Quick Review a=bc/d 2. b=ad/c 3. c=ad/b 4. d=bc/a 7 sin L sin 23 9 sin

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1 Section 5.5 The Lw of Sines 9 () The sine regression curve through the oints defined y L nd L is y4.656 sin(0.05x-0.85) This is firly good fit, ut not relly s good s one might exect from dt generted y sinusoidl hysicl model. [ 30, 370] y [ 60, 60] (c) Using the formul L-Y(L) (where Y is the sine regression curve), the residul list is: {3.64, 7.56, 3.35, 5.94, 9.35, 3.90, 5., 9.43, 3.90, 4.57, 9.7, 3.}. (d) The following is sctter lot of the dys st Jnury s x-coordintes (L) nd the residuls (the difference etween the ctul numer of minutes (L) nd the numer of minutes redicted y the regression curve (Y)) s y-coordintes (L3) for the time of dy tht stronomicl twilight egn in northestern Mli in 005. The sine regression curve through the oints defined y L nd L3 is y8.856 sin(0.0346x ) (Note: Round L3 to deciml lces to otin this nswer.) This is nother firly good fit, which indictes tht the residuls re not due to chnce. There is eriodic vrition tht is most roly due to hysicl cuses. [ 30, 370] y [ 5, 5] (e) The first regression indictes tht the dt re eriodic nd nerly sinusoidl. The second regression indictes tht the vrition of the dt round the redicted vlues is lso eriodic nd nerly sinusoidl. Periodic vrition round eriodic models is redictle consequence of odies oriting odies, ut ncient stronomers hd difficult time reconciling the dt with their simler models of the universe. Section 5.5 The Lw of Sines Exlortion. If BC AB, the segment will not rech from oint B to the dotted line. On the other hnd, if BC 7 AB, then circle of rdius BC will intersect the dotted line in unique oint. (Note tht the line only extends to the left of oint A.). A circle of rdius BC will e tngent to the dotted line t C if BCh, thus determining unique tringle. It will miss the dotted line entirely if BC 6 h, thus determining zero tringles. 3. The second oint (C ) is the reflection of the first oint (C ) on the other side of the ltitude. 4. sin (-C )sin cos C -cos. 5. If BC AB, then BC cn only extend to the right of the ltitude, thus determining unique tringle. Quick Review 5.5. c/d. d/c 3. cd/ 4. dc/ 7 sin L 3.34 sin 3 9 sin 6. L sin 4 7. xsin x80 -sin x80 -sin ( 0.7) x360 +sin ( 0.7) Section 5.5 Exercises. Given: 3.7, B45, A60 n AAS cse. C80 -(A+B)75 ; c 3.7 sin 60 L 4.5; sin 45. Given: c7, B5, C0 n AAS cse. A80 -(B+C)45 ; c 7 sin 45 L 3.9; c sin 0 c 7 sin 5 L 5. c sin 0 3. Given: A00, C35, n AAS cse. B80 -(A+C)45 ; L 5.8; 4. Given: A8, B40, 9 n AAS cse. C80 -(A+B)59 ; c L 4.4; 5. Given: A40, B30, 0 n AAS cse. C80 -(A+B)0 ; c L.9; 6. Given: A50, B6, 4 n AAS cse. C80 -(A+B)68 ; c c sin 45 sin 00 sin 35 sin 00 9 sin 8 sin 40 9 sin 59 sin 40 0 sin 40 sin 30 0 sin 0 sin 30 4 sin 6 sin 50 4 sin 68 sin 50 c L.8 L.7 L 8.8 L 4.6; L sin 75 sin 45 L 5.

