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Higher
Mathematics
Trigonometry
Contents
Trigonometry
1
2
3
4
5
6
7
8
9
10
11
12
13
Radians
Exact Values
Solving Trigonometric Equations
Trigonometry in Three Dimensions
Compound Angles
Double-Angle Formulae
Further Trigonometric Equations
Expressing pcosx + qsinx in the form kcos(x – a)
Expressing pcosx + qsinx in other forms
Multiple Angles
Maximum and Minimum Values
Solving Equations
Sketching Graphs of y = pcosx + qsinx
1
EF
EF
RC
EF
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EF
RC
EF
EF
EF
EF
RC
EF
CfE Edition
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Higher Mathematics
Trigonometry
Trigonometry
1
Radians
EF
Degrees are not the only units used to measure angles. The radian (RAD on
the calculator) is a measurement also used.
Degrees and radians bear the relationship:
π radians
= 180°.
The other equivalences that you should become familiar with are:
30° = π6 radians
45° = π4 radians
60° = π3 radians
90° = π2 radians
135° = 34π radians
360° =2π radians.
Converting between degrees and radians is straightforward.
• To convert from degrees to radians, multiply by π
× π
and divide by 180.
180
Degrees
• To convert from radians to degrees, multiply by 180
× 180
π
and divide by π .
Radians
π = 5 π radians .
For example, 50°= 50 × 180
18
2
Exact Values
EF
The following exact values must be known. You can do this by either
memorising the two triangles involved, or memorising the table.
DEG RAD
1
1
hsn.uk.net
2
1
sin x cos x tan x
0
0
0
1
0
30
π
6
π
4
3
2
1
2
1
3
45
60
π
3
1
2
3
90
π
2
1
2
1
2
3
2
1
0
–
Page 1
1
Tip
You’ll probably find it
easier to remember the
triangles.
CfE Edition
Higher Mathematics
3
Trigonometry
Solving Trigonometric Equations
RC
You should already be familiar with solving some trigonometric equations.
EXAMPLES
1. Solve sin x° = 12 for 0 < x < 360 .
sin x° = 12
180° − x °
x°
 S A
T C 360° − x °
180° + x °
Since
sin x°
Remember
The exact value triangle:
positive
First quadrant solution:
2 30° 3
x = sin −1 ( 12 )
=
x 30
x = 30
= 30.
or 180 − 30
or 150.
is
60°
1
2. Solve cos x° = − 1 for 0 < x < 360 .
5
cos x° = − 1
5
180° − x °
x°
S A
 T C 360° − x °
180° + x °
negative
Since
cos x°
is
( 5)
x = cos −1 1
= 63·435 (to 3 d.p.).
180 − 63·435 or 180 + 63·435
x=
x = 116·565 or 243·435.
3. Solve sin x° =3 for 0 < x < 360 .
There are no solutions since −1 ≤ sin x ° ≤ 1 .
Note that −1 ≤ cos x ° ≤ 1 , so cos x° =3 also has no solutions.
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Page 2
CfE Edition
Higher Mathematics
Trigonometry
4. Solve tan x° = −5 for 0 < x < 360 .
180° − x °
x°
tan x° = −5
S A
T C
180° + x °
360° − x °
Since
tan x°
is
negative
x = tan −1 ( 5 )
= 78·690 (to 3 d.p.).
180 − 78·690 or 360 − 78·690
x=
x = 101·310 or 281·310.
Note
All trigonometric equations we will meet can be reduced to problems like
those above. The only differences are:

the solutions could be required in radians – in this case, the question will not
have a degree symbol, e.g. “Solve 3tan x = 1 ” rather than “ 3tan x° =1 ”;

exact value solutions could be required in the non-calculator paper – you
will be expected to know the exact values for 0, 30, 45, 60 and 90 degrees.
Questions can be worked through in degrees or radians, but make sure the
final answer is given in the units asked for in the question.
EXAMPLES
5. Solve 2sin 2 x° − 1 =0 where 0 ≤ x ≤ 360 .
180° − 2 x °
2x°
2sin 2 x ° =1
 S A
sin 2 x ° = 12
T C 360° − 2 x °
180° + 2 x °
2 x = sin −1 ( 12 )
= 30.
