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January 27, 2005 11:58     l24-appa-sv        Sheet number 1 Page number 524                            black



                                                               APPENDIX A
                                            Trigonometry Review

             EXERCISE SET A

              1. (a) 5/12                    (b) 13/6                                (c) /9                          (d) 23/30

              2. (a) 7/3                     (b) /12                                 (c) 5/4                         (d) 11/12

              3. (a) 12                      (b) (270/)                             (c) 288                         (d) 540

              4. (a) 18                      (b) (360/)                             (c) 72                          (d) 210


              5.          sin 慮     cos 慮   tan 慮       csc 慮        sec 慮 cot 慮
                                                                          
                   (a)     21/5  2/5          21/2     5/ 21          5/2 2/ 21
                                                                        
                   (b)    3/4     7/4       3/ 7        4/3          4/ 7    7/3
                                                                  
                   (c)   3/ 10 1/ 10              3      10/3          10   1/3


              6.         sin 慮     cos 慮    tan 慮     csc 慮        sec 慮       cot 慮
                                                                 
                   (a)   1/ 2      1/ 2       1          2            2            1
                   (b)   3/5       4/5       3/4      5/3        5/4           4/3
                                                                            
                   (c)   1/4        15/4    1/ 15      4        4/ 15            15

                                                                                                                
              7. sin 慮 = 3/ 10, cos 慮 = 1/ 10                                      8. sin 慮 =       5/3, tan 慮 =       5/2
                                                                                                                
              9. tan 慮 =       21/2, csc 慮 = 5/ 21                             10. cot 慮 =           15, sec 慮 = 4/ 15

             11. Let x be the length of the side adjacent to 慮, then cos 慮 = x/6 = 0.3, x = 1.8.

             12. Let x be the length of the hypotenuse, then sin 慮 = 2.4/x = 0.8, x = 2.4/0.8 = 3.

             13.           慮        sin 慮  cos 慮 tan 慮  csc 慮  sec 慮  cot 慮
                                                              
                   (a)   225      1/ 2 1/ 2     1     2     2      1
                                                                     
                   (b)   210      1/2    3/2 1/ 3     2   2/ 3    3
                                                                      
                   (c)   5/3       3/2    1/2   3   2/ 3     2   1/ 3
                   (d)   3/2        1           0                           1                       0


             14.           慮        sin 慮    cos 慮     tan 慮  csc 慮 sec 慮                       cot 慮
                                                                                               
                   (a)   330       1/2       3/2    1/ 3    2   2/ 3                         3
                                                                                               
                   (b)   120      3/2     1/2         3  2/ 3   2                         1/ 3
                                                                  
                   (c)   9/4      1/ 2      1/ 2        1       2     2                         1
                   (d)   3          0       1           0                          1        

                                                                           524
January 27, 2005 11:58         l24-appa-sv           Sheet number 2 Page number 525             black



              Exercise Set A                                                                                                525


              15.          sin 慮        cos 慮       tan 慮   csc 慮       sec 慮    cot 慮
                    (a)    4/5          3/5          4/3    5/4         5/3       3/4
                    (b)    4/5  3/5   4/3 5/4  5/3 3/4
                                                    
                    (c)     1/2  3/2 1/ 3   2  2 3  3
                                                    
                    (d)    1/2   3/2 1/ 3 2   2/ 3  3
                                               
                    (e)    1/ 2 1/ 2    1      2    2  1
                                                
                    (f )   1/ 2 1/ 2   1     2  2   1


              16.              sin 慮cos 慮  tan 慮                     csc 慮        sec 慮     cot 慮
                                                                                         
                    (a)      1/4     15/4 1/ 15                         4        4/ 15        15
                                                                                          
                    (b)      1/4    15/4 1/ 15                        4        4/ 15      15
                                                                               
                    (c)    3/ 10   1/ 10     3                        10/3          10        1/3
                                                                                
                    (d)    3/ 10 1/ 10     3                       10/3         10       1/3
                                                                                           
                    (e)       21/5  2/5   21/2                    5/ 21          5/2    2/ 21
                                                                                           
                    (f )    21/5   2/5    21/2                    5/ 21         5/2    2/ 21


              17. (a) x = 3 sin 25  1.2679                                        (b) x = 3/ tan(2/9)  3.5753

              18. (a) x = 2/ sin 20  5.8476                                       (b) x = 3/ cos(3/11)  4.5811

