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

3600 eg (i) 3cos 4sin 2  

3. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
let tan
2
t



4. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
           

let tan
2
t



5. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
           

let tan
2
t


2 2
3 3 8 2 2t t t   

6. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
           

let tan
2
t


2 2
3 3 8 2 2t t t   
2
5 8 1 0t t  

7. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
           

let tan
2
t


2 2
3 3 8 2 2t t t   
2
5 8 1 0t t  
8 84
10
t



8. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
   
        

let tan
2
t


2 2
3 3 8 2 2t t t   
2
5 8 1 0t t  
8 84
10
t


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

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
   
        

let tan
2
t


2 2
3 3 8 2 2t t t   
2
5 8 1 0t t  
8 84
10
t


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

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
   
        

let tan
2
t


2 2
3 3 8 2 2t t t   
2
5 8 1 0t t  
8 84
10
t


4 21 4 21
tan or tan
2 5 2 5
  
 
Q2
21 4
tan
5
6 39




 

11. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
   
        

let tan
2
t


2 2
3 3 8 2 2t t t   
2
5 8 1 0t t  
8 84
10
t


4 21 4 21
tan or tan
2 5 2 5
  
 
Q2
21 4
tan
5
6 39




 
173 21
2
346 42







12. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
   
        

let tan
2
t


2 2
3 3 8 2 2t t t   
2
5 8 1 0t t  
8 84
10
t


4 21 4 21
tan or tan
2 5 2 5
  
 
Q2
21 4
tan
5
6 39




 
173 21
2
346 42






Q1

13. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
   
        

let tan
2
t


2 2
3 3 8 2 2t t t   
2
5 8 1 0t t  
8 84
10
t


4 21 4 21
tan or tan
2 5 2 5
  
 
Q2
21 4
tan
5
6 39




 
173 21
2
346 42






Q1
4 21
tan
5
59 47




 

14. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
   
        

let tan
2
t


2 2
3 3 8 2 2t t t   
2
5 8 1 0t t  
8 84
10
t


4 21 4 21
tan or tan
2 5 2 5
  
 
Q2
21 4
tan
5
6 39




 
173 21
2
346 42






Q1
4 21
tan
5
59 47




 
59 47
2
119 33







15. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
   
        

let tan
2
t


2 2
3 3 8 2 2t t t   
2
5 8 1 0t t  
8 84
10
t


4 21 4 21
tan or tan
2 5 2 5
  
 
Q2
21 4
tan
5
6 39




 
173 21
2
346 42






Q1
4 21
tan
5
59 47




 
59 47
2
119 33







180: Test

16. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
   
        

let tan
2
t


2 2
3 3 8 2 2t t t   
2
5 8 1 0t t  
8 84
10
t


4 21 4 21
tan or tan
2 5 2 5
  
 
Q2
21 4
tan
5
6 39




 
173 21
2
346 42






Q1
4 21
tan
5
59 47




 
59 47
2
119 33







180: Test
3cos180 4sin180
4 2

  
 

17. Equations of the form asinx + bcosx = c
Method 1: Using the t results

3600 eg (i) 3cos 4sin 2  
2
2 2
1 2
3 4 2 0 180
1 1 2
t t
t t
   
        

let tan
2
t


2 2
3 3 8 2 2t t t   
2
5 8 1 0t t  
8 84
10
t


4 21 4 21
tan or tan
2 5 2 5
  
 
Q2
21 4
tan
5
6 39




 
173 21
2
346 42






Q1
4 21
tan
5
59 47




 
59 47
2
119 33







180: Test
3cos180 4sin180
4 2

  
 
119 33 ,346 42     

18. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 

19. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  

20. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
 sincoscossin 

21. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
 sincoscossin 


22. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
 sincoscossin 

sin corresponds to 3, so
3 goes on the opposite
side
3

23. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
 sincoscossin 

3

24. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
 sincoscossin 

3
cos corresponds to 4, so
4 goes on the adjacent
side
4

25. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
 sincoscossin 

3
4
by Pythagoras the
hypotenuse is 5
5

26. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
 sincoscossin 

3
4
by Pythagoras the
hypotenuse is 5
5
3 4
5 cos sin 2
5 5
     
 

27. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
 sincoscossin 

3
4
by Pythagoras the
hypotenuse is 5
5
 5sin 2  
3 4
5 cos sin 2
5 5
     
 

28. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
 sincoscossin 

3
4
by Pythagoras the
hypotenuse is 5
5
 5sin 2  
The hypotenuse becomes the
coefficient of the trig function
3 4
5 cos sin 2
5 5
     
 

29. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  

3
4
5
 5sin 2  
3 4
5 cos sin 2
5 5
     
  3
tan
4
 
 
2
sin
5
  

30. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  

3
4
5
 5sin 2  
3 4
5 cos sin 2
5 5
     
  3
tan
4
 
36 52  
 
2
sin
5
  

31. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  

3
4
5
 5sin 2  
3 4
5 cos sin 2
5 5
     
  3
tan
4
 
36 52  
 
2
sin
5
  
Q1, Q2

32. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  

3
4
5
 5sin 2  
3 4
5 cos sin 2
5 5
     
  3
tan
4
 
36 52  
 
2
sin
5
  
Q1, Q2
2
sin
5
 

33. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  

3
4
5
 5sin 2  
3 4
5 cos sin 2
5 5
     
  3
tan
4
 
36 52  
 
2
sin
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   
3600 
3cos 4sin 2  

3
4
5
 5sin 2  
3 4
5 cos sin 2
5 5
     
  3
tan
4
 
36 52  
 
2
sin
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   
3600 
3cos 4sin 2  

3
4
5
 5sin 2  
3 4
5 cos sin 2
5 5
     
  3
tan
4
 
36 52  
 
2
sin
5
  
Q1, Q2
2
sin
5
 
23 35  
36 52 23 35 ,156 25     
1317 ,119 33     

36. Method 2: Auxiliary Angle Method
(i) Change into a sine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  

3
4
5
 5sin 2  
3 4
5 cos sin 2
5 5
     
  3
tan
4
 
36 52  
 
2
sin
5
  
Q1, Q2
2
sin
5
 
23 35  
36 52 23 35 ,156 25     
119 33 ,346 43     
1317 ,119 33     

37. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 

38. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  

39. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

40. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   


41. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

cos corresponds to 3, so
3 goes on the adjacent
side
3

42. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

3

43. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

3
cos corresponds to 4, so
4 goes on the adjacent
side
4

44. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

3
4
by Pythagoras the
hypotenuse is 5
5

45. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

3
4
by Pythagoras the
hypotenuse is 5
5
3 4
5 cos sin 2
5 5
     
 

46. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

3
4
by Pythagoras the
hypotenuse is 5
5
 5cos 2  
3 4
5 cos sin 2
5 5
     
 

47. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

3
4
by Pythagoras the
hypotenuse is 5
5
 5cos 2  
The hypotenuse becomes the
coefficient of the trig function
3 4
5 cos sin 2
5 5
     
 

48. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

3
4
5
 5cos 2  
3 4
5 cos sin 2
5 5
     
  4
tan
3
 
 
2
cos
5
  

49. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

3
4
5
 5cos 2  
3 4
5 cos sin 2
5 5
     
  4
tan
3
 
53 8  
 
2
cos
5
  

50. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

3
4
5
 5cos 2  
3 4
5 cos sin 2
5 5
     
  4
tan
3
 
53 8  
 
2
cos
5
  
Q1, Q4

51. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

3
4
5
 5cos 2  
3 4
5 cos sin 2
5 5
     
  4
tan
3
 
53 8  
 
2
cos
5
  
Q1, Q4
2
cos
5
 

52. Method 2: Auxiliary Angle Method
(ii) Change into a cosine function
eg (i) 3cos 4sin 2   
3600 
3cos 4sin 2  
cos cos sin sin   

