1sin 4x 2 40 Note that starting with the rightside is easier this time Indeed observe that cos 2x sin 2xcos 2x sin 2x cos2 2x sin2 2x cos2 2x cos 4x 855 Chapter 7 41 Before graphing observe that 7 sin2x 7 2sinxcosx 7 2sinxcosx licos 2x 17cos2x7sin2x cos2xsi117 x7cos7 x7 sinzx 7 2sinxcosx 7 cosx cotx 2 sin2 x sin x So a snapshot of the graph is as follows 42 Before graphing observe that 2 tan x 2 tan x 2 tan x tan2x 27sec2x 271tan2x litanzx So the period ofthis function is Z and the Vertical asymptotes occur When cos 2x 0 namely When 2x 2nr171 and so x 21111 The graph is as follows 856 Section 74 43 Before graphing observe that cosx sinx cos2xsin2 x 1 icotxtanx7 cosxi 39 sx icos2xsin2xi 1 7 cotxrtanx i coszxisinzx coszxrsinzx 0st sin x cosx was 2x So the period of this mction is 27 Ir The graph is as follows 44 Before graphing observe that L y tanccotxsecxcscxl cchx 2 T 2 cosx sin x sin 2 So the period of this mction is 27 Ir The graph is as follows Chapter 7 45 Before graphing observe that as follows sin2x cos2x 7sin4x The graph is 46 Before graphing observe that 3sin3x cosr3x 2sin3x cos3x 7 7sin6x The graph is as follows a a Before graphing observe that o 1 7 secxcscx 7 cos x sin x 17 sin x cos x 17 sin2x The graph is as follows 48 Before graphing observe that 3 7 2 sec 2 2 273tan2x csc 2x an The graph is as follows 858 Section 74 2sinx73cos2x The graph is as follows 50 Before graphing observe that sin 2 sin 2 cos 225inxoes M 2 73c052x2 3c052x cosx x i 3 cos 2 x 2Zsinx73coszx The graph is as follows a 7 5 5 l 3 2 51 Observe that F 500sin60 2002 7501405120 375 51311201 sooggjwlwolktal fl 250 3 22500000 37 500 m 22565385 lbs 859 Chapter 7 52 Observe that F 500sin 75 20027501 cosl50 375sinl50 You can immediately enter this into the calculator to get the estimate indicated below However we simplify the various quantities using addition formulas rst in order to provide the exact not just approximate answer To this end observe that sin 75 sin 30 45 sin 30 cos 45 cos 30 sin 45 1 Q 6 4 2 2 2 2 4 cos 75 cos 30 45 cos 30 cos 45 sin 30 sin 45 1 JE 2 2 2 2 2 4 As such we use this information to now compute the following cosl50 cos2 75 cos2 75 sin2 75 xg sEZ JEJ 2 TT 16 16 sinl50 sin2 75 2sin 75 cos 75 6 2JE2 62JE2 4 5 43 7 4 4 l6 4 So we have F 500 5 gl2002 7501 3751H 4 2 2 4 12s 1257375000007315018750 m 28009614 lbs 860 Section 74 53 Consider the following diagram 3ft 1ft I y 3 Observe that tan23 i 1 tanBi 2 litan B y Substituting 2 into 1 yields the following equation that we solve for y 2 L L L 2 LL23 2 L123 2 37121 2 2 176 y 13 y 2 y yy 21 y 2 2y24y224 2 42y2 2 y22 2 yJ ft 54 Consider the following diagram Observe that 7 7 stB 3 2schosB x x 4 3 Since smB g and cosB g we have 2g 2 723 2 24x 175 2 sin 2x sin 2x 55 Should use sinx 2 since we 56 N0 that tan 2x 110t 3 cos 2x cosx are assuming that sin x lt 0 861 Chapter 7 58 False Let A 7239 Observe that cos 2 7239 1 sin 2 sin 1 1 cos47r1 4 2 Since 1 1 at l the statement is false sin 4 j sin7r 0 Since 1 1 0 the statement is false 57 False Let A Observe that 59 F 1 N t th t t 2 2tanx 60 False Note that sin2x 