              19.               sin 慮               cos 慮      tan 慮               csc 慮       sec 慮             cot 慮
                                                                                                         
                    (a)       a/3       9  a2 /3            a/ 9  a2              3/a     3/ 9  a2           9  a2 /a
                                                                                         
                    (b)    a/ a2 + 25 5/ a2 + 25                  a/5             a2 + 25/a   a2 + 25/5        5/a
                                                                                                           
                    (c)      a2  1/a     1/a                     a2  1        a/ a2  1        a          1/ a2  1


              20. (a) 慮 = 3/4 賊 2n and 慮 = 5/4 賊 2n, n = 0, 1, 2, . . .
                  (b) 慮 = 5/4 賊 2n and 慮 = 7/4 賊 2n, n = 0, 1, 2, . . .

              21. (a) 慮 = 3/4 賊 n, n = 0, 1, 2, . . .
                  (b) 慮 = /3 賊 2n and 慮 = 5/3 賊 2n, n = 0, 1, 2, . . .

              22. (a) 慮 = 7/6 賊 2n and 慮 = 11/6 賊 2n, n = 0, 1, 2, . . .
                  (b) 慮 = /3 賊 n, n = 0, 1, 2, . . .

              23. (a) 慮 = /6 賊 n, n = 0, 1, 2, . . .
                  (b) 慮 = 4/3 賊 2n and 慮 = 5/3 賊 2n, n = 0, 1, 2, . . .

              24. (a) 慮 = 3/2 賊 2n, n = 0, 1, 2, . . .                            (b) 慮 =  賊 2n, n = 0, 1, 2, . . .

              25. (a) 慮 = 3/4 賊 n, n = 0, 1, 2, . . .                             (b) 慮 = /6 賊 n, n = 0, 1, 2, . . .

              26. (a) 慮 = 2/3 賊 2n and 慮 = 4/3 賊 2n, n = 0, 1, 2, . . .
                  (b) 慮 = 7/6 賊 2n and 慮 = 11/6 賊 2n, n = 0, 1, 2, . . .
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             526                                                                                                             Appendix A


             27. (a) 慮 = /3 賊 2n and 慮 = 2/3 賊 2n, n = 0, 1, 2, . . .
                 (b) 慮 = /6 賊 2n and 慮 = 11/6 賊 2n, n = 0, 1, 2, . . .

             28. sin 慮 = 3/5, cos 慮 = 4/5, tan 慮 = 3/4, csc 慮 = 5/3, sec 慮 = 5/4, cot 慮 = 4/3
                                                                                                 
             29. sin 慮 = 2/5, cos 慮 =  21/5, tan 慮 = 2/ 21, csc 慮 = 5/2, sec 慮 = 5/ 21, cot 慮 =  21/2

             30. (a) 慮 = /2 賊 2n, n = 0, 1, 2, . . .                 (b) 慮 = 賊2n, n = 0, 1, 2, . . .
                 (c) 慮 = /4 賊 n, n = 0, 1, 2, . . .                  (d) 慮 = /2 賊 2n, n = 0, 1, 2, . . .
                 (e) 慮 = 賊2n, n = 0, 1, 2, . . .                      (f ) 慮 = /4 賊 n, n = 0, 1, 2, . . .

             31. (a) 慮 = 賊n, n = 0, 1, 2, . . .                       (b) 慮 = /2 賊 n, n = 0, 1, 2, . . .
                 (c) 慮 = 賊n, n = 0, 1, 2, . . .                       (d) 慮 = 賊n, n = 0, 1, 2, . . .
                 (e) 慮 = /2 賊 n, n = 0, 1, 2, . . .                  (f ) 慮 = 賊n, n = 0, 1, 2, . . .

             32. Construct a right triangle with one angle equal to 17 , measure the lengths of the sides and
                 hypotenuse and use formula (6) for sin 慮 and cos 慮 to approximate sin 17 and cos 17 .

             33. (a) s = r慮 = 4(/6) = 2/3 cm                         (b) s = r慮 = 4(5/6) = 10/3 cm

             34. r = s/慮 = 7/(/3) = 21/                         35. 慮 = s/r = 2/5

                                  1 2    1          1
             36. 慮 = s/r so A =     r 慮 = r2 (s/r) = rs
                                  2      2          2
                                               2  慮
             37. (a) 2r = R(2  慮), r =             R
                                                 2                        
                                                                               4慮  慮2
                   (b) h =    R2  r 2 =    R2  (2  慮)2 R2 /(4 2 ) =                R
                                                                                 2
             38. The circumference of the circular base is 2r. When cut and 鍖attened, the cone becomes a circular
                 sector of radius L. If 慮 is the central angle that subtends the arc of length 2r, then 慮 = (2r)/L
                 so the area S of the sector is S = (1/2)L2 (2r/L) = rL which is the lateral surface area of the
                 cone.

