Chapter 7 Analytic Trigonometry - Mr. Tower
Chapter 7Analytic TrigonometryChapter 7 Mixed Review Worksheets1. sin 1cot θ 3,12θ Find the angle θ , 1.21sin θ ,2πθ 6ππ θ , whose sine22Find the angle θ , 0 θ π, whose secantequals 1 .sec θ ( 1) , 0 θ πθ πThus, sec 1 ( 1) π . 2 3 6. csc 1 3 Find the angle θ , 0 θ π, whose cosecant2. tan 1 1ππ θ , whose tangent22equals 1.θ equals ππ θ 22π4Thus, tan 1 (1) π.4Thus, sec 1 2 π2 θ π2π.4 π 7. sin 1 sin follows the form of the 8 3.23cos θ , 0 θ π2πθ 6 3 π .Thus, cos 1 2 6 ()()equation f 1 f ( x ) sin 1 sin ( x ) x . Sinceequals(2 3.3 2 3 csc θ ,3 πθ 3 3 3. cos 1 2 Find the angle θ , 0 θ π, whose cosine4. cot 1 35π.65. sec 1 ( 1) 1 πThus, sin 1 . 2 6tan θ 1,)5π6(ππ θ 22Find the angle θ , ππ θ 22Thus, cot 1 3 equals π π is in the interval , , we can apply8 2 2 the equation directly and get π πsin 1 sin .88 π()8. cos cos 1 ( 0.3) follows the form of the)()()equation f f 1 ( x ) cos cos 1 ( x ) x .Find the angle θ , 0 θ π , whose cotangentequals 3 .1Copyright 2013 Pearson Education, Inc.
Chapter 7: Analytic Trigonometry514. cos csc 1 3 Since 0.3 is in the interval 1,1 , we canapply the equation directly and get(5ππSince cscθ , θ , let r 5 and y 3 .322Solve for x: x 2 9 25)cos cos 1 ( 0.3) 0.3 .9. Since there is no angle θ such that sin θ 1.6 ,x 2 16 1the quantity sin 1.6 is not defined. Thus,(x 4Since θ is in quadrant I, x 4 .5x 4Thus, cos csc 1 cosθ .3r5 )sin sin 1 1.6 is not defined.7π π 1 110. tan tan tan ( 1) 44 3 15. tan cos 1 5 3Since cos θ , 0 θ π , let x 3 and5 7π 3 5π 11. cos 1 cos cos 1 6 2 6r 5 . Solve for y: 9 y 2 25y 2 16y 4Since θ is in quadrant II, y 4 . 1 12. tan cos 1 2 Find the angle θ , 0 θ π, whose cosine1equals .21cos θ , 0 θ π22πθ 3 1 2πSo, cos 1 2 3 1 2π Thus, tan cos 1 tan 3 .2 3 y 44 3 Thus, tan cos 1 tan θ 533 x 16.y tan ( 2 x 3) 1x tan ( 2 y 3) 1x 1 tan ( 2 y 3)2 y 3 tan 1 ( x 1)2 y tan 1 ( x 1) 3 3 13. csc sin 1 2 Find the angle θ , f ( x ) tan ( 2 x 3) 113tan 1 ( x 1) f 1 ( x )22The domain of f ( x ) is all real numbers suchy ππ θ , whose sine equals223.2that x ( 2k 1) π 3where k is an integer. To42find the domain of f 1 ( x ) we note that the3ππ, θ 222πθ 33 π .So, sin 123 1 3 π 2 3Thus, csc sin. csc 23 3 sin θ argument of the inverse tangent function can beany real number. Thus, the domain of f 1 ( x ) isall real numbers, or ( , ) in interval notation.Recall that the domain of a function is the rangeof its inverse and the domain of the inverse is therange of the function. Therefore, the range off ( x ) is ( , ) and the range of f 1 ( x ) is all2Copyright 2013 Pearson Education, Inc.