2sinxcosx 39 a se39 0 e a an x 1tan2 x 39 For instance there exist x values for which Take any xe 1 For suchx sinxgt0andcosxlt 0namely gltxlt 7239 4 2 Th f th 1 39 2 lt0 tanxgtlsothat l tan2xlt0 So for us or esexvaues sm x these values ofx tan 2x lt 0 61 True Observe that 62 False Observe that 0ltcosxltsinxlt12 0ltcoszxltsin2xltl sin2x 2sinxcosxgt 0 ifand only if Hence Since COS 2x COSZ xsin2 x we sinx and cosx have the same sign This conclude that cos 2x lt 0 occurs In Q1 and QIH39 63 True 64 False cos 2 3x cos2 3x sin2 3x 1 2 sin2 1 sin2 sin2 cos2 s1n2 cos 2 65 False sin 2x 2 sin x cos x 2 sin x cos x 2 sin xcos x sin 2x and so it cannot be even 66 Ztanx 4tanx 2 2 tan4xtan22x Ztanzzx 1 tan x 2 ltazn x l tan 2x 1 2tanx 1tan2x 4tan2x 2 1 tan x 1tan2x2 4tan xl tan2 x 1 tan2 x2 4tan2 x 862 Section 74 67 Using the odd identity for tangent togetherwith Exercise 66wefln that 2 2 A gt A 4tanx17tan x 7 4tanxtan x71 2 2 litanzx 4th litanzx 74mm lhe Jucuul JUI ml 68 One A 39 lhe identity fUl mi 69 One A 70 Consider the following graphs Observe that y dotted gloom agood Observe that y dotted graph ls a good approxlmatlon of y2 0n 711 approxlmanon of y2 on 711 72 Consider the following graphs 39 119 i 9n enuminn i no an idemitv Chapter 7 3973 Consider the following graphs m From the picture it seems as though the given equation is an identity But one cannot prove an identity by inspection alone The formal proof is as follows 1725quot x 7 25inx 905x sin2 x cos2 x7 25in2 x 7 25inx cosx 25inxcosxcoszxisin2x 25inxcosxcoszxisin2x cos2 x 75in x7 25inx cost 7 25inx costcos2 xisin2 x 7 cos 2x 7 sin Zr sin 2x cos 2x 3974 To the right csc 2x sec 2x cos 2x 75in 2x are the graphs of the following functions cos 2 solid y2 2cosx dashed y 2cos2 x71 Note that the graphs of y and y are the same Semun7 5 75 To me nght are the graphs of me followmg funeuons y 5m 2x solid y 25m x dashed 25m xcosx V3 Note thatthe graphs of y and y are the Secdnn 1 Use sm 139 2 mm 1430 to obtam sm15 sm3 2 Use cos 2 thh 2445 to obtam J2 a a 1 ess2zsoe05 45 e 2 2 2 2 4 3 Use cos i M wnh 1441 750 obtam 2 2 6 1m 1 1m 1 J3 1m 7 57 A 7 2J 7 J2J V 2 V 2 V 4 2 w Chapter 7 4 Use sin V w withA to obtain 2 2 4 J5 1 7 242 J2 JE 2 4 2 39 V H 5 oo l II V H 5 NJgtll O o T H 5 Use cos j with A150 to obtain J5 o o 1 750m 2 J2 2 2 2 4 2 6 Use sin J Jw with A150 to obtain J3 o 0 1 Sin7sosin 150 1 cos150 l 2 f2J 2 I 2 2 2 4 2 Alternatively you could use Exercise 5 together with the identity sinA 1 cos2 A with A 75 to obtain 2 sin75 l cos275 1 quot2 E 1 LJ g 7 Use tan i W with A 135 to obtain 2 1 cosA tan675 tan 135 2 1 cosl35 Since cosl35 7 cos 45 7g the above further simpli es to 2x2 25 25 44J22 27 2 5 472 3242 L 27 Z tan675 866 Section 75 8 Use tan i E with A2025 to obtain 2 lcosA 4050 if licos405O 2 7 l1cos405 39 2 tan2025O tanE Since cos 405 005360quot 45 cos45 7 the above further simpli es to o 27J2 242 Lb22 tan 2025 7 7 7 7 7372 2 2J2 242 472 9 Observe that 1 1 1 Also note that cos cos 277 cos Hence using the formula S 2 97r J5 9 1co 1 7 cos w1thA9 glves c0 7 7 4 7 2 7V2 5 2 2 4 2 2 2 1 7 Using this in 1 