             39. Let h be the altitude as shown in the 鍖gure, then
                                              1                
                 h = 3 sin 60 = 3 3/2 so A = (3 3/2)(7) = 21 3/4.                      3             h
                                               2
                                                                                            60属
                                                                                                           7

             40. Draw the perpendicular from vertex C as shown in the 鍖gure,                                             C
                 then                                   
                 h = 9 sin 30 = 9/2, a = h/ sin 45 = 9 2/2,
                                                                                                           9        h              a
                 c1 = 9 cos 30 = 9 3/2, c2 = a cos 45 = 9/2,
                              
                                                                                                                              45属
                                                                                                          30属
                 c1 + c2 = 9( 3 + 1)/2, angle C = 180  (30 + 45 ) = 105                      A
                                                                                                                c1             c2
                                                                                                                                        B




             41. Let x be the distance above the ground, then x = 10 sin 67  9.2 ft.

             42. Let x be the height of the building, then x = 120 tan 76  481 ft.
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              Exercise Set A                                                                                                 527


              43. From the 鍖gure, h = x  y but x = d tan 硫,
                  y = d tan 留 so h = d(tan 硫  tan 留).


                                                                                                  h

                                                                                              x

                                                                                                  y
                                                                                                            硫 留

                                                                                                            d




              44. From the 鍖gure, d = x  y but x = h cot 留,
                  y = h cot 硫 so d = h(cot 留  cot 硫),
                             d                                                                                         h
                  h=                .
                      cot 留  cot 硫                                                                     硫
                                                                                                  留
                                                                                                  d             y
                                                                                                        x

                                                              
              45. (a) sin 2慮 = 2 sin 慮 cos 慮 = 2( 5/3)(2/3) = 4 5/9
                  (b) cos 2慮 = 2 cos2 慮  1 = 2(2/3)2  1 = 1/9
                                                                                            
              46. (a) sin(留  硫) = sin 留 cos 硫  cos 留 sin 硫 = (3/5)(1/ 5)  (4/5)(2/ 5) = 1/ 5
                                                                                              
                  (b) cos(留 + 硫) = cos 留 cos 硫  sin 留 sin 硫 = (4/5)(1/ 5)  (3/5)(2/ 5) = 2/(5 5)

              47. sin 3慮 = sin(2慮 + 慮) = sin 2慮 cos 慮 + cos 2慮 sin 慮 = (2 sin 慮 cos 慮) cos 慮 + (cos2 慮  sin2 慮) sin 慮
                  = 2 sin 慮 cos2 慮 + sin 慮 cos2 慮  sin3 慮 = 3 sin 慮 cos2 慮  sin3 慮; similarly, cos 3慮 = cos3 慮  3 sin2 慮 cos 慮

                    cos 慮 sec 慮   cos 慮 sec 慮   cos 慮     cos 慮
              48.               =             =       =            = cos2 慮
                    1 + tan2 慮      sec2 慮      sec 慮   (1/ cos 慮)

                    cos 慮 tan 慮 + sin 慮   cos 慮(sin 慮/ cos 慮) + sin 慮
              49.                       =                             = 2 cos 慮
                           tan 慮                  sin 慮/ cos 慮

                                   2            2             1        1
              50. 2 csc 2慮 =            =               =                    = csc 慮 sec 慮
                                 sin 2慮   2 sin 慮 cos 慮     sin 慮    cos 慮

                                      sin 慮   cos 慮   sin2 慮 + cos2 慮        1              2           2
              51. tan 慮 + cot 慮 =           +       =                 =             =               =        = 2 csc 2慮
                                      cos 慮   sin 慮      sin 慮 cos 慮    sin 慮 cos 慮   2 sin 慮 cos 慮   sin 2慮
                    sin 2慮 cos 2慮    sin 2慮 cos 慮  cos 2慮 sin 慮      sin 慮
              52.                 =                             =             = sec 慮
                     sin 慮   cos 慮           sin 慮 cos 慮           sin 慮 cos 慮

                    sin 慮 + cos 2慮  1   sin 慮 + (1  2 sin2 慮)  1   sin 慮(1  2 sin 慮)
              53.                      =                            =                    = tan 慮
                      cos 慮  sin 2慮        cos 慮  2 sin 慮 cos 慮     cos 慮(1  2 sin 慮)