=
2 x 30 or 180 − 30
or 360 + 30 or 360 + 180 − 30
or 360 + 360 + 30
Remember
The exact value triangle:
2 30° 3
60°
1
2 x = 30 or 150 or 390 or 510
x = 15 or 75 or 195 or 255.
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0 ≤ x ≤ 360
0 ≤ 2 x ≤ 720
Page 3
Note
There are more solutions
every 360°, since
sin(30°) = sin(30°+360°) = …
So keep adding 360 until
2x > 720.
CfE Edition
Higher Mathematics
6. Solve
Trigonometry
2 cos 2 x = 1 where 0 ≤ x ≤ π .
cos 2x = 1
2
2x
π − 2x
S A

π + 2 x T C 2π − 2 x
0≤ x ≤π
0 ≤ 2 x ≤ 2π
( 2)
Remember
The exact value triangle:
2 x = cos −1 1
= π4 .
=
2 x π4 or 2π − π4
2
π
4
π
4 1
1
or 2π + π4
2 x = π4 or 74π
x = π8 or 78π .
7. Solve 4cos 2 x = 3 where 0 < x < 2π .
( cos x )2 = 34
cos x = ± 34
cos x = ± 23
x = π6
or
x = π6
 S A  Since cos x can be positive or negative
 T C
Remember
The exact value triangle:
( )
x = cos −1 23
= π6 .
or π − π6 or π + π6 or 2π − π6
2π + π6
or 56π or 76π or 116π .
π
3
1
8. Solve 3tan ( 3 x° − 20° ) =5 where 0 ≤ x ≤ 360 .
3tan ( 3 x ° − 20° ) =5
tan ( 3 x ° − 20° ) = 53
S A
T C
0 ≤ x ≤ 360
0 ≤ 3 x ≤ 1080
−20 ≤ 3 x − 20 ≤ 1060
( )
3 x − 20 =
tan −1 53
= 59·036 (to 3 d.p.)
=
3 x − 20 59·036 or 180 + 59·036
or 360 + 59·036 or 360 + 180 + 59·036
or 360 + 360 + 59·036 or 360 + 360 + 180 + 59·036
or 360 + 360 + 360 + 59·036.
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π
6
2
Page 4
CfE Edition
3
Higher Mathematics
Trigonometry
3 x − 20 =
59·036 or 239·036 or 419·036
or 599·036 or 779·036 or 959·036
3 x = 79·036 or 259·036 or 439·036
or 619·036 or 799·036 or 979·036
x = 26·35 or 86·35 or 146·35 or 206·35 or 266·35 or 326·35.
(
)
9. Solve cos 2 x + π3 =
0·812 for 0 < x < 2π .
(
)
cos 2 x + π3 =
0·812
S A
T C
0 < x < 2π
0 < 2 x < 4π
π < 2 x + π < 4π + π
3
3
3
π
1·047 < 2 x + 3 < 13·614 (to 3 d.p.)
Remember
2 x + π3 =
cos −1 ( 0·812 )
Make sure your
= 0·623 (to 3 d.p.).
calculator uses radians
2 x + π3 =
0·623 or 2π − 0·623
or 2π + 0·623 or 2π + 2π − 0·623
or 2π + 2π + 0·623 or 2π + 2π + 2π − 0·623
2 x + π3 =
5.660 or 6.906 or 11.943 or 13.189
2 x = 4.613 or 5.859 or 10.896 or 12.142
x = 2.307 or 2.930 or 5.448 or 6.071.
4
Trigonometry in Three Dimensions
EF
It is possible to solve trigonometric problems in three dimensions using
techniques we already know from two dimensions. The use of sketches is
often helpful.
The angle between a line and a plane
The angle a between the plane P and the line ST is calculated by adding a line
perpendicular to the plane and then using basic trigonometry.
T
a
P
S
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Page 5
CfE Edition
Higher Mathematics
Trigonometry
EXAMPLE
1. The triangular prism ABCDEF is shown below.
3 cm
F
B
E
C
6 cm
A
10 cm
Calculate the acute angle between:
(a) The line AF and the plane ABCD.
(b) AE and ABCD.
