it is about math

1. 1
ACTIVE MATHS 4 BOOK 2
Learning Outcomes
In this chapter you have learned to:
• Solve right-angled triangles using the theorem of Pythagoras
• Define tan x
• Find trigonometric ratios in surd form for angles 30°, 45° and 60°
• Calculate the area of a triangle
• Use the trigonometric ratios sin, cos and tan to solve problems
• Use the Sine and Cosine Rules to solve problems
• Calculate the area of a sector of a circle and the length of an arc of a circle
• Define sin x and cos x for all values of x
• Graph the trigonometric functions sine, cosine, tangent
• Graph trigonometric functions of type a sin nƟ, a cos nƟ for a,n ∈ N
• Solve trigonometric equations
• Derive the trigonometric formulae 1, 2, 3, 4, 5, 6, 7, 9
• Apply the trigonometric formulae 1–24
Chapter 7 – Trigonometry
07

2. 2
ACTIVE MATHS 4 BOOK 2
Pythagoras’ theorem: In a right-angled triangle, the area of
the square on the hypotenuse is equal to the sum of the
areas of the squares on the other two sides.
Formula
𝑐2 = 𝑎2 + 𝑏2 (F and T: P16)
𝑎2
𝑐2
𝑏2
90°
A mast on a sailing boat is held in place by steel wires called
stays. The mast is 12 m tall. The stay is 13 m long.
What is the width of the deck between the base of the mast
and the stay?
Let the distance between the base and the stay be 𝑥.
132 = 122 + 𝑥2
169 = 144 + 𝑥2
25 = 𝑥2
𝑥 = 5 m
169 − 144 = 𝑥2
Right-Angled Triangles and Pythagoras’ Theorem
07 Trigonometry

3. 3
ACTIVE MATHS 4 BOOK 2
90°
In a right-angled triangle, we have the following special ratios:
A
opposite
adjacent
sin 𝐴 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝐴 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
tan 𝐴 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
(F and T: P16)
In the following right-angled triangle, write down the value of each of the following ratios:
sin A, cos A and tan A; also sin B, cos B and tan B.
90°
B
A
8
6
10
sin 𝐴 =
8
10
=
4
5
cos 𝐴 =
6
10
=
3
5
tan 𝐴 =
8
6
=
4
3
sin 𝐵 =
6
10
=
3
5
cos 𝐵 =
8
10
=
4
5
tan 𝐵 =
6
8
=
3
4
Right-Angled Triangles and The Trigonometric Ratios
07 Trigonometry

4. 4
ACTIVE MATHS 4 BOOK 2
Compass Directions
N
S
E
W
Ɵ
N Ɵ° E
S Ɵ° E
Ɵ
S Ɵ° W
Ɵ
Ɵ
N Ɵ° W
Angles of Elevation and Depression
The angle of elevation is
the angle above the horizontal.
The angle of depression is the angle below the horizontal.
John is standing on a cliff top and observes a boat drifting towards the base of the cliff.
He decides to call the emergency services and give them the position of the boat. He
measures the angle of depression of the boat from the cliff top to be 30°, and he knows
the cliff top is 200 m above sea level. How far is the boat from the base of the cliff?
30°
200 m
x m
tan 30° =
200
𝑥
⇒
1
3
=
200
𝑥
⇒ 𝑥 = 200 3
⇒ 𝑥 = 346∙41 m
Using Trigonometry to Solve Practical Problems
07 Trigonometry

5. 5
ACTIVE MATHS 4 BOOK 2
Definition One radian is the measure of the angle at the centre of a circle when
the length of the arc is equal to the length of the radius.
𝜃
2𝑥
𝑥
If you marked an arc of length l = 2x cm on a circle
with radius length r = x cm, the angle 𝜃 that is
created is two radians in measure.
𝜃
𝑙
𝑙 𝑙 = 𝑟𝜃(when 𝜃 is in radians)
Formula (F and T: P9)
NOTE: 1° =
𝜋
180°
≈ 0∙01745 radians 𝜋 radians = 180° (F and T: P13)
Convert 135° to radians.
180° = 𝜋 radians
1° =
𝜋
180
radians
135° =
135𝜋
180
radians
135° =
3𝜋
4
radians
Convert
𝟕𝝅
𝟗
radians to degrees.
180° = 𝜋 radians
7𝜋
9
radians =
7(180°)
9
=
1,260°
9
= 140°
Degrees and Radians
07 Trigonometry