Chapter 7 Mixed Review Worksheetsreal numbers such that y ( 2k 1) π 34219. Let θ csc 1 u so that csc θ u , where(f ( x ) 2sin ( x 1))sin csc 1 u sin θ y 2sin ( x 1)x 2sin ( y 1)x sin ( y 1)2 x y 1 sin 1 2 x y sin 1 1 2 x y 1 sin 1 2 The domain of f ( x ) is the set of all real(23.sin θ1 cos θ sin 2 θ (1 cos θ )2 1 cos θsin θsin θ (1 cos θ )sin 2 θ 1 2 cos θ cos 2 θ sin θ (1 cos θ )1 2cos θ 1sin θ (1 cos θ )2 2 cos θ sin θ (1 cos θ )2(1 cos θ ) sin θ (1 cos θ )2 sin θ 2 csc θx 12 2 x 2The domain of f 1 ( x ) is { x 2 x 2} , or 1 2, 2 in interval notation.Recall that the domain of a function is the rangeof its inverse and the domain of the inverse is therange of the function. Therefore, the range off ( x ) is 2, 2 and the range of f 1 ( x ) is24. 1 ( , ) .π2() sin 2 θ1 cosθ sin 2 θ 1 cosθ1 cosθ1 cosθ 1 cos 2 θ 1 cosθcosθ cos 2 θ 1 cosθcosθ (1 cosθ ) 1 cosθ cosθ(,u 1 . Then,cos csc 1 u cos θ cos θ ) 4 4 cos 2 θ 2 cos 2 θ 4 2 cos 2 θxand2that it must lie in the interval 1,1 . That is, θ 1 sin 2 θsin θ22. 4 sin 2 θ 2 cos 2 θ 4 1 cos 2 θ 2 cos 2 θargument of the inverse sine function is2221. (1 sin 2 θ )(1 tan 2 θ ) cos 2 θ sec 2 θ1 cos 2 θ cos 2 θ 1find the domain of f 1 ( x ) we note that the18. Let θ csc u so that csc θ u , π 1 sin 2 θ cos 2 θnumbers, or ( , ) in interval notation. Toπ θ 11 csc θ u20. sin θ csc θ sin 2 θ sin θ 12and θ 0 , u 1 . Then,k is an integer.17.πsin θ cot θ sin θsin θcot θcot 2 θcsc2 θ 1 csc θcsc θcsc θu2 1u3Copyright 2013 Pearson Education, Inc.)
Chapter 7: Analytic Trigonometry1cos θ cos θ1cos θcos θ1 cos θ 11 cos θ 1 cos θ 11 cos θ1 cos 2 θ 1 cos θsin 2 θ 1 cos θ1 sec θ 25.sec θ1 29.30.1csc θθ 1 cos θsin 26.1 cos θ 1 cos θ 1 cos θ1 1 cos θ sin θ 1 cos 2 θ1 cos θ sin θ sin 2 θ1 cos θ sin 3 θ( 2sinθ 1)33.())([ cos(2θ )]cos ( 2θ ))34.)() 1 2 cos θ 1 1 2 cos θ 2tan ( 3θ )tan ( 3θ )tan θ tan ( 3θ )tan ( 3θ )tan θtan θcos(2θ ) cos(10θ ) 2θ 10θ 2θ 10θ 2 sin sin 22 2sin(6θ )sin( 4θ ) 2sin(6θ )sin(4θ ) 2 sin ( 2 2θ ) sin ( 4θ )sin ( 2θ ) sin ( 4θ )tan θ2θ 4θ2 cos θ 4θ 2 sin 2 2 tan ( 3θ ) 2θ 4θ tan θ 2θ 4θ 2sin cos 22 2sin ( 3θ ) cos ( θ ) tan ( 3θ ) 2sin ( θ ) cos ( 3θ )tan θ 0 1 cos 2 θ sin 2 θ 1 cos ( 2θ ) cos ( 2θ ))sin ( 2θ ) sin ( 4θ ) 22 1 cos θ 1 cos θsin θ tan ( 3θ ) cot ( θ ) 2 2 sin θ (((θ32. 