(a) Start with a sketch:
F
tan a
=
3 cm
A
a
10 cm
D
D
Opposite 3
=
Adjacent 10
( )
3
a = tan −1 10
= 16.699° (or 0.291 radians) (to 3 d.p.).
Note
Since the angle is in a right-angled triangle, it must be acute so there is no
need for a CAST diagram.
(b) Again, make a sketch:
E
3 cm
C
b
angle to be
A
calculated
We need to calculate the length of AC first using Pythagoras’s Theorem:
C
=
AC
102 + 62
6 cm
= 136
A
D
10 cm
Therefore:
E
Opposite
tan b = 3
=
3 cm
Adjacent 136
C
b
A
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b = tan −1
( 1363 )
= 14.426° (or 0.252 radians) (to 3 d.p.).
136 cm
Page 6
CfE Edition
Higher Mathematics
Trigonometry
The angle between two planes
The angle a between planes P and Q is calculated by adding a line
perpendicular to Q and then using basic trigonometry.
P
a
Q
EXAMPLE
2. ABCDEFGH is a cuboid with dimensions 12 × 8 × 8 cm as shown below.
H
G
F
E
8 cm
D
C
8 cm
A
12 cm
B
(a) Calculate the size of the angle between the planes AFGD and ABCD.
(b) Calculate the size of the acute angle between the diagonal planes
AFGD and BCHE.
(c) Start with a sketch:
F
tan a
=
8 cm
A
a
12 cm
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a = tan −1 23
= 33.690° (or 0.588 radians) (to 3 d.p.).
Let AF and BE intersect at T
 TBA

AB
33.690°
=
 ATB is isosceles, so T=
= 180° − ( 33.690° + 33.690° )
ATB
Note
= 112.620°.
T
•
( )
Note
Angle GDC is the same
size as angle FAB.
B
(d) Again, make a sketch:
E
F
A
Opposite 8
=
Adjacent 12
•
The angle could also
B
So the acute angle is:
have been calculated


.
BTF
= ATE
= 180° − 112 620° using rectangle DCGH.
= 67.380° (or 1.176 radians) (to 3 d.p.).
Page 7
CfE Edition
Higher Mathematics
5
Trigonometry
Compound Angles
EF
When we add or subtract angles, the result is a compound angle.
For example, 45° + 30° is a compound angle. Using a calculator, we find:
• sin ( 45° + 30
=
° ) sin ( 75
=
° ) 0.966
• sin ( 45° ) + sin ( 30° ) =
1.207 (both to 3 d.p.).
This shows that sin ( A + B ) is not equal to sin A + sin B . Instead, we can use
the following identities:
sin ( A=
+ B ) sin A cos B + cos A sin B
sin ( A=
− B ) sin A cos B − cos A sin B
cos ( A=
+ B ) cos A cos B − sin A sin B
cos ( A=
− B ) cos A cos B + sin A sin B .
These are given in the exam in a condensed form:
sin ( A=
± B ) sin A cos B ± cos A sin B
cos ( A ± B ) =
cos A cos B  sin A sin B .
EXAMPLES
1. Expand and simplify cos ( x ° + 60° ) .
cos ( x ° + 60=
° ) cos x ° cos 60° − sin x ° sin 60°
Remember
The exact value triangle:
= 12 cos x ° − 23 sin x °.
2 30° 3
60°
1
2. Show that sin ( a=
+ b ) sin a cos b + cos a sin b for a = π6 and b = π3 .
LHS
= sin ( a + b )
(
= sin π6 + π3
= sin ( π2 )
= 1.
)
=
RHS sin a cos b + cos a sin b
= sin π6 cos π3 + cos π6 sin π3
(
= ( 12 × 12 ) + 23 × 23
= 14 + 34 = 1.
)
Since LHS = RHS , the claim is true for a = π6 and b = π3 .
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Page 8
Remember
The exact value triangle:
π
6
2
π
3
1
CfE Edition
3
Higher Mathematics
Trigonometry
3. Find the exact value of sin 75° .
sin=
75° sin ( 45° + 30° )
= sin 45° cos30° + cos 45° sin30°
(
) (
= 1 × 23 + 1 × 12
2
2
)
3 + 1
2 2 2 2
= 3 +1
2 2
= 6 +4 2 .