6. 6
ACTIVE MATHS 4 BOOK 2
The unit circle has its centre at (0,0) and has a radius length of 1 unit.
(0,1)
(1,0)
(0,−1)
(−1,0)
(0,0)
(x,y)
1
Ɵ x
y
1
Ɵ
cos 𝜃 =
𝑥
1
⇒ 𝑥 = cos 𝜃
sin 𝜃 =
𝑦
1
⇒ 𝑦 = sin 𝜃
tan 𝜃 =
𝑦
𝑥
⇒ tan 𝜃 =
sin 𝜃
cos 𝜃
(cos Ɵ,sin Ɵ)
1
(0,1)
(1,0)
(0,−1)
(−1,0)
(0,0)
Ɵ cos 0°=1, sin 0°= 0
cos 90°=0, sin 90°= 1
cos 180°= −1, sin 180°= 0
cos 270°=0, sin 270°= −1
cos 360°=1, sin 360°= 0
The Unit Circle
07 Trigonometry

7. 7
ACTIVE MATHS 4 BOOK 2
60°
60°
60°
2
2
2
90°
30°
30°
90° 60°
2
1
3 cos 30° =
3
2
sin 30° =
1
2
tan 30° =
1
3
sin 60° =
3
2
cos 60° =
1
2
tan 60° =
3
1
90°
45°
45°
1
1
2
cos 45° =
1
2
sin 45° =
1
2
tan 45° =
1
1
These ratios appear in (F and T: P13)
Special Angles 30°, 45° and 60°
07 Trigonometry
When asked to give an answer in surd form, the sine, cosine and tangent ratios can
be read easily for the angles 30°, 45° and 60°.

8. 8
ACTIVE MATHS 4 BOOK 2
In the first quadrant, all three ratios are positive.
In the second quadrant, sin is positive; cos and tan are negative.
In the third quadrant, tan is positive; sin and cos are negative.
In the fourth quadrant, cos is positive; sin and tan are negative.
The diagram summarises this.
CAST
C
(cos +)
A
(all +)
S
(sin +)
T
(tan +)
|∠AOB| = 140°.
140°
Reference angle
O
B
A
180°
40°
|∠AOB| = 250°.
250°
B
O
A
180°
= 40°
= 180° − 140°
Reference angle
= 70°
= 250° − 180°
70°
|∠AOB| = 330°.
B
O
A
Reference angle
= 30°
= 360° − 330°
330°
30°
The Sign of the Ratios in Each Quadrant
07 Trigonometry
Reference Angles

9. 9
ACTIVE MATHS 4 BOOK 2
Write in surd form: (i) cos 225° (ii) tan 330° (iii) sin 135°
Step 1
(i) cos 225°
Draw an angle of 225°.
Step 2
180°
cos is negative (CAST).
Third quadrant.
225°
Reference angle = 225° – 180°
Step 3
= 45°
Step 4
225°
cos 45° =
1
2
∴ cos 225° = −
1
2
45°
Step 1
(ii) tan 330°
Draw an angle of 330°.
Step 2
180°
tan is negative (CAST).
Fourth quadrant.
330°
Reference angle = 360° – 330°
Step 3
= 30°
Step 4
330°
tan 30° =
1
3
∴ tan 330° = −
1
3
30°
Step 1
(iii) sin 135°
Draw an angle of 135°.
Step 2
180°
sin is positive(CAST).
Second quadrant.
135°
Reference angle = 180° – 135°
Step 3
= 45°
Step 4
135°
sin 45° =
1
2
∴ sin135° =
1
2
45°
07 Trigonometry

10. 10
ACTIVE MATHS 4 BOOK 2
0°
270°
360°
90°
180° π/2 π 3π/2 2π 5π/2 3π 7π/2 4π
−2
−1
1
2
x
y
Graphing the Sine Function (y = sin Ɵ)
The graph repeats itself every 2π radians, so it is a periodic function.
Period = 2π Range = [−1,1]
Graphing Trigonometric Functions
07 Trigonometry

