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 θ + 4sin θ = 2 0 ≤ θ ≤ 360

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

4. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0≤ ≤ 180
3 2
2
2 1+ t  1+ t  2

5. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0≤ ≤ 180
3 2
2
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 θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0≤ ≤ 180
3 2
2
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 θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0≤ ≤ 180
3 2
2
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 θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0≤ ≤ 180
3 2
2
2 1+ t  1+ t  2
3 − 3t 2 + 8t = 2 + 2t 2
5t 2 − 8t − 1 = 0
8 ± 84
t=
10
θ 4 − 21 θ 4 + 21
tan = tan =
or
2 5 2 5

9. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0≤ ≤ 180
3 2
2
2 1+ t  1+ t  2
3 − 3t 2 + 8t = 2 + 2t 2
5t 2 − 8t − 1 = 0
8 ± 84
t=
10
θ 4 − 21 θ 4 + 21
tan = tan =
or
2 5 2 5
Q2

10. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0≤ ≤ 180
3 2
2
2 1+ t  1+ t  2
3 − 3t 2 + 8t = 2 + 2t 2
5t 2 − 8t − 1 = 0
8 ± 84
t=
10
θ 4 − 21 θ 4 + 21
tan = tan =
or
2 5 2 5
21 − 4
Q2 tan α =
5
α = 639′

11. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0≤ ≤ 180
3 2
2
2 1+ t  1+ t  2
3 − 3t 2 + 8t = 2 + 2t 2
5t 2 − 8t − 1 = 0
8 ± 84
t=
10
θ 4 − 21 θ 4 + 21
tan = tan =
or
2 5 2 5
21 − 4
Q2 tan α =
5
α = 639′
θ
= 173 21′
2
θ = 346 42′

12. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0≤ ≤ 180
3 2
2
2 1+ t  1+ t  2
3 − 3t 2 + 8t = 2 + 2t 2
5t 2 − 8t − 1 = 0
8 ± 84
t=
10
θ 4 − 21 θ 4 + 21
tan = tan =
or
2 5 2 5
21 − 4
Q2 tan α = Q1
5
α = 639′
θ
= 173 21′
2
θ = 346 42′

13. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0≤ ≤ 180
3 2
2
2 1+ t  1+ t  2
3 − 3t 2 + 8t = 2 + 2t 2
5t 2 − 8t − 1 = 0
8 ± 84
t=
10
θ 4 − 21 θ 4 + 21
tan = tan =
or
2 5 2 5
21 − 4 4 + 21
Q2 tan α = Q1 tan α =
5 5
α = 639′ α = 59 47′
θ
= 173 21′
2
θ = 346 42′

14. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0≤ ≤ 180
3 2
2
2 1+ t  1+ t  2
3 − 3t 2 + 8t = 2 + 2t 2
5t 2 − 8t − 1 = 0
8 ± 84
t=
10
θ 4 − 21 θ 4 + 21
tan = tan =
or
2 5 2 5
21 − 4 4 + 21
Q2 tan α = Q1 tan α =
5 5
α = 639′ α = 59 47′
θ
θ
= 59 47′
′
= 173 21

2
2
θ = 11933′
θ = 346 42′

15. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0 ≤ ≤ 180
3 
1+ t 2  1+ t 2 
2 2
 
3 − 3t 2 + 8t = 2 + 2t 2
5t 2 − 8t − 1 = 0
8 ± 84
t=
10
θ 4 − 21 θ 4 + 21
tan = tan =
or
Test : θ = 180
2 5 2 5
21 − 4 4 + 21
Q2 tan α = Q1 tan α =
5 5
α = 639′ α = 59 47′
θ
θ
= 59 47′
′
= 173 21

2
2
θ = 11933′
θ = 346 42′

16. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0 ≤ ≤ 180
3 
1+ t 2  1+ t 2 
2 2
 
3 − 3t 2 + 8t = 2 + 2t 2
5t 2 − 8t − 1 = 0
8 ± 84
t=
10
θ 4 − 21 θ 4 + 21
tan = tan =
or
Test : θ = 180
2 5 2 5
21 − 4 4 + 21
Q2 tan α = Q1 tan α = 3cos180 + 4sin180
5 5
= −4 ≠ 2
α = 639′ α = 59 47′
θ
θ
= 59 47′
′
= 173 21

2
2
θ = 11933′
θ = 346 42′

17. Equations of the form asinx + bcosx = c
Method 1: Using the t results
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
θ θ
 1 − t 2   2t 
let t = tan + 4 =2 0 ≤ ≤ 180
3 
1+ t 2  1+ t 2 
2 2
 