=
Finding Trigonometric Ratios
You should already be familiar with the following formulae (SOH CAH TOA).
Hypotenuse
a
Opposite
Adjacent
sin a =
Opposite
Hypotenuse
cos a =
Adjacent
Hypotenuse
tan a =
Opposite
.
Adjacent
p
If we have sin a = q where 0 < a < π2 , then we can form a right-angled
triangle to represent this ratio.
Opposite
p
Since sin a = q then:
=
Hypotenuse
q
p
 the side opposite a has length p;
a
 the hypotenuse has length q.
The length of the unknown side can be found using Pythagoras’s Theorem.
Once the length of each side is known, we can find cos a and tan a using
SOH CAH TOA.
The method is similar if we know cos a and want to find sin a or tan a .
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Page 9
CfE Edition
Higher Mathematics
Trigonometry
EXAMPLES
5 . Show that
4. Acute angles p and q are such that sin p = 54 and sinq = 13
63 .
sin ( p + q ) =
65
5
sin p = 54
4
13
q
12
cos p = 35
p
3
5
5
sinq = 13
cosq = 12
13
+ q ) sin p cos q + cos p sin q
sin ( p=
(
) (
3 5
= 54 × 12
13 + 5 × 13
)
Note
Since “Show that” is
used in the question, all
of this working is
required.
48 + 15
= 65
65
63
= 65 .
Using compound angle formulae to confirm identities
EXAMPLES
5. Show that sin ( x − π2 ) =
− cos x .
sin ( x − π2 )
= sin x cos π2 − cos x sin π2
= sin x × 0 − cos x × 1
= − cos x .
sin ( s + t )
6. Show that
= tan s + tan t for cos s ≠ 0 and cos t ≠ 0 .
cos s cos t
sin ( s + t ) sin s cos t + cos s sin t
=
cos s cos t
cos s cos t
sin s cos t cos s sin t
=
+
cos s cos t cos s cos t
Remember
sin x
sin s sin t
 tan x .
=
+
cos x
cos s cos t
= tan s + tan t .
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Page 10
CfE Edition
Higher Mathematics
6
Trigonometry
Double-Angle Formulae
EF
Using the compound angle identities with A = B , we obtain expressions for
sin 2A and cos 2A . These are called double-angle formulae.
sin 2 A = 2sin A cos A
cos
=
2 A cos 2 A − sin 2 A
= 2cos 2 A − 1
= 1 − 2sin 2 A.
Note that these are given in the exam.
EXAMPLES
1. Given that tanθ = 34 , where 0 < θ < π2 , find the exact value of sin2θ and
cos2θ.
5
θ
sin 2θ = 2sin θ cos θ
sinθ = 34
4
cos
2θ cos 2 θ − sin 2 θ
=
= 2 × 54 × 35
cosθ = 35
= 24
25 .
3
( ) ( )
3 2− 4 2
5
5
9 − 16 Note
= 25
25
Any of the cos2A
7 .
= − 25
formulae could have
been used here.
=
5 , where 0 < x < π , find the exact values of sin x
2. Given that cos 2x = 13
and cos x .
Since cos 2 x = 1 − 2 sin 2 x ,
5
1 − 2sin 2 x =
13
8
2sin 2 x = 13
8
sin 2 x = 26
4
= 13
sin x = ±
2 .
13
We are told that 0 < x < π , so only sin x =
13
x
2
a=
2
13 − 2 2 =
2
13
is possible.
13 − 4 =
9 = 3.
a
So cos x =
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3 .
13
Page 11
CfE Edition
Higher Mathematics
7
Trigonometry
Further Trigonometric Equations
RC
We will now consider trigonometric equations where double-angle formulae
can be used to find solutions. These equations will involve:
 sin2x and either sin x or cos x ;
Remember
The double-angle
 cos2x and cos x ;
formulae are given in
 cos2x and sin x .
the exam.