11. 11
ACTIVE MATHS 4 BOOK 2
0°
270°
360°
90°
180° π/2 π 3π/2 2π 5π/2 3π 7π/2 4π
−2
−1
1
2
x
y
Graphing the Cosine Function (y = cos Ɵ)
The graph repeats itself every 2π radians so it is a periodic function.
Period = 2π Range = [−1,1]
07 Trigonometry

12. 12
ACTIVE MATHS 4 BOOK 2
0°
270°
360°
180°
45°
90°
135°
225°
315°
π/2 π 3π/2 2π 5π/2 3π 7π/2 4π
−2
−1
1
2
x
y
Graphing the Tangent Function (y = tan Ɵ)
The graph repeats itself every π radians so it is a periodic function.
Period = π
There are asymptotes at 𝜃 = ±
𝜋
2
, ±
3𝜋
2
, ±
5𝜋
2
……………
07 Trigonometry

13. 13
ACTIVE MATHS 4 BOOK 2
Range = [−4,4] Period =
2𝜋
2
= 𝜋
As the range is [−4,4] the minimum point is −4 and the maximum point is 4.
As the period is 2𝜋, mark four intervals on the x axis and then continue to 2𝜋.
As it is a sin graph it starts at (0,0).
State the period and range the functions: y = 4 sin 2𝜽, and sketch its graph
in the domain 𝟎 ≤ 𝜽 ≤ 𝟐𝝅
The range of a graph a sin n𝜃 or a cos n𝜃 = [−a,a]
The period of a graph a sin n𝜃 or a cos n𝜃 =
2𝜋
𝑛
y = 4 sin 2𝜃
x
y
− 2
2
−4
−3
− 1
1
3
4
𝝅
𝝅
𝟐
𝝅
𝟒
𝟑𝝅
𝟒
𝟐𝝅
𝟓𝝅
𝟒
𝟔𝝅
𝟒
𝟕𝝅
𝟒
Graphing Functions of the Form a sin nɵ and a cos nɵ for a, n ∈ N
07 Trigonometry

14. 14
ACTIVE MATHS 4 BOOK 2
Solve for Ɵ if cos Ɵ = −
𝟑
𝟐
, 𝐟𝐨𝐫 𝟎° ≤ 𝜽 ≤ 𝟑𝟔𝟎°
cos Ɵ = −
3
2
, for 0° ≤ 𝜃 ≤ 360°
Find the reference angle. Ignore the minus sign and find cos-1 3
2
.
Step 1
150°
210°
30°
30°
√
√
Answer: 150°, 210°
Ɵ = cos-1 3
2
⇒ 𝜃 = 30°
Establish where cos is negative, i.e. in the second and third quadrants.
Step 2
Thus, Ɵ = 180° − 30° = 150° and Ɵ = 180° + 30° = 210°.
Solving Trigonometric Equations
07 Trigonometry

15. 15
ACTIVE MATHS 4 BOOK 2
Solve for Ɵ if tan 3Ɵ = 𝟏, 𝐟𝐨𝐫 𝟎° ≤ 𝜽 ≤ 𝟑𝟔𝟎°
tan 3Ɵ = 1, for 0° ≤ 𝜃 ≤ 360°
Find the reference angle.
Step 1
45°
225°
√
√
Ɵ = tan-1 1 ⇒ 𝜃 = 45°
Establish where tan is positive, i.e. in the first and third quadrants.
Step 2
45°
45°
If 0° ≤ Ɵ ≤ 360°
Step 3
3 Ɵ = 45° and 3 Ɵ = 225°
0° × 3 ≤ 3 Ɵ ≤ 360° × 3 0° ≤ 3Ɵ ≤ 1080°
Solutions for 3 Ɵ must be listed up to 1,080°:
3Ɵ = 45° + 360° = 405°
3Ɵ = 405 + 360° = 765°
3Ɵ = 765° + 360° = 1,125° (outside range)
3Ɵ = 225° + 360° = 585°
3Ɵ = 585° + 360° = 945°
So, 3Ɵ = 45°, 225°, 405°, 585°, 765°, 945°
∴ Ɵ = 15°, 75°, 135°, 195°, 255°, 315°
07 Trigonometry