then yields sec 7 8 7 2 J2 xl2 J2 2 10 First note that using the formula cos l with A 97 yields 97 J5 l 9n 9 4 Host 4 41 2 2 COS COS 8 2 V 2 V 2 2 Now using sin A 11 cos2 A With A 9 subsequently yields 97z 1 1 71 71 71 Cm sing Alexi W275i Chapter 7 11 First note that using tan w with A13 we obtain 2 1 cos A 4 1 1 1 1 cot B l 8 we 2 lcos13 Next observe that cos 13 cos 277 5 cos 5 g Using this in 1 subsequently yields Bit 7 a 1 1 1 m 7 7 2J 7 2J 2J 7 44J 27 32J 275 2745 2J 472 Note Recall that there are two other formulae equivalent to tan V and 2 1 cos A either one could be used in place of this one to obtain an equivalent but perhaps different looking result For instance using tan smA with A 13 yields 2 1 cos A 4 137 l l l cot 2 8 tan 137 13 sm13 8 tan 2 f 7 1 cos Since mewcoslt2n57zgtcosltw snowsin2nssm we can use these in 2 to obtain 137 1 245 245 J3 2 54 7 g 7J5 1 2 Even though the forms of the two answers are different they are equivalent 868 Section 75 12 First note that using tan 4 with A 7 we obtain 2 1 cosA 4 0071 1 1 1 8 tan tan 1 cos7 2 1 cos7 Using this in 1 subsequently yields Z 1 1245 2 6245 25 25 2 5 Note Recall that there are two other formulae equivalent to tan 2 l cos A and either one could be used in place of this one to obtain an equivalent but perhaps different looking result For instance using tan with A 7 yields 2 l cos A 4 cot 1 1 1 2 cosmgtcosltm smmrswmr we can use these in 2 to obtain MEL 1 2 2 L22 71 5 8 7 Since ii aE 7J5 J5 2 J5 17 2 Even though the forms of the two answers are different they are equivalent 869 Chapter 7 13 First note that using the formula cos f with A 57 yields 57 J5 1 s 5 1 ml 4 1 2 12 5 COS COS 8 2 V 2 V 2 2 Then we have sec5 1 72 8 cOS i M 2 7J5 2 14 First note that using the formula sin J y with A 5T iyields 5 J5 15 Observe that a 1 1 1 cot 135 cot 2 tan tanz720 1COS270 1cos270 10 16 Observe that 00M1 1 1 12 J 2 5 1cos210 1 2 l IS 17 Since cos x 13 and sin x lt 0 the terminal side ofx lies in QIV Hence the terminal side of 3 lies in Q11 and so sin gt 0 Therefore we have 5 x l cosx E 13 5 8 2 2l3 gm 2 2 2 26 26 l3 l3 870 Section 7 5 13 Since cosr 7ahd sin lt0 the terminal side ofr lies in QIH Hence the terminal side of lies in Q11 and so cos lt0 Therefore we have cos x 7 l1cosx7 8 7 2 2 2 26 J13 w f th di tht 19 Since tanx and7rltxlt37 we em mquot e agmm 2 cosxri Soywehave know that smx lt0 and cosxlt0 Also 13 note that since lt lt3 sin 1 gt0 2 2 4 2 Consider the following diagram Chapter 7 20 Since tmx2 and ltxlt3lvwe We seefromthe diagramthat know that sinxlt 0 and cosxlt 0 Also m i 39 8039 we have I x 37 note that since 7ltEltTy cosgjdy x 1 cosx Consider the followin dia ram cos 7 T 7 21 Since secx J5 we know that we see from me diagram 31315 cosx This together with sinxgt 0 So we have implies that the terminal side ofx is in Q1 an si 2 Hence the terminal side of g is also in 2 1 co 1 5 Q1 Thusm g gt0 7 2J511 5 71 cons der me fallowquot d a mm Alternatively we could have used one of the other two equivalent formulae for an Speci cally the following 2 5 computation is also valid Section 7 5 