Unit 7A: Trigonometry

Classwork/Homework
Activity 7A.1. P.10
1. If 𝑝 = 49° and 𝑞 = 32°, use a calculator to determine whether the
following statements are true or false.
1.1. sin 𝑝 + 3 sin 𝑝 = 4 sin 𝑝
1.2. cos(𝑝 − 𝑞) = cos 𝑝 − cos 𝑞
1.3. sin(2𝑝) = 2 sin 𝑝 cos 𝑝

2. Determine the following angles (correct to one decimal place):

2.1. cos 𝛼 = 0,64
1
2.2. cos 2𝛽 = 0,3
2
2.3. 2 sin 3𝛽 + 1 = 2,6
2.4. sin 𝜃 2
=4
cos 𝜃 3

3. In ∆𝐴𝐵𝐶, 𝐴𝐶̂ 𝐵 = 30°, 𝐴𝐶 = 20 cm and 𝐵𝐶 = 22 cm. The perpendicular

line from 𝐴 intersects 𝐵𝐶 at 𝑇.

Determine:
3.1. the length 𝑇𝐶 (Leave your answer in simplest surd form)
3.2. the length 𝐴𝑇
3.3. the angle 𝐵𝐴̂𝑇 (Round your answer off to 2 decimal places)

4. A rhombus has a perimeter of 40 cm and one of the internal angles is 30°.

Round your answer off to 2 decimal places.
4.1. Determine the lengths of the sides.
4.2. Determine the lengths of the diagonals.
4.3. Calculate the area of the rhombus.

5. Simplify the following without using a calculator:

5.1. 2 sin 45° × 2 cos 45°
5.2. cos2 30° − sin2 60°
5.3. 4 sin 60° cos 30° − 2 tan 45° + tan 60° − 2 sin 60°
5.4. sin 60° × √2 tan 45° + 1 − sin 30°
6. Given the diagram below.

B(2; 2√3)

Determine the following without using a calculator:

6.1. 𝛽
6.2. cos 𝛽
6.3. cos2 𝛽 + sin2 𝛽

7. The 10 m ladder of a fire truck leans against the wall of a burning building
at an angle of 60°. The height of an open window is 9 m from the ground.
Will the ladder reach the window?

Answers:

1. 𝑝 = 49° and 𝑞 = 32°

1.1. sin 𝑝 + 3 sin 𝑝 = 4 sin 𝑝

𝐿𝐻𝑆 = sin 49° + 3 sin 49°

𝐿𝐻𝑆 = 0,75470958 + 3(0,75470958)
𝐿𝐻𝑆 = 0,75470958 + 2,264128741
𝐿𝐻𝑆 = 3,018838321
𝐿𝐻𝑆 = 3,02

𝑅𝐻𝑆 = 4 sin 49°

𝑅𝐻𝑆 = 4(0,75470958)
𝑅𝐻𝑆 = 3,018838321
𝑅𝐻𝑆 = 3,02

∴ True, 𝐿𝐻𝑆 = 𝑅𝐻𝑆

1.2. cos(𝑝 − 𝑞) = cos 𝑝 − cos 𝑞

𝐿𝐻𝑆 = cos(49° − 32°)

𝐿𝐻𝑆 = cos 17°
𝐿𝐻𝑆 = 0,956304756
𝐿𝐻𝑆 = 0,96
 𝑅𝐻𝑆 = cos 49° − cos 32°
𝑅𝐻𝑆 = 0,656059029 − 0,848048096
𝑅𝐻𝑆 = −0,191989067
𝑅𝐻𝑆 = −0,19

∴ False, 𝐿𝐻𝑆 ≠ 𝑅𝐻𝑆

1.3. sin(2𝑝) = 2 sin 𝑝 cos 𝑝

𝐿𝐻𝑆 = sin 2(49°)

𝐿𝐻𝑆 = sin 98°
𝐿𝐻𝑆 = 0,990268068
𝐿𝐻𝑆 = 0,99

𝑅𝐻𝑆 = 2 sin 49° cos 49°

𝑅𝐻𝑆 = 2(0,75470958)(0,656059029)
𝑅𝐻𝑆 = 0,990268068
𝑅𝐻𝑆 = 0,99

∴ True, 𝐿𝐻𝑆 = 𝑅𝐻𝑆

1
2.1. cos 𝛼 = 0,64 2.2. cos 2𝛽 = 0,3
2
𝛼 = cos −1 0,64 cos 2𝛽 = 0,6
𝛼 = 50,2081805° 2𝛽 = cos −1 0,6
𝛼 = 50,2° 2𝛽 = 53,13010235°
𝛽 = 26,56505118°
𝛽 = 26,6°

2.3. 2 sin 3𝛽 + 1 = 2,6 2.4. sin 𝜃 2

=4
cos 𝜃 3
2
2 sin 3𝛽 = 1,6 tan 𝜃 = 4
3
2
sin 3𝛽 = 0,8 𝜃= tan−1 4
3
3𝛽 = sin−1 0,8 𝜃 = 77,90524292°
3𝛽 = 53,13010235° 𝜃 = 77,9°
𝛽 = 17,71003412°
𝛽 = 17,7°

𝑎𝑑𝑗 𝑜𝑝𝑝
3.1 cos 𝜃 = 3.2 sin 𝜃 =
ℎ𝑦𝑝 ℎ𝑦𝑝
𝐶𝑇 𝐴𝑇
cos 30° = sin 30° =
20 20
𝐶𝑇 = 20 cos 30° 𝐴𝑇 = 20 sin 30°
√3 1
𝐶𝑇 = 20 ( ) 𝐴𝑇 = 20 ( )
2 2
 𝐶𝑇 = 10√3 𝐴𝑇 = 10 cm
𝐶𝑇 = 17,32050808
𝐶𝑇 = 17,32 cm

3.3. 𝐶𝐵 = 𝐶𝑇 + 𝑇𝐵 4.1. 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 4 × 𝑙

22 = 17,32 + 𝑇𝐵 40 = 4 × 𝑙
𝑇𝐵 = 22 − 17,32 𝑙 = 10 𝑐𝑚
𝑇𝐵 = 4,68 cm

𝑜𝑝𝑝
tan 𝛼 =
𝑎𝑑𝑗
4,68
tan 𝐵𝐴̂𝑇 =
10
tan 𝐵𝐴̂𝑇 = 0,468
𝐵𝐴̂𝑇 = tan−1 0,468
𝐵𝐴̂𝑇 = 25,07959402°
𝐵𝐴̂𝑇 = 25,08°

4.2. Determine the lengths of the diagonals.

• Two diagonals are perpendicular

• Diagonals bisect opposite angles
• Diagonals bisect each other

In ∆𝐸𝐷𝐶:

𝐷𝐸
⇝ 𝐷𝐸: sin 15° =
𝐷𝐶
𝐷𝐸
𝐷𝐸: sin 15° =
10
𝐷𝐸: 𝐷𝐸 = 10 sin 15°
𝐷𝐸: 𝐷𝐸 = 10(0,258819045)
𝐷𝐸: 𝐷𝐸 = 2,588190451
𝐷𝐸: 𝐷𝐸 = 2,59 cm
∴ 𝐷𝐵 = 2𝐷𝐸
∴ 𝐷𝐵 = 2(2,59)
∴ 𝐷𝐵 = 5,18 cm

𝐸𝐶
⇝ 𝐷𝐸: cos 15° =
𝐷𝐶
𝐸𝐶
𝐷𝐸: cos 15° =
10
 𝐷𝐸: 𝐸𝐶 = 10 cos 15°
𝐷𝐸: 𝐸𝐶 = 10(0,965925826)
𝐷𝐸: 𝐸𝐶 = 9,659258263
𝐷𝐸: 𝐸𝐶 = 9,66 cm

