“Teach A Level Maths”
Vol. 2: A2 Core Modules
32: The function
a cos x  b sin x
© Christine Crisp
The function a cos x  b sin x
Module C3
Module C4
Edexcel
AQA
OCR
MEI/OCR
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The function a cos x  b sin x
If you have either Autograph or a graphical
calculator, draw the graph of
y  4 cos x  3 sin x
On a calculator choose
 180  x  360 and  6  y  6
You will have the following graph:
The function a cos x  b sin x
y  4 cos x  3 sin x
It is clearly the shape of a sin or cos function but
it has been transformed.
Can you describe, giving approximate values, the
transformations from y  cos x that give this curve?
ANS:
a stretch of s.f. 5 parallel to the y-axis, and
40 
a translation of approx.  
 0 
The function a cos x  b sin x
y  4 cos x  3 sin x
The equation of the curve is approximately
y  5 cos( x  40)

( As the cosine curve keeps repeating we could
translate much further, for example ( 40  360 
but there’s no point doing this. )
400)

The function a cos x  b sin x
We need an exact method of finding constants R and
a so that
4 cos x  3 sin x  R cos( x  a )       (1)
(R>0)
Using the addition formula
cos( A  B )  cos A cos B  sin A sin B
the r.h.s. of (1) becomes
R cos( x  a )  R(cos x cos a  sin x sina )
Substitute in (1):
4 cos x  3 sin x  R(cos x cos a  sin x sina )
 4 cos x  3 sin x  R cos x cosa  R sin x sina
The function a cos x  b sin x
4 cos x 3sin x  R cos x cos a  R sin x sina
Since this is an identity, the l.h.s. and the r.h.s.
must be exactly the same.
So, the
cos x term must be the same on both sides
The function a cos x  b sin x
4 cos x 3sin x  R cos x cos a  R sin x sina
Since this is an identity, the l.h.s. and the r.h.s.
must be exactly the same.
So, the
cos x term must be the same on both sides
The function a cos x  b sin x
4 cos x 3sin x  R cos x cos a  R sin x sina
Since this is an identity, the l.h.s. and the r.h.s.
must be exactly the same.
So, the
and the
cos x term must be the same on both sides
sin x term on both sides must be the same.
The function a cos x  b sin x
4 cos x 3sin x  R cos x cos a  R sin x sina
Since this is an identity, the l.h.s. and the r.h.s.
must be exactly the same.
So, the
and the
cos x term must be the same on both sides
sin x term on both sides must be the same.
The function a cos x  b sin x
4 cos x 3sin x  R cos x cos a  R sin x sina
Since this is an identity, the l.h.s. and the r.h.s.
must be exactly the same.
So, the
and the
cos x term must be the same on both sides
sin x term on both sides must be the same.
So, we can equate the coefficients
Coef. of
cos x:
The function a cos x  b sin x
4 cos x 3sin x  R cos x cos a  R sin x sina
Since this is an identity, the l.h.s. and the r.h.s.
must be exactly the same.
So, the
and the
cos x term must be the same on both sides
sin x term on both sides must be the same.
So, we can equate the coefficients
Coef. of
Coef. of
cos x: 4  R cosa
sin x :
The function a cos x  b sin x
4 cos x 3sin x  R cos x cos a  R sin x sina
Since this is an identity, the l.h.s. and the r.h.s.
must be exactly the same.
So, the
and the
cos x term must be the same on both sides
sin x term on both sides must be the same.
So, we can equate the coefficients
Coef. of
Coef. of
cos x: 4  R cosa
sin x : 3  R sina
Be careful to
check the signs
The function a cos x  b sin x
4 cos x 3sin x  R cos x cos a  R sin x sina
Since this is an identity, the l.h.s. and the r.h.s.
must be exactly the same.
So, the
and the
cos x term must be the same on both sides
sin x term on both sides must be the same.
So, we can equate the coefficients
Coef. of
Coef. of
cos x: 4  R cosa
sin x : 3  R sina
      (1)
      ( 2)
We can now solve to find R and
a
The function a cos x  b sin x
4 cos x 3sin x  R cos x cos a  R sin x sina
4  R cosa       (1)
3  R sina       (2)
The easiest way to find a is unusual.
3 R sin a
Divide (2) by (1):