22 Since secx J3quot we know that cost This together with sin lt 0 implies that the tenninal side of is in QIV Hence the teiminal side of is in Q11 Ihustan lt 0 Consider the following diagram We see from the diagram that 2 sinxr sowehave J3 J7 ms393 45 2 1cosgtc 11 J31 3 42 434 dis571 51 J34 371 Altematively we couldhave used one of o equivalent foimulae for tan Speci cally the following computation is also valid Chapter 7 23 Since cscx 3 we know that 5 1 We see from the diagram that 3 This together with cosx lt 0 implies that cosx ii So we have the terminal side ofx is in QII Hence the 3 terminal side of i is in Q1 So 5nigt0 2 2 Consider the following diagram sin I 24 Since cscx 73 we know that We see from the diagram that 1 5mg 7 This together With cosx gt 0 my 232 I SO we have implies that the terminal side ofx is in QIV Hence the terminal side of is in 1 amp 2 cos x if 1cosx77 3 QII Thus COSEJlt 0 2 2 7 Consider the following diagram Section 75 25 Since cscx lt 0 it follows that We see from the diagram that sinx lt0 Thistogetherwith GOSPJ 15 4 sinx7T Sowehave implies that the terminal side ofx is in QIII Hence the terminal side of is in QII Thus co 1 lt 0 2 Consider the following diagram 26 Since cotx lt 0 and cosx i it we see from me diagram that 15 follows that sin x lt 0 So the terminal 5 39 50 We have side ofx is in QIV and the terminal side of x 1 1 1 39 39 csc 1 In Q1 So mg and mg 2 Sin W i Consider the following diagram 2 2 4 Chapter 7 24 7 We see from the diagram that and 7ltxlt7rwe 27 Since cotx cosrr Sowehave know that smx gt 0 and Cost lt 0 Also J601 notethat since 0 ltilt1 cos 1 gt0 1 2 2 Con der the follo g agram We see from the dia am that 23 Since cotxr and ltxlt7rwe 24 y 5 2 cosxr Sowehave know that smxgt0 and C0516 lt0 Also 4601 notethatsince olt lt1 sin 1 gt0 2 2 I Consider the followin dia ram 7 Section 75 29 Since secx gt 0 it follows that We see from the diagram that cosxgt0 This together with cosx 091 smwe have sinx 703 implies that the terminal side ofxisianV Hence the terminal side of anx 17 05 2 1cosx i isinQII Thustan 5 lt0 2 2 Consider the following diagram JD 91 30 Since secx lt 0 it follows that We see from the diagram that 905xlt0 This ogemer With cosx7l091 Sowe have sin x 70 3 implies that the terminal side x 1 1 ofx is in QIII Hence the terminal side of 90 x 1 m 7 7 7 cos x gisinQII Thuscot lt0 2 JHcosx Consider the following diagram 1 1 091 17 091 il 113 x 1 877 Chapter 7 31 Since sec x 25 we know that We have cost 04 This together with cot 1 1 25 2 t x tanx gt 0 implies that the terminal side of E x is in Q1 Hence the terminal side of x is In Q1 Thuscot3 gt 0 Consider the following diagram 25 J5 25 1 32 Since sec x 73 we know that We have cost 7 This together with cotx lt 0 3 tan lrcosx implies that the terminal side ofx is in Q11 2 1 cost Hence the terminal side of is in Q1 1 Thustan 1 gt0 2 Consider the following diagram N 3 x 1 Section 75 33 Using the formula cos J with A 5 yields 57239 lcos 5n f 6cos cos 5 2 2 12 35 Using the formula tan with A 150 yields 2 1 cosA Sm150 O tan 150 tan75 l cos 150 2 36 