Upcoming SlideShare
×

# 11 X1 T04 13 Asinx + Bcosx = C

5,477 views

Published on

Published in: Education, Technology, Travel
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
5,477
On SlideShare
0
From Embeds
0
Number of Embeds
66
Actions
Shares
0
61
0
Likes
0
Embeds 0
No embeds

No notes for slide

### 11 X1 T04 13 Asinx + Bcosx = C

1. 1. Equations of the form asinx + bcosx = c Method 1: Using the t results
2. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 18. Method 2: Auxiliary Angle Method (i) Change into a sine function eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
19. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 37. Method 2: Auxiliary Angle Method (ii) Change into a cosine function eg (i) 3cos θ + 4sin θ = 2 0 ≤ θ ≤ 360
38. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 55. 2005 Extension 1 HSC Q5c) (i) (ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α )
56. 56. 2005 Extension 1 HSC Q5c) (i) (ii) Express 3sin 3t − cos3t in the form R sin ( 3t − α ) 3sin 3t − cos3t
57. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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