∴ 𝐴𝐶 = 2𝐸𝐶
∴ 𝐴𝐶 = 2(9,66)
∴ 𝐴𝐶 = 19,32 cm

4.3. 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = 1 (𝑏𝑎𝑠𝑒)(𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ℎ𝑒𝑖𝑔ℎ𝑡)

2
1
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = (19,32)(2,59)
2
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = 25,0194 cm2

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟ℎ𝑜𝑚𝑏𝑢𝑠 = 2(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒)

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑜𝑚𝑏𝑢𝑠 = 2(25,0194 )
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑜𝑚𝑏𝑢𝑠 = 50,0388
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑜𝑚𝑏𝑢𝑠 = 50,04 cm2

OR

⏊ℎ
⏊ℎ𝑒𝑖𝑔ℎ𝑡: sin 30° =
10
⏊ℎ𝑒𝑖𝑔ℎ𝑡: ⏊ℎ = 10 sin 30°
1
⏊ℎ𝑒𝑖𝑔ℎ𝑡: ⏊ℎ = 10 ( )
2
⏊ℎ𝑒𝑖𝑔ℎ𝑡: ⏊ℎ = 5 cm

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟ℎ𝑜𝑚𝑏𝑢𝑠 = (𝑏𝑎𝑠𝑒)(⏊ℎ𝑒𝑖𝑔ℎ𝑡 )

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑜𝑚𝑏𝑢𝑠 = 10(5)
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑜𝑚𝑏𝑢𝑠 = 50 cm2
5.

5.1. 2 sin 45° × 2 cos 45° 5.2. cos2 30° − sin2 60°
1 1 2 2
√3 √3
= 2( ) × 2( ) =( ) −( )
√2 √2 2 2
 2 2 3 3
= × = −
√2 √2 4 4
4
= =0
2
=2

5.3. 4 sin 60° cos 30° − 2 tan 45° + tan 60° − 2 sin 60°
√3 √3 √3
= 4 ( ) ( ) − 2(1) + (√3) − 2 ( )
2 2 2
3
= 4 ( ) − 2 + √3 − √3
4
=3−2
=1

5.4. sin 60° × √2 tan 45° + 1 − sin 30°

√3 1
= ( ) × √2(1) + 1 − ( )
2 2
√3 1
= ( ) × √3 −
2 2
3 1
= −
2 2
=1

6.1. tan 𝛽 = 𝑜𝑝𝑝 6.2. 𝑟 2 = 𝑥 2 + 𝑦 2

𝑎𝑑𝑗
2√3
tan 𝛽 =
2
𝑟 = √(2)2 + (2√3)

tan 𝛽 = √3 𝑟 = √4 + 4.3
𝛽 = tan−1 √3 𝑟 = √4 + 12
𝛽 = 60° 𝑟 = √16
𝑟 =4

𝑥
cos 𝛽 =
𝑟
2
cos 𝛽 =
4
1
cos 𝛽 =
2

6.3. cos2 𝛽 + sin2 𝛽 7.

2
1 2 2√3
=( ) +( )
2 4
1 4.3
= +
4 16
1 12
= +
4 16
1 3
= +
4 4
=1
 𝐻𝑒𝑖𝑔ℎ𝑡
sin 60° =
10
𝐻𝑒𝑖𝑔ℎ𝑡 = 10 sin 60°
√3
𝐻𝑒𝑖𝑔ℎ𝑡 = 10 ( )
2
𝐻𝑒𝑖𝑔ℎ𝑡 = 5√3
𝐻𝑒𝑖𝑔ℎ𝑡 = 8,66 m

∴ The ladder is too short

Classwork/Homework
Activity 7A.2. P.16
1. Reduce the following to one trigonometric ratio:

1.1. sin 𝛼
tan 𝛼
1.2. cos 𝜃 tan2 𝜃 + tan2 𝜃 sin2 𝜃
2

1.3. 1 − sin 𝜃 cos 𝜃 tan 𝜃

1.4. 1−cos2 𝛽
( ) − tan2 𝛽
cos2 𝛽

2. Prove the following identities:

2.1. 1+sin 𝜃 cos 𝜃

=
cos 𝜃 1−sin 𝜃
2.2. sin 𝛼 + (cos 𝛼 − tan 𝛼)(cos 𝛼 + 𝑡𝑎𝑛𝛼) = 1 − tan2 𝛼
2

2.3. 1 cos 𝜃 tan2 𝜃

− = cos 𝜃
cos 𝜃 1
2.4. 2 sin 𝜃 cos 𝜃 1
= sin 𝜃 + cos 𝜃 −
sin 𝜃+cos 𝜃 sin 𝜃+cos 𝜃
2.5. cos 𝛽 1
( + tan 𝛽) 𝑐𝑜𝑠𝛽 =
sin 𝛽 sin 𝛽
2.6. 1 1 2 tan 𝜃
+ =
1+sin 𝜃 1−sin 𝜃 sin 𝜃 cos 𝜃
2.7. (1+tan2 𝛼) cos 𝛼 1
=
(1−tan 𝛼) cos 𝛼−sin 𝛼

Answers:

1.1. sin 𝛼
tan 𝛼
sin 𝛼 sin 𝛼
= sin 𝛼 = tan 𝛼
cos 𝛼 cos 𝛼
cos 𝛼
= sin 𝛼 ×
sin 𝛼
= cos 𝛼
1.2. cos2 𝜃 tan2 𝜃 + tan2 𝜃 sin2 𝜃
= tan2 𝜃 (cos2 𝜃 + sin2 𝜃)
= tan2 𝜃 (1) sin2 𝜃 + cos 2 𝜃 = 1
= tan2 𝜃

1.3. 1 − sin 𝜃 cos 𝜃 tan 𝜃

sin 𝜃 sin 𝜃
= 1 − sin 𝜃 cos 𝜃 = tan 𝜃
cos 𝜃 cos 𝜃
= 1 − sin 𝜃 sin 𝜃
= 1 − sin2 𝜃
= cos2 𝜃 cos2 𝜃 = 1 − sin2 𝜃

1.4. (1−cos2 𝛽) − tan2 𝛽

2
cos 𝛽
sin2 𝛽
=( ) − tan2 𝛽 sin2 𝛽 = 1 − cos2 𝛽
cos2 𝛽
sin2 𝛽
=( ) − tan2 𝛽
cos2 𝛽
sin2 𝛽
= tan2 𝛽 − tan2 𝛽 = tan2 𝛽
cos2 𝛽
=0

2.1. 1+sin 𝜃 cos 𝜃

=
cos 𝜃 1−sin 𝜃

1+sin 𝜃
𝐿𝐻𝑆 =
cos 𝜃
(1+sin 𝜃) (1−sin 𝜃)
𝐿𝐻𝑆 = ×
cos 𝜃 (1−sin 𝜃)
1−sin2 𝜃
𝐿𝐻𝑆 =
cos 𝜃(1−sin 𝜃)
cos2 𝜃 = 1 − sin2 𝜃
cos2 𝜃
𝐿𝐻𝑆 =
cos 𝜃(1−sin 𝜃)
cos 𝜃 cos 𝜃
𝐿𝐻𝑆 =
cos 𝜃(1−sin 𝜃)
cos 𝜃
𝐿𝐻𝑆 =
1−sin 𝜃

cos 𝜃
𝑅𝐻𝑆 =
1−sin 𝜃

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

2.2. sin2 𝛼 + (cos 𝛼 − tan 𝛼)(cos 𝛼 + 𝑡𝑎𝑛𝛼) = 1 − tan2 𝛼

𝐿𝐻𝑆 = sin2 𝛼 + (cos 𝛼 − tan 𝛼)(cos 𝛼 + 𝑡𝑎𝑛𝛼 )