Observe that l cos150 l cos150 lcos150 l cos2150o sin150 sin150 1cos150 sin150 1cos150 sin2150 sin150 sin150 1cos150 lcos150 Now using the formula tan i with A 150 yields 2 1 cosA L500tan 150 tan75 lcos150 2 l cos5 5 7 l cos15 a 37 J tan tan5T 38 Jmtan tan 39 Use the known Pythagorean identity sin2 9 cos2 9 1 with 9 g to conclude that the given equation is in fact an identity 40 Observe that cos2 sin2 cos 2 J cos x 4 p t Observe that 2sin cos sin2 sin x sin x 879 Chapter 7 42 Observe that 2 2 43 Observethat tan2 i tan i i 1cosx 1cosx 2 2 1cosx 1cosx 44 Note that starting with the rightside is easier Indeed observe that cscxcotx2 1 cbssz 1 cosx2 l cosx1 cosx s1nx s1nx sin2 x 1 cos2 x 1 x 1 005x l cosx Jeni 2x tan tan 1 x 1cosx 1COSX 2 2 45 Observe that A A sinA 1 sinA 1cosA tan cot 2 2 1cosA SlIlA 1cosA sinA 1cosA sinZA1cosA2 1 cos2 A1cosA2 sinA1cosA sinA1cosA 1 cosA1cosA1cosA2 sinA1cosA 2 2cscA sinAW sinA 46 Observe that A A 1 sinA 1cosA sinA cot tan 2 2 511114 1cosA sinA 1cosA 1cosA 1cosA2 sin2A 1cosA2 1 cos2 A sinA1cosA sinA1cosA 1cosA2 1 cosA1cosA sinA1 cos A W1cosA lCOSA 2c0sA 2 A 2 t gnaw sinA Co 880 Section 75 4 7 Observe that A 1 1 2 CSC 2 2 sin 5 160 1405A 2 2 2 1cosA721cosA 21cosA I COSAl i COSAi licoszA sinzA 48 Observe that 2 2 cos 1 1cosA 1COSA 2 2 2 1405A 7 217cosA7 217cosA 1cosA17cosA7 licoszA 7 siniA 49 Using Exercise 47 we see that csc i wit 214 qicsc2 A22cosA ilcscAlJZJchosl U 5111 50 Using Exercise 48 we see that 2 17 A sec r j r csc2 A272cosA rlcscAHZi 2cosA sin 51 Before graphing observe that 2 y4c052 i 4 4 22cosx 2 ll 2 2 39 i 7 L rquot 4 39 2 L 39 quot and the vertical shift is up 2 units So 39 39 Chapter 6 L L L 39 follows 2 A 1 1 2 SEC 2 Chapter 7 52 Before graphing observe that 2 licosx 76sin2 i 76 7 y 2 ill 2 y L rr 4 3 shift is down 3 units 0 f 39 L L39 as follows 733cosx 2 39 and the Vertical L L L is L Chapter 6 2 x 17w 1cosx717cosx 717m EJ7171C05X 7 W iZcosxi 1mn2 j 117cosx 1cosx17cosx 2 2 W So the graph is as follows Section 75 u a Before graphing observe that wtmgjwostaj 17wansmowsowosza m 21 2 So the graph is that of y sinx re ected over the x axis as shown below 55 Before graphing observe that y 45in2 Q71 4 The graph is as shown below Chapter 7 56 Before graphing observe that 1cos 21 27 r cosx2r cosx we The graph is as shown below The graph is as shown below Section 75 58 Before graphing observe that y l 3c0s3 f3 The graph is as shown below S 7 5 4 3 2 1 1 1 2 3 4 5 6 7 S 9 1G 72 3 4 59 Observe that 60 Consider the following triangle A 17h 6 6 asm a cos 2 2 aZl 2sin2cos2 7 7 2 2 2 a2 sin 6 2 Since the sum of the interior angles in a triangle must bel80 we see that 6 30 So using Exercise 41 the area ofthis triangle is 7 I Z A m sin30 inz 2 4 37r 7r x 37r x 61 If 7r lt x lt 7 then 3 lt3 lt7 so that sm 3 1s pos1trve not negative sinZ i sinZ i 62 Note that tanZ not 2 z x cosZ x I cos 2 Chapter 7 63 False Use A 7239 for instance Indeed observe that sin jsin j 2 while sin72390 64 False Use A 7239 for instance Indeed observe that cos gcos 0 while cos7r l 65 False Use x 57 for instance Observe that l gt 0 while tan lt 0 