3 − 3t 2 + 8t = 2 + 2t 2
5t 2 − 8t − 1 = 0
8 ± 84
t=
10
θ 4 − 21 θ 4 + 21
tan = tan =
or
Test : θ = 180
2 5 2 5
21 − 4 4 + 21
Q2 tan α = Q1 tan α = 3cos180 + 4sin180
5 5
= −4 ≠ 2
α = 639′ α = 59 47′
θ
θ
= 59 47′
′
= 173 21

2
2 ∴θ = 11933′,346 42′
θ = 11933′
θ = 346 42′

18. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360

19. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
3cos θ + 4sin θ = 2

20. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
sin α cosθ + cos α sin θ
3cos θ + 4sin θ = 2

21. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
sin α cosθ + cos α sin θ
3cos θ + 4sin θ = 2
α

22. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
sin α cosθ + cos α sin θ
3
3cos θ + 4sin θ = 2
α
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 θ + 4sin θ = 2 0 ≤ θ ≤ 360
sin α cosθ + cos α sin θ
3
3cos θ + 4sin θ = 2
α

24. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
sin α cosθ + cos α sin θ
3
3cos θ + 4sin θ = 2
α
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 θ + 4sin θ = 2 0 ≤ θ ≤ 360
sin α cosθ + cos α sin θ
5
3
3cos θ + 4sin θ = 2
α
4
by Pythagoras the
hypotenuse is 5

26. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
sin α cosθ + cos α sin θ
5
3
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 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 θ + 4sin θ = 2 0 ≤ θ ≤ 360
sin α cosθ + cos α sin θ
5
3
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  4

 
5 5
5sin ( α + θ ) = 2 by Pythagoras the
hypotenuse is 5

28. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
sin α cosθ + cos α sin θ
5
3
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  4

 
5 5
5sin ( α + θ ) = 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 θ + 4sin θ = 2 0 ≤ θ ≤ 360
5
3
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  4

 
5 5 3
tan α =
5sin ( α + θ ) = 2 4
2
sin ( α + θ ) =
5

30. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
5
3
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  4

 
5 5 3
tan α =
5sin ( α + θ ) = 2 4
2
sin ( α + θ ) = α = 3652′
5

31. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
5
3
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  4

 
5 5 3
tan α =
5sin ( α + θ ) = 2 4
2
sin ( α + θ ) = α = 3652′
5
Q1, Q2

32. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
5
3
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  4

 
5 5 3
tan α =
5sin ( α + θ ) = 2 4
2
sin ( α + θ ) = α = 3652′
5
Q1, Q2
2
sin β =
5

33. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
5
3
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  4

 
5 5 3
tan α =
5sin ( α + θ ) = 2 4
2
sin ( α + θ ) = α = 3652′
5
Q1, Q2
2
sin β =
5
β = 2335′

34. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
5
3
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  4

 
5 5 3
tan α =
5sin ( α + θ ) = 2 4
2
sin ( α + θ ) = α = 3652′
5
Q1, Q2
2
sin β =
5
β = 2335′
3652′ + θ = 2335′,156 25′

35. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
5
3
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  4

 
5 5 3
tan α =
5sin ( α + θ ) = 2 4
2
sin ( α + θ ) = α = 3652′
5
Q1, Q2
2
sin β =
5
β = 2335′
3652′ + θ = 2335′,156 25′
θ = −1317′,11933′

36. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
5
3
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  4

 
5 5 3
tan α =
5sin ( α + θ ) = 2 4
2
sin ( α + θ ) = α = 3652′
5
Q1, Q2
2
sin β =
5
β = 2335′
∴θ = 11933′,346 43′
36 52′ + θ = 23 35′,156 25′
  
θ = −1317′,11933′

37. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360

38. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
3cos θ + 4sin θ = 2

39. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
3cos θ + 4sin θ = 2

40. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
3cos θ + 4sin θ = 2
α

41. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
3cos θ + 4sin θ = 2
α
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 θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
3cos θ + 4sin θ = 2
α
3

43. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
4
3cos θ + 4sin θ = 2
α
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 θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
5
4
3cos θ + 4sin θ = 2
α
3
by Pythagoras the
hypotenuse is 5

45. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
5
4
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 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 θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
5
4
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  3

 
5 5
5cos ( θ − α ) = 2 by Pythagoras the
hypotenuse is 5

47. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
5
4
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  3

 
5 5
5cos ( θ − α ) = 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 θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
5
4
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  3

 
5 5 4
tan α =
5cos ( θ − α ) = 2 3
2
cos ( θ − α ) =
5

49. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
5
4
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  3

 
5 5 4
tan α =
5cos ( θ − α ) = 2 3
2
cos ( θ − α ) = α = 538′
5

50. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
5
4
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  3

 
5 5 4
tan α =
5cos ( θ − α ) = 2 3
2
cos ( θ − α ) = α = 538′
5
Q1, Q4

51. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
5
4
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  3

 
5 5 4
tan α =
5cos ( θ − α ) = 2 3
2
cos ( θ − α ) = α = 538′
5
Q1, Q4
2
cos β =
5

52. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
5
4
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  3

 
5 5 4
tan α =
5cos ( θ − α ) = 2 3
2
cos ( θ − α ) = α = 538′
5
Q1, Q4
2
cos β =
5
β = 66 25′

53. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
5
4
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  3

 
5 5 4
tan α =
5cos ( θ − α ) = 2 3
2
cos ( θ − α ) = α = 538′
5
Q1, Q4
2
cos β =
5
β = 66 25′
θ − 538′ = 66 25′, 29335′

54. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
cos α cos θ + sin α sin θ
5
4
3cos θ + 4sin θ = 2
α
 3 cos θ + 4 sin θ  = 2
5×  3

 
5 5 4
tan α =
5cos ( θ − α ) = 2 3
2
cos ( θ − α ) = α = 538′
5
Q1, Q4
2
cos β =
5
β = 66 25′
θ − 538′ = 66 25′, 29335′
∴θ = 11933′,346 43′

55. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )

56. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
3sin 3t − cos3t

57. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
( sin 3t cos α − cos3t sin α )
3sin 3t − cos3t

58. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
( sin 3t cos α − cos3t sin α ) 2 1
3sin 3t − cos3t
α
3

59. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
( sin 3t cos α − cos3t sin α ) 2 1
3sin 3t − cos3t = 2sin ( 3t − α )
α
3

60. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
( sin 3t cos α − cos3t sin α ) 2 1
3sin 3t − cos3t = 2sin ( 3t − α )
α
3
1
tan α =
3
α = 30

61. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
( sin 3t cos α − cos3t sin α ) 2 1
3sin 3t − cos3t = 2sin ( 3t − α )
= 2sin ( 3t − 30 ) α
3
1
tan α =
3
α = 30

62. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
( sin 3t cos α − cos3t sin α ) 2 1
3sin 3t − cos3t = 2sin ( 3t − α )
= 2sin ( 3t − 30 ) α
3
1
tan α =
3
α = 30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx − cos x in the form R cos ( x + α )

63. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
( sin 3t cos α − cos3t sin α ) 2 1
3sin 3t − cos3t = 2sin ( 3t − α )
= 2sin ( 3t − 30 ) α
3
1
tan α =
3
α = 30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx − cos x in the form R cos ( x + α )
sin x − cos x

64. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
( sin 3t cos α − cos3t sin α ) 2 1
3sin 3t − cos3t = 2sin ( 3t − α )
= 2sin ( 3t − 30 ) α
3
1
tan α =
3
α = 30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx − cos x in the form R cos ( x + α )
( cos x cos α − sin x sin α )
sin x − cos x

65. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
( sin 3t cos α − cos3t sin α ) 2 1
3sin 3t − cos3t = 2sin ( 3t − α )
= 2sin ( 3t − 30 ) α
3
1
tan α =
3
α = 30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx − cos x in the form R cos ( x + α )
( cos x cos α − sin x sin α )
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 − α )
( sin 3t cos α − cos3t sin α ) 2 1
3sin 3t − cos3t = 2sin ( 3t − α )
= 2sin ( 3t − 30 ) α
3
1
tan α =
3
α = 30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx − cos x in the form R cos ( x + α )
( cos x cos α − sin x sin α ) 2 1
sin x − cos x = − ( cos x − sin x )
α
1

67. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
( sin 3t cos α − cos3t sin α ) 2 1
3sin 3t − cos3t = 2sin ( 3t − α )
= 2sin ( 3t − 30 ) α
3
1
tan α =
3
α = 30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx − cos x in the form R cos ( x + α )
( cos x cos α − sin x sin α ) 2 1
sin x − cos x = − ( cos x − sin x )
α
= − 2 cos ( x − α )
1

68. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
( sin 3t cos α − cos3t sin α ) 2 1
3sin 3t − cos3t = 2sin ( 3t − α )
= 2sin ( 3t − 30 ) α
3
1
tan α =
3
α = 30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx − cos x in the form R cos ( x + α )
( cos x cos α − sin x sin α ) 2 1
sin x − cos x = − ( cos x − sin x )
α
= − 2 cos ( x − α )
1
tan α = 1
α = 45

69. 2005 Extension 1 HSC Q5c) (i)
(ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
( sin 3t cos α − cos3t sin α ) 2 1
3sin 3t − cos3t = 2sin ( 3t − α )
= 2sin ( 3t − 30 ) α
3
1
tan α =
3
α = 30
2003 Extension 1 HSC Q2e) (i)
(iii) Express sinx − cos x in the form R cos ( x + α )
( cos x cos α − sin x sin α ) 2 1
sin x − cos x = − ( cos x − sin x )
α
= − 2 cos ( x − α )
= − 2 cos ( x − 45 ) 1
tan α = 1
α = 45

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