              54. Using (47), 2 sin 2慮 cos 慮 = 2(1/2)(sin 慮 + sin 3慮) = sin 慮 + sin 3慮

              55. Using (47), 2 cos 2慮 sin 慮 = 2(1/2)[sin(慮) + sin 3慮] = sin 3慮  sin 慮

                                   sin(慮/2)        2 sin2 (慮/2)     1  cos 慮
              56. tan(慮/2) =                =                     =
                                   cos(慮/2)   2 sin(慮/2) cos(慮/2)     sin 慮
January 27, 2005 11:58     l24-appa-sv      Sheet number 5 Page number 528           black



             528                                                                                                      Appendix A


                               sin(慮/2)   2 sin(慮/2) cos(慮/2)     sin 慮
             57. tan(慮/2) =             =           2 (慮/2)
                                                              =
                               cos(慮/2)       2 cos             1 + cos 慮


             58. From (52), cos(/3 + 慮) + cos(/3  慮) = 2 cos(/3) cos 慮 = 2(1/2) cos 慮 = cos 慮

                                                                                                     C
                                          1
             59. From the 鍖gures, area = hc but h = b sin A
                                          2
                            1                                                                                     a
                 so area = bc sin A. The formulas                                                b       h
                            2
                         1                     1
                 area = ac sin B and area = ab sin C
                         2                     2                                            A
                                                                                                             c
                                                                                                                       B
                 follow by drawing altitudes from vertices B and C, respectively.

             60. From right triangles ADC and BDC,                                                   C
                 h1 = b sin A = a sin B so a/ sin A = b/ sin B.
                 From right triangles AEB and CEB,
                                                                                                     h1           a
                 h2 = c sin A = a sin C so a/ sin A = c/ sin C                               b
                 thus a/ sin A = b/ sin B = c/ sin C.                                            E           h2

                                                                                                     D
                                                                                        A                                  B
                                                                                                             c



             61. (a) sin(/2 + 慮) = sin(/2) cos 慮 + cos(/2) sin 慮 = (1) cos 慮 + (0) sin 慮 = cos 慮
                   (b) cos(/2 + 慮) = cos(/2) cos 慮  sin(/2) sin 慮 = (0) cos 慮  (1) sin 慮 =  sin 慮
                   (c) sin(3/2  慮) = sin(3/2) cos 慮  cos(3/2) sin 慮 = (1) cos 慮  (0) sin 慮 =  cos 慮
                   (d) cos(3/2 + 慮) = cos(3/2) cos 慮  sin(3/2) sin 慮 = (0) cos 慮  (1) sin 慮 = sin 慮


                                  sin(留 + 硫)   sin 留 cos 硫 + cos 留 sin 硫
             62. tan(留 + 硫) =                =                           , divide numerator and denominator by
                                  cos(留 + 硫)   cos 留 cos 硫  sin 留 sin 硫
                                               sin 留                sin 硫
                   cos 留 cos 硫 and use tan 留 =        and tan 硫 =          to get (38);
                                               cos 留               cos 硫
                                                    tan 留 + tan(硫)         tan 留  tan 硫
                   tan(留  硫) = tan(留 + (硫)) =                         =                  because
                                                   1  tan 留 tan(硫)       1 + tan 留 tan 硫
                   tan(硫) =  tan 硫.


             63. (a) Add (34) and (36) to get sin(留  硫) + sin(留 + 硫) = 2 sin 留 cos 硫 so
                     sin 留 cos 硫 = (1/2)[sin(留  硫) + sin(留 + 硫)].
                   (b) Subtract (35) from (37).                         (c) Add (35) and (37).