4 R cos a
3
 tan a 
 a  36  9
4
( 3 s.f. )
Coef. of
cos x :
Coef. of sin x :
This equation has an infinite number of solutions, but
find us
R the
we can
square (1)
and cosine
(2) andcurve.
add them.
aTogives
translation
of the
We
can take the principal value which will translate the
Why does this give R ?
curve by the least amount.
The function a cos x  b sin x
4 cos x 3sin x  R cos x cos a  R sin x sina
Coef. of
cos x :
Coef. of sin x :
3
tan a 
4
(1) 2  ( 2) 2 : 4 2  3 2
4  R cosa       (1)
3  R sina       (2)
 a  36  9 ( 3 s.f. )
 R 2 cos 2 a  R 2 sin 2 a
 4 2  3 2  R 2 (cos 2 a  sin 2 a )
2
2
But, cos A  sin A  1
 42  32  R2  R  4 2  3 2
R is positive because it gives
from y  cos x
 the
R stretch
5
So,
4 cos x  3 sin x  R cos( x  a )
The function a cos x  b sin x
4 cos x 3sin x  R cos x cos a  R sin x sina
Coef. of
cos x :
Coef. of sin x :
3
tan a 
4
(1) 2  ( 2) 2 : 4 2  3 2
4  R cosa       (1)
3  R sina       (2)
 a  36  9 ( 3 s.f. )
 R 2 cos 2 a  R 2 sin 2 a
 4 2  3 2  R 2 (cos 2 a  sin 2 a )
2
2
But, cos A  sin A  1
 42  32  R2  R  4 2  3 2
 R5
So,
4 cos x  3 sin x  R cos( x  a )
The function a cos x  b sin x
4 cos x 3sin x  R cos x cos a  R sin x sina
Coef. of
cos x :
Coef. of sin x :
3
tan a 
4
(1) 2  ( 2) 2 : 4 2  3 2
4  R cosa       (1)
3  R sina       (2)
 a  36  9 ( 3 s.f. )
 R 2 cos 2 a  R 2 sin 2 a
 4 2  3 2  R 2 (cos 2 a  sin 2 a )
2
2
But, cos A  sin A  1
 42  32  R2  R  4 2  3 2
 R5
So,
4 cos x  3 sin x  5 cos( x  36  9  ) ( 3 s.f. )
The function a cos x  b sin x
SUMMARY
To express
4 cos x  3 sin x in the form R cos( x  a ):
• Expand cos( x  a ) using the addition formula
cos( x  a )  cos x cos a  sin x sina
•
Write
•
Equate the coefficients of
(1) cos x and ( 2) sin x
•
Divide the equations to find
•
Square and add the equations to get R 2  4 2  3 2
(or get this directly from the given expression)
•
Choose the value of R > 0.
4 cos x  3 sin x  R cos x cosa  R sin x sina
tan a and solve for a
The function a cos x  b sin x
We’ve used 4 cos x  3 sin x as an example of the
more general function
a cos x  b sin x
where a and b are constants and can be positive or
negative.
5 cos x  12 sin x .
5 cos x  12 sin x  R cos( x  a )
5 cos x  12 sin x  R cos x cosa  R sin x sina
Suppose we have
Let

By choosing this addition formula, we have matched
the signs on the l.h.s. and the r.h.s. So, cos a and
sina are both positive and a is an acute angle. The
translation from cos x is less than 90
The function a cos x  b sin x
We’ve used 4 cos x  3 sin x as an example of the
more general function
a cos x  b sin x
where a and b are constants and can be positive or
negative.
5 cos x  12 sin x .
5 cos x  12 sin x  R cos( x  a )
5 cos x  12 sin x  R cos x cosa  R sin x sina
Suppose we have
Let

Can you complete this to find
and R ?
a correct to 3 d.p.
The function a cos x  b sin x
5 cos x  12 sin x  R cos( x  a )
5 cos x  12 sin x  R cos x cosa  R sin x sina

Coef. of cos x :
5  R cosa       (1)
Coef. of
sin x :

( 2)
(1)
So,
 12   R sina
12  R sina
12 R sin a

5 R cos a
      ( 2)
12
 tan a 
5
a  67  4
R 2  5 2  12 2  R  13
5 cos x  12 sin x  13 cos( x  67  4) 
Tip: If you have a graphical calculator, check by
drawing both forms. They should give one curve.
The function a cos x  b sin x
To express
we’ve used
a cos x  b sin x as a single trig ratio,
R cos( x  a ) and R cos( x  a )
Since we can translate Rsin x instead of Rcos x
to get curves of the same form, we can also use
R sin( x  a ) or R sin( x  a )
If you can choose which form to use, it’s better to
choose the version which, when expanded, gives the
same signs for the corresponding terms as the
original expression.
Just look in the formula book to see which of the 4
addition formulae match the expression.
The function a cos x  b sin x
Exercise
For each of questions (1) to (4), select the
expression that would be easiest to use from the 4
below
R cos( x  a ) , R sin( x  a )
3 sin x  4 cos x choose R sin( x  a )
(2) For 3 cos x  4 sin x choose R cos( x  a )
(1) For
(3) For
3 cos x  4 sin x choose R cos( x  a )
(4) For
3 sin x  4 cos x choose R sin( x  a )
For (3) and (4) the terms could be switched so that
either R cos( x  a ) or R sin( x  a ) could be used.
The function a cos x  b sin x
One reason for expressing
of the forms
a cos x  b sin x in one
R cos( x  a ) or R sin( x  a )
is that the stretch from cos x or sin x is obvious.
e.g.
5 cos x  12 sin x  13 cos( x  67  4) 
so the max. is 13 and the min is 13.
The function a cos x  b sin x
Exercise
For the following
(i) express f ( x ) in the given form where R > 0 and