𝐿𝐻𝑆 = sin2 𝛼 + cos 2 𝛼 − tan2 𝛼
 𝐿𝐻𝑆 = 1 − tan2 𝛼 sin2 𝛼 + cos 2 𝛼 = 1

𝑅𝐻𝑆 = 1 − tan2 𝛼

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

2.3. 1 cos 𝜃 tan2 𝜃

− = cos 𝜃
cos 𝜃 1

1 cos 𝜃 tan2 𝜃
𝐿𝐻𝑆 = −
cos 𝜃 1
sin2 𝜃
cos 𝜃 sin2 𝜃
𝐿𝐻𝑆 =
1
− cos2 𝜃 = tan2 𝜃
cos 𝜃 1 cos2 𝜃
sin2 𝜃
1 cos 𝜃
cos 𝜃 cos 𝜃
𝐿𝐻𝑆 = −
cos 𝜃 1
1 sin2 𝜃
𝐿𝐻𝑆 = −
cos 𝜃 cos 𝜃
1−sin2 𝜃
𝐿𝐻𝑆 = cos2 𝜃 = 1 − sin2 𝜃
cos 𝜃
cos2 𝜃
𝐿𝐻𝑆 =
cos 𝜃
cos 𝜃 cos 𝜃
𝐿𝐻𝑆 =
cos 𝜃
𝐿𝐻𝑆 = cos 𝜃

𝑅𝐻𝑆 = cos 𝜃

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

2.4. 2 sin 𝜃 cos 𝜃 1

= sin 𝜃 + cos 𝜃 −
sin 𝜃+cos 𝜃 sin 𝜃+cos 𝜃

1
𝑅𝐻𝑆 = sin 𝜃 + cos 𝜃 −
sin 𝜃+cos 𝜃
sin 𝜃(sin 𝜃+cos 𝜃) cos 𝜃(sin 𝜃+cos 𝜃) 1
𝑅𝐻𝑆 = + −
sin 𝜃+cos 𝜃 sin 𝜃+cos 𝜃 sin 𝜃+cos 𝜃
sin2 𝜃+sin 𝜃 cos 𝜃+sin 𝜃 cos 𝜃+cos2 𝜃−1
𝑅𝐻𝑆 =
sin 𝜃+cos 𝜃
sin2 𝜃+cos2 𝜃+2sin 𝜃 cos 𝜃−1
𝑅𝐻𝑆 = sin2 𝜃 + cos 2 𝜃 = 1
sin 𝜃+cos 𝜃
1+2sin 𝜃 cos 𝜃−1
𝑅𝐻𝑆 =
sin 𝜃+cos 𝜃
2sin 𝜃 cos 𝜃
𝑅𝐻𝑆 =
sin 𝜃+cos 𝜃

2 sin 𝜃 cos 𝜃
𝐿𝐻𝑆 =
sin 𝜃+cos 𝜃

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆
2.5. (cos 𝛽 + tan 𝛽) 𝑐𝑜𝑠𝛽 = 1
sin 𝛽 sin 𝛽

cos 𝛽
𝐿𝐻𝑆 = ( + tan 𝛽) 𝑐𝑜𝑠𝛽
sin 𝛽
cos 𝛽 sin 𝛽 sin 𝛽
𝐿𝐻𝑆 = ( + ) 𝑐𝑜𝑠𝛽 = tan 𝛽
sin 𝛽 cos 𝛽 cos 𝛽
cos2 𝛽 sin 𝛽𝑐𝑜𝑠𝛽
𝐿𝐻𝑆 = +
sin 𝛽 cos 𝛽
cos2 𝛽
𝐿𝐻𝑆 = + sin 𝛽
sin 𝛽
cos2 𝛽+sin2 𝛽
𝐿𝐻𝑆 =
sin 𝛽
sin2 𝛽 + cos2 𝛽 = 1
1
𝐿𝐻𝑆 =
sin 𝛽

1
𝑅𝐻𝑆 =
sin 𝛽

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

2.6. 1 1 2 tan 𝜃
+ =
1+sin 𝜃 1−sin 𝜃 sin 𝜃 cos 𝜃

1 1
𝐿𝐻𝑆 = +
1+sin 𝜃 1−sin 𝜃
1(1−sin 𝜃)
𝐿𝐻𝑆 = +
(1+sin 𝜃)(1−sin 𝜃)
1(1+sin 𝜃)
(1−sin 𝜃)(1+sin 𝜃)
1−sin 𝜃+1+sin 𝜃
𝐿𝐻𝑆 =
1−sin2 𝜃
2
𝐿𝐻𝑆 =
cos2 𝜃
cos2 𝜃 = 1 − sin2 𝜃

2 tan 𝜃
𝑅𝐻𝑆 =
sin 𝜃 cos 𝜃
sin 𝜃
2 sin 𝜃
𝑅𝐻𝑆 = cos 𝜃 = tan 𝜃
sin 𝜃 cos 𝜃 cos 𝜃
2 sin 𝜃 1
𝑅𝐻𝑆 = ×
cos 𝜃 sin 𝜃 cos 𝜃
2
𝑅𝐻𝑆 =
cos2 𝜃

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆
2.7. (1+tan2 𝛼) cos 𝛼 1
=
(1−tan 𝛼) cos 𝛼−sin 𝛼

(1+tan2 𝛼) cos 𝛼
𝐿𝐻𝑆 =
(1−tan 𝛼)
sin2 𝛼
(1+ ) cos 𝛼 sin2 𝛼
cos2 𝛼
𝐿𝐻𝑆 = sin 𝛼 = tan 𝛼
(1− ) cos2 𝛼
cos 𝛼
2
cos 𝛼+sin 𝛼 2
( ) cos 𝛼 sin 𝛼
cos2 𝛼
𝐿𝐻𝑆 = cos 𝛼−sin 𝛼
= tan 𝛼
( ) cos 𝛼
cos 𝛼
2
cos 𝛼+sin 𝛼 2
( ) cos 𝛼
𝐿𝐻𝑆 =
cos 𝛼 cos 𝛼
cos 𝛼−sin 𝛼
sin2 𝛼 + cos 2 𝛼 = 1
( )
cos 𝛼
1
( )
cos 𝛼
𝐿𝐻𝑆 = cos 𝛼−sin 𝛼
( )
cos 𝛼
1 cos 𝛼
𝐿𝐻𝑆 = ×
cos 𝛼 cos 𝛼−sin 𝛼
1
𝐿𝐻𝑆 =
cos 𝛼−sin 𝛼

1
𝑅𝐻𝑆 =
cos 𝛼−sin 𝛼

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

Classwork/Homework
Activity 7A.3. P.23
1. Rewrite each ratio as a ratio of an acute angle:
1.1. cos 156°
1.2. sin 166°
1.3. cos 99°
1.4. cos 225°
1.5. tan 209°
1.6. sin 216°
1.7. cos 300°
1.8. sin 302°
1.9. tan 322°

2. Rewrite each ratio as a ratio of a positive acute angle:

2.1. cos(−125°)
2.2. sin(−106°)
2.3. sin(−204°)
2.4. tan (−189°)
2.5. cos(−251°)
2.6. cos(−292°)
2.7. tan(−286°)

3. Determine the value of the following expressions without using a

calculator:
3.1. tan 150° sin 30° − cos 210°
3.2. (1 + cos 120°)(1 − sin2 240°)
3.3. cos2 140° + sin2 220°