                                      A+B         AB      1
             64. (a) From (47), sin           cos       = (sin B + sin A) so
                                         2         2       2
                                             A+B       AB
                       sin A + sin B = 2 sin       cos       .
                                               2         2
                   (b) Use (49)                                    (c) Use (48)


                                          留硫       留+硫
             65. sin 留 + sin(硫) = 2 sin        cos       , but sin(硫) =  sin 硫 so
                                            2         2
                                         留+硫      留硫
                   sin 留  sin 硫 = 2 cos      sin       .
                                          2         2
January 27, 2005 11:58         l24-appa-sv   Sheet number 6 Page number 529         black



              Exercise Set A                                                                                        529


              66. (a) From (34), C sin(留 + ) = C sin 留 cos  + C cos 留 sin  so C cos  = 3 and C sin  = 5, 
                                                                                            
                      square and add to get C 2 (cos2  + sin2 ) = 9 + 25, C 2 = 34. If C = 34 then cos  = 3/ 34
                                      
                      and sin  = 5/ 34  is the 鍖rst-quadrant angle for which tan  = 5/3.
                                          so
                      3 sin 留 + 5 cos 留 = 34 sin(留 + ).
                                                                                                         
                  (b) Follow the procedure of part (a) to get C cos  = A and C sin  = B, C = A2 + B 2 ,
                      tan  = B/A where the quadrant in which  lies is determined by the signs of A and B because
                                                                                
                      cos  = A/C and sin  = B/C, so A sin 留 + B cos 留 = A2 + B 2 sin(留 + ).

              67. Consider the triangle having a, b, and d as sides. The angle formed by sides a and b is   慮 so
                                                                                                   
                  from the law of cosines, d2 = a2 + b2  2ab cos(  慮) = a2 + b2 + 2ab cos 慮, d = a2 + b2 + 2ab cos 慮.