2 0 Chter 5 Anlytic Trigonometry 7. Given: A33, B70, 7 n AAS cse. C80 -(A+B)77 ; c L 4.; 8. Given: B6, C03, c n AAS cse. A80 -(B+C)6 ; c c 7 sin 33 sin 70 7 sin 77 sin 70 sin 6 sin 03 sin 6 sin 03 L 7.3 L 0.8; L Given: A3, 7, n SSA cse. h 5.8; h<<, so there is one tringle. B sin - sin L 0. C80 -(A+B) 7.9 ; c 7 sin 7.9 L 5.3 sin 3 0. Given: A49, 3, 8 n SSA cse. h.; h<<, so there is one tringle. B sin - sin L 4.3 C80 -(A+B)89.7 ; c 3 sin 89.7 L 4.4 sin 49. Given: B70, 4, c9 n SSA cse. hc 8.5; h<c<, so there is one tringle. C sin - c sin L 37. A80 -(B+C) 7.8 ; 4 sin 7.8 L 4. sin 70. Given: C03, 46, c6 n SSA cse. h 44.8; h<<c, so there is one tringle. B sin - sin L 47.3 c A80 -(B+C)9.7 ; c 6 sin 9.7 L 3.0 sin Given: A36,, 7. h 4.; <h, so no tringle is formed. 4. Given: B8, 7, c5. hc 4.9; h<c<, so there is one tringle. 5. Given: C36, 7, c6. h 0.0; h<c<, so there re two tringles. 6. Given: A73, 4, 8. h 6.8; <h, so no tringle is formed. 7. Given: C30, 8, c9. h 9; hc, so there is one tringle. 8. Given: B88, 4, c6. hc 6.0; <h, so no tringle is formed. 9. Given: A64, 6, 7. h 5.3; h<<, so there re two tringles. B sin - sin L 7.7 C 80 -(A+B ) 43.3 ; c 6 sin 43.3 L. sin 64 Or (with B otuse): B 80 -B 07.3 ; C 80 -(A+B ) 8.7 ; c L.7 0. Given: B38,, c5. hc 5.4; h<<c, so there re two tringles. C sin - c sin L 47. A 80 -(B+C ) 94.9 ; sin 94.9 L 34.0 sin 38 Or (with C otuse): C 80 -C 3.9 ; A 80 -(B+C ) 9. ; L 5.4. Given: C68, 9, c8. h 7.6; h<c<, so there re two tringles. A sin - sin L 78. c B 80 -(A+C) 33.8 ; c 8 sin 33.8 L 0.8 sin 68 Or (with A otuse): A 80 -A 0.8 ; B 80 -(A +C) 0. ; c L 3.4. Given: B57,, 0. h 9.; h<<, so there re two tringles. A sin - sin L 67.3 C 80 -(A +B) 55.7 ; c 0 sin 55.7 L 9.9 sin 57 Or (with A otuse): A 80 -A.7 ; C 80 -(A +B) 0.3 ; c L. 3. h0 sin , so: () () 6.69 or 0. (c) h sin , so: () c 6. () c or c. (c) c () No: this is n SAS cse () No: only two ieces of informtion given.

3 Section 5.5 The Lw of Sines 6. () Yes: this is n AAS cse. B80 -(A+C)3 ; 8 sin 3 L 88.5; sin 9 8 sin 9 c L 46. sin 9 () No: this is n SAS cse. 7. Given: A6, 8, n SSA cse. h 8.4; <h, so no tringle is formed. 8. Given: B47, 8, n SSA cse. h 5.9; h<<, so there is one tringle. A sin - sin L 6. C80 -(A+B)6.8 ; c sin 6.8 L 5.6 sin Given: A36, 5, 8 n SSA cse. h 9.5; <h, so no tringle is formed. 30. Given: C5,, c7 n SSA cse. h 0.9; c<h, so no tringle is formed. 3. Given: B4, c8, C39 n AAS cse. A80 -(B+C)99 ; c c L 8.3; 3. Given: A9,, B47 n AAS cse. C80 -(A+B)4 ; c 8 sin 99 sin 39 8 sin 4 sin 39 sin 9 sin 47 sin 4 sin 47 L 9.8; 33. Given: C75, 49, c48. n SSA cse. h 47.3; h<c<, so there re two tringles. B sin - sin L 80.4 c A 80 -(B+C) 4.6 ; c 48 sin 4.6 L 0.7 sin 75 Or (with B otuse): B 80 -B 99.6 ; A 80 -(B +C) 5.4 ; c L Given: A54, 3, 5. n SSA cse. h.; h<<, so there re two tringles. B sin - sin L 69.0 C 80 -(A+B ) 57.0 ; c 3 sin 57.0 L 3.5 sin 54 Or (with B otuse): B 80 -B.0 ; C 80 -(A+B ) 5.0 ; c L 4. L 9. L Cnnot e solved y lw of sines (n SAS cse). 36. Cnnot e solved y lw of sines (n SAS cse). 