4. Write the following in terms of a single trigonometric ratio:

4.1. (tan 180° − 𝜃) × sin(180° + 𝜃)
4.2. tan(180°+𝜃) cos(180°−𝜃)
sin(180°−𝜃)

5. If 𝑡 = tan 40°, express the following in terms of 𝑡:

5.1. tan 140° + 3 tan 220°
5.2. cos 220°
sin 140°

6. Simplify the following:

6.1. tan(180°−𝜃) sin(360°−𝜃)
cos(180°+𝜃) tan(360°−𝜃)
2(
6.2. cos 360° + 𝜃 + cos(180° + 𝜃) tan(360° − 𝜃) sin(360° + 𝜃)
)
6.3. 𝑠𝑖𝑛(360°+𝛼) tan(180°+𝛼)
cos(360°−𝛼) tan2 (360°+𝛼)

7. Write the following in terms of cos 𝛽:

cos(360°−𝛽) cos(−𝛽)−1
sin(360°+𝛽) tan(360°−𝛽)

8. Simplify the following without using a calculator:

8.1. cos 300° tan 150°
sin 225°cos (−45°)
8.2. 3 tan 405° + 2 tan 330° cos 750°
8.3. cos 315° cos 405°+sin 45° sin 135°
sin 750°
8.4. tan 150° cos 390° − 2 sin 510°
8.5. 2 sin 120°+3 cos 765°−2 sin 240°−3 cos 45°
5 sin 300°+3 tan 225°−6 cos 60°

9. Simplify the following:

9.1. cos(90°+𝜃) sin(𝜃+90°)
sin (−𝜃)
9.2. 2 sin(90°−𝑥)+sin (90°+𝑥)
sin(90°−𝑥)+cos (180°+𝑥)
 10. Given cos 36° = 𝑝, express the following in terms of 𝑝:
10.1. sin 54°
10.2. sin 36°
10.3. tan 126°
10.4. cos 324°

11. Write 𝐴 and 𝐵 as a single trigonometric ratio:

11.1. 𝐴 = sin(360° − 𝜃) cos(180° − 𝜃)tan (360° + 𝜃)
11.2. cos(360°+𝜃)cos (−𝜃)sin (−𝜃)
𝐵=
cos (90°+𝜃)
11.3. Hence, determine:
11.3.1. 𝐴 + 𝐵 =. ….
11.3.2. 𝐴 =. ….
𝐵

12. Write the following as a function of an acute angle:

12.1. sin 163°
12.2. cos 327°
12.3. tan 248°
12.4 cos(−213°)

13. Determine the value of the following, without using a calculator:

13.1. sin (−30°)
+ cos 330°
tan(150°)
13.2. tan 300° cos 120°
13.3. (1 − cos 30°)(1 − cos 210°)
13.4. cos 780° − (sin 315°)(cos 405°)

14. Prove that the following identity is true and state any restrictions:
sin(180°+𝛼) tan(360°+𝛼) cos 𝛼
= sin 𝛼
cos (90°−𝛼)

Answers:

1.1. cos 156°

= cos(180° − 24°)
= − cos 24°

1.2. sin 166°

= sin(180° − 14°)
= sin 14°

1.3. cos 99°

= cos(180° − 81°)
= − cos 81°
1.4. cos 225°
= cos(180° + 45°)
= − cos 45°
1.5. tan 209°
= tan(180° + 29°)
= tan 29°

1.6. sin 216°

= sin(180° + 36°)
= −sin 36°

1.7. cos 300°

= cos(360° − 60°)
= cos 60°

1.8. sin 302°

= sin(360° − 58°)
= −sin 58°

1.9. tan 322°

= tan(360° − 38°)
= − tan 38°

2.1. cos(−125°)
= cos(125°) cos(−𝜃) = cos 𝜃
= cos(180° − 55°)
= cos 55°

2.2. sin(−106°)
= −sin(106°) sin(−𝜃) = − sin 𝜃
= −sin(180° − 74°)
= − sin 74°

2.3. sin(−204°)
= −sin(204°) sin(−𝜃) = − sin 𝜃
= −sin(180° + 24°)
= −(− sin 24°)
= sin 24°

2.4. tan (−189°)

= −tan(189°) tan(−𝜃) = − tan 𝜃
= −tan(180° + 9°)
= − tan 9°
2.5. cos(−251°)
= cos(251°) cos(−𝜃) = cos 𝜃
= cos(180° + 71°)
= −cos 71°

2.6. cos(−292°)
= cos(292°) cos(−𝜃) = cos 𝜃
= cos(360° − 68°)
= cos 68°

2.7. tan(−286°)
= −tan(286°) tan(−𝜃) = − tan 𝜃
= −tan(360° − 74°)
= −(− tan 74°)
= tan 74°

3.1. tan 150° sin 30° − cos 210°

= tan(180° − 30°) sin 30° − cos(180° + 30°)
= (− tan 30°)(sin 30°) − (− cos 30°)
1 1 √3
= (− )( ) + ( )
√3 2 2
−1 √3
= +
2√3 2
−1+√3×√3
=
2√3
−1+3
=
2√3
2
=
2√3
1
=
√3

3.2. (1 + cos 120°)(1 − sin2 240°)

= (1 + cos(180° − 60°))(1 − sin2 (180° + 60°))
= (1 + (− cos 60°))(1 − (− sin 60°)2 )
= (1 − cos 60°)(1 − (sin 60°)2
2
1 √3
= (1 − ) (1 − ( ) )
2 2
1 3
= ( ) (1 − )
2 4
1 1
= ( )( )
2 4
1
=
8
3.3. cos2 140° + sin2 220°
= cos2 (180° − 40°) + sin2 (180° + 40°)
= (− cos 40°)2 + (− sin 40°)2
= cos2 40° + sin2 40°
=1

4.1. (tan 180° − 𝜃) × sin(180° + 𝜃)

= (− tan 𝜃) × (− sin 𝜃)
sin 𝜃
= (− ) (− sin 𝜃)
cos 𝜃
sin2 𝜃
=
cos 𝜃
1−cos2 𝜃
=
cos 𝜃

4.2. tan(180°+𝜃) cos(180°−𝜃)

sin(180°−𝜃)
(tan 𝜃)(− cos 𝜃)
=
sin 𝜃
sin 𝜃
( )(− cos 𝜃)
𝑐𝑜𝑠𝜃
=
sin 𝜃
− sin 𝜃
=
sin 𝜃
= −1

5. 𝑡 = tan 40°
5.1. tan 140° + 3 tan 220°
= tan(180° − 40°) + 3 tan(180° + 40°)
= (− tan 40°) + 3 tan 40°
= 2 tan 40°
= 2𝑡

5.2. cos 220°

sin 140°
cos(180°+40°)
=
sin(180°−40°)
− cos 40°
=
sin 40°
sin 40° −1
= −( )
cos 40°
= −(𝑡𝑎𝑛40°)−1
= −(𝑡 )−1
−1
=
𝑡
6.1. tan(180°−𝜃) sin(360°−𝜃)
cos(180°+𝜃) tan(360°−𝜃)
(− tan 𝜃)(− sin 𝜃)
=
(− cos 𝜃)(− tan 𝜃)
−sin 𝜃
=
− cos 𝜃
= tan 𝜃

6.2. cos2 (360° + 𝜃) + cos(180° + 𝜃) tan(360° − 𝜃) sin(360° + 𝜃)

