Grade 12
Mathematics
Maths and Science Infinity
CAPS Syllabus
Trigonometry
Trigonometric Identities
Solutions
Felix Pagona
Asitandile Yanxa
Thulani Mjikwa
Senior Facilitator
Lead Facilitator
Senior Facilitator
Activity 1
No.
1.1
Solution
tan 480° . sin 300° . cos 14° . sin(−135°) 3
=
sin 104° cos 225°
2
Marks
(6)
LHS
=
tan 300° . sin(360° − 60°) . cos 14° . sin(−135° + 360°)
sin(90° + 14°). cos(180° + 45°)
=
tan(360° − 60°) . (−)sin 60° . cos 14° . sin(180° + 45°)
(−) cos 14° . (−) cos 45°
=
(−) tan 60° . (−)sin 60° . cos 14° . (−) sin 45°
cos 14° . (−) cos 45°
= √3.
=
1.2
√3
2
3
2
LHS = RHS
√2(√3 − 1)
cos 75° =
4
(4)
LHS
cos 75°
= cos(45° + 30°)
= cos 45° cos 30° − sin 45° sin 30°
=(
√2 √3
√2 1
)( ) − ( )( )
2
2
2
2
=
√6 √2
−
4
4
=
√2. √3 − √2
4
=
√2(√3 − 1)
4
2
LHS = RHS
1.3
cos(90° − 2𝑥) . tan(180° + 𝑥) + sin2 (360° − 𝑥) = 3 sin2 𝑥
(6)
LHS
sin 2𝑥 . tan 𝑥 + sin2 𝑥
= 2 sin 𝑥 cos 𝑥 .
sin 𝑥
+ sin2 𝑥
cos 𝑥
= 2𝑠𝑖𝑛2 𝑥 + 𝑠𝑖𝑛2 𝑥
= 3𝑠𝑖𝑛2 𝑥
1.4
LHS = RHS
(tan 𝑥 − 1)(sin 2𝑥 − 2cos 2 𝑥) = 2(1 − 2 sin 𝑥 cos 𝑥)
(5)
LHS
=(
=
sin 𝑥
− 1) (2 sin 𝑥 cos 𝑥 − 2cos2 𝑥)
cos 𝑥
sin 𝑥
sin 𝑥
× 2 sin 𝑥 cos 𝑥 −
× 2cos2 𝑥 − 2 sin 𝑥 cos 𝑥 + 2cos2 𝑥
cos 𝑥
cos 𝑥
= 2sin2 𝑥 − 2 sin 𝑥 cos 𝑥 − 2 sin 𝑥 cos 𝑥 + 2cos2 𝑥
= 2(sin2 𝑥 + cos2 𝑥) − 4 sin 𝑥 cos 𝑥
= 2(1) − 4 sin 𝑥 cos 𝑥
= 2(2 − 2 sin 𝑥 cos 𝑥)
1.5
LHS = RHS
sin 190° . cos 225° . tan 390°
1
=−
cos 100° . sin 135°
√3
(7)
LHS
=
sin(180° + 10°) . cos(180° + 45°) . tan 210°
cos(90° + 10°) . sin(180° − 45°)
=
(−)sin 10° . (−) cos 45° . tan(180° + 30°)
(−) sin 10° . sin 45°
3
= (−) tan 30°
=−
1.6
1
√3
LHS = RHS
cos(360° − 𝑥) . tan2 𝑥
1
=
sin(𝑥 − 180°) . cos(90° + 𝑥) cos 𝑥
(5)
LHS
cos 𝑥 . tan2 𝑥
=
sin(𝑥 − 180° + 360°) . (−) sin 𝑥
=
cos 𝑥 . tan2 𝑥
sin(180° + 𝑥) . (−) sin 𝑥
sin2 𝑥
cos 2 𝑥
=
(−) sin 𝑥 . (−) sin 𝑥
cos 𝑥 .