2 sin ( 2θ ) 1 2 sin 2 θ 2sin ( 2θ ) cos ( 2θ ) 1 2sin 2 θ 28. sin 4 θ cos 4 θsin 2 θ cos 2 θ sin 2 θ cos 2 θ2cos(α β ) cos α cos β sin α sin β sin α cos βsin α cos βcos α cos β sin α sin β sin α cos β sin α cos βcos α sin β sin α cos β cot α tan β31. sin θ tan1 cos θ1 cos θ27. sin θ1 cos θ 1 cos θsin θcsc θ cot θ csc θ cot θ csc θ cot θ csc θ cot θ(csc θ cot θ ) 2 csc2 θ cot 2 θ(csc θ cot θ ) 2 1 (csc θ cot θ )22sin(α β ) sin α cos β cos α sin β sin α cos βsin α cos βsin α cos β cos α sin β sin α cos β sin α cos β 1 cot α tan βsin ( 4θ )cos ( 4θ )sin ( 4θ )cos ( 4θ ) 2 sin ( 6θ ) cos ( 4θ )1 2 sin ( 6θ 4θ ) sin ( 6θ 4θ ) 2 tan ( 4θ ) sin (10θ ) sin ( 2θ ) tan ( 4θ ) sin ( 2θ ) sin (10θ ) 4Copyright 2013 Pearson Education, Inc.
Chapter 7 Mixed Review Worksheets35. sin17π 15π 2π sin 12 12 12 5π5πππ sin cos cos sin46462 3 2 1 2 2 2 216 2 4(36. cos)7π 4π 3π cos 12 12 12 ππππ cos cos sin sin34341 23 2 2 22 21 2 64(a.sin(α β ) sin α cos β cos α sin β 3 5 4 12 5 13 5 13 15 48 6533 65b.cos(α β ) cos α cos β sin α sin β 4 5 3 12 5 13 5 13 20 36 6556 65c.sin(α β ) sin α cos β cos α sin β3 5 4 12 5 13 5 1315 48 6563 65)37. tan 285º tan (135º 150º )tan135º tan150º 1 tan135º tan150º 3 1 3 3 1 1 3 d.tan(α β ) 3 12 4 5 3 12 1 4 5 33 20 20 56 202033 563 3 33 3 3 31 3 1 38. costan α tan β1 tan α tan βππ13π13π 13π π cos sin sin cos 248248 24 8 2π cos31 2e. 3 4 24sin(2α ) 2sin α cos α 2 5 5 25f.cos(2β ) cos 2 β sin 2 β2225 144119 5 12 169 13 13 169 1694π5π39. cos α , 0 α ; cos β , β 05213 2331212sin α , tan α , sin β , tan β ,54135α ππ β0 , 02 44 25Copyright 2013 Pearson Education, Inc.
Chapter 7: Analytic Trigonometryg.h.sincosβ1 cos β2251 13 28422 1313 2131313α2 d.tan(α β ) 4 12 3 5 4 12 1 3 5 56 56 15 5615 33 15 33 33 151 cos α2491 5 5 9 3 3 10 22101010 e.24 4 3 sin(2α ) 2 sin α cos α 2 25 5 5 2f.4 π5 π40. sin α , α 0; cos β , β π5 213 2341212cos α , tan α , sin β , tan β ,53135π απ β π 0, 4 24 2 2a. sin(α β ) sin α cos β cos α sin β 5 12 cos(2β ) cos 2 β sin 2 β 13 13 25 144 169 169119 169βg.sinh.cos2 4 5 3 12 5 13 5 1320 36 65 6556 65b.c.tan α tan β1 tan α tan βcos(α β ) cos α cos β sin α sin β 3 5 4 12 5 13 5 13 15 48 65 6533 65sin(α β ) sin α cos β cos α sin β 4 5 3 12 5 13 5 13 20 36 65 6516 65α221 cos β2 5 1 13 218933 13 13 2131313 1 cos α231 5 28422 5 5 2555 4 π123π41. tan α , α π; cot β , π β 3 25243512sin α , cos α , sin β , cos β ,551313π α π π β 3π , 4 2 2 2 246Copyright 2013 Pearson Education, Inc.