= cos2 𝜃 + (− cos 𝜃)(− tan 𝜃)(sin 𝜃)
sin 𝜃
= cos2 𝜃 + (cos 𝜃) ( ) (sin 𝜃)
cos 𝜃
= cos2 𝜃 + sin2 𝜃
=1

6.3. 𝑠𝑖𝑛(360°+𝛼) tan(180°+𝛼)

cos(360°−𝛼) tan2 (360°+𝛼)
(sin 𝛼)(tan 𝛼)
= (cos
𝛼)(tan2 𝛼)
(sin 𝛼)(tan 𝛼)
= (cos
𝛼)(tan 𝛼)(tan 𝛼)
(sin 𝛼)
= (cos
𝛼)(tan 𝛼)
(sin 𝛼)
= sin 𝛼
(cos 𝛼)( )
cos 𝛼
sin 𝛼
=
sin 𝛼
=1

7. cos(360°−𝛽) cos(−𝛽)−1
sin(360°+𝛽) tan(360°−𝛽)
(cos 𝛽)(cos 𝛽)−1
=
(sin 𝛽)(− tan 𝛽)
cos2 𝛽−(sin2 𝛽+cos2 𝛽)
= sin 𝛽
sin 𝛽(− )
cos 𝛽
cos2 𝛽−sin2 𝛽−cos2 𝛽
= sin2 𝛽
(− )
cos 𝛽
− sin2 𝛽
= sin2 𝛽
(− )
cos 𝛽
− sin2 𝛽 cos 𝛽
= ×
1 − sin2 𝛽
= cos 𝛽
8.1. cos 300° tan 150°
sin 225°cos (−45°)
cos(360°−60°) tan(180°−30°)
=
sin(180°−45°) cos 45°
(cos 60°)(− tan 30°)
=
(sin 45°)(cos 45°)
1 1
( )(− )
2 √3
= 1 1
( )( )
√2 √2
1 1
( )(− )
2 √3
= 1
( )
2
1
=−
√3

8.2. 3 tan 405° + 2 tan 330° cos 750°

= 3 tan(360° + 45°) + 2 tan(360° − 30°) cos(2(360°) + 30°)
= 3 tan 45° + 2(− tan 30°)(cos 30°)
1 √3
= 3(1) + 2 (− )( )
√3 2
1
= 3 + 2 (− )
2
=3−1
=2

8.3. cos 315° cos 405°+sin 45° sin 135°

sin 750°
cos(360°−45°) cos(360°+45°)+sin 45° sin(180°−45°)
=
sin(2(360°)+30°)
(cos 45°)(cos 45°)+(sin 45°)(sin 45°)
=
sin 30°
1 1 1 1
( )( )+( )( )
√2 √2 √2 √2
= 1
( )
2
1 1
+
2 2
= 1
2
1
= 1
2
=2

8.4. tan 150° cos 390° − 2 sin 510°

= tan(180° − 30°) cos(360° + 30°) − 2sin (150°)
= tan(180° − 30°) cos(360° + 30°) − 2sin (180° − 30°)
= (− tan 30°)(cos 30°) − 2 sin 30°
1 √3 1
= (− )( ) − 2( )
√3 2 2
1
=− −1
2
3
=−
2
8.5. 2 sin 120°+3 cos 765°−2 sin 240°−3 cos 45°
5 sin 300°+3 tan 225°−6 cos 60°
2 sin(180°−60°)+3 cos(2(360°)+45°)−2 sin(180°+60°)−3 cos 45°
=
5 sin(360°−60°)+3 tan(180°+45°)−6 cos 60°
2 sin 60°+3 cos 45°−2(− sin 60°)−3 cos 45°
=
5(− sin 60°)+3 tan 45°−6 cos 60°
√3 1 √3 1
2( )+3( )+2( )−3( )
2 √2 2 √2
= √3 1
5(− )+3(1)−6( )
2 2
√3
4( )
2
= √3
−5( )+3−3
2
√3
4( )
2
= √3
−5( )
2
4
=−
5

9.1. cos(90°+𝜃) sin(𝜃+90°)

sin (−𝜃)
− sin 𝜃.𝑐𝑜𝑠𝜃
=
− sin 𝜃
= cos 𝜃

9.2. 2 sin(90°−𝑥)+sin (90°+𝑥)

sin(90°−𝑥)+cos (180°+𝑥)
2 cos 𝑥+cos 𝑥
=
cos 𝑥+(− cos 𝑥)
3 cos 𝑥
=
0
∴ undefined

10. cos 36° = 𝑝

10.1. sin 54°
= sin(90° − 36°)
= cos 36°
=𝑝

10.2. sin 36°

sin2 36° = 1 − cos2 36°
sin2 36° = 1 − 𝑝2
√sin2 36° = √1 − 𝑝2
∴ sin 36° = √1 − 𝑝2
10.3. tan 126°
sin 126°
=
cos 126°
sin(90°+36°)
=
cos(90°+36°)
cos 36°
=
− sin 36°
𝑝
=
−√1−𝑝2

10.4. cos 324°

= cos(360° − 36°)
= cos 36°
=𝑝

11.1. 𝐴 = sin(360° − 𝜃 ) cos(180° − 𝜃)tan (360° + 𝜃)

𝐴 = (− sin 𝜃)(− cos 𝜃)(tan 𝜃)
sin 𝜃
𝐴 = (− sin 𝜃)(− cos 𝜃) ( )
cos 𝜃
𝐴 = sin2 𝜃

11.2. cos(360°+𝜃)cos (−𝜃)sin (−𝜃)

𝐵=
cos (90°+𝜃)
(cos 𝜃)(cos 𝜃)(− sin 𝜃)
𝐵=
(− sin 𝜃)
2
𝐵 = cos 𝜃

11.3.1. 𝐴 + 𝐵 =. ….
= sin2 𝜃 + cos 2 𝜃
=1

𝐴
11.3.2. =. ….
𝐵
sin2 𝜃
=
cos2 𝜃
2
= tan 𝜃

12.1. sin 163°

= sin(180° − 17°)
= sin 17°

12.2. cos 327°

= cos(360° − 33°)
= cos 33°

12.3 tan 248°

= tan (180° + 68°)
= tan 68°
12.4. cos(−213°)
= cos(−(213°))
= cos 213°
= cos(180° + 33°)
= − cos 33°

13.1. sin (−30°)

+ cos 330°
tan(150°)
− sin 30°
= + cos(360° − 30°)
tan (180°−30°)
− sin 30°
= + cos 30°
− tan 30°
1
(− ) √3
2
= 1 +( )
(− ) 2
√3
√3 √3
= +
2 2
2√3
=
2
= √3

13.2. tan 300° cos 120°

= tan(360° − 60°) cos(180° − 60°)
= (− tan 60°) (− cos 60°)
1
= (−√3) (− )
2
√3
=
2

13.3. (1 − cos 30°)(1 − cos 210°)

= (1 − cos 30°)(1 − cos(180° + 30°))
= (1 − cos 30°)(1 − (− cos 30°))
= (1 − cos 30°)(1 + cos 30°)
= 1 − cos2 30°
2
√3
=1−( )
2
3
=1−
4
1
=
4

13.4. cos 780° − (sin 315°)(cos 405°)

= cos(2(360°) + 60°) − sin(360° − 45°) cos(360° + 45°)
= cos 60° − (− sin 45°)(cos 45°)
1 1 1
= + ( )( )
2 √2 √2
1 1
= +
2 2
=1
14. sin(180°−𝛼) tan(360°+𝛼) cos 𝛼
= sin 𝛼
cos (90°−𝛼)