since the terminal side of 5 is in QII 66 False Use x 450 for instance Observe that o sin 450 sin 360 90 sin90 lgt 0 while sin450 J sin225 lt 0 since the terminal side of 225 is in QIII 886 Section 75 67 Observe that 2 tan A 2i l cosA 2i l cosAd cosA 2i lcosA 2 lcosA lcosA l cosA l coszA l cosA2 l cosAl cosA lcosA V sinZA sinA I sinA lcosA l coszA sinZA sinA I sinAlcosA sinAlcosA lcosA Case 1 The terminal side of g is in Q1 or QIII Here we use tan Further for such angles A both sinA gt 0 and 2 1 cos A lcosAZO Hence gt0 so thatwe have tan lcosA 2 lcosA lcosA Case 2 The terminal side of g is in QII or QIV sin A A Here we use tan lcosA 2 Further for such angles A both sinA lt 0 and lcosAZO Hence lt 0 so that we have lcosA A sinA sinA sinA an 2 lcosA lcosA lcosA I So in either case we conclude that tan smA l cos A I One cannot evaluate the identity at A 7239 since the denominator on the rightside would be 0 hence the fraction is not wellde ned and g is not in the domain of tangent 887 Chapter 7 68 Observe that 2 tan A 2i l cosA 2i l cosAl cosA 2i lcosA 2 lJlcosA VlcosA l cosA l coszA l cosA2 2i 2 sm A 4 l cosA sinA Case 1 The terminal side of g is in Q1 or QIII Here we use tan Further for such angles A both sinA gt 0 and 2 s1nA l cosAZ 0 Hence gt 0 so thatwe have tan 1ciA s1nA 2 s1nA s1nA Case 2 The terminal side of g is in QII or QIV l cosA Here we use tan Further for such angles A both sinA lt 0 and sinA l cosA Z 0 Hence w lt 0 so that we have s1nA A l cosA l cosA tan 2 A sinA sinA So in either case we conclude that tan 1 cos A sinA 1 cos A sin A One cannot evaluate the identity at A 7239 since the denominator on the rightside would be 0 hence the fraction is not wellde ned and g is not in the domain of tangent 69 Observe that 1 1 1 lcos t A i co 4 tan tan 1cos 1 cos lcos 888 Section 7 5 70 Observe that 7 seccscrtan Meadows 7 1 chcA 71 observe that tan gt0 when 0 lt4 lt7 or 7 lt lt 3 2 So the desired values ofx in 017 for which is true are 72 observe that cot g g 0 when l lt A lt 7 So the desired values ofx in 027r for 272 whichistme are 73 Consider the graphs of the following functions a s i i 1 L L a y 2 3 5 y 2 E 5 4 Observe that y dotted graph is a good 3 approximation of y2 on 711 2 1 123455 439 Chapter 7 74 Consider the graphs of the following functions Xj Observe that yl dotted graph is a good approximation of y2 on 711 75 Consider the graphs of the following functions csc2 i sec2 i 4cch x y 2 2 yz n a E 7 E 5 4 3 2 1 43757473724 123455 Since the graphs coincide this suggests that the giVen equation is an identity The proof ofthis fact is as follows ml wtai 1 1 2 3sm2 21 130036 WSW wwowmpchmwwamk 890 Section 75 y2 72cotz xsecx 76 Consider the graphs ofthe following functions x x tan2 cot2 y 2 2 L L 39 39 the i en nua nni not anidemic 77 To the right are the graphs ofthe followingfuncti s 2 y cos solid 5 3 2 3 3 2 Note that the graphs of y and y arethe same 73 To the right are the graphs ofthe following functions yfsihe solid y2 7sih dashed e and y arethe 2 Note that the graphs of y same

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