16. 16
ACTIVE MATHS 4 BOOK 2
Solve for Ɵ if cos Ɵ = −
𝟏
𝟐
, 𝐟𝐨𝐫 𝜽 ∈ 𝑹, 𝐰𝐡𝐞𝐫𝐞 𝜽 𝐢𝐬 𝐢𝐧 𝐫𝐚𝐝𝐢𝐚𝐧𝐬
cos Ɵ = −
1
2
Find the reference angle.
Step 1
2𝜋
3
4𝜋
3
√
√
𝜃 = cos-1 1
2
⇒ 𝜃 =
𝜋
3
Establish where cos is negative, i.e. in the second and third quadrants.
Step 2
𝜋
3
𝜋
3
Step 3
NOTE: In this question, the range of values for Ɵ is not specified,
so it is necessary to give a general solution.
With every full rotation, the same ratio is given.
For the general solution add 2𝑛𝜋.
∴ Ɵ =
2𝜋
3
+ 2𝑛𝜋 or Ɵ =
2𝜋
3
+ 2𝑛𝜋, where 𝑛 ∈ 𝑍.
07 Trigonometry

17. 17
ACTIVE MATHS 4 BOOK 2
cos2
𝐴 + sin2
𝐴 = 1
Formula 1
Proof: (cos A,sin A) is a point on the unit circle.
(cos A,sin A)
1
A
(0,0)
Hence, the distance from (cos A, sin A) to (0,0) is one unit.
∴ (cos 𝐴 − 0)2 + (sin 𝐴 − 0)2 = 1
∴ (cos 𝐴 − 0)2
+ (sin 𝐴 − 0)2
= 1 (squaring both sides)
cos2
𝐴 + sin2
𝐴 = 1
Derivation of Trigonometric Formulae 1, 2 and 3
07 Trigonometry

18. 18
ACTIVE MATHS 4 BOOK 2
Formula 2
There are two cases: 1. Acute-angled triangles 2. Obtuse-angled triangles
Δ ABC with sides a, b and c as shown below.
Given:
Sine Formula:
𝑎
sin 𝐴
=
𝑏
sin 𝐵
=
𝑐
sin 𝐶
(i) ∠C acute (ii) ∠C obtuse
To prove:
𝑎
sin 𝐴
=
𝑏
sin 𝐵
=
𝑐
sin 𝐶
In each triangle, draw BD perpendicular to AC.
Construction:
Proof: Acute case
|𝐵𝐷|
𝑎
= sin 𝐶 ……1
|𝐵𝐷|
𝑐
= sin 𝐴 ……2
∴ By dividing (1) by (2), we get:
𝑎
𝑐
=
sin 𝐴
sin 𝐶
.
(i)
A
B
C
a
b
c
(ii) B
C
A
a
c
b
D
90°
D
90°
Obtuse case
|𝐵𝐷|
𝑐
= sin 𝐴 ……1
|𝐵𝐷|
𝑎
= sin (180° − 𝐶) …… 2
= sin 𝐶
Similarly, by drawing a perpendicular from C to AB, we can prove that:
𝑐
sin 𝐶
Multiplying both sides by we get
𝑎
sin 𝐴
=
𝑐
sin 𝐶
.
𝑎
sin 𝐴
=
𝑏
sin 𝐵
∴
𝑎
sin 𝐴
=
𝑏
sin 𝐵
=
𝑐
sin 𝐶
.
(F and T: P16)
07 Trigonometry

19. 19
ACTIVE MATHS 4 BOOK 2
Formula 3 Cosine Formula: 𝑎2
= 𝑏2
+ 𝑐2
− 2𝑏𝑐 cos 𝐴
Proof: Increase the unit circle by a factor of b.
∴ The radius is b, and any point on the
circle can be described as (b cos A,b sin A),
where A is the angle between the x-axis
and the radius to the point.
Based on the co-ordinate geometry distance
formula, the distance between the vertices
containing angles B and C is:
𝑑 =∴ (𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2
∴ 𝑎 = (𝑏 cos 𝐴 − 𝑐)2 + (𝑏 sin 𝐴 − 0)2
𝑎2
= 𝑏2
𝑐𝑜𝑠2
𝐴 − 2𝑏𝑐 𝑐𝑜𝑠 𝐴 + 𝑐2
+ 𝑏2
𝑠𝑖𝑛2
𝐴
𝑎2 = 𝑏2(𝑐𝑜𝑠2 𝐴 + 𝑠𝑖𝑛2 𝐴) + 𝑐2 − 2𝑏𝑐 𝑐𝑜𝑠 𝐴
𝑎2
= 𝑏2
+ 𝑐2
− 2𝑏𝑐 𝑐𝑜𝑠 𝐴
As 𝑐𝑜𝑠2
𝐴 + 𝑠𝑖𝑛2
𝐴 = 1
(F and T: P16)
07 Trigonometry