Solving equations involving sin2x and either sinx or cosx
EXAMPLE
1. Solve sin 2 x ° = − sin x ° for 0 ≤ x < 360 .
Replace sin2x using the double angle
formula
2sin x ° cos x ° + sin x ° =0
 Take all terms to one side, making the
sin x ° ( 2cos x ° + 1) =0
equation equal to zero
 Factorise the expression and solve
2cos x ° + 1 =0
sin x ° =0
S A
T C
cos x ° = − 12
x = 0 or 180 or 360
−1 1
x=
180 − 60 or 180 + 60 x = cos ( 2 )

2sin x ° cos x ° = − sin x °
= 60.
= 120 or 240.
So x = 0 or 120 or 180 or 240 .
Solving equations involving cos2x and cosx
EXAMPLE
2. Solve cos2 x = cos x for 0 ≤ x ≤ 2π .
cos 2 x = cos x
2cos 2 x − 1 =
cos x
2cos 2 x − cos x − 1 =
0
0
( 2cos x + 1)( cos x − 1) =
2cos x + 1 =
0
cos x = − 12
x=
π − π3 or π + π3
= 23π
or 43π
Replace cos2x by 2 cos 2 x − 1
 Take all terms to one side, making a
quadratic equation in cos x
 Solve the quadratic equation (using
factorisation or the quadratic formula)

S A
T C
x = cos −1 ( 12 )
cos x − 1 =
0
cos x = 1
x = 0 or 2π.
= π3
So x = 0 or 23π or 43π or 2π .
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Page 12
CfE Edition
Higher Mathematics
Trigonometry
Solving equations involving cos2x and sinx
EXAMPLE
3. Solve cos2 x = sin x for 0 < x < 2π .
1 − 2sin 2 x =
sin x
2sin 2 x + sin x − 1 =
0
0
( 2sin x − 1)( sin x + 1) =
2sin x − 1 =
0
sin x = 12
=
x π6 or π − π6
= π6 or
5π
6
Replace cos2x by 1 − 2 sin 2 x
 Take all terms to one side,
making a quadratic equation in

cos 2 x = sin x
sin x

Solve the quadratic equation
(using factorisation or the
quadratic formula)
 S A
T C
x = sin
= π6
−1
sin x + 1 =
0
sin x = −1
( )
1
2
x = 32π .
So x = π6 or 56π or 32π .
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Page 13
CfE Edition
Higher Mathematics
8
Trigonometry
Expressing pcosx + qsinx in the form kcos(x – a)
EF
An expression of the form p cos x + q sin x can be written in the form
k cos ( x − a ) where:
k sin a
=
k
p 2 + q 2 and tan a =
.
k cos a
The following example shows how to achieve this.
EXAMPLES
1. Write 5cos x ° + 12sin x ° in the form k cos ( x ° − a ° ) where 0 ≤ a < 360 .
Step 1
Expand k cos ( x − a ) using the
compound angle formula.
Step 2
Rearrange to compare with
p cos x + q sin x .
Step 3
Compare the coefficients of cos x
and sin x with p cos x + q sin x .
Step 4
Mark the quadrants on a CAST
diagram, according to the signs
of k cos a and k sin a .
Step 5
Find k and a using the formulae
above (a lies in the quadrant
marked twice in Step 4).
Step 6
State p cos x + q sin x in the form
k cos ( x − a ) using these values.
hsn.uk.net
k cos ( x ° − a ° )
= k cos x ° cos a ° + k sin x ° sin a °
= k cos a ° cos x ° + k sin a ° sin x °
= k
cos a ° cos x ° + k
sin
a ° sin x °
12
5
k cos a ° =5
k sin a ° =12
180° − a °
a°
 S A 
T C
180° + a °
360° − a °
=
k
52 + 122
= 169
= 13
tan a ° =
k sin a °
k cos a °
= 12
5
( )
a = tan −1 12
5
.
= 67 4 (to 1 d.p.)
5cos x ° + 12sin
=
x ° 13cos ( x ° − 67.4° )
Page 14
CfE Edition
Higher Mathematics
Trigonometry
2. Write 5cos x − 3sin x in the form k cos ( x − a ) where 0 ≤ a < 2π .
5cos x − 3sin x = k cos ( x − a )
= k cos x cos a + k sin x sin a
=
( k cos a ) cos x + ( k sin a ) sin x
k sin a = −3
π−a
tan a =
= 34
( )
a
Note
tan −1 35
Make sure your
= 0.540 (to 3 d.p.). calculator is in radian
mode.