Restriction: denominator may not equal to 0

∴ cos(90° − 𝛼) ≠ 0
∴ sin 𝛼 ≠ 0
𝛼 ≠ 0°; 180°; 360°; 𝑒𝑡𝑐

sin(180°+𝛼) tan(360°+𝛼) cos 𝛼

𝐿𝐻𝑆 =
cos (90°−𝛼)
(sin 𝛼) (tan 𝛼)(cos 𝛼)
𝐿𝐻𝑆 =
sin 𝛼
sin 𝛼
(sin 𝛼)( )(cos 𝛼)
cos 𝛼
𝐿𝐻𝑆 =
sin 𝛼
𝐿𝐻𝑆 = sin 𝛼

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

Classwork/Homework
Activity 7A.4. P.26
1. Given tan 𝑥 = √5 where 𝑥 ∈ [90°; 270°] and √2 cos 𝑦 + 1 = 0 where
sin 𝑦 > 0. Determine the following using relevant diagrams.
1.1. sin2 𝑥
cos2 𝑦
1.2. cos(180° + 𝑥)

2. If 2 sin 𝑥 − √3 = 0 and cos 𝑥 < 0, find the following using a relevant

diagram:
2.1. cos2 𝑥
2.2. sin(180°+𝑥)
tan(360°−𝑥)
2.3. cos(90° − 𝑥) . cos 30°

𝑝
3. If tan 𝐴 = and 𝐴 < 90°, find the following in terms of 𝑝 and 𝑘.
𝑘
3.1. sin 𝐴
1
3.2. cos 𝐴
𝑘

−3
4. If cos 𝑃 = and sin 𝑃 > 0, use a sketch (no calculator) to find the value
4
of:
√7 cos(90° + 𝑃)
Answers:

1. tan 𝑥 = √5 where 𝑥 ∈ [90°; 270°] √2 cos 𝑦 + 1 = 0 where sin 𝑦 > 0

−√5 𝑦 −1 𝑥
tan 𝑥 = = cos 𝑦 = =
−1 𝑥 √2 𝑟

√2

𝑥 𝑦
−1
−1
−√5

𝑟2 = 𝑥2 + 𝑦2 𝑦2 = 𝑟2 − 𝑥2
2 2
𝑟 = √(−1)2 + (−√5) 𝑦 = √(√2) − (−1)2
𝑟 = √1 + 5 𝑦 = √4 − 1
𝑟 = √6 𝑦 = √3

1.1. sin2 𝑥
cos2 𝑦
2
−√5
( )
√6
= −1 2
( )
√2
5
6
= 1
2
5 2
= ×
6 1
10 5 2
= = =1
6 3 3

1.2. cos(180° + 𝑥)
= − cos 𝑥
−1
=
√6

2. 2 sin 𝑥 − √3 = 0 and cos 𝑥 < 0

√3 𝑦
sin 𝑥 = =
2 𝑟
 2
√3
𝑥

𝑥2 = 𝑟2 − 𝑦2
2
𝑥 = √(2)2 − (√3)
𝑥 = √4 − 3
𝑥 = ±√1
𝑥 = ±1
But 2nd quadrant 𝑥 is negative
∴ 𝑥 = −1

2.1. cos2 𝑥
−1 2
=( )
2
1
=
4

2.2. sin(180°+𝑥)
tan(360°−𝑥)
√3
−
2
= √3
−
1
−√3 −1
= ×
2 √3
1
=
2

2.3. cos(90° − 𝑥) . cos 30°

= sin 𝑥 . cos 30°
√3 √3
= ×
2 2
3
=
4

𝑝
3. tan 𝐴 = and 𝐴 < 90°
𝑘
𝑝 𝑦
tan 𝐴 = =
𝑘 𝑥
 𝑝
𝐴
𝑘

𝑟2 = 𝑥2 + 𝑦2
𝑟 = √(𝑘 )2 + (𝑝)2
𝑟 = √𝑘 2 + 𝑝2

3.1. sin 𝐴
𝑝
= 2
√𝑘 +𝑝2

1
3.2. cos 𝐴
𝑘
1 𝑘
= ×
𝑘 √𝑘 2 +𝑝2
1
=
√𝑘 2 +𝑝2
−3
4. cos 𝑃 = and sin 𝑃 > 0
4
−3 𝑥
cos 𝑃 = =
4 𝑟

𝑃
−3

𝑟2 = 𝑥2 + 𝑦2
𝑦 = √(4)2 − (−3)2
𝑟 = √16 − 9
𝑟 = ±√7
But 𝑟 is always positive
∴ 𝑟 = √7
√7 cos(90° + 𝑃)
= √7(− sin 𝑃)
√7
= −√7 ( )
4
7 3
= − = −1 = −1,75
4 4
 Classwork/Homework
Activity 7A.5. P.30
1. Determine the value of 𝛼 for 𝛼 ∈ [0°; 360°] if:
1.1. 4 cos 𝛼 = 2
1.2. sin 𝛼 + 3,65 = 3
1
1.3. tan 𝛼 = 5 4
1.4. cos 𝛼 + 0,939 = 0
1.5. 5 sin 𝛼 = 3
1
1.6. tan 𝛼 = −1,4
2

2. Determine the value of 𝜃 for 𝜃 ∈ [−180°; 180°] if:

2.1. sin 𝜃 = 0,6
3
2.2. cos 𝜃 + 4 = 0
2.3. 3 tan 𝜃 = 20
2.4. sin 𝜃 = cos 180°
4
2.5. 2 cos 𝜃 = 5

Answers:

1.1. 4 cos 𝛼 = 2
2
cos 𝛼 = 4
1
cos 𝛼 = 2

ref ∠ for 𝛼 = cos −1 0,5

ref ∠ for 𝛼 = 60°

1st Quadrant

𝛼 = 60°

4th Quadrant

𝛼 = 360° − 60°
𝛼 = 300°

∴ 𝛼 = 60° 𝑜𝑟 300°

1.2. sin 𝛼 + 3,65 = 3

sin 𝛼 = 3 − 3,65 m
sin 𝛼 = 0,65

ref ∠ for 𝛼 = sin−1 0,65

ref ∠ for 𝛼 = 40,54160187°

1st Quadrant
 𝛼 = 40,54°

2nd Quadrant

𝛼 = 180° − 40,54160187°
𝛼 = 139,46°

∴ 𝛼 = 40,54° 𝑜𝑟 139,46°

1
1.3. tan 𝛼 = 5 4

ref ∠ for 𝛼 = tan−1 5,25

ref ∠ for 𝛼 = 79,21570213°

1st Quadrant
m
𝛼 = 79,22°

3rd Quadrant

𝛼 = 180° + 79,21570213°
𝛼 = 259,22°

∴ 𝛼 = 79,22° 𝑜𝑟 259,22° °

1.4. cos 𝛼 + 0,939 = 0

cos 𝛼 = −0,939

ref ∠ for 𝛼 = cos −1 0,939

ref ∠ for 𝛼 = 20,11570805°

2nd Quadrant

𝛼 = 180° − 20,11570805°
𝛼 = 159,88°

3rd Quadrant

𝛼 = 180° + 20,11570805°
𝛼 = 200,12°

∴ 𝛼 = 159,88° 𝑜𝑟 200,12°

1.5. 5 sin 𝛼 = 3
3
sin 𝛼 = 5 m

ref ∠ for 𝛼 = sin−1 0,6

ref ∠ for 𝛼 = 36,86989765°

1st Quadrant

𝛼 = 36,87°

2nd Quadrant
 𝛼 = 180° − 36,86989765°
𝛼 = 143,13°

∴ 𝛼 = 36,87° 𝑜𝑟 143,13°
1
1.6. tan 𝛼 = −1,4
2
tan 𝛼 = −2,8

ref ∠ for 𝛼 = tan−1 2,8

ref ∠ for 𝛼 = 70,34617594°

2nd Quadrant m

𝛼 = 180° − 70,34617594°
𝛼 = 109,65°

4th Quadrant

𝛼 = 360° − 70,34617594°
𝛼 = 289,65°

∴ 𝛼 = 109,65° 𝑜𝑟 289,65°

2.1. sin 𝜃 = 0,6

ref ∠ for 𝜃 = sin−1 0,6 m

ref ∠ for 𝜃 = 36,86989765°

1st Quadrant

𝜃 = 36,87°

2nd Quadrant

𝜃 = 180° − 36,86989765°
𝜃 = 143,13°

∴ 𝜃 = 36,87° 𝑜𝑟 143,13°

3
2.2. cos 𝜃 + 4 = 0
3
cos 𝜃 = − 4
3
ref ∠ for 𝜃 = cos −1 4
ref ∠ for 𝜃 = 41,40962211°