20. 20
ACTIVE MATHS 4 BOOK 2
Find the values of the unknown sides and angles correct to one decimal place.
75∙5°
48°
5∙6 cm
A C
B
𝐵𝐴𝐶 = 180° − 48° − 75∙5° = 56∙5°
|𝐴𝐶|
sin 75∙5°
=
5∙6
sin 56∙5°
𝐴𝐶 =
5∙6sin 75∙5°
sin 56∙5°
𝐴𝐶 = 6∙5 cm
|𝐴𝐵|
sin 48°
=
5∙6
sin 56∙5°
|𝐴𝐵| =
5∙6 sin 48°
sin 56∙5°
|𝐴𝐵| = 4∙9906
𝐴𝐵 = 5 cm
𝐴𝐶 = 6∙5016
Sine Rule:
𝑎
sin 𝐴
=
𝑏
sin 𝐵
=
𝑐
sin 𝐶
(F and T: P16)
Using the Sine Rule
07 Trigonometry

21. 21
ACTIVE MATHS 4 BOOK 2
Cosine Rule: 𝑎2
= 𝑏2
+ 𝑐2
− 2𝑏𝑐 cos 𝐴 (F and T: P16)
NOTE: If the lengths of two sides and the angle between these sides are known, use Cosine Rule.
NOTE: If the lengths of all three sides are known, use Cosine Rule.
In the triangle given find |AB|.
A
B C
6
8
x
40°
𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴
𝑥2
= 62
+ 82
− 2 6 (8) cos 40°
𝑥2 = 26∙4597
𝑥 = 5∙1439
𝑥 = 5∙14
Calculate the measure of the angle A.
A
8
5
7
𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴
72
= 52
+ 82
− 2(5)(8) cos 𝐴
49 = 25 + 64 − 80 cos 𝐴
80 cos 𝐴 = 89 − 49
cos 𝐴 =
1
2
∴ 𝐴 = 60°
Using the Cosine Rule
07 Trigonometry

22. 22
ACTIVE MATHS 4 BOOK 2
Formula Area of a triangle =
1
2
𝑎𝑏 sin 𝐶 (F and T: P16)
c
b
a
C
NOTE: To use this formula we need the lengths
of two sides and the angle between them.
Find the area of Δ ABC.
A
B
C
12 cm
8 cm
30°
Area =
1
2
𝑎𝑏 sin 𝐶
Area =
1
2
(12)(8) sin 30°
Area =
1
2
(12)(8)
1
2
Area = 24 𝑐𝑚2
In the given triangle the area is 13.6 cm2.
Find the measure of the angle A to the
nearest degree.
A
8
7
1
2
8 7 sin 𝐴 = 13∙6
28 sin 𝐴 = 13∙6
sin 𝐴 =
13∙6
28
⇒ 𝐴 = 𝑠𝑖𝑛−1
13∙6
28
𝐴 = 29° 𝑂𝑅 𝐴 = 151°
From the diagram, A is clearly acute. ⇒ 𝐴 = 29°.
Area of a Triangle
07 Trigonometry

23. 23
ACTIVE MATHS 4 BOOK 2
Formula 𝜃
𝑙
𝐴
𝐵
Formula
Length of an arc 𝑙 = 𝑟𝜃 where 𝜃 is in radians. (F and T: P9)
Area of a sector A =
1
2
𝑟2
𝜃 where 𝜃 is in radians. (F and T: P9)
The arc AB subtends an angle
𝝅
𝟒
at the centre, O,
of a circle radius 8 cm.
(i) Find the length of the arc AB.
(ii) Find the area of the sector OAB.
𝜋
4
𝑙
𝐴
𝐵
8 cm
O
(i) Length of the arc AB.
𝑙 = 𝑟𝜃
𝑙 = (8)
𝜋
4
Length of the arc 2𝜋 cm.
(ii) Area of the sector OAB.
Area =
1
2
𝑟2
𝜃
=
1
2
(8)2
𝜋
4
= 32
𝜋
4
= 8𝜋 cm2
.
Area of a Sector and Length of an Arc
07 Trigonometry