So =
a 2π − 0.540
= 5.743 (to 1 d.p.).
S A

 T C
π+a
2π − a
Hence a is in the
fourth quadrant.
34 cos ( x − 5.743 ) .
Hence 5cos x − 3sin=
x
9
k sin a
= − 35
k cos a
First quadrant answer is:
52 + ( −3 )2
=
k
k cos a = 5
Expressing pcosx + qsinx in other forms
EF
An expression in the form p cos x + q sin x can also be written in any of the
following forms using a similar method:
k cos ( x + a ) ,
k sin ( x + a ) .
k sin ( x − a ) ,
EXAMPLES
1. Write 4cos x ° + 3sin x ° in the form k sin ( x ° + a ° ) where 0 ≤ a < 360 .
4cos x ° + 3sin
=
x ° k sin ( x ° + a ° )
= k sin x ° cos a ° + k cos x ° sin a °
=
( k cos a ° ) sin x ° + ( k sin a ° ) cos x °.
=
k
k cos a ° =3
k sin a ° =4
180° − a °
4 2 + 32
= 25
a°
 S A 
T C
180° + a °
360° − a °
=5
tan a °
=
So:
k sin a ° 4
=
k cos a ° 3
( )
a = tan −1 34
= 53.1 (to 1 d.p.).
Hence a is in the
first quadrant.
Hence 4cos x ° + 3sin
=
x ° 5sin ( x ° + 53.1° ) .
hsn.uk.net
Page 15
CfE Edition
Higher Mathematics
Trigonometry
2. Write cos x − 3 sin x in the form k cos ( x + a ) where 0 ≤ a < 2π .
cos x − 3 sin x =k cos ( x + a )
= k cos x cos a − k sin x sin a
=
( k cos a ) cos x − ( k sin a ) sin x .
k cos a = 1
k=
k sin a = 3
12 + ( − 3 )
=
1+ 3
π−a
2
So:
a = tan −1
= 4
=2
a
 S A 
T C
π+a
2π − a
k sin a
=
k cos a
=
tan a
= π3 .
3
( 3)
Hence a is in the
first quadrant.
(
)
Hence cos x − 3 sin x =2cos x + π3 .
10 Multiple Angles
EF
We can use the same method with expressions involving the same multiple
angle, i.e. p cos (nx ) + q sin (nx ) , where n is a constant.
EXAMPLE
Write 5cos 2 x ° + 12sin 2 x ° in the form k sin ( 2 x ° + a ° ) where 0 ≤ a < 360 .
5cos 2 x ° + 12sin
2 x ° k sin ( 2 x ° + a ° )
=
= k sin 2 x ° cos a ° + k cos 2 x ° sin a °
=
=
k
k cos a ° =12
k sin a ° =5
180° − a °
( k cos a ° ) sin 2 x ° + ( k sin a ° ) cos 2 x °.
a°
 S A 
T C
180° + a °
360° − a °
122 + 52
= 169
= 13
=
tan a °
So:
k sin a ° 5
=
k cos a ° 12
( )
5
a = tan −1 12
= 22.6 (to 1 d.p.).
Hence a is in the
first quadrant.
Hence 5cos 2 x ° + 12sin
=
2 x ° 13sin ( 2 x ° + 22.6° ) .
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CfE Edition
Higher Mathematics
Trigonometry
11 Maximum and Minimum Values
EF
To work out the maximum or minimum values of p cos x + q sin x , we can
rewrite it as a single trigonometric function, e.g. k cos ( x − a ) .
Recall that the maximum value of the sine and cosine functions is 1, and their
minimum is –1.
y
y
y = cos x
y = sin x
max. = 1
max. = 1
1
1
x
min. = –1
O
–1
x
min. = –1
O
–1
EXAMPLE
Write 4sin x + cos x in the form k cos ( x − a ) where 0 ≤ a ≤ 2π and state:
(i) the maximum value and the value of 0 ≤ x < 2π at which it occurs
(ii) the minimum value and the value of 0 ≤ x < 2π at which it occurs.