1.7
=
sin2 𝑥
÷ sin2 𝑥
cos 𝑥
=
sin2 𝑥
1
× 2
cos 𝑥 sin 𝑥
=
1
cos 𝑥
LHS = RHS
sin4 𝑥+sin2 𝑥.cos2 𝑥
= 1 − cos 𝑥
1+cos 𝑥
(4)
LHS
=
sin4 𝑥 + sin2 𝑥. cos 2 𝑥
1 + cos 𝑥
=
sin2 𝑥. sin2 𝑥 + sin2 𝑥. cos2 𝑥
1 + cos 𝑥
4
=
sin2 𝑥(sin2 𝑥 + cos 2 𝑥)
1 + cos 𝑥
sin2 𝑥(1)
=
1 + cos 𝑥
1 − cos 2 𝑥
=
1 + cos 𝑥
=
(1 − cos 𝑥)(1 + cos 𝑥)
(1 + cos 𝑥)
= 1 − cos 𝑥
1.8
LHS = RHS
4 sin θ cos θ cos 2θ sin 15° √6 − √2
=
sin 2θ (tan 225° − 2sin2 θ)
2
(6)
LHS
=
2 × 2 sin θ cos θ cos 2θ sin 15°
sin 2θ (tan(180° + 45°) − 2sin2 θ)
=
2 sin 2𝜃 . cos 2θ sin 15°
sin 2θ (tan 45° − 2sin2 θ)
=
2 cos 2θ sin 15°
(1 − 2sin2 θ)
=
2 cos 2θ sin 15°
cos 2θ
= 2[sin(45° − 30°)]
= 2(sin 45° cos 30° − sin 30° cos 45°)
1 √2
√2 √3
= 2 [( ) ( ) − ( ) ( )]
2
2
2
2
√2. √3 √2
= 2(
− )
4
4
5
√2. √3 − √2
= 2(
)
2×2
=
1.9
√6 − √2
2
LHS = RHS
cos 2𝑥
cos 𝑥 − sin 𝑥
=
1 + sin 2𝑥 sin 𝑥 + cos 𝑥
(4)
LHS
=
cos2 𝑥 − sin2 𝑥
sin2 𝑥 + cos2 𝑥 + 2 sin 𝑥 cos 𝑥
=
(cos 𝑥 − sin 𝑥)(cos 𝑥 + sin 𝑥)
(sin 𝑥 + cos 𝑥)(sin 𝑥 + cos 𝑥)
=
(cos 𝑥 − sin 𝑥)
(sin 𝑥 + cos 𝑥)
LHS = RHS
Activity 2
No.
2.1
Solution
tan(180° + 𝑥) . cos(360° − 𝑥)
sin(180 − 𝑥) . cos(90° + 𝑥) + cos(540° + 𝑥) . cos(−𝑥)
=
tan 𝑥 . cos 𝑥
sin 𝑥 . (−) sin 𝑥 + cos(540° + 𝑥 − 360°) . cos 𝑥
sin 𝑥
cos
𝑥 × cos 𝑥
=
(−)sin2 𝑥 + cos(180° + 𝑥) . cos 𝑥
=
=
sin 𝑥
(−)sin2 𝑥 − cos2 𝑥
sin 𝑥
+ cos2 𝑥)
−1(sin2 𝑥
6
Marks
(8)
=
sin 𝑥
−1(1)
= − sin 𝑥
2.2
2.3
1 − cos 2𝑥 − sin 𝑥
sin 2𝑥 − cos 𝑥
=
1 − (1 − 2sin2 𝑥) − sin 𝑥
2 sin 𝑥 cos 𝑥 − cos 𝑥
=
2sin2 𝑥 − sin 𝑥
cos 𝑥 (2 sin 𝑥 − 1)
=
sin 𝑥 (2 sin 𝑥 − 1)
cos 𝑥 (2 sin 𝑥 − 1)
(5)
= tan 𝑥
sin(90° − 𝑥) . cos(180° − 𝑥) + tan 𝑥 . cos(−𝑥) . sin(180° + 𝑥)