Chapter 7 Mixed Review Worksheetsa.sin(α β ) sin α cos β cos α sin β 4 12 3 5 5 13 5 13 48 15 6533 65b.cos(α β ) cos α cos β sin α sin β 3 12 4 5 5 13 5 13 36 20 6556 65c.d.sin(α β ) sin α cos β cos α sin β 4 12 3 5 5 13 5 13 48 15 6563 65f.cos(2β ) cos 2 β sin 2 β2 12 5 13 13 144 25 169 169119 169β1 cos β22 12 1 13 2252555 26 13 2262626h.cosα 1 cos α22 3 1 5 2 25 1 1 52555ππ α π; sec β 3, β π22133sin α , cos α , tan α ,2232 21sin β , cos β , tan β 2 2,33π α π π β π , 4 2 2 4 2 2a. sin(α β ) sin α cos β cos α sin β3 2 2 1 1 2 3 2 3 1 2 6 6tan α tan β1 tan α tan β4 5 3 12 4 5 1 3 12 11 12563611 36 12 5633 56e.sin42. csc α 2,tan(α β ) 24 4 3 sin(2α ) 2sin α cos α 2 25 5 5 g.2b.cos(α β ) cos α cos β sin α sin β 3 1 1 2 2 2 3 2 3 3 2 2 6c.sin(α β ) sin α cos β cos α sin β1 1 3 2 2 2 3 2 3 1 2 6 67Copyright 2013 Pearson Education, Inc.
Chapter 7: Analytic Trigonometryd.tan α tan β1 tan α tan β3 2 23 3 1 2 2 3 ππ α π; cot β 2, β π221211tan β , sin α , cos α ,sin β ,2555π α π π β π2 , cos β ,5 4 2 2 4 2 2a. sin(α β ) sin α cos β cos α sin β 2 2 1 1 5 5 5 5 4 1 5 55 5 1tan(α β ) (43. tan α 2,)() 3 6 23 3 2 63 3 6 2 3 2 6 3 2 6 3 2 6 27 3 24 2 159 3 8 2 5e.sin(2α ) 2 sin α cos α3 3 1 2 222 f.cos(2β ) cos 2 β sin 2 β2 1 2 2 3 3 1 87 9 99g.h.sincosβ1 cos β22 1 1 3 242 3 23α23 cos(α β ) cos α cos β sin α sin β 1 2 2 1 5 5 5 5 2 2 5 5 0c.sin(α β ) sin α cos β cos α sin β 2 2 1 1 5 5 5 5 4 1 5 53 5d.tan(α β ) e. 2 1 4sin(2α ) 2sin α cos α 2 555 f.cos(2β ) cos 2 β sin 2 β2 2b.33 631 cos α2 3 1 2 2 2 32 2tan α tan β1 tan α tan β 1 2 2 1 1 ( 2) 2 5 2 ; undefined022 3 424 1 3 2 1 55 55 5 2 328Copyright 2013 Pearson Education, Inc.
Chapter 7 Mixed Review Worksheetsg. h.54 sin cos 1 sin 1 sin (α β )135 sin α cos β cos α sin β3 5 4 2 5 3 5 3 3 5 8 8 3 5 15 1515 2 1 β1 cos β5 sin 2225 2525 212 45. cos tan 1 3 sin 1 13 2 5 5 2 510 10 5 2 510 12 Let α tan 1 ( 3) and β sin 1 . α is in 13 quadrant I; β is in quadrant I. Thentan α 3, 0 α 1 1 α1 cos α5 cos 222 0 β 5 152.sin α 1 cos 2 α2 521 1 1 1 24 5 51033 42cos β 1 sin 2 β10 5 5 102144255 12 1 1 1316916913 4 2 44. sin cos 1 sin 1 5 3 4 2 Let α cos 1 and β sin 1 . α is in5 3 0 α 2sec α 1 tan 2 α 1 ( 3) 2 4 21cos α 25 1quadrant I; β is in quadrant IV. Then cos α ππ12, and sin β ,213 4 cos tan 1 ( 1) sin 1 cos (α β ) 5 cos α cos β sin α sin β 1 5 3 12 2 13 2 13 5 12 3 26265 12 3 264,5ππ 2 , and sin β , β 0 .22 3 sin α 1 cos 2 α21693 4 1 1 525255 5 46. sin 2sin 1 13 5 Let α sin 1 . α is in quadrant IV. 13 cos β 1 sin 2 β2455 2 1 1 3993 9Copyright 2013 Pearson Education, Inc.