37. Given: cab56, A7, B53 n ASA cse, so C80 -(A+B)55 c 56 sin 53 () AC L 54.6 ft. in 55 () h ( ) 5.9 ft. 38. Given: c5, A , B n ASA cse, so C80 -(A+B)9 nd c 5 sin 5 L 9.7 mi. sin 9 c 5 sin 37 L 5.0 mi, sin 9 nd finlly h.9 mi. 39. Given: c6, C90-68, B n AAS cse. A80 -(B+C)47, so c 6 sin 47 L 4.9 ft. sin Given: c.3, A8, B37 n ASA cse. C80 -(A+B)5 ; c.3 sin 8 L. mi. sin 5 c.3 sin 37 L.5 mi. sin 5 Therefore, the ltitude is h (.5) sin mi or (.) sin 37 mi 0.7 mi ft 8 0 The length of the rce is the leg of the lrger tringle. sin 8 x, so x.9 ft B ft x C The center of the wheel (A) nd two djcent chirs 360 (B nd C) form tringle with 5.5, A 6.5, nd BC This is n ASA cse, so 5.5 sin the rdius is c L 39.7 ft. sin.5 Alterntively, let D e the midoint of BC, nd consider right ^ABD, with mjbad.5 nd BD7.75 ft; then r is the hyotenuse of this tringle, so 7.75 r L 39.7 ft. sin.5

4 Chter 5 Anlytic Trigonometry 43. Consider the tringle with vertices t the to of the flgole (A) nd the two oservers (B nd C). Then 600, B9, nd C (n ASA cse), so A80 -(B+C)40 ; 600 sin 9 L 303.9; sin sin c L sin 40 nd finlly h c 08.9 ft. 44. Consider the tringle with vertices t the to of the tree (A) nd the two oservers (B nd C). Then 400, B5, nd C0 (n ASA cse), so A80 -(B+C)45 ; 400 sin 5 L 80.5; sin sin 0 c L 38.5; sin 45 nd finlly h c 6.7 ft. 45. Given: c0, B5, C33 n AAS cse. A80 -(B+C)95, so c 0 sin 95 L 36.6 mi, nd sin 33 c 0 sin 5 L 8.9 mi. sin We use the men (verge) mesurements for A, B, nd AB, which re 79.7, 83.9, nd 5.9 feet, resectively. This gives 6.4 for ngle C. By the Lw of Sines, 5.9 sin 83.9 AC L 9. feet. sin True. By the lw of sines,, which is equivlent to (since, Z 0). 48. Flse. By the lw of sines, the third side of the tringle 0 sin 00 mesures, which is out 5.3 inches. Tht sin 40 mkes the erimeter out , which is less thn 36 inches. 49. The third ngle is 3. By the Lw of Sines, sin 3 sin 53, which cn e solved for x..0 x The nswer is C. 50. With SSA, the known side oosite the known ngle sometimes hs two different ossile ositions. The nswer is D. 5. The longest side is oosite the lrgest ngle, while the shortest side is oosite the smllest ngle. By the Lw of sin 50 sin 70 Sines,, which cn e solved for x. 9.0 x The nswer is A. 5. Becuse BC>AB, only one tringle is ossile. The nswer is B. 53. () Given ny tringle with side lengths,, nd c, the lw of sines sys tht. c But we cn lso find nother tringle(using ASA) with two ngles the sme s the first (in which cse the third ngle is lso the sme) nd different side length sy,. Suose tht k for some constnt k. Then for this new tringle, we hve. Since c k #, we cn see tht #, k k so tht k nd similrly, c kc. So for ny choice of ositive constnt k, we cn crete tringle with ngles A, B, nd C. () Possile nswers:, 3, c (or ny set of three numers roortionl to these). (c) Any set of three identicl numers. 54. In ech roof, ssume tht sides,, nd c re oosite ngles A, B, nd C, nd tht c is the hyotenuse. sin 90 () c c c o hy sin 90 () c cos> - B c cos A c dj hy (c) cos A tn A o dj 55. () hab () BC 6 AB (c) BC AB or BCAB (d) AB 6 BC 6 AB 56. Drwing the line suggested in the hint, nd extending to meet tht line t, sy, D, gives right ^ADC nd right ^ADB. A 8 C 5 B D BC Then AD8 sin 3.0 nd DC8 cos 7.4, so DBDC-5 nd cab AD + DB L 3.9. Finlly, A(90 - )-sin DB nd AB L 9. B80 -A-C 8.9.