24. 24
ACTIVE MATHS 4 BOOK 2
To solve a three-dimensional (3D) problem, it is important to be able to identify the
right-angled triangles in the question. Then redraw these right-angled triangles and use
appropriate trigonometric ratios or Pythagoras’ theorem to solve them.
A vertical tower stands 91 m tall on a horizontal plane. A tourist is standing at
point C. From C, the angle of elevation to the top of the tower, A, is 32°.
The tourist’s husband is standing at a point which is 235 m
from the base of the tower. Given |∠CBD| = 49°, calculate
(to two decimal places): (i) |BC| (ii) |CD|
90°
32°
91 m
49°
235 m
A
B
C
D
(i) tan 32° =
91
|𝐵𝐶|
𝐵𝐶 =
91
tan 32°
= 145∙63044
= 145∙63 m
49°
235 m
145∙63 m
B
C
D
Using the Cosine Rule: 𝑎2
= 𝑏2
+ 𝑐2
− 2𝑏𝑐 cos 𝐴
|𝐶𝐷|2 = (145∙63)2+(235)2−2(145∙63)(235) cos 49°
|𝐶𝐷|2
= 31,528∙415
|𝐶𝐷| = 177∙5624
(ii)
𝐶𝐷 = 177∙56 m
Three-Dimensional (3D) Problems
07 Trigonometry

25. 25
ACTIVE MATHS 4 BOOK 2
Formula 4
Let P (cos A,sin A) and Q (cos B,sin B) be two points on the unit circle.
Q (cos B,sin B)
P (cos A,sin A)
B
A
cos 𝐴 − 𝐵 = cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵 (F and T: P14)
1
1
A-B
Q (cos B,sin B)
P (cos A,sin A)
1
1
Using the Cosine Rule:
|𝑃𝑄|2 = 12 + 12 − 2 1 (1) cos (𝐴 − 𝐵)
Proof:
|𝑃𝑄|2
= 2 − 2 cos 𝐴 − 𝐵 ❶
Using the distance formula:
𝑃𝑄 = (cos 𝐴 − cos 𝐵)2 + (sin 𝐴 − sin 𝐵)2
|𝑃𝑄|2
= (cos 𝐴 − cos 𝐵)2
+ (sin 𝐴 − sin 𝐵)2
|𝑃𝑄|2
= cos2
𝐴 − 2cos 𝐴 cos 𝐵 + cos2
𝐵 + sin2
𝐴 − 2sin 𝐴 sin 𝐵 + sin2
𝐴
|𝑃𝑄|2
= cos2
𝐴 + sin2
𝐴 + cos2
𝐵 + sin2
𝐴 − 2cos 𝐴 cos 𝐵 − 2sin 𝐴 sin 𝐵
|𝑃𝑄|2
= 1 + 1 − 2cos 𝐴 cos 𝐵 − 2sin 𝐴 sin 𝐵
|𝑃𝑄|2
= 2 − 2cos 𝐴 cos 𝐵 − 2sin 𝐴 sin 𝐵 ❷
❶ = ❷ ⇒ 2 − 2 cos 𝐴 − 𝐵 = 2 − 2cos 𝐴 cos 𝐵 − 2sin 𝐴 sin 𝐵 Divide by −2
cos 𝐴 − 𝐵 = cos 𝐴 cos 𝐵 − 2sin 𝐴 sin 𝐵
Derivation of Trigonometric Formulae 4, 5, 6, 7 and 9
07 Trigonometry