(7)
= cos 𝑥 . (−) cos 𝑥 + tan 𝑥 . cos 𝑥 . (−) sin 𝑥
= (−)cos2 𝑥 +
sin 𝑥
. cos 𝑥 . (−) sin 𝑥
cos 𝑥
= −cos 2 𝑥 − sin2 𝑥
= −1(cos2 𝑥 + sin2 𝑥)
= −1
2.4
sin2 θ
sin(180° − θ) . cos(90° + θ) + tan 45°
=
sin2 θ
sin 𝜃 . (−) sin 𝜃 + 1
=
sin2 θ
1 − sin2 θ
(5)
sin2 θ
=
cos 2 θ
= tan2 𝑥
7
2.5
2.6
2.7
sin 104° (2cos2 15° − 1)
tan 38° . sin2 412°
=
sin(90° + 14°) cos 30°
tan 38° sin2 (412° − 360°)
=
cos 14°
sin 38°
2
cos 38° . sin 52°
=
sin 76°
sin 38°. cos 38°
=
sin 2(38°)
sin 38°. cos 38°
=
2 sin 38°. cos 38°
sin 38°. cos 38°
(8)
=2
sin(90° + θ) + cos(180° + θ) sin(−θ)
sin 180° − tan 135°
=
cos 𝜃 + (−) cos 𝜃 . (−) sin 𝜃
0 − tan(180° − 45°)
=
cos 𝜃 + cos 𝜃 . sin 𝜃
0 − (−)tan 45°
=
cos 𝜃 (1 + sin 𝜃)
1
(5)
= cos 𝜃 (1 + sin 𝜃)
sin(450° − 𝑥). tan(𝑥 − 180°) sin 23° cos 23°
cos 44° . sin(−𝑥)
=
sin(450° − 𝑥 − 360°). tan(𝑥 − 180° + 180°) sin 23° cos 23°
cos 44° . (−)sin 𝑥
=
sin(90° − 𝑥). tan 𝑥 . sin 23° cos 23°
cos 44° . (−)sin 𝑥
=
cos 𝑥. tan 𝑥 . sin 23° cos 23°
2 sin 23° cos 23° . (−)sin 𝑥
8
(6)
=
sin 𝑥
cos 𝑥 . cos 𝑥
−2 sin 𝑥
=−
2.8
1
2
sin 130° . tan 60°
cos 540° . tan 230° . sin 400°
(7)
=
sin(180° − 50°) tan 60°
cos(540° − 360°) tan(180° + 50°) sin(400° − 360°)
=
sin 50° tan 60°
cos 180° tan 50° sin 40°
=
sin 50° . √3
sin 50°
(−1)
× sin 40°
cos 50°
= −√3
2.9
(4)
(1 − √2 sin 75°)(√2 sin 75° + 1)
= (1 − √2 sin 75°)(1 + √2 sin 75°)
= 1 − 2sin2 75°
= sin 2(75°)
= sin 150°
= sin(180° − 30°)
= sin 30°
=
1
2
2.10 tan(360°−𝑥).sin(90°+𝑥)
sin(−𝑥)
=
(5)
(−) tan 𝑥 . cos 𝑥
sin(−𝑥 + 360°)
9
=−
(−) tan 𝑥 . cos 𝑥
(−) sin 𝑥
sin 𝑥
× cos 𝑥
= cos 𝑥
sin 𝑥
=1
Activity 3
No.