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Appendix a page_524

  • 1. January 27, 2005 11:58 l24-appa-sv Sheet number 1 Page number 524 black APPENDIX A Trigonometry Review EXERCISE SET A 1. (a) 5/12 (b) 13/6 (c) /9 (d) 23/30 2. (a) 7/3 (b) /12 (c) 5/4 (d) 11/12 3. (a) 12 (b) (270/) (c) 288 (d) 540 4. (a) 18 (b) (360/) (c) 72 (d) 210 5. sin 慮 cos 慮 tan 慮 csc 慮 sec 慮 cot 慮 (a) 21/5 2/5 21/2 5/ 21 5/2 2/ 21 (b) 3/4 7/4 3/ 7 4/3 4/ 7 7/3 (c) 3/ 10 1/ 10 3 10/3 10 1/3 6. sin 慮 cos 慮 tan 慮 csc 慮 sec 慮 cot 慮 (a) 1/ 2 1/ 2 1 2 2 1 (b) 3/5 4/5 3/4 5/3 5/4 4/3 (c) 1/4 15/4 1/ 15 4 4/ 15 15 7. sin 慮 = 3/ 10, cos 慮 = 1/ 10 8. sin 慮 = 5/3, tan 慮 = 5/2 9. tan 慮 = 21/2, csc 慮 = 5/ 21 10. cot 慮 = 15, sec 慮 = 4/ 15 11. Let x be the length of the side adjacent to 慮, then cos 慮 = x/6 = 0.3, x = 1.8. 12. Let x be the length of the hypotenuse, then sin 慮 = 2.4/x = 0.8, x = 2.4/0.8 = 3. 13. 慮 sin 慮 cos 慮 tan 慮 csc 慮 sec 慮 cot 慮 (a) 225 1/ 2 1/ 2 1 2 2 1 (b) 210 1/2 3/2 1/ 3 2 2/ 3 3 (c) 5/3 3/2 1/2 3 2/ 3 2 1/ 3 (d) 3/2 1 0 1 0 14. 慮 sin 慮 cos 慮 tan 慮 csc 慮 sec 慮 cot 慮 (a) 330 1/2 3/2 1/ 3 2 2/ 3 3 (b) 120 3/2 1/2 3 2/ 3 2 1/ 3 (c) 9/4 1/ 2 1/ 2 1 2 2 1 (d) 3 0 1 0 1 524
  • 2. January 27, 2005 11:58 l24-appa-sv Sheet number 2 Page number 525 black Exercise Set A 525 15. sin 慮 cos 慮 tan 慮 csc 慮 sec 慮 cot 慮 (a) 4/5 3/5 4/3 5/4 5/3 3/4 (b) 4/5 3/5 4/3 5/4 5/3 3/4 (c) 1/2 3/2 1/ 3 2 2 3 3 (d) 1/2 3/2 1/ 3 2 2/ 3 3 (e) 1/ 2 1/ 2 1 2 2 1 (f ) 1/ 2 1/ 2 1 2 2 1 16. sin 慮cos 慮 tan 慮 csc 慮 sec 慮 cot 慮 (a) 1/4 15/4 1/ 15 4 4/ 15 15 (b) 1/4 15/4 1/ 15 4 4/ 15 15 (c) 3/ 10 1/ 10 3 10/3 10 1/3 (d) 3/ 10 1/ 10 3 10/3 10 1/3 (e) 21/5 2/5 21/2 5/ 21 5/2 2/ 21 (f ) 21/5 2/5 21/2 5/ 21 5/2 2/ 21 17. (a) x = 3 sin 25 1.2679 (b) x = 3/ tan(2/9) 3.5753 18. (a) x = 2/ sin 20 5.8476 (b) x = 3/ cos(3/11) 4.5811 19. sin 慮 cos 慮 tan 慮 csc 慮 sec 慮 cot 慮 (a) a/3 9 a2 /3 a/ 9 a2 3/a 3/ 9 a2 9 a2 /a (b) a/ a2 + 25 5/ a2 + 25 a/5 a2 + 25/a a2 + 25/5 5/a (c) a2 1/a 1/a a2 1 a/ a2 1 a 1/ a2 1 20. (a) 慮 = 3/4 賊 2n and 慮 = 5/4 賊 2n, n = 0, 1, 2, . . . (b) 慮 = 5/4 賊 2n and 慮 = 7/4 賊 2n, n = 0, 1, 2, . . . 21. (a) 慮 = 3/4 賊 n, n = 0, 1, 2, . . . (b) 慮 = /3 賊 2n and 慮 = 5/3 賊 2n, n = 0, 1, 2, . . . 22. (a) 慮 = 7/6 賊 2n and 慮 = 11/6 賊 2n, n = 0, 1, 2, . . . (b) 慮 = /3 賊 n, n = 0, 1, 2, . . . 23. (a) 慮 = /6 賊 n, n = 0, 1, 2, . . . (b) 慮 = 4/3 賊 2n and 慮 = 5/3 賊 2n, n = 0, 1, 2, . . . 24. (a) 慮 = 3/2 賊 2n, n = 0, 1, 2, . . . (b) 慮 = 賊 2n, n = 0, 1, 2, . . . 25. (a) 慮 = 3/4 賊 n, n = 0, 1, 2, . . . (b) 慮 = /6 賊 n, n = 0, 1, 2, . . . 26. (a) 慮 = 2/3 賊 2n and 慮 = 4/3 賊 2n, n = 0, 1, 2, . . . (b) 慮 = 7/6 賊 2n and 慮 = 11/6 賊 2n, n = 0, 1, 2, . . .