0  a  90 , giving a correct to 1 d.p.
(ii) write down the minimum and maximum value of f ( x )
1.
f ( x )  3 cos x  sin x
in the form R cos( x  a )
2.
f ( x )  3 sin x  4 cos x
in the form
R sin( x  a )
3. f ( x )  7 sin x  24 cos x in the form R sin( x  a )
The function a cos x  b sin x
1. f ( x) 
Solution:
3 cos x  sinx in the form R cos( x  a )
3 cos x  sin x  R cos( x  a )

3 cos x  sin x  R cos x cosa  R sin x sina
Coef. of cos x :
3  R cos a       (1)
Coef. of sin x :
 1   R sina
1  R sina       (2)
( 2)
1
R sin a
1

 tan a 
(1)
3 R cos a
3

a

30
R 2  ( 3 ) 2  12  R  2
So,
3 cos x  sin x  2 cos( x  30) 
The max. is 2 and the min is 2.

The function a cos x  b sin x
f ( x )  3 sin x  4 cos x in the form R sin( x  a )
Solution: 3 sin x  4 cos x  R sin( x  a )
3 sin x  4 cos x  R sin x cosa  R cos x sina

Coef. of sin x :
3  R cosa       (1)
2.
Coef. of
cos x :

( 2)
(1)
 4   R sina
4  R sina
4 R sin a

3 R cos a
      ( 2)

4
tan a 
3
a  53  1
R2  32  42  R  5
So,
3 sin x  4 cos x  5 sin( x  53  1) 
The max. is 5 and the min is 5.
The function a cos x  b sin x
3. f ( x )  7 sin x  24 cos x in the form R sin( x  a )
Solution: 7 sin x  24 cos x  R sin( x  a )
7 sin x  24 cos x  R sin x cosa  R cos x sina
Coef. of sin x :
7  R cosa       (1)
Coef. of cos x :
24  R sina       (2)

( 2)
(1)
So,
24 R sin a

7 R cos a
24
 tan a 
7

a

73

7
2
2
2
R  7  24  R  25
7 sin x  24 cos x  25 sin( x  73  7) 
The max. is 25 and the min is 25.
The function a cos x  b sin x
The function a cos x  b sin x
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
The function a cos x  b sin x
SUMMARY
To express
4 cos x  3 sin x in the form R cos( x  a ):
• Expand cos( x  a ) using the addition formula
cos( x  a )  cos x cos a  sin x sina
•
Write
•
Equate the coefficients of
(1) cos x and ( 2) sin x
•
Divide the equations to find
•
Square and add the equations to get R 2  4 2  3 2
(or get this directly from the given expression)
•
Choose the value of R > 0.
4 cos x  3 sin x  R cos x cosa  R sin x sina
tan a and solve for a
The function a cos x  b sin x
e.g. Express
4 cos x  3 sin x in the form R cos( x  a )
Solution:
cos( x  a )  cos x cos a  sin x sina

4 cos x  3 sin x  R(cos x cos a  sin x sina )
Coef. of cos x :
4  R cosa       (1)
Coef. of sin x :
3  R sina       (2)
3
( 2)

:
( 3 s.f. )
tan a 
a

36

9

(1)
4
(1) 2  ( 2) 2 : 4 2  3 2  R 2 cos 2 a  R 2 sin 2 a
 4 2  3 2  R 2 (cos 2 a  sin 2 a )
But,
cos 2 A  sin 2 A  1
 42  32  R2  R  4 2  3 2  R  5
So,
4 cos x  3 sin x  5 cos( x  36  9  ) ( 3 s.f. )
The function a cos x  b sin x
We’ve used 4 cos x  3 sin x as an example of the
more general function
a cos x  b sin x
where a and b are constants and can be positive or
negative.
Suppose we have
5 cos x  12 sin x .
It is easier to solve for a if
positive so we choose to use

cos a and sina are
5 cos x  12 sin x  R cos( x  a )
5 cos x  12 sin x  R cos x cosa  R sin x sina
The function a cos x  b sin x
5 cos x  12 sin x  R cos x cosa  R sin x sina
Coef. of
cos x :
Coef. of sin x :

( 2)
(1)
So,
5  R cosa       (1)
 12   R sina
12  R sina
12 R sin a

5 R cos a
      ( 2)
12
 tan a 
5
a  67  4
R 2  5 2  12 2  R  13
5 cos x  12 sin x  13 cos( x  67  4) 
Tip: If you have a graphical calculator, check by
drawing both forms. They should give one curve.
The function a cos x  b sin x
To express
we’ve used
a cos x  b sin x as a single trig ratio,
R cos( x  a ) and R cos( x  a )
Since we can translate Rsin x instead of Rcos x
to get curves of the same form, we can also use
R sin( x  a ) or R sin( x  a )
If can choose which form to use, it’s better to
choose the version which, when expanded, gives the
same signs for the corresponding terms as the
original expression.
Just look in the formula book to see which of the 4
addition formulae match the expression.
The function a cos x  b sin x
One reason for expressing
of the forms
a cos x  b sin x in one
R cos( x  a ) or R sin( x  a )
is that the stretch from cos x or sin x is obvious.
e.g.
5 cos x  12 sin x  13 cos( x  67  4) 
so the max. is 13 and the min is 13.