1. Equations of the form asinx + bcosx = c
Method 1: Using the t results

2. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360

3. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360

let t  tan
2

4. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3 2 
 4 2 
2 0  180
2 1 t  1 t  2

5. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3 2 
 4 2 
2 0  180
2 1 t  1 t  2
3  3t 2  8t  2  2t 2

6. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3 2 
 4 2 
2 0  180
2 1 t  1 t  2
3  3t 2  8t  2  2t 2
5t 2  8t  1  0

7. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3 2 
 4 2 
2 0  180
2 1 t  1 t  2
3  3t 2  8t  2  2t 2
5t 2  8t  1  0
8  84
t
10

8. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3 2 
 4 2 
2 0  180
2 1 t  1 t  2
3  3t 2  8t  2  2t 2
5t 2  8t  1  0
8  84
t
10
 4  21  4  21
tan  or tan 
2 5 2 5

9. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3 2 
 4 2 
2 0  180
2 1 t  1 t  2
3  3t 2  8t  2  2t 2
5t 2  8t  1  0
8  84
t
10
 4  21  4  21
tan  or tan 
2 5 2 5
Q2

10. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3 2 
 4 2 
2 0  180
2 1 t  1 t  2
3  3t 2  8t  2  2t 2
5t 2  8t  1  0
8  84
t
10
 4  21  4  21
tan  or tan 
2 5 2 5
21  4
Q2 tan  
5
  639

11. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3 2 
 4 2 
2 0  180
2 1 t  1 t  2
3  3t 2  8t  2  2t 2
5t 2  8t  1  0
8  84
t
10
 4  21  4  21
tan  or tan 
2 5 2 5
21  4
Q2 tan  
5
  639

 173 21
2
  346 42

12. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3 2 
 4 2 
2 0  180
2 1 t  1 t  2
3  3t 2  8t  2  2t 2
5t 2  8t  1  0
8  84
t
10
 4  21  4  21
tan  or tan 
2 5 2 5
21  4
Q2 tan   Q1
5
  639

 173 21
2
  346 42

13. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3 2 
 4 2 
2 0  180
2 1 t  1 t  2
3  3t 2  8t  2  2t 2
5t 2  8t  1  0
8  84
t
10
 4  21  4  21
tan  or tan 
2 5 2 5
21  4 4  21
Q2 tan   Q1 tan  
5 5
  639   59 47

 173 21
2
  346 42

14. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3 2 
 4 2 
2 0  180
2 1 t  1 t  2
3  3t 2  8t  2  2t 2
5t 2  8t  1  0
8  84
t
10
 4  21  4  21
tan  or tan 
2 5 2 5
21  4 4  21
Q2 tan   Q1 tan  
5 5
  639   59 47
 
 173 21  59 47
2 2
  346 42   11933

15. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3  4  2 0   180
2  1 t 2  1 t 2 
 2
3  3t 2  8t  2  2t 2
5t 2  8t  1  0
8  84
t
10
 4  21  4  21
tan  or tan 
2 5 2 5 Test :   180
21  4 4  21
Q2 tan   Q1 tan  
5 5
  639   59 47
 
 173 21  59 47
2 2
  346 42   11933

16. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3  4  2 0   180
2  1 t 2  1 t 2 
 2
3  3t 2  8t  2  2t 2
5t 2  8t  1  0
8  84
t
10
 4  21  4  21
tan  or tan 
2 5 2 5 Test :   180
21  4 4  21
Q2 tan   Q1 tan   3cos180  4sin180
5 5
  639   59 47  4  2
 
 173 21  59 47
2 2
  346 42   11933

17. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos   4sin   2 0    360
  1  t 2   2t  
let t  tan 3  4  2 0   180
2  1 t 2  1 t 2 
 2
3  3t 2  8t  2  2t 2
5t 2  8t  1  0
8  84
t
10
 4  21  4  21
tan  or tan 
2 5 2 5 Test :   180
21  4 4  21
Q2 tan   Q1 tan   3cos180  4sin180
5 5
  639   59 47  4  2
 