3
4
5
 5cos 2  
3 4
5 cos sin 2
5 5
     
  4
tan
3
 
53 8  
 
2
cos
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   
3600 
3cos 4sin 2  
cos cos sin sin   

3
4
5
 5cos 2  
3 4
5 cos sin 2
5 5
     
  4
tan
3
 
53 8  
 
2
cos
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   
3600 
3cos 4sin 2  
cos cos sin sin   

3
4
5
 5cos 2  
3 4
5 cos sin 2
5 5
     
  4
tan
3
 
53 8  
 
2
cos
5
  
Q1, Q4
2
cos
5
 
66 25  
53 8 66 25 ,293 35      
119 33 ,346 43     

55.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)

56.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t

57.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

58.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12

59.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12
 2sin 3t  

60.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12
 2sin 3t  
1
tan
3
30



 

61.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12
 2sin 3t  
1
tan
3
30



 
 2sin 3 30t  

62.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12
 2sin 3t  
1
tan
3
30



 
 2sin 3 30t  
 (iii) Express sin cos in the form cosx x R x  
2003 Extension 1 HSC Q2e) (i)

63.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12
 2sin 3t  
1
tan
3
30



 
 2sin 3 30t  
 (iii) Express sin cos in the form cosx x R x  
2003 Extension 1 HSC Q2e) (i)
sin cosx x

64.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12
 2sin 3t  
1
tan
3
30



 
 2sin 3 30t  
 (iii) Express sin cos in the form cosx x R x  
2003 Extension 1 HSC Q2e) (i)
sin cosx x
 cos cos sin sinx x 

65.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12
 2sin 3t  
1
tan
3
30



 
 2sin 3 30t  
 (iii) Express sin cos in the form cosx x R x  
2003 Extension 1 HSC Q2e) (i)
sin cosx x
 cos cos sin sinx x 
 cos sinx x  

66.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12
 2sin 3t  
1
tan
3
30



 
 2sin 3 30t  
 (iii) Express sin cos in the form cosx x R x  
2003 Extension 1 HSC Q2e) (i)
sin cosx x
 cos cos sin sinx x 

1
12
 cos sinx x  

67.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12
 2sin 3t  
1
tan
3
30



 
 2sin 3 30t  
 (iii) Express sin cos in the form cosx x R x  
2003 Extension 1 HSC Q2e) (i)
sin cosx x
 cos cos sin sinx x 

1
12
 2cos x   
 cos sinx x  

68.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12
 2sin 3t  
1
tan
3
30



 
 2sin 3 30t  
 (iii) Express sin cos in the form cosx x R x  
2003 Extension 1 HSC Q2e) (i)
sin cosx x
 cos cos sin sinx x 

1
12
 2cos x   
tan 1
45



 
 cos sinx x  

69.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12
 2sin 3t  
1
tan
3
30



 
 2sin 3 30t  
 (iii) Express sin cos in the form cosx x R x  
2003 Extension 1 HSC Q2e) (i)
sin cosx x
 cos cos sin sinx x 

1
12
 2cos x   
tan 1
45



 
 2cos 45x   
 cos sinx x  

70.  (ii) Express 3sin3 cos3 in the form sin 3t t R t  
2005 Extension 1 HSC Q5c) (i)
3sin3 cos3t t
 sin3 cos cos3 sint t 

3
12
 2sin 3t  
1
tan
3
30



 
 2sin 3 30t  
 (iii) Express sin cos in the form cosx x R x  
2003 Extension 1 HSC Q2e) (i)
sin cosx x
 cos cos sin sinx x 

1
12
 2cos x   
tan 1
45



 
 2cos 45x   
 cos sinx x  
Exercise 2E;
6, 7, 10bd, 11, 13, 14, 16ac, 20a, 23