3.1
Solution
cos 2𝑥 = 1 − 3 cos 𝑥
()
2cos2 𝑥 − 1 + 3 cos 𝑥 − 1 = 0
2cos2 𝑥 + 3 cos 𝑥 − 2 = 0
(2 cos 𝑥 − 1)(cos 𝑥 + 2) = 0
1
cos 𝑥 = 2
or
𝑥𝑟𝑒𝑓 = 60°
or
cos 𝑥 = −2
No solution
First quadrant: 𝑥 = 60° + 360°𝑘, 𝑘 ∈ ℤ
First quadrant: 𝑥 = −300° + 360°𝑘, 𝑘 ∈ ℤ
Fourth quadrant: 𝑥 = 300° + 360°𝑘, 𝑘 ∈ ℤ
Fourth quadrant: 𝑥 = −60° + 360°𝑘, 𝑘 ∈ ℤ
3.2
Marks
tan 𝑥 − 1
= −3
2
tan 𝑥 − 1 = −6
tan 𝑥 = −5
𝑥𝑟𝑒𝑓 = 78,69°
Second quadrant: 𝑥 = 101,31° + 180°𝑘, 𝑘 ∈ ℤ
Second quadrant: 𝑥 = −258,69° + 180°𝑘, 𝑘 ∈ ℤ
Fourth quadrant: 𝑥 = 281,31° + 180°𝑘, 𝑘 ∈ ℤ
10
Fourth quadrant: 𝑥 = −78,69° + 180°𝑘, 𝑘 ∈ ℤ
3.3
sin 𝑥 + 2cos2 𝑥 = 1
sin 𝑥 + 2(1 − sin2 𝑥) − 1 = 0
−2sin2 𝑥 + sin 𝑥 + 2 − 1 = 0
2sin2 𝑥 − sin 𝑥 − 1 = 0
(2 sin 𝑥 + 1)(sin 𝑥 − 1) = 0
1
sin 𝑥 = − 2
sin 𝑥 = 1
or
1
2
= 30°
sin 𝑥 = 1
sin 𝑥 = −
𝑥𝑟𝑒𝑓
Third quadrant:
𝑥 = 210° + 360°𝑘,
𝑥𝑟𝑒𝑓 = 90°
𝑥 = 90° + 360°𝑘,
𝑘∈ℤ
𝑥 = −150° + 360°𝑘,
Fourth quadrant:
3.4
𝑘∈ℤ
𝑥 = −270° + 360°𝑘,
𝑥 = 330° + 360°𝑘,
𝑘∈ℤ
𝑥 = −30° + 360°𝑘,
𝑘∈ℤ
6 cos 𝑥 − 5 =
4
cos 𝑥
; cos 𝑥 ≠ 0
6cos2 𝑥 − 5 cos 𝑥 − 4 = 0
(2 cos 𝑥 + 1)(3 cos 𝑥 − 4) = 0
1
𝑘∈ℤ
4
cos 𝑥 = − 2
or
cos 𝑥 = 3
𝑥𝑟𝑒𝑓 = 60°
or
No solution
Second quadrant: 𝑥 = 120° + 360°𝑘, 𝑘 ∈ ℤ
Second quadrant: 𝑥 = −240° + 360°𝑘, 𝑘 ∈ ℤ
Third quadrant: 𝑥 = 240° + 360°𝑘, 𝑘 ∈ ℤ
Third quadrant: 𝑥 = −120° + 360°𝑘, 𝑘 ∈ ℤ
11
𝑘∈ℤ
3.6
1 + 4sin2 𝑥 − 5 sin 𝑥 + cos 2𝑥 = 0
1 + 4sin2 𝑥 − 5 sin 𝑥 + 1 − 2sin2 𝑥 = 0
2sin2 𝑥 − 5 sin 𝑥 + 2 = 0
(2 sin 𝑥 − 1)(sin 𝑥 − 2) = 0
1
3.7
sin 𝑥 = − 2
or
sin 𝑥 = 2
𝑥𝑟𝑒𝑓 = 30°
or
No solution
Third quadrant:
𝑥 = 210° + 360°𝑘, 𝑘 ∈ ℤ
Third quadrant:
Fourth quadrant:
𝑥 = −150° + 360°𝑘, 𝑘 ∈ ℤ
𝑥 = 330° + 360°𝑘, 𝑘 ∈ ℤ
Fourth quadrant:
𝑥 = −30° + 360°𝑘, 𝑘 ∈ ℤ
sin2 𝑥 + cos 2𝑥 − cos 𝑥 = 0
1 − cos2 𝑥 + 2cos 2 𝑥 − 1 − cos 𝑥 = 0