5 Section 5.6 The Lw of Cosines Given: c4., B5, C An AAS cse: A80 -(B+C)43.5, so c 4. sin 5 AC L 8.7 mi, nd sin.5 c 4. sin 43.5 BC L. mi. sin.5 The height is h sin 5 sin mi. Section 5.6 The Lw of Cosines Exlortion. The semierimeters re 54 nd 50. A ces. 4, squre feet squre miles cres 5. The estimte of little over n cre seems questionle, ut the roughness of their mesurement system does not rovide firm evidence tht it is incorrect. If Jim nd Brr wish to mke n issue of it with the owner, they would e well-dvised to get some more relile dt. 6. Yes. In fct, ny olygonl region cn e sudivided into tringles. Quick Review 5.6. Acos Ccos ( 0.3) Acos ( 0.68) Ccos () () Acos 6. () cos A 8 - x - y -xy () Acos x + y - 8 xy cos A y - x x - y x + y - 8. xy x - y One nswer: (x-)(x-)x -3x+. Generlly: (x-)(x-)x -(+)x+ for ny two ositive numers nd. 8. One nswer: (x-)(x+)x -. Generlly, (x-)(x+)x -(-)x- for ny two ositive numers nd. 9. One nswer: (x-i)(x+i)x + 0. One nswer: (x-) x -x+. Generlly: (x-) x -x+ for ny ositive numer. Section 5.6 Exercises. Given: B3, c8, 3 n SAS cse. + c - c cos B L L 9.; C cos c L cos L 8.3 ; A 80 - B + C L Given: C4,, 4 n SAS cse. c + - cos C L L 9.5; A cos - + c - L cos L 80.3 ; c B 80 - A + C L Given: 7, 9, c4 n SSS cse. A cos - + c - L cos L 76.8 ; c B cos - + c - L cos L 43. ; c C 80 - A + B L Given: 8, 35, c7 n SSS cse. A cos - + c - L cos L 5. ; c B cos - + c - L cos L 99. ; c C 80 - A + B L Given: A55,, c7 n SAS cse. + c - c cos A L L 9.8; B cos - + c - L cos L 89.3 ; c C 80 - A + B L Given: B35, 43, c9 n SAS cse. + c - c cos B L L 9.5; C cos c L cos L.7 ; A 80 - B + C L Given:,, C95 n SAS cse. c + - cos C L L 5.; A cos - + c - L cos L 8.5 ; c B 80 - A + C L Given:, c3, A8 n SAS cse. + c - c cos A L L 35.4; B cos - + c - L cos L 37.9 ; c C 80 - A + B L No tringles ossile (+c) 0. No tringles ossile (+<c). Given: 3., 7.6, c6.4 n SSS cse. A cos - + c - L cos L 4.6 ; c B cos - + c - L cos L 99. ; c C 80 - A + B L 56..

6 4 Chter 5 Anlytic Trigonometry. No tringles ossile (+<c) Exercises 3 6 re SSA cses, nd cn e solved with either the Lw of Sines or the Lw of Cosines. The lw of cosines solution is shown. 3. Given: A4, 7, 0 n SSA cse. Solve the qudrtic eqution 7 0 +c -(0)c cos 4, or c -(4.86 )c+50; there re two ositive + c - solutions: L9.487 or Since cos B : c c 9.487, B cos (0.94) 7.9, nd C 80 -(A+B ) 65., or c 5.376, B cos ( 0.94) 07., nd C 80 -(A+B ) Given: A57,, 0 n SSA cse. Solve the qudrtic eqution 0 +c -(0)C cos 57, or c -(0.893)c-0; there is one ositive + c - solution c.564. Since cos B, c B cos (0.647) 49.7 nd C80 -(A+B) Given: A63, 8.6,. n SSA cse. Solve the qudrtic eqution c -(.)c cos 63, or c -(0.079)c+49.50; there re no rel solutions, so there is no tringle. 6. Given: A7, 9.3, 8.5 n SSA cse. Solve the qudrtic eqution c -(8.5)c cos 7, or c -(5.535)c-4.40; there is one ositive + c - solution: c Since cos B, c B cos (0.503) 59.8 nd C80 -(A+B) Given: A47, 3, c9 n SAS cse. + c - c cos A L L 3.573, so Are L L.33 ft (using Heron s formul). Or, use A c. 8. Given: A5, 4, c n SAS cse. + c - c cos A L L 6.583, so Are L L 5.84 m (using Heron s formul). Or, use A c. 9. Given: B0, 0, c n SAS cse. + c - c cos B L L 5.845, so Are L L cm (using Heron s formul). Or, use A c. 0. Given: C,.8, 5. n SAS cse. c + - cos C L 36.8 L 6.0, so Are L 8. L 4.6 in. (using Heron s formul). Or, use A. For # 8, tringle cn e formed if +<c, +c<, nd +c<. 7. s ; Are L 8.8. s ; Are L No tringle is formed (+c). 4. s7; Are, L ; Are 46, L No tringle is formed (+<c) 7. s4.; Are 98, L s3.8; Are0,69.4 L Let 4, 5, nd c6. The lrgest ngle is oosite the lrgest side, so we cll it C. Since cos C + - c, C cos - 8 L rdins. 30. The shorter digonl slits the rllelogrm into two (congruent) tringles with 6, B39, nd c8. The digonl hs length + c - c cos B 7.59 L 6.5 ft. 3. Following the method of Exmle 3, divide the hexgon into 6 tringles. Ech hs two -inch sides tht form 60 ngle. 3. Following the method of Exmle 3, divide the nongon into 9 tringles. Ech hs two 0-inch sides tht form 40 ngle * sin L 374. squre inches 9 * 00sin 40 L 89.3 squre inches 30 s s In the figure, nd so s sec The re of the hexgon is 6 * 8383sin L squre inches s s In the figure, 0 nd so s0 sec 0. The re of the nongon is 9 * 0 sec 0 0 sec 0 sin 40 L 37.6 squre inches.