  • 3. January 27, 2005 11:58 l24-appa-sv Sheet number 3 Page number 526 black 526 Appendix A 27. (a) 慮 = /3 賊 2n and 慮 = 2/3 賊 2n, n = 0, 1, 2, . . . (b) 慮 = /6 賊 2n and 慮 = 11/6 賊 2n, n = 0, 1, 2, . . . 28. sin 慮 = 3/5, cos 慮 = 4/5, tan 慮 = 3/4, csc 慮 = 5/3, sec 慮 = 5/4, cot 慮 = 4/3 29. sin 慮 = 2/5, cos 慮 = 21/5, tan 慮 = 2/ 21, csc 慮 = 5/2, sec 慮 = 5/ 21, cot 慮 = 21/2 30. (a) 慮 = /2 賊 2n, n = 0, 1, 2, . . . (b) 慮 = 賊2n, n = 0, 1, 2, . . . (c) 慮 = /4 賊 n, n = 0, 1, 2, . . . (d) 慮 = /2 賊 2n, n = 0, 1, 2, . . . (e) 慮 = 賊2n, n = 0, 1, 2, . . . (f ) 慮 = /4 賊 n, n = 0, 1, 2, . . . 31. (a) 慮 = 賊n, n = 0, 1, 2, . . . (b) 慮 = /2 賊 n, n = 0, 1, 2, . . . (c) 慮 = 賊n, n = 0, 1, 2, . . . (d) 慮 = 賊n, n = 0, 1, 2, . . . (e) 慮 = /2 賊 n, n = 0, 1, 2, . . . (f ) 慮 = 賊n, n = 0, 1, 2, . . . 32. Construct a right triangle with one angle equal to 17 , measure the lengths of the sides and hypotenuse and use formula (6) for sin 慮 and cos 慮 to approximate sin 17 and cos 17 . 33. (a) s = r慮 = 4(/6) = 2/3 cm (b) s = r慮 = 4(5/6) = 10/3 cm 34. r = s/慮 = 7/(/3) = 21/ 35. 慮 = s/r = 2/5 1 2 1 1 36. 慮 = s/r so A = r 慮 = r2 (s/r) = rs 2 2 2 2 慮 37. (a) 2r = R(2 慮), r = R 2 4慮 慮2 (b) h = R2 r 2 = R2 (2 慮)2 R2 /(4 2 ) = R 2 38. The circumference of the circular base is 2r. When cut and 鍖attened, the cone becomes a circular sector of radius L. If 慮 is the central angle that subtends the arc of length 2r, then 慮 = (2r)/L so the area S of the sector is S = (1/2)L2 (2r/L) = rL which is the lateral surface area of the cone. 39. Let h be the altitude as shown in the 鍖gure, then 1 h = 3 sin 60 = 3 3/2 so A = (3 3/2)(7) = 21 3/4. 3 h 2 60属 7 40. Draw the perpendicular from vertex C as shown in the 鍖gure, C then h = 9 sin 30 = 9/2, a = h/ sin 45 = 9 2/2, 9 h a c1 = 9 cos 30 = 9 3/2, c2 = a cos 45 = 9/2, 45属 30属 c1 + c2 = 9( 3 + 1)/2, angle C = 180 (30 + 45 ) = 105 A c1 c2 B 41. Let x be the distance above the ground, then x = 10 sin 67 9.2 ft. 42. Let x be the height of the building, then x = 120 tan 76 481 ft.
  • 4. January 27, 2005 11:58 l24-appa-sv Sheet number 4 Page number 527 black Exercise Set A 527 43. From the 鍖gure, h = x y but x = d tan 硫, y = d tan 留 so h = d(tan 硫 tan 留). h x y 硫 留 d 44. From the 鍖gure, d = x y but x = h cot 留, y = h cot 硫 so d = h(cot 留 cot 硫), d h h= . cot 留 cot 硫 硫 留 d y x 45. (a) sin 2慮 = 2 sin 慮 cos 慮 = 2( 5/3)(2/3) = 4 5/9 (b) cos 2慮 = 2 cos2 慮 1 = 2(2/3)2 1 = 1/9 46. (a) sin(留 硫) = sin 留 cos 硫 cos 留 sin 硫 = (3/5)(1/ 5) (4/5)(2/ 5) = 1/ 5 (b) cos(留 + 硫) = cos 留 cos 硫 sin 留 sin 硫 = (4/5)(1/ 5) (3/5)(2/ 5) = 2/(5 5) 47. sin 3慮 = sin(2慮 + 慮) = sin 2慮 cos 慮 + cos 2慮 sin 慮 = (2 sin 慮 cos 慮) cos 慮 + (cos2 慮 sin2 慮) sin 慮 = 2 sin 慮 cos2 慮 + sin 慮 cos2 慮 sin3 慮 = 3 sin 慮 cos2 慮 sin3 慮; similarly, cos 3慮 = cos3 慮 3 sin2 慮 cos 慮 cos 慮 sec 慮 cos 慮 sec 慮 cos 慮 cos 慮 48. = = = = cos2 慮 1 + tan2 慮 sec2 慮 sec 慮 (1/ cos 慮) cos 慮 tan 慮 + sin 慮 cos 慮(sin 慮/ cos 慮) + sin 慮 49. = = 2 cos 慮 tan 慮 sin 慮/ cos 慮 2 2 1 1 50. 2 csc 2慮 = = = = csc 慮 sec 慮 sin 2慮 2 sin 慮 cos 慮 sin 慮 cos 慮 sin 慮 cos 慮 sin2 慮 + cos2 慮 1 2 2 51. tan 慮 + cot 慮 = + = = = = = 2 csc 2慮 cos 慮 sin 慮 sin 慮 cos 慮 sin 慮 cos 慮 2 sin 慮 cos 慮 sin 2慮 sin 2慮 cos 2慮 sin 2慮 cos 慮 cos 2慮 sin 慮 sin 慮 52. = = = sec 慮 sin 慮 cos 慮 sin 慮 cos 慮 sin 慮 cos 慮 sin 慮 + cos 2慮 1 sin 慮 + (1 2 sin2 慮) 1 sin 慮(1 2 sin 慮) 53. = = = tan 慮 cos 慮 sin 2慮 cos 慮 2 sin 慮 cos 慮 cos 慮(1 2 sin 慮) 54. Using (47), 2 sin 2慮 cos 慮 = 2(1/2)(sin 慮 + sin 3慮) = sin 慮 + sin 3慮 55. Using (47), 2 cos 2慮 sin 慮 = 2(1/2)[sin(慮) + sin 3慮] = sin 3慮 sin 慮 sin(慮/2) 2 sin2 (慮/2) 1 cos 慮 56. tan(慮/2) = = = cos(慮/2) 2 sin(慮/2) cos(慮/2) sin 慮