cos2 𝑥 − cos 𝑥 = 0
cos 𝑥 (cos 𝑥 − 1) = 0
cos 𝑥 = 0
or
cos 𝑥 = 1
cos 𝑥 = 0
𝑥𝑟𝑒𝑓 = 90°
𝑥 = 90° + 360°𝑘,
3.8
cos 𝑥 = 1
𝑘∈ℤ
𝑥𝑟𝑒𝑓 = 0°
𝑥 = 0° + 360°𝑘,
𝑥 = −270° + 360°𝑘,
𝑥 = 270° + 360°𝑘,
𝑘∈ℤ
𝑘∈ℤ
𝑥 = 360° + 360°𝑘,
𝑥 = −90° + 360°𝑘,
𝑘∈ℤ
𝑥 = −360° + 360°𝑘,
cos 2𝑥 − 7 cos 𝑥 − 3 = 0
2cos2 𝑥 − 1 − 7 cos 𝑥 − 3 = 0
2cos2 𝑥 − 7 cos 𝑥 − 4 = 0
(2 cos 𝑥 + 1)(cos 𝑥 − 4) = 0
12
𝑘∈ℤ
𝑘∈ℤ
𝑘∈ℤ
1
cos 𝑥 = − 2
𝑥𝑟𝑒𝑓 = 60°
cos 𝑥 = 4
or
or
No solution
Second quadrant: 𝑥 = 120° + 360°𝑘, 𝑘 ∈ ℤ
Second quadrant: 𝑥 = −240° + 360°𝑘, 𝑘 ∈ ℤ
Third quadrant: 𝑥 = 240° + 360°𝑘, 𝑘 ∈ ℤ
Third quadrant: 𝑥 = −120° + 360°𝑘, 𝑘 ∈ ℤ
3.9
cos 2𝑥 = sin 𝑥 − 2
1 − 2sin2 𝑥 − sin 𝑥 + 2 = 0
2sin2 𝑥 + sin 𝑥 − 3 = 0
(2 sin 𝑥 + 3)(sin 𝑥 − 1) = 0
3
sin 𝑥 = 1
or
sin 𝑥 = − 2
𝑥𝑟𝑒𝑓 = 90°
or
No solution
𝑥 = 90° + 360°𝑘,
𝑘∈ℤ
𝑥 = −270° + 360°𝑘,
𝑘∈ℤ
3.10 sin 𝑥 = 2cos 2 15° − 1
sin 𝑥 = cos 30°
sin 𝑥 =
√3
2
𝑥𝑟𝑒𝑓 = 60°
First quadrant:
𝑥 = 60° + 360°𝑘, 𝑘 ∈ ℤ
First quadrant:
𝑥 = −300° + 360°𝑘, 𝑘 ∈ ℤ
Second quadrant:
𝑥 = 120° + 360°𝑘, 𝑘 ∈ ℤ
Second quadrant:
𝑥 = −240° + 360°𝑘, 𝑘 ∈ ℤ
13
Activity 4
No.
4.1.1
Solution
Marks
(4)
√5 − 𝑝2
tan 𝛽 = −
𝑝
4.1.2
cos 2𝛽
(3)
= 2cos2 𝛽 − 1
= 2 (−
=
4.2.1
2
) −1
√5
2𝑝2
−1
5
cos 𝛼
=−
4.2.2
𝑝
(3)
12
13
cos(𝛼 + 𝛽)
(5)
= cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽
= (−
=
4.3.1
33
65
tan 𝛼
=−
4.3.2
12
4
5 3
) (− ) − ( ) ( )
13
5
13 5
(3)
8
15
sin(90° + 𝛼)
(2)
= cos 𝛼
=−
15
17
14
4.3.3
cos 2𝛼
(3)
= 1 − 2sin2 𝑥
8 2
= 1 − 2( )
17
=
4.4.1
sin 𝛼
=
4.4.2
161
289
(3)
3
5
cos2 (90° − 𝛼) − 1
(2)
= sin2 𝑥 − 1
3 2
=( ) −1
5
=−
4.4.3
16
25
1 − 2 sin 2𝛼
(3)
= 1 − 2.2 sin 𝛼 cos 𝛼
3 4
= 1 − 4( )( )
5 5
=−
4.5.1
23
25
sin θ + cos θ
(4)
3 4
=− −
5 5
=−
4.5.2
7
5
tan 2θ
=
(5)
sin 2𝜃
cos 2𝜃
15
=