2nd Quadrant

𝜃 = 180° − 41,40962211°
𝜃 = 138,59°

3rd Quadrant

𝜃 = 180° + 41,40962211°
𝜃 = 221,41°

∴ 𝜃 = −138,59° 𝑜𝑟 138,59°
2.3. 3 tan 𝜃 = 20
20
tan 𝜃 = 3

20
ref ∠ for 𝜃 = tan−1 3
ref ∠ for 𝜃 = 81,46923439°

1st Quadrant
m
𝜃 = 81,47°

3rd Quadrant

𝜃 = 180° + 81,46923439°
𝜃 = 261,47°

∴ 𝜃 = −98,53° 𝑜𝑟 81,47°

2.4. sin 𝜃 = cos 180°

sin 𝜃 = −1

ref ∠ for 𝜃 = sin−1 1

ref ∠ for 𝜃 = 90°

3rd Quadrant m
𝜃 = 180° + 90°
𝜃 = 270°

4th Quadrant

𝛼 = 360° − 90°
𝛼 = 270°

∴ 𝜃 = −90°

4
2.5. 2 cos 𝜃 = 5
2
cos 𝜃 = 5

ref ∠ for 𝜃 = cos −1 0,4

ref ∠ for 𝜃 = 66,42182152°

1st Quadrant

𝜃 = 66,42°

4th Quadrant

𝜃 = 360° − 66,42182152°
𝜃 = 293,58°

∴ 𝜃 = −66,42° 𝑜𝑟 66,42°
 Classwork/Homework
Activity 7A.6. P.34
1. Find the general solution for each equation.
1.1. cos(𝜃 + 20°) = 0 1.4. cos(𝛼 − 25°) = 0,707
1.2. 3𝜃
sin 3𝛼 = −1 1.5. 2 sin = −1
2
1.3. 5
tan 4𝛽 = 0,866 1.6. 5 tan(𝛽 + 15°) = −
√3

2. • Find the general solution for each equation.

• Hence, find all the solutions in the interval [−180°; 180°].
2.1. cos(𝜃 + 25°) = 0,231 2.6 cos 𝜃 = −1
2.2. sin 2𝛼 = −0,327 2.7. 4 sin 𝜃 = 0
2.3. 𝜃
2 tan 𝛽 = −2,68 2.8. tan 2 = 0,9
2.4. cos 𝛼 = 1 2.9. 2 sin 2𝜃 = −
√3
2
2.5. 4 cos 𝜃 + 3 = 1

Answers:

1.1. cos(𝜃 + 20°) = 0

ref ∠ for 𝜃 + 20° = cos −1 0

ref ∠ for 𝜃 + 20° = 90°

1st Quadrant

𝜃 + 20° = 90° + 𝑛. 360° 𝑛∈ℤ

𝜃 = 70° + 𝑛. 360°

4th Quadrant

𝜃 + 20° = 360° − 90° + 𝑛. 360°

𝜃 = 250° + 𝑛. 360°

∴ 𝜃 = 70° + 𝑛. 360° 𝑜𝑟 250° + 𝑛. 360°

1.2. sin 3𝛼 = −1

ref ∠ for 3𝛼 = sin−1 1

ref ∠ for 3𝛼 = 90°

3rd Quadrant

3𝛼 = 180° + 90° + 𝑛. 360° 𝑛∈ℤ

3𝛼 = 270° + 𝑛. 360°
3𝛼 270° 𝑛.360°
= +
3 3 3
𝛼 = 90° + 𝑛. 120°
 4th Quadrant

3𝛼 = 360° − 90° + 𝑛. 360°

3𝛼 = 270° + 𝑛. 360°
3𝛼 270° 𝑛.360°
= +
3 3 3
𝛼 = 90° + 𝑛. 120°

∴ 𝛼 = 90° + 𝑛. 120°

1.3. tan 4𝛽 = 0,866

ref ∠ for 4𝛽 = tan−1 0,866

ref ∠ for 4𝛽 = 40,89256291°

1st Quadrant

4𝛽 = 40,89256291° + 𝑛. 180° 𝑛∈ℤ

4𝛽 40,89256291° 𝑛. 180°
= +
4 4 4
𝛽 = 10,22° + 𝑛. 45°
3rd Quadrant

4𝛽 = 180° + 40,89256291° + 𝑛. 180°

4𝛽 = 220,89256291° + 𝑛. 180°
4𝛽 220,89256291° 𝑛.180°
= +
4 4 4
𝛽 = 55,22° + 𝑛. 45°

∴ 𝛽 = 10,22° + 𝑛. 45° 𝑜𝑟 55,22° + 𝑛. 45°

1.4. cos(𝛼 − 25°) = 0,707

ref ∠ for 𝛼 − 25° = cos −1 0,707

ref ∠ for 𝛼 − 25° = 45,00865166°

1st Quadrant

𝛼 − 25° = 45,00865166° + 𝑛. 360° 𝑛∈ℤ

𝛼 = 70,01° + 𝑛. 360°

4th Quadrant

𝛼 − 25° = 360° − 45,00865166° + 𝑛. 360°

𝛼 − 25° = 314,9913483° + 𝑛. 360°
𝛼 = 339,99° + 𝑛. 360°

∴ 𝛼 = 70,01° + 𝑛. 360° 𝑜𝑟 339,99° + 𝑛. 360°

1.5. 3𝜃
2 sin = −1
2
3𝜃 −1
sin =
2 2
 3𝜃
ref ∠ for = sin−1 0,5
2
3𝜃
ref ∠ for = 30°
2

3rd Quadrant
3𝜃
= 180° + 30° + 𝑛. 360° 𝑛∈ℤ
2
3𝜃
= 210° + 𝑛. 360°
2
3𝜃° = 420° + 𝑛. 720°
3𝜃 420° 𝑛.720°
= +
3 3 3
𝜃 = 140° + 𝑛. 240°

4th Quadrant
3𝜃
= 360° − 30° + 𝑛. 360°
2
3𝜃
= 330° + 𝑛. 360°
2
3𝜃° = 660° + 𝑛. 720°
3𝜃 660° 𝑛.720°
= +
3 3 3
𝜃 = 220° + 𝑛. 240°