Chapter 7: Analytic TrigonometryThen sin α 5π, α 0.13249. 2 cos θ 2 02 cos θ 2cosα 1 sin 2 αcos θ 225144 12 5 1 1 169169 13 13 3π5π 2kπ or θ 2kπ , k is any integer44 3π 5π On 0 θ 2π , the solution set is , .4 4θ 3 sin 2cos 1 sin 2α 5 2sin α cosα120 5 12 2 169 13 13 50. cos ( 2θ ) 0π3π 2k π or 2θ 2k π22π3πθ kπθ k π,44where k is any integerOn the interval 0 θ 2π , the solution set is π 3π 5π 7 π , , , . 4 4 4 4 2θ 1 1 47. cos tan 1 2 2 1 Let α tan 1 . α is in quadrant IV. Then 2 1πtan α , α 0 .22sec α tan 2 α 12155 1 1 1 2442 51. sin ( 3θ ) 4 3sin ( 3θ ) 122 5cos α 55 1 1 1 cos tan 1 cos α 2 2 21 cos α 2 48. sin θ 221 2 55 2π 2k π2π 2k π, where k is any integerθ 63On the interval 0 θ 2π , the solution set is π 5π 3π , . , 6 6 2 3θ 5 2 552 5 2 55 2 5 1010 10 5 2 51053. 0.9cos(2θ ) 0.7Find the intersection of Y1 0.9 cos(2θ ) andY2 0.7 :324π5π 2kπ or θ 2kπ , k is any integer33 4π 5π On 0 θ 2π , the solution set is , .3 3θ 10Copyright 2013 Pearson Education, Inc.
Chapter 7 Mixed Review WorksheetsOn the interval 0 θ 2π , x 0.34 orx 2.80 or x 3.48 or x 5.94The solution set is {0.34, 2.80,3.48,5.94} .57. 8 12 sin 2 θ 4 cos 2 θ(8 12 sin θ 4 4 sin θ2cos ( 2θ ) sin θ54.24 8sin 2 θ1 2sin θ sin θ2sin 2 θ 2sin θ sin θ 1 02(2sin θ 1)(sin θ 1) 02sin θ 1 0or sin θ 1 01sin θ 1sin θ 23πθ π 5π2θ ,6 6 π 5π 3π On 0 θ 2π , the solution set is , , . 6 6 2 55.θ 1212 22π 3π 5π 7πθ , , ,4 4 4 4 π 3π 5π 7π On 0 θ 2π , the solution set is , , , . 4 458.3,θ 3(cos θ 0π2θ 3,or 2 cos θ 3 0π 3π2,cos θ 2 2 659. sin θ 3 cos θ 2Divide each side byθ π35π πOn 0 θ 2π , the solution set is , π , . 32:13sin θ cos θ 122Rewrite using the difference of two anglesπ1formula where φ , so cos φ and23cos θ 1π 5π)325π 7πθ ,6 6 π 5π 7π 3π On 0 θ 2π , the solution set is , , , .θ or cos θ 1 01cos θ 21 3 cos θ cos ( 2θ ) 0cos θ 2 cos θ 3 02 cos 2 θ cos θ 1 0(2 cos θ 1)(cos θ 1) 02 cos θ 1 04 2 cos 2 (θ ) 3 cos θ 0 π π 5π On 0 θ 2π , the solution set is , , . 3 2 3 56.41 3 cos θ 2 cos 2 (θ ) 1 0or sin θ 1π 5π4 1 8 2sin θ sin(2θ ) sin θ 2cos θ 1 02sin θ cos θ sin θ 2cos θ 1 0sin θ (2cos θ 1) 1(2 cos θ 1) 0(2cos θ 1)(sin θ 1) 0cos θ )8 12 sin 2 θ 4 1 sin 2 θ3.2sin θ cos φ cos θ sin φ 1sin (θ φ ) 1sin φ 3 11Copyright 2013 Pearson Education, Inc.62
