1. Expressing asinx + bcosx in the forms
Rsin(x ± a) or Rcos(x ± a)
The graph below is y = 3cosx + 4sinx.
This can be considered as either a sine or a
cosine graph which has been translated horizontally and stretched vertically.

2. If it is considered to be a cosine curve then it has
been translated horizontally and stretched
vertically by a factor of
530 5
The object is to find the horizontal translation and
vertical stretch.

3. If it is considered to be a sine curve then it has
been translated horizontally and stretched
vertically by a factor of
-370 5
The object is to find the horizontal translation and
vertical stretch.

4. If the curve is taken to be a translated cosine curve then its equation will be of the form
3cosx + 4sinx = Rcos(x - a)
Where a is the horizontal translation
And R is the vertical stretch
Note the question contains a PLUS in the middle and the translated equation contains a MINUS because the curve is translated in a positive x direction.

5. 3
cosx + sinx = Rcos(x - a)
= R(cosxcosa + sinxsina) Using cos(A - B)
cosx
Matching up the left and right hand side then
Rcosa = 3
Rsina = 4 4
= (Rcosa)
+ (Rsina)
sinx

6. Rsina = 4
Rcosa = 3
Dividing these two equations
a = 53.1o

7. Rsina = 4
Rcosa = 3
Squaring these two equations
R2 sin2a = and R2 cos2a = 9
Adding these two equations
R2 sin2a + R2cos2a = = 25
R2(sin2a + cos2a) = 25
R = 5 as sin2a + cos2a = 1

8. Hence 3cosx + 4sinx = Rcos(x - a)
This is a cosine graph which has been translated horizontally and stretched vertically by a factor of 5
This is evident from the graph on the right

9. If the curve is taken to be a translated sine curve then its equation will be of the form
3cosx + 4sinx = Rsin(x + a)
Where a is the horizontal translation
And R is the vertical stretch
Note the question contains a PLUS in the middle and the translated equation contains a PLUS because the curve is translated in a negative x direction.

10. 3
cosx + sinx = Rsin(x + a)
= R(sinxcosa + cosxsina) Using sin(A + B)
sinx
Matching up the left and right hand side then
Rcosa = 4
Rsina = 3 4
= (Rcosa)
+ (Rsina)
cosx

11. Rsina = 3
Rcosa = 4
Dividing these two equations
a = 36.9o

12. Rsina = 3
Rcosa = 4
Squaring these two equations
R2 sin2a = 9 and R2 cos2a = 16
Adding these two equations
R2 sin2a + R2cos2a = = 25
R2(sin2a + cos2a) = 25
R = 5 as sin2a + cos2a = 1

13. Hence 3cosx + 4sinx = Rsin(x + a)
This is a sine graph which has been translated horizontally –36.90 and stretched vertically by a factor of 5
This is evident from the graph on the right

14. Using the Rsin(x ± a) or Rcos(x ± a) form
to solve equations of the form acosx + bsinx = c
Solve cosx + 4sinx = 4
3cosx + 4sinx = 5cos(x ) Shown previously
So 5cos(x – 53.1) = 4
Let y = x – 53.1
So cosy = 
y = cos-1  = 36.8, -36.8, 323.1
x – 53.1 = 36.8, -36.8, 323.1
find 1st two answers and add 360
x = 89.9, 16.3, add 53.1 to both sides