  • 5. January 27, 2005 11:58 l24-appa-sv Sheet number 5 Page number 528 black 528 Appendix A sin(慮/2) 2 sin(慮/2) cos(慮/2) sin 慮 57. tan(慮/2) = = 2 (慮/2) = cos(慮/2) 2 cos 1 + cos 慮 58. From (52), cos(/3 + 慮) + cos(/3 慮) = 2 cos(/3) cos 慮 = 2(1/2) cos 慮 = cos 慮 C 1 59. From the 鍖gures, area = hc but h = b sin A 2 1 a so area = bc sin A. The formulas b h 2 1 1 area = ac sin B and area = ab sin C 2 2 A c B follow by drawing altitudes from vertices B and C, respectively. 60. From right triangles ADC and BDC, C h1 = b sin A = a sin B so a/ sin A = b/ sin B. From right triangles AEB and CEB, h1 a h2 = c sin A = a sin C so a/ sin A = c/ sin C b thus a/ sin A = b/ sin B = c/ sin C. E h2 D A B c 61. (a) sin(/2 + 慮) = sin(/2) cos 慮 + cos(/2) sin 慮 = (1) cos 慮 + (0) sin 慮 = cos 慮 (b) cos(/2 + 慮) = cos(/2) cos 慮 sin(/2) sin 慮 = (0) cos 慮 (1) sin 慮 = sin 慮 (c) sin(3/2 慮) = sin(3/2) cos 慮 cos(3/2) sin 慮 = (1) cos 慮 (0) sin 慮 = cos 慮 (d) cos(3/2 + 慮) = cos(3/2) cos 慮 sin(3/2) sin 慮 = (0) cos 慮 (1) sin 慮 = sin 慮 sin(留 + 硫) sin 留 cos 硫 + cos 留 sin 硫 62. tan(留 + 硫) = = , divide numerator and denominator by cos(留 + 硫) cos 留 cos 硫 sin 留 sin 硫 sin 留 sin 硫 cos 留 cos 硫 and use tan 留 = and tan 硫 = to get (38); cos 留 cos 硫 tan 留 + tan(硫) tan 留 tan 硫 tan(留 硫) = tan(留 + (硫)) = = because 1 tan 留 tan(硫) 1 + tan 留 tan 硫 tan(硫) = tan 硫. 63. (a) Add (34) and (36) to get sin(留 硫) + sin(留 + 硫) = 2 sin 留 cos 硫 so sin 留 cos 硫 = (1/2)[sin(留 硫) + sin(留 + 硫)]. (b) Subtract (35) from (37). (c) Add (35) and (37). A+B AB 1 64. (a) From (47), sin cos = (sin B + sin A) so 2 2 2 A+B AB sin A + sin B = 2 sin cos . 2 2 (b) Use (49) (c) Use (48) 留硫 留+硫 65. sin 留 + sin(硫) = 2 sin cos , but sin(硫) = sin 硫 so 2 2 留+硫 留硫 sin 留 sin 硫 = 2 cos sin . 2 2
  • 6. January 27, 2005 11:58 l24-appa-sv Sheet number 6 Page number 529 black Exercise Set A 529 66. (a) From (34), C sin(留 + ) = C sin 留 cos + C cos 留 sin so C cos = 3 and C sin = 5, square and add to get C 2 (cos2 + sin2 ) = 9 + 25, C 2 = 34. If C = 34 then cos = 3/ 34 and sin = 5/ 34 is the 鍖rst-quadrant angle for which tan = 5/3. so 3 sin 留 + 5 cos 留 = 34 sin(留 + ). (b) Follow the procedure of part (a) to get C cos = A and C sin = B, C = A2 + B 2 , tan = B/A where the quadrant in which lies is determined by the signs of A and B because cos = A/C and sin = B/C, so A sin 留 + B cos 留 = A2 + B 2 sin(留 + ). 67. Consider the triangle having a, b, and d as sides. The angle formed by sides a and b is 慮 so from the law of cosines, d2 = a2 + b2 2ab cos( 慮) = a2 + b2 + 2ab cos 慮, d = a2 + b2 + 2ab cos 慮.