 173 21  59 47
2 2   11933,346 42
  346 42   11933

18. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360

19. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
3cos   4sin   2

20. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
sin  cos  cos  sin 
3cos   4sin   2

21. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
sin  cos  cos  sin 
3cos   4sin   2


22. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
sin  cos  cos  sin 
3
3cos   4sin   2

sin corresponds to 3, so
3 goes on the opposite
side

23. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
sin  cos  cos  sin 
3
3cos   4sin   2


24. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
sin  cos  cos  sin 
3
3cos   4sin   2

4
cos corresponds to 4, so
4 goes on the adjacent
side

25. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
sin  cos  cos  sin 
5
3
3cos   4sin   2

4
by Pythagoras the
hypotenuse is 5

26. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
sin  cos  cos  sin 
5
3
3cos   4sin   2

 3 cos   4 sin    2
5   4
 5 5 
by Pythagoras the
hypotenuse is 5

27. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
sin  cos  cos  sin 
5
3
3cos   4sin   2

 3 cos   4 sin    2
5   4
 5 5 
5sin      2 by Pythagoras the
hypotenuse is 5

28. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
sin  cos  cos  sin 
5
3
3cos   4sin   2

 3 cos   4 sin    2
5   4
 5 5 
5sin      2 by Pythagoras the
hypotenuse is 5
The hypotenuse becomes the
coefficient of the trig function

29. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
5
3
3cos   4sin   2

 3 cos   4 sin    2
5   4
 5 5  3
tan  
5sin      2 4
2
sin     
5

30. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
5
3
3cos   4sin   2

 3 cos   4 sin    2
5   4
 5 5  3
tan  
5sin      2 4
2
sin        3652
5

31. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
5
3
3cos   4sin   2

 3 cos   4 sin    2
5   4
 5 5  3
tan  
5sin      2 4
2
sin        3652
5
Q1, Q2

32. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
5
3
3cos   4sin   2

 3 cos   4 sin    2
5   4
 5 5  3
tan  
5sin      2 4
2
sin        3652
5
Q1, Q2
2
sin  
5

33. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
5
3
3cos   4sin   2

 3 cos   4 sin    2
5   4
 5 5  3
tan  
5sin      2 4
2
sin        3652
5
Q1, Q2
2
sin  
5
  2335

34. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
5
3
3cos   4sin   2

 3 cos   4 sin    2
5   4
 5 5  3
tan  
5sin      2 4
2
sin        3652
5
Q1, Q2
2
sin  
5
  2335
3652    2335,156 25

35. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
5
3
3cos   4sin   2

 3 cos   4 sin    2
5   4
 5 5  3
tan  
5sin      2 4
2
sin        3652
5
Q1, Q2
2
sin  
5
  2335
3652    2335,156 25
  1317,11933

36. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos   4sin   2 0    360
5
3
3cos   4sin   2

 3 cos   4 sin    2
5   4
 5 5  3
tan  
5sin      2 4
2
sin        3652
5
Q1, Q2
2
sin  
5
  2335
36 52    2335,156 25

  11933,346 43
  1317,11933

37. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360

38. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
3cos   4sin   2

39. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
3cos   4sin   2

40. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
3cos   4sin   2


41. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
3cos   4sin   2

3
cos corresponds to 3, so
3 goes on the adjacent
side

42. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
3cos   4sin   2

3

43. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
4
3cos   4sin   2

3
cos corresponds to 4, so
4 goes on the adjacent
side

44. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
5
4
3cos   4sin   2

3
by Pythagoras the
hypotenuse is 5

45. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
5
4
3cos   4sin   2

 3 cos   4 sin    2
5   3
 5 5 
by Pythagoras the
hypotenuse is 5

46. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
5
4
3cos   4sin   2

 3 cos   4 sin    2
5   3
 5 5 
5cos      2 by Pythagoras the
hypotenuse is 5

47. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
5
4
3cos   4sin   2

 3 cos   4 sin    2
5   3
 5 5 
5cos      2 by Pythagoras the
hypotenuse is 5
The hypotenuse becomes the
coefficient of the trig function

48. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
5
4
3cos   4sin   2

 3 cos   4 sin    2
5   3
 5 5  4
tan  
5cos      2 3
2
cos     
5

49. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
5
4
3cos   4sin   2

 3 cos   4 sin    2
5   3
 5 5  4
tan  
5cos      2 3
2
cos        538
5

50. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
5
4
3cos   4sin   2

 3 cos   4 sin    2
5   3
 5 5  4
tan  
5cos      2 3
2
cos        538
5
Q1, Q4

51. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
5
4
3cos   4sin   2

 3 cos   4 sin    2
5   3
 5 5  4
tan  
5cos      2 3
2
cos        538
5
Q1, Q4
2
cos  
5

52. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
5
4
3cos   4sin   2

 3 cos   4 sin    2
5   3
 5 5  4
tan  
5cos      2 3
2
cos        538
5
Q1, Q4
2
cos  
5
  66 25

53. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
5
4
3cos   4sin   2

 3 cos   4 sin    2
5   3
 5 5  4
tan  
5cos      2 3
2
cos        538
5
Q1, Q4
2
cos  
5
  66 25
  538  66 25, 29335

54. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos   4sin   2 0    360
cos  cos   sin  sin 
5
4
3cos   4sin   2

 3 cos   4 sin    2
5   3
 5 5  4
tan  
5cos      2 3
2
cos        538
5
Q1, Q4
2
cos  
5
  66 25
  538  66 25, 29335
  11933,346 43

55. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   

56. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
3sin 3t  cos3t

57. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin  
3sin 3t  cos3t

58. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t

3

59. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t  2sin  3t   

3

60. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t  2sin  3t   

3
1
tan  
3
  30

61. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t  2sin  3t   
 2sin  3t  30  
3
1
tan  
3
  30

62. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t  2sin  3t   
 2sin  3t  30  
3
1
tan  
3
  30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx  cos x in the form R cos  x   

63. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t  2sin  3t   
 2sin  3t  30  
3
1
tan  
3
  30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx  cos x in the form R cos  x   
sin x  cos x

64. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t  2sin  3t   
 2sin  3t  30  
3
1
tan  
3
  30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx  cos x in the form R cos  x   
 cos x cos   sin x sin  
sin x  cos x

65. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t  2sin  3t   
 2sin  3t  30  
3
1
tan  
3
  30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx  cos x in the form R cos  x   
 cos x cos   sin x sin  
sin x  cos x    cos x  sin x 

66. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t  2sin  3t   
 2sin  3t  30  
3
1
tan  
3
  30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx  cos x in the form R cos  x   
 cos x cos   sin x sin   2
sin x  cos x    cos x  sin x 
1

1

67. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t  2sin  3t   
 2sin  3t  30  
3
1
tan  
3
  30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx  cos x in the form R cos  x   
 cos x cos   sin x sin   2
sin x  cos x    cos x  sin x 
1
  2 cos  x    
1

68. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t  2sin  3t   
 2sin  3t  30  
3
1
tan  
3
  30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx  cos x in the form R cos  x   
 cos x cos   sin x sin   2
sin x  cos x    cos x  sin x 
1
  2 cos  x    
1
tan   1
  45

69. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t  2sin  3t   
 2sin  3t  30  
3
1
tan  
3
  30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx  cos x in the form R cos  x   
 cos x cos   sin x sin   2
sin x  cos x    cos x  sin x 
1
  2 cos  x    
  2 cos  x  45  1
tan   1
  45

70. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t  cos3t in the form R sin  3t   
 sin 3t cos   cos3t sin   2 1
3sin 3t  cos3t  2sin  3t   
 2sin  3t  30  
3
1
tan  
Exercise 2E; 3
6, 7, 10bd, 11, 13, 14, 16ac, 20a, 23   30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx  cos x in the form R cos  x   
 cos x cos   sin x sin   2
sin x  cos x    cos x  sin x 
1
  2 cos  x    
  2 cos  x  45  1
tan   1
  45