∴ 𝜃 = 140° + 𝑛. 240° 𝑜𝑟 220° + 𝑛. 240°

1.6. 5
5 tan(𝛽 + 15°) = −
√3
1
tan(𝛽 + 15°) = −
√3

1
ref ∠ for 𝛽 + 15° = tan−1
√3
ref ∠ for 𝛽 + 15° = 30°

2nd Quadrant

𝛽 + 15° = 180° − 30° + 𝑛. 180° 𝑛∈ℤ

𝛽 + 15° = 150° + 𝑛. 180°
𝛽 = 135° + 𝑛. 180°

4th Quadrant

𝛽 + 15° = 360° − 30° + 𝑛. 180°

𝛽 + 15° = 330° + 𝑛. 180°
𝛽 = 315° + 𝑛. 180°

∴ 𝛽 = 135° + 𝑛. 180° 𝑜𝑟 315° + 𝑛. 180°

2.1. cos(𝜃 + 25°) = 0,231 for −180° ≤ 𝑥 ≤ 180°

ref ∠ for 𝜃 + 25° = cos −1 0,231

ref ∠ for 𝜃 + 25° = 76,64404693°

1st Quadrant

𝜃 + 25° = 76,64404693° + 𝑛. 360° 𝑛∈ℤ

𝜃 = 51,64° + 𝑛. 360°

4th Quadrant

𝜃 + 25° = 360° − 76,64404693° + 𝑛. 360°

𝜃 + 25° = 283,3559531° + 𝑛. 360°
𝜃 = 258,36° + 𝑛. 360°

∴ 𝜃 = 51,64° + 𝑛. 360° 𝑜𝑟 258,36° + 𝑛. 360°

𝜃 = 51,64° + 𝑛. 360° 𝜃 = 258,36° + 𝑛. 360°

𝑛 = −2 −668,36  −461,64° 
𝑛 = −1 −308,36°  −101,64° ✓
𝑛=0 51,64° ✓ 258,36° ✓
𝑛=1 411,64°  618,36° 
∴ 𝜃 = −101,64°; 51,64°; 258,36°

2.2. sin 2𝛼 = −0,327 for −180° ≤ 𝑥 ≤ 180°

ref ∠ for 2𝛼 = sin−1 0,327

ref ∠ for 2𝛼 = 19,0867885°

3rd Quadrant

2𝛼 = 180° + 19,0867885° + 𝑛. 360° 𝑛∈ℤ

2𝛼 = 199,0867885° + 𝑛. 360°
2𝛼 199,0867885° 𝑛.360°
= +
2 2 2
𝛼 = 99,54° + 𝑛. 180°

4th Quadrant

2𝛼 = 360° − 19,0867885° + 𝑛. 360°

2𝛼 = 340,9132115° + 𝑛. 360°
2𝛼 340,9132115° 𝑛.360°
= +
2 2 2
𝛼 = 170,46° + 𝑛. 180°

∴ 𝛼 = 99,54° + 𝑛. 180° 𝑜𝑟 170,46° + 𝑛. 180°

𝛼 = 99,54° + 𝑛. 180° 𝛼 = 170,46° + 𝑛. 180°

𝑛 = −2 −260,46°  −189,54° 
𝑛 = −1 −80,46° ✓ −9,54° ✓
 𝑛=0 99,54° ✓ 170,46° ✓
𝑛=1 279,54°  350,46° 
∴ 𝑥 = −80,46°; −9,54°; 99,54°; 170,46°

2.3. 2 tan 𝛽 = −2,68 for −180° ≤ 𝑥 ≤ 180°

tan 𝛽 = −1,34

ref ∠ for 𝛽 = tan−1 1,34

ref ∠ for 𝛽 = 53,26717334°

2nd Quadrant

𝛽 = 180° − 53,26717334° + 𝑛. 180° 𝑛∈ℤ

𝛽 = 126,73° + 𝑛. 180°

4th Quadrant

𝛽 = 360° − 53,26717334° + 𝑛. 180°

𝛽 = 306,73° + 𝑛. 180°

∴ 𝛽 = 126,73° + 𝑛. 180° 𝑜𝑟 306,73° + 𝑛. 180°

𝛽 = 126,73° + 𝑛. 180° 𝛽 = 306,73° + 𝑛. 180°

𝑛 = −2 −233,27°  −53,27° ✓
𝑛 = −1 −53,27° ✓ 126,73° ✓
𝑛=0 126,73° ✓ 306,73° 
𝑛=1 306,73°  486,73° 
∴ 𝛽 = −53,27°; 126,73°

2.4. cos 𝛼 = 1 for −180° ≤ 𝑥 ≤ 180°

ref ∠ for 𝛼 = cos −1 1

ref ∠ for 𝛼 = 0°

1st Quadrant

𝛼 = 0° + 𝑛. 360° 𝑛∈ℤ

4th Quadrant

𝛼 = 360° − 0° + 𝑛. 360°
𝛼 = 360° + 𝑛. 360°

∴ 𝛼 = 0° + 𝑛. 360° 𝑜𝑟 360° + 𝑛. 360°

𝛼 = 0° + 𝑛. 360° 𝛼 = 360° + 𝑛. 360°

𝑛 = −1 −360°  0° ✓
 𝑛=0 0° ✓ 360° 
𝑛=1 360°  720° 
∴ 𝛼 = 0°

2.5. 4 cos 𝜃 + 3 = 1 for −180° ≤ 𝑥 ≤ 180°

4 cos 𝜃 = −2
1
cos 𝜃 = − 2
1
ref ∠ for 𝜃 = cos −1 2
ref ∠ for 𝜃 = 60°

2nd Quadrant

𝜃 = 180° − 60° + 𝑛. 360° 𝑛∈ℤ

𝜃 = 120° + 𝑛. 360°
3rd Quadrant

𝜃 = 180° + 60° + 𝑛. 360°

𝜃 = 240° + 𝑛. 360°

∴ 𝜃 = 120° + 𝑛. 360° 𝑜𝑟 240° + 𝑛. 360°

𝜃 = 120° + 𝑛. 360° 𝜃 = 240° + 𝑛. 360°

𝑛 = −2 −600°  −480° 
𝑛 = −1 −240°  −120° ✓
𝑛=0 120° ✓ 240° 
𝑛=1 480°  600° 
∴ 𝜃 = −120°; 120°

2.6. cos 𝜃 = −1 for −180° ≤ 𝑥 ≤ 180°

ref ∠ for 𝜃 = cos −1 1

ref ∠ for 𝜃 = 0°

2nd Quadrant

𝜃 = 180° − 0° + 𝑛. 360° 𝑛∈ℤ

𝜃 = 180° + 𝑛. 360°

3rd Quadrant

𝜃 = 180° + 0° + 𝑛. 360°
𝜃 = 180° + 𝑛. 360°

∴ 𝜃 = 180° + 𝑛. 360° 𝑜𝑟 180° + 𝑛. 360°

𝜃 = 180° + 𝑛. 360° 𝜃 = 180° + 𝑛. 360°

𝑛 = −1 −180°  −180° 
 𝑛=0 180° ✓ 180° ✓
𝑛=1 540°  540° 
∴ 𝜃 = −180°; 180°

2.7. 4 sin 𝜃 = 0 for −180° ≤ 𝑥 ≤ 180°

sin 𝜃 = 0

ref ∠ for 𝜃 = sin−1 0

ref ∠ for 𝜃 = 0°

1st Quadrant

𝜃 = 0° + 𝑛. 360° 𝑛∈ℤ

2nd Quadrant

𝜃 = 180° − 0° + 𝑛. 360°
𝜃 = 180° + 𝑛. 360°

∴ 𝜃 = 0° + 𝑛. 360° 𝑜𝑟 180° + 𝑛. 360°

𝜃 = 0° + 𝑛. 360° 𝜃 = 180° + 𝑛. 360°

𝑛 = −1 −360°  −180° ✓
𝑛=0 0° ✓ 180° ✓
𝑛=1 360°  540° 
∴ 