Chapter 7: Analytic Trigonometryπ2π πθ 3 25πθ 611. Now sin θ , so θ lies in551quadrant I. The calculator yields sin 1 0.20 ,5which is an angle in quadrant I, socsc 1 ( 5 ) 0.20 .θ φ equals 5π On 0 θ 2π , the solution set is . 6 60. sin 1 ( 0.4 ) 0.4165. 2 x 5sin xFind the intersection of Y1 2 x and Y2 5sin x :6 2 61. cos 1 0.84 3 62π 2π 2π 62π 6On the interval 0 x 2π , x 0 or x 2.13 .The solution set is {0, 2.13} .66. 3cos x x sin xFind the intersection of Y1 3cos x x andY2 sin x : 1 62. cot 1 ( 3) tan 1 3 We seek the angle θ , 0 θ π , whose tangent611equals . Now tan θ , so θ lies in33quadrant II. The calculator yields2π 2π 2π 1 tan 1 0.32 , which is an angle in 3 quadrant IV. Since θ lies in quadrant II,θ 0.32 π 2.82 . Therefore,6 62π 6On the interval 0 x 2π , x 1.89 orx 3.07 .The solution set is {1.89,3.07} .cot 1 ( 4 ) 2.90 .67. sin x e xFind the intersection of Y1 sin x and Y2 e x :6 1We seek the angle θ , 2π 2π 2π 1 64. csc ( 5 ) sin 5 1π2 θ π2 6, whose sine612Copyright 2013 Pearson Education, Inc.2π 6
Chapter 7 Mixed Review WorksheetsOn the interval 0 x 2π , x 0.59 or x 3.10 .The solution set is {0.59,3.10} .68. 2 tan 1 x 3 032 3 x tan 2 ?The solution set is {?} .tan 1 x 13Copyright 2013 Pearson Education, Inc.
Analytic Trigonometry Chapter 7 Mixed Review Worksheets 11. 1 sin 2 Find the angle ,, 22 θθ ππ whose sine equals 1 2. 1 sin , 22 2 6 θθ θ ππ π Thus, 1 1 sin 26 π . 2. 1tan 1 Find the angle ,, 22 θθ ππ whose tangent equals 1. ta
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
(Answers for Chapter 5: Analytic Trigonometry) A.5.1 CHAPTER 5: Analytic Trigonometry SECTION 5.1: FUNDAMENTAL TRIGONOMETRIC IDENTITIES 1) Left Side Right Side Type of Identity (ID) csc(x) 1 sin(x) Reciprocal ID tan(x) 1
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
CHAPTER 1 1 Angles and Applications 1.1 Introduction Trigonometry is the branch of mathematics concerned with the measurement of the parts, sides, and angles of a triangle. Plane trigonometry, which is the topic of this book, is restricted to triangles lying in a plane. Trigonometry is based on certain ratio
Trigonometry Standard 12.0 Students use trigonometry to determine unknown sides or angles in right triangles. Trigonometry Standard 5.0 Students know the definitions of the tangent and cotangent functions and can graph them. Trigonometry Standard 6.0 Students know the definitions of the secant and cosecant functions and can graph them.
Analytic Continuation of Functions. 2 We define analytic continuation as the process of continuing a function off of the real axis and into the complex plane such that the resulting function is analytic. More generally, analytic continuation extends the representation of a function
COMPLEX ANALYTIC GEOMETRY AND ANALYTIC-GEOMETRIC CATEGORIES YA’ACOV PETERZIL AND SERGEI STARCHENKO Abstract. The notion of a analytic-geometric category was introduced by v.d. Dries and Miller in [4]. It is a category of subsets of real analytic manifolds which extends the c