4sin x + cos x= k cos ( x − a )
= k cos x cos a + k sin x sin a
=
( k cos a ) cos x + ( k sin a ) sin x .
k = ( −1)2 + 42
k cos a = 1
k sin a = 4
π−a
= 17
tan a
=
k sin a
= 4
k cos a
So:
a = tan −1 ( 4 )
= 1.326 (to 3 d.p.).
a
 S A 
T C
π+a
2π − a
Hence a is in the
first quadrant.
Hence 4sin x + cos
x
=
17 cos ( x − 1.326 ) .
The minimum value of − 17
occurs when:
cos ( x − 1.326 ) =
−1
The maximum value of 17
occurs when:
cos ( x − 1.326 ) =
1
cos −1 (1)
x − 1.326 =
0
x − 1.326 =
x 1.326 (to 3 d.p.).
=
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x − 1.326 = cos −1 ( −1)
x − 1.326 =
π
x 4.468 (to 3 d.p.).
=
CfE Edition
Higher Mathematics
Trigonometry
12 Solving Equations
RC
The method of writing two trigonometric terms as one can be used to help
solve equations involving both a sin ( nx ) and a cos ( nx ) term.
EXAMPLES
1. Solve 5cos x ° + sin x ° = 2 where 0 ≤ x < 360 .
First, we write 5cos x ° + sin x ° in the form k cos ( x ° − a ° ) :
=
x ° k cos ( x ° − a ° )
5cos x ° + sin
= k cos x ° cos a ° + k sin x ° sin a °
=
( k cos a ° ) cos x ° + ( k sin a ° ) sin x °.
=
k
k cos a ° =5
k sin a ° =1
180° − a °
52 + 12
=
tan a °
= 26
a°
So:
k sin a ° 1
=
k cos a ° 5
()
a = tan −1 15
= 11.3 (to 1 d.p.).
 S A 
T C
180° + a °
360° − a °
Hence a is in the
first quadrant.
Hence 5cos x °=
+ sin x °
26 cos ( x ° − 11.3° ) .
Now we use this to help solve the equation:
x°
5cos x ° + sin x ° = 2
180° − x °
S A
26 cos ( x ° − 11.3° ) = 2

180° + x ° T C 360° − x °
cos ( x ° − 11.3° ) = 2
26
x − 11.3 =
cos −1 2
( 26 )
=
x − 11.3 66.9 or 360 − 66.9
x − 11.3 =
66.9 or 293.1
x = 78.2 or 304.4.
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= 66.9 (to 1d.p.).
CfE Edition
Higher Mathematics
Trigonometry
2. Solve 2cos 2 x + 3sin 2 x =
1 where 0 ≤ x < 2π .
First, we write 2cos 2 x + 3sin 2 x in the form k cos ( 2 x − a ) :
2cos 2 x + 3sin 2 x = k cos ( 2 x − a )
= k cos 2 x cos a + k sin 2 x sin a
=
( k cos a ) cos 2 x + ( k sin a ) sin 2 x .
k cos a = 2
=
k
22 + ( −3 )2
=
tan a
k sin a = 3
=
4+9
So:
π−a
a
 S A 
T C
π+a
2π − a
= 13
k sin a 3
=
k cos a 2
( )
a = tan −1 32
= 0.983 (to 3 d.p.).
Hence a is in the
first quadrant.
Hence 2cos 2 x + 3sin 2=
x
13 cos ( 2 x − 0.983 ) .
Now we use this to help solve the equation:
2x
2cos 2 x + 3sin 2 x =
1
π − 2x
0 < x < 2π
S A
13 cos ( 2 x − 0.983 ) =
1

0 < 2 x < 4π
π + 2 x T C 2π − 2 x
1
.
cos ( 2 x − 0 983 ) =
13
2 x − 0.983 =
cos −1 1
( 13 )
= 1.290 (to 3 d.p.).
2 x −=
0.983 1.290 or 2π − 1.290
or 2π + 1.290 or 2π + 2π − 1.290
or 2π + 2π + 1.290
2 x − 0.983 =
1.290 or 4.993 or 7.573 or 11.276
2 x = 2.273 or 5.976 or 8.556 or 12.259
x = 1.137 or 2.988 or 4.278 or 6.130.