2 sin 𝜃 cos 𝜃
2cos 2 − 1
3
4
2 (− ) (− )
5
5
=
2
4
2 (− ) − 1
5
=
4.6.1
24
7
sin 203°
(2)
= sin(180° + 23°)
= − sin 23°
= −√𝑘
4.6.2
cos 23°
(3)
= √1 − 𝑘
4.6.3
tan(−23°)
(2)
= tan(−23° + 180°)
= (−)tan(23°)
=−
4.7.1
√𝑘
√1 − 𝑘
cos 113°
(2)
= cos(90° + 23°)
= − sin 23°
= −𝑝
4.7.2
cos 23°
(2)
= √1 − 𝑝2
16
4.7.3
sin 46°
(2)
= 2 sin 23° cos 23°
= 2(𝑝) (√1 − 𝑝2 )
= 2𝑝√1 − 𝑝2
4.8.1
sin 48°
(3)
= sin(36° + 12°)
= sin 36° cos 12° + cos 36° sin 12°
=𝑝+𝑞
4.8.2
sin 24°
(3)
= sin(36° − 12°)
= sin 36° cos 12° − cos 36° sin 12°
=𝑝−𝑞
4.8.3
cos 24°
(3)
2 sin 24° cos 24° = sin 48°
2(𝑝 − 𝑞) cos 24° = 𝑝 + 𝑞
cos 24° =
4.9.1
𝑝+𝑞
2(𝑝 − 𝑞)
cos 28°
(2)
= √1 − 𝑎2
4.9.2
cos 64°
(3)
= cos 2(32°)
= 2cos2 32° − 1
= 2𝑏 2 − 1
17
4.9.3
sin 4°
(4)
= sin(32° − 28°)
= sin 32° cos 28° − sin 28° cos 32°
= (√1 − 𝑏 2 ) (√1 − 𝑎2 ) − (𝑎)(𝑏)
= √(1 − 𝑏 2 )(1 − 𝑎2 ) − 𝑎𝑏
4.10.1 sin 28°
(2)
=𝑚
4.10.2 cos 362°
(4)
= cos 2°
= cos(62° − 60°)
= cos 62° cos 60° + sin 62° sin 60°
1
√3
= (𝑚) ( ) + (√1 − 𝑚2 ) ( )
2
2
=
𝑚 + √3(1 − 𝑚2 )
2
4.11.1 sin 241°
(2)
= sin(180° + 61°)
= − sin 61°
= −√𝑝
4.11.2 cos 61°
(2)
= √1 − 𝑝
4.11.3 cos 122°
(3)
= cos 2(61°)
18
= 2cos2 61° − 1
2
= 2(√1 − 𝑝) − 1
= 2 − 2𝑝 − 1
= 1 − 2𝑝
4.12.1 cos 214°
(2)
= cos(180° + 34°)
= − cos 34°
= −𝑝
4.12.2 cos 68°
(2)
= cos 2(34°)
= 2cos2 34° − 1
= 2𝑝2 − 1
4.12.3 tan 56°
(4)
𝑟2 = 𝑥2 + 𝑦2
12 = 𝑝2 + 𝑦 2
𝑦 = √1 − 𝑝2
=
√1 − 𝑝2
𝑝
4.13.1 sin 58°
=
(2)
1
√1 + 𝑚2
4.13.2 sin 296°
(3)
= sin(360° − 64°)
= − sin 64°
19
= − sin 2(32)°
= −2 sin 32° cos 32°
= −2 (
=−
𝑚
1
)(
)
√1 + 𝑚2 √1 + 𝑚2
2𝑚
1 + 𝑚2
4.13.3 cos 2°
(3)
= cos(32° − 30°)
= cos 32° cos 30° + sin 32° sin 30°
=(
=
=
1
𝑚
1
√3
)( ) + (
)( )
2
√1 + 𝑚2
√1 + 𝑚2 2
√3
2√1 + 𝑚2
+
𝑚
2√1 + 𝑚2
√3 + 𝑚
2√1 + 𝑚2
20