26. 26
ACTIVE MATHS 4 BOOK 2
Formula 5 cos 𝐴 + 𝐵 = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 (F and T: P14)
Use the same proof as for Formula 4 above. Then replace B with −B on both sides:
cos 𝐴 − (−𝐵 = cos 𝐴 cos −𝐵 − sin 𝐴 sin (−𝐵)
As cos (−B) = cos B
⇒ cos 𝐴 + 𝐵 = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
and sin (−B) = −sin B
Formula 6
Proof:
Proof:
cos 𝐴 + 𝐵 = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
cos 𝐴 + 𝐴 = cos 𝐴 cos 𝐴 − sin 𝐴 sin 𝐴
cos 2𝐴 = co𝑠2
𝐴 − 𝑠𝑖𝑛2
𝐴
cos 2𝐴 = co𝑠2
𝐴 − 𝑠𝑖𝑛2
𝐴 (F and T: P14)
07 Trigonometry

27. 27
ACTIVE MATHS 4 BOOK 2
Formula 7 sin A + B = sin A cos B + cos A sin B (F and T: P14)
Proof:
Given: cos 𝐴 − 𝐵 = cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵
Replace A with
𝜋
2
− 𝐴 on both sides:
cos
𝜋
2
− 𝐴 − 𝐵 = cos
𝜋
2
− 𝐴 cos 𝐵 − sin
𝜋
2
− 𝐴 sin 𝐵
But cos
𝜋
2
− 𝐴 = sin 𝐴 ……… proof not needed for this: not needed for this course.
Also sin
𝜋
2
− 𝐴 = cos 𝐴
∴ cos
𝜋
2
− 𝐴 − 𝐵 = sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵
cos
𝜋
2
− (𝐴 + 𝐵) = sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵
⇒ sin 𝐴 + 𝐵 = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
07 Trigonometry

28. 28
ACTIVE MATHS 4 BOOK 2
Formula 9 tan 𝐴 + 𝐵 =
tan 𝐴 + tan 𝐵
1 − tan 𝐴 tan 𝐵
(F and T: P14)
Proof:
Given: tan 𝐴 =
sin 𝐴
cos 𝐴
tan (𝐴 + 𝐵) =
sin (𝐴 + 𝐵)
cos (𝐴 + 𝐵)
tan (𝐴 + 𝐵) =
sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
Divide above and below by cos A cos B:
tan (𝐴 + 𝐵) =
sin 𝐴 cos 𝐵
cos 𝐴 cos 𝐵 +
cos 𝐴 sin 𝐵
cos 𝐴 cos 𝐵
cos 𝐴 cos 𝐵
cos 𝐴 cos 𝐵 −
sin 𝐴 sin 𝐵
cos 𝐴 cos 𝐵
tan (𝐴 + 𝐵) =
tan 𝐴 + tan 𝐵
1 − tan 𝐴 tan 𝐵
tan (𝐴 + 𝐵) =
sin 𝐴
cos 𝐴 +
sin 𝐵
cos 𝐵
1 −
sin 𝐴 sin 𝐵
cos 𝐴 cos 𝐵
07 Trigonometry

29. 29
ACTIVE MATHS 4 BOOK 2
Express, in surd form: cos 15°.
cos 15° = cos (45° − 30) = cos 45° cos 30° + sin 45° sin 45°
=
1
2
×
3
2
+
1
2
×
1
2
=
3 + 1
2 2
Application of Formulae 1–24
07 Trigonometry

30. 30
ACTIVE MATHS 4 BOOK 2
If tan (A + B) = 4, and tan B = 3, find the value of tan A.
tan 𝐴 + 𝐵 =
tan 𝐴 + tan 𝐵
1 − tan 𝐴 tan 𝐵
Let tan 𝐴 = 𝑡.
Substitute what you know: 4 =
𝑡 + 3
1 − 3𝑡
Multiply both sides by (1 − 3𝑡) 4 1 − 3𝑡 = 𝑡 + 3
4 − 12𝑡 = 𝑡 + 3
1 = 13𝑡
𝑡 =
1
13
tan 𝐴 =
1
13
Write sin 6x + sin 4x as a product.
sin 𝐴 + sin 𝐵 = 2sin
𝐴 + 𝐵
2
cos
𝐴 − 𝐵
2
sin 6𝑥 + sin 4𝑥 = 2sin
6𝑥 + 4𝑥
2
cos
6𝑥 − 4𝑥
2
= 2sin 5𝑥 cos𝑥
07 Trigonometry