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CfE Edition
Higher Mathematics
Trigonometry
13 Sketching Graphs of y = pcosx + qsinx
EF
Expressing p cos x + q sin x in the form k cos ( x − a ) enables us to sketch the
graph=
of y p cos x + q sin x .
EXAMPLES
1. (a) Write 7 cos x ° + 6sin x ° in the form k cos ( x ° − a ° ) , 0 ≤ a < 360 .
(b) Hence sketch the graph =
of y 7 cos x ° + 6sin x ° for 0 ≤ x ≤ 360 .
(a) First, we write 7 cos x ° + 6sin x ° in the form k cos ( x ° − a ° ) :
=
x ° k cos ( x ° − a ° )
7 cos x ° + 6sin
= k cos x ° cos a ° + k sin x ° sin a °
=
( k cos a ° ) cos x ° + ( k sin a ° ) sin x °.
k cos a ° =7
=
k
62 + 7 2
=
tan a °
k sin a ° =6
=
36 + 49
So:
180° − a °
= 85
a°
 S A 
T C
180° + a °
360° − a °
k sin a ° 6
=
k cos a ° 7
( )
a = tan −1 76
= 40.6 (to 1 d.p.).
Hence a is in the
first quadrant.
Hence 7 cos x ° +=
6sin x °
85 cos ( x ° − 40.6° ) .
(b) Now we can sketch the graph =
of y 7 cos x ° + 6sin x ° :
y y 7 cos x ° + 6 sin x °
=
85
x
O
− 85
40.6
hsn.uk.net
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Page 20
CfE Edition
Higher Mathematics
Trigonometry
2. Sketch the graph of=
y sin x ° + 3 cos x ° for 0 ≤ x ≤ 360 .
First, we write sin x ° + 3 cos x ° in the form k cos ( x ° − a ° ) :
sin x ° + 3 cos
=
x ° k cos ( x ° − a ° )
= k cos x ° cos a ° + k sin x ° sin a °
=
( k cos a ° ) cos x ° + ( k sin a ° ) sin x °.
=
k
k cos a ° = 3
k sin a ° =1
180° − a °
12 + 3
=
2
1+ 3
=2
a°
 S A 
T C
180° + a °
360° − a °
=
tan a °
k sin a °
=
k cos a °
So:
1
3
( 3)
a = tan −1 1
= 30.
Hence a is in the
first quadrant.
Hence sin x ° + 3 cos
=
x ° 2cos ( x ° − 30° ) .
Now we can sketch the graph of=
y sin x ° + 3 cos x ° :
y
=
y sin x ° + 3 cos x °
2
O
−2
x
30
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360
Page 21
CfE Edition
Higher Mathematics
Trigonometry
3. (a) Write 5sin x ° − 11cos x ° in the form k sin ( x ° − a ° ) , 0 ≤ a < 360 .
(b) Hence sketch the graph of
=
y 5sin x ° − 11cos x ° + 2 , 0 ≤ x ≤ 360 .
(a) First, we write 5sin x ° − 11cos x ° in the form k sin ( x ° − a ° ) :
=
x ° k sin ( x ° − a ° )
5sin x ° − 11cos
= k sin x ° cos a ° − k cos x ° sin a °
=
( k cos a ° ) sin x ° − ( k sin a ° ) cos x °.
k cos a ° =5
=
k
52 + 11
k sin a ° = 11
=
25 + 11
180° − a °
a°
 S A 
T C
180° + a °
360° − a °
= 36
=6
2
=
tan a °
So:
k sin a °
=
k cos a °
11
5
( )
a = tan −1 11
5
= 33.6 (to 1 d.p.).
Hence a is in the
first quadrant.
=
x ° 6sin ( x ° − 33.6° ) .
Hence 5sin x ° − 11cos
(b) We can now sketch the graph of
=
y 5sin x ° − 11cos x °=
+ 2 6sin ( x ° − 33.6° ) + 2 :
y
6
4
y 5sin x ° − 11cos x ° + 2
=
2
x
O
−4
−6
33.6
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CfE Edition