7 Section 5.6 The Lw of Cosines Given: C54, BC60, AC0 n SAS cse. AB c + - cos C 7, L 30.4 ft. 36. () The home-to-second segment is the hyotenuse of right tringle, so the distnce from the itcher s ruer to second se is L 66.8 ft. This is it more thn c cos L 63.7 ft. () B cos - + c - L cos c () c cos L 4.5 ft. () The home-to-second segment is the hyotenuse of right tringle, so the distnce from the itcher s ruer to second se is L 44.9 ft. (c) B cos - + c - L cos c Given: 75, 860, nd C78. An SAS cse, so ABc + - cos C L 707, ft. 39. () Using right ACE, mjcae tn tn - 3 L () Using A L 8.435, we hve n SAS cse, so DF cos A L ft. (c) EF cos A L ft. 40. After two hours, the lnes hve trveled 700 nd 760 miles, nd the ngle etween them is.5, so the distnce is cos.5 84,59.77 L 90.8 mi. 4. AB cos L.5 yd. 4. mjhab 35,so HB cos L 37.0 ft. Note tht AB is the hyotenuse of n equilterl right 0 tringle with leg length, nd HC is the 0 hyotenuse of n equilterl right tringle with leg length 0 + 0, so HC L 48.3 ft. Finlly, using right HAD with leg lengths HA 0 ft nd AD HC L 48.3 ft, we hve HD HA + AD L 5.3 ft. 43. AB c + 3 3, AC + 3 0, nd BC + 5, so A cos - + c - mjcab c 9 cos ABC is right tringle (C90 ), with BC + nd AC, so ABc nd Bm ABCsin + 3 j True. By the Lw of Cosines, +c -c cos A, which is ositive numer. Since +c -c cos A>0, it follows tht +c >c cos A. 46. True. The digonl oosite ngle slits the rllelogrm into two congruent tringles, ech with re sin. θ 47. Following the method of Exmle 3, divide the dodecgon into tringles. Ech hs two -inch sides tht form 30 ngle. * sin The nswer is B. 48. The semierimeter is s(7+8+9)/. Then y Heron s Formul, A The nswer is B. 49. After 30 minutes, the first ot hs trveled miles nd the second hs trveled 6 miles. By the Lw of Cosines, the two ots re cos 0 L 3.05 miles rt. The nswer is C. 50. By the Lw of Cosines, (7)(5) cos, so cos The nswer is E. 5. Consider tht n-sided regulr olygon inscried within circle cn divide into n equilterl tringles, ech with r 360 equl re of sin. (The two equl sides of the n equilterl tringle re of length r, the rdius of the circle.) Then, the re of the olygon is exctly nr 360 sin. n + c - 5. () + c - + c - c cos A c c Lw of Cosines c cos A c cos A () The identity in () hs two other equivlent forms: cos B + c - c cos C + - c c c

8 6 Chter 5 Anlytic Trigonometry () Shi A: 5. knots; hr Shi B:.4 knots hrs () We use them ll in the roof: cos A + cos B + cos C c + c - c cos A + c - c A 35.8 (c) c + - cos C (49.6) +(60.4) -(49.6)(60.4) cos (35.8 ).04, so the ots re 34.8 nuticl miles rt t noon. 54. Use the re formul nd the Lw of Sines: A sin qlw of Sines r 55. Let P e the center of the circle. Then, cos P , so P The re 55 of the segment is The re of the tringle, however, is 55 sin in, so the re of the shded region is rox. 6.9 in. Chter 5 Review. sin 00 cos 00 sin c - c c c + c c c c + + c c r # L 5 # 0.47 L 9.39 in. tn 40. tn 80 - tn ; the exression simlifies to (cos ) +( sin cos ) (cos ) +(sin ). 