This PowerPoint helps students to consider the concept of infinity.
11 X1 T04 13 Asinx + Bcosx = C
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 = 639′
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 = 639′
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 = 639′
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 = 639′ 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 = 639′ α = 59 47′
θ
θ
= 59 47′
′
= 173 21
2
2
θ = 11933′
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 = 639′ α = 59 47′
θ
θ
= 59 47′
′
= 173 21
2
2
θ = 11933′
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 = 639′ α = 59 47′
θ
θ
= 59 47′
′
= 173 21
2
2
θ = 11933′
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 = 639′ α = 59 47′
θ
θ
= 59 47′
′
= 173 21
2
2 ∴θ = 11933′,346 42′
θ = 11933′
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 = 3652′
5
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 = 538′
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 = 538′
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 = 538′
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 = 538′
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 = 538′
5
Q1, Q4
2
cos β =
5
β = 66 25′
θ − 538′ = 66 25′, 29335′
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 = 538′
5
Q1, Q4
2
cos β =
5
β = 66 25′
θ − 538′ = 66 25′, 29335′
∴θ = 11933′,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