4. cos x; the exression cn e rewritten -( sin x cos x) -(sin x) cos x. 5. cos 3xcos(x+x)cos x cos x-sin x sin x (cos x-sin x) cos x-( sin x cos x) sin x cos 3 x-3 sin x cos x cos 3 x-3(-cos x)cos x cos 3 x-3 cos x+3 cos 3 x 4 cos 3 x-3 cos x 6. cos x-cos x(-sin x)-(-sin x) sin x-sin x 7. tn x-sin xsin x - cos x cos x sin x # sin x sin x tn x cos x 8. sin cos 3 + sin 3 cos ( sin cos )(cos +sin ) ( sin cos )()sin. 9. csc x-cos x cot x sin x - cos x # cos x sin x - cos x sin x sin x sin x sin x tn u + sin u 0. + cos u + cos u ; tn u A cos u + tn u. Recll tht tn cot. - tn u + + cot u - cot u + tn u - cot u + + cot u - tn u - tn u - cot u + tn u - cot u cot u - tn u - - tn u - cot u tn u - cot u. sin 3 sin( + )sin cos +cos sin sin cos +(cos -sin ) sin 3 sin cos -sin 3 3. cos t c; B + cos t d + cos t + cos t sec t sec t + sec t sec t tn 3 g - cot 3 g 4. tn g + csc g 5. tn g - cot gtn g + tn g cot g + cot g tn g + csc g tn g - cot gtn g + + cot g tn g + csc g tn g - cot gtn g + csc g tn g - cot g tn g + csc g cos f - tn f + sin f - cot f cos f - tn f cos f cos f + sin f - cot f sin f sin f cos f sin f sin f - cos f cos f - sin f cos f - sin f + cos f - sin f cos f + sin f cos -z cos -z 6. sec -z + tn -z 3 + sin -z4>cos -z cos -z cos z + sin -z - sin z - sin z - sin z + sin z

9 Chter 5 Review 7 - cos y 7. B + cos y - cos y B + cos y - cos y - cos y B - cos y - cos y B sin y 0 - cos y 0 - cos y since -cos y 0, 0 sin y 0 0 sin y 0 we cn dro tht solute vlue. - sin g 8. B + sin g - sin g + sin g B + sin g - sin g B + sin g cos g B + sin g 0 cos g 0 0 cos g 0 since +sin 0, 0 + sin g 0 + sin g we cn dro tht solute vlue. 9. tn u tn u + tn3>4 - tn u tn3>4 tn u + - tn u - - tn u - + tn u 0. sin 4 sin ( ) ( sin cos ) ( sin cos )(cos -sin ) sin cos 3 -cos sin 3. tn - cos sin sin - cos csc - cot sin. Let rctn t, so tht tn t. Then tn u tn. Note lso tht since - tn u t - t <t<, -, nd therefore -. 6 u u 6 4 Tht mens tht is in the rnge of the rctn function, t nd so rctn, or equivlently - t t rctn nd of course, rctn t. - t 3. Yes: sec x-sin x tn x cos x - sin x cos x - sin x cos x. cos x cos x cos x 4. Yes: (sin Å-cos )(tn Å+) (sin Å-cos Å)(sec Å) sin - cos sin cos cos - tn Mny nswers re ossile, for exmle, sin 3x+cos 3x (3 sin x-4 sin 3 x)+(4 cos 3 x-3 cos x) 3(sin x-cos x)-4(sin 3 x-cos 3 x) (sin x-cos x)[3-4(sin x+sin x cos x+cos x)] (sin x-cos x) (3-4-4 sin x cos x) (cos x-sin x)(+4 sin x cos x). Check other nswers with grher. 6. Mny nswers re ossile, for exmle, sin x+cos 3x sin x cos x+4 cos 3 x-3 cos x cos x( sin x+4 cos x-3) cos x( sin x+-4 sin x). Check other nswers with grher. 7. Mny nswers re ossile, for exmle, cos x-sin x-sin x-sin x -4 sin x cos x- sin x cos x. Check other nswers with grher. 8. Mny nswers re ossile, for exmle (using Review Exercise #), sin 3x-3 sin x 3 cos x sin-sin 3 x-6 sin x cos x sin x(3 cos x-sin x-6 cos x) sin x(4 cos x--6 cos x). Check other nswers with grher. In #9 33, n reresents ny integer sin x0.5 when x +n or x +n, so x +n or x +n cos x when x ; +n 6 3. tn x when x - +n 4 3. If sin x, then xsin. 33. If tn x, then xtn. 34. cos x cos x cos x- cos 3 x 4 3 cos x 5 So x or x for n ny integer. 6 + n 6 + n 35. x. or x 5.6 [0, ] y [ 4, 4] 36. x 0.4 or x 3.79 [0, ] y [ 3, ]

10 8 Chter 5 Anlytic Trigonometry 37. x.5 [0, ] y [ 3, ] 38. x.85 or x 3.59 [0, ] y [ 3, ] cos x, so x or x sin 3x(sin x)(4 cos x-). This eqution ecomes (sin x)(4 cos x-)sin x,or (sin x)( cos x-)0, so sin x0 or 3 cos x ; x0, x, x, x, x, or x The left side fctors to (sin x-3)(sin x+)0; 3 only sin x is ossile, so x. 4. cos t-cos t, or cos t-cos t-0, or ( cos t+)(cos t-)0. Then cos t 4 or cos t: t0, t or t sin(cos x) only if cos x. No choice of + n n gives vlue in [, ], so there re no solutions. 44. cos(x)+5 cos x- cos x-+5 cos x- cos x+5 cos x-30. ( cos x-)(cos x+3) 0, so cos(x) nd cos(x) 3. The ltter is extrneous so x e 3, 5 3 f For #45 48, use grhs to suggest the intervls. To find the endoints of the intervls, tret the inequlities s equtions nd solve. 45. cos x hs solutions x 6, x 5 6, x 7 6, nd x in intervl [0, ). The solution 6 set for the inequlity is, 0 x or or 6 x 6 ; 6 6 x tht is, c 0, , 7 6 6, 46. sin x cos x cos x is equivlent to (cos x)(sin x-)0, so the solutions in (0, ] re x nd x 3. The solution set for the inequlity is ; tht is,., 3 6 x cos x hs solutions x nd x 5 in the 3 3 intervl [0, ). The solution set for the inequlity is ; tht is,. 3, x tn xsin x is equivlent to (sin x)(cos x-)0, so the only solution in - is x0. The solution, set for the inequlity is - ; tht is, -. 6 x 6 0, y5 sin (3x+cos (3/5)) 5 sin (3x+0.973) 50. y3 sin (x-cos (5/3)) 3 sin (x-.76) 5. Given: A79, B33, 7 n AAS cse. C 80 - A + B 68 7 sin 33 L 3.9; sin 79 c 5. Given: 5, 8, B0 n SSA cse. Using the Lw of Sines: h 4.7; h<<, so there is one tringle. A sin - L sin L 36.0 ; C 80 - A + B L 34.0 ; c 8 sin 34.0 L L 4.8. sin 0 Using Lw of Cosines: Solve the qudrtic eqution 8 5 +c -(5)c cos 0, or c +(3.40)c-390; there is one ositive solution: + c - c 4.8. Since cos A : c A cos (0.809) 36.0 nd C80 -(A+B) Given: 8, 3, B30 n SSA cse. Using the Lw of Sines: h 4; <h, so no tringle is formed. Using the Lw of Cosines: Solve the qudrtic eqution 3 8 +c -(8)c cos 30, or c - 83c ; there re no rel solutions. 54. Given: 4.7, A9.3, C33 n AAS cse. B80 -(A+C)7.7, nd c 55. Given: A34, B74, c5 n ASA cse. C80 -(A+B)7 ; c c 7 sin 68 sin sin 7.7 sin sin 33 sin sin 34 sin 7 5 sin 74 sin 7 L 6.6. L 6.4. L.9; L 5.. L 6.6;

11 Chter 5 Review Given: c4, A.9, C55. n AAS cse. B80 -(A+C)0 ; c c 57. Given: 5, 7, c6 n SSS cse. A cos - + c - L cos L 44.4 ; c B cos - + c - L cos - 0. L 78.5 ; c C 80 - A + B L Given: A85, 6, 4 n SSA cse. Using the Lw of Sines: h 4.0; h<<, so there is one tringle. B sin - L sin L 4.6 ; C 80 - A + B L 53.4 ; c 4 sin.9 sin sin 0 sin sin 53.4 sin 85 L 9.5; L L 4.8. Using the Lw of Cosines: Solve the qudrtic eqution 6 4 +c -(4)c cos 85, or c -(0.697)c-00; there is one ositive solution: + c - c 4.8. Since cos B : c B cos (0.747) 4.6 nd C80 -(A+B) s (3+5+6)7; Are ss - s - s - c L c 7.67 so re 58.4 L 3.0 (using Heron s formul). Or, use A. 6. h sin 8 5.6, so: () 5.6<<. () 5.6 or. (c) < () C80 -(A+B)45, so c 80 sin 65 AC L 0.5 ft. sin 45 () The distnce cross the cnyon is 96.4 ft. 63. Given: c.75, A33, B37 n ASA cse, so C80 -(A+B)0 ; c.75 sin 33 L.0; sin 0 c.75 sin 37 L., sin 0 nd finlly, the height is h sin 0.6 mi. 64. Given: C70, 5, 900 n SAS cse, so ABc + - cos C 7,06.84 L ft. 65. Let 8, 9, nd c0. The lrgest ngle is oosite the lrgest side, so we cll it C. + - c Since cos C, Ccos ,.53 rd. 66. The shorter digonl slits the rllelogrm into two (congruent) tringles with 5, B40, nd c4. The shorter digonl hs length + - c cos B L L ft. Since djcent ngles re sulementry, the other ngle is 40. The longer digonl slits the rllelogrm into (two) congruent tringles with 5, B40, nd c4, so the longer digonl length is + c - c cos B L L ft. 67. () The oint (x, y) hs coordintes (cos, sin ), so the ottom is units wide, the to is x cos units wide, nd the height is hysin units. Either use the formul for the re of trezoid, A ( + )h, or notice tht the trezoid cn e slit into two tringles nd rectngle. Either wy: A( )sin +sin cos sin (+cos ) sin + sin. () The mximizing ngle is 60 ; the mximum 3 3 re is.30 squre units. 4 3 L 68. () Sustituting the vlues of nd : Su cot u + 3 # csc u cos u sin u [0, 9.4] y [ 6., 6.] () Considering only ngles etween 0 nd, the minimum occurs when (c) The minimum vlue of S is roximtely S(0.96) 7.7 in. 69. () Slit the qudrilterl in hlf to leve two (identicl) right tringles, with one leg 4000 mi, hyotenuse 4000+h mi, nd one cute ngle /. u Then cos ; solve for h to leve h h sec u miles. cosu> () cos u, so u cos - 0 L 0.6 L

12 30 Chter 5 Anlytic Trigonometry 70. Rewrite the left side of the eqution s follows: sin x-sin x+sin 3x sin x- sin x cos x+sin(x+x) sin x- sin x cos x+sin x cos x+cos x sin x sin x- sin x cos x+ sin x cos x+ (cos x-sin x) sin x sin x- sin x cos x+ sin x cos x+ sin x cos x-sin 3 x sin x- sin x cos x+3 sin x cos x-sin 3 x sin x (- cos x+3 cos x-sin x) sin x ((-sin x)- cos x+3 cos x) sin x (cos x- cos x+3 cos x) sin x (4 cos x- cos x) sin x cos x ( cos x-) sin x ( cos x-) So sin x or. So or x ; 0 cosx - 0 x n, n n integer. 3 + n 7. The hexgon is mde u of 6 equilterl tringles; using Heron s formul (or some other method), we find tht ech tringle hs re , The hexgon s re is therefore, 3843 cm, nd the rdius of the circle is 6 cm, so the re of the circle is 56 cm, nd the re outside the hexgon is L cm. 7. The entgon is mde u of 5 tringles with se length 6 cm nd height cm, so its re is out tn 36 L cm. The rdius of the circle is the height of those tringles, so the desired re is out (8.58) 33.5 cm. 73. The volume of cylinder with rdius r nd height h is V r h, so the wheel of cheese hs volume (9 )(5)405 cm 3 ; 5 wedge would hve frction of tht volume, or L 53.0 cm () (cos(u-v)-cos(u+v)) (cos u cos v+sin u sin v -(cos u cos v-sin u sin v)) ( sin u sin v) sin u sin v () (cos(u-v)+cos(u+v)) (cos u cos v+sin u sin v+cos u cos v -sin u sin v) ( cos u cos v) cos u cos v (c) (sin(u+v)+sin(u-v)) (sin u cos v+cos u sin v+sin u cos v -cos u sin v) ( sin u cos v) sin u cos v 75. () By the roduct-to-sum formul in 74 (c), u + v u - v sin cos # sin u + v + u - v u + v - u - v + sin sin u+sin v () By the roduct-to-sum formul in 74 (c), u - v u + v sin cos # sin u - v + u + v u - v - u + v + sin sin u+sin( v) sin u-sin v (c) By the roduct-to-sum formul in 74 (), u + v u - v cos cos # u + v - u - v cos + cos u + v + u - v cos v+cos u cos u+cos v (d) By the roduct-to-sum formul in 74 (), u + v u - v sin sin - # u + v - u - v cos -cos u + v + u - v (cos v-cos u) cos u-cos v 76. Pt fked the dt. The Lw of Cosines cn e solved to show tht x. Only Crmen s vlues B - cos u re consistent with the formul. 77. () Any inscried ngle tht intercets n rc of 80 is right ngle. () Two inscried ngles tht intercet the sme rc re congruent. (c) In right A BC, o hy d. (d) Becuse ja nd ja re congruent, >d d. (e) Of course. They oth equl y the Lw of Sines.

13 Chter 5 Review 3 Chter 5 Project. [, 34] y [ 0.,.]. We set the mlitude s hlf the difference etween the mximum vlue,.00, nd the minimum vlue, 0.00, so 0.5. And we set the verge vlue s the verge of the mximum nd minimum vlues, so k0.5. Since cos ((x-h)) tkes on its mximum vlue t h, we set h9. Exerimenting with the grh suggests tht should e out /30.5. So the eqution is y L 0.5 cos x Using the identities from Questions 3 nd 4 together, y L 0.5 cos x cos 9 - x sin x sin x , 30.5 which is equivlent to y L 0.5 sin 0.x Sinusoidl regression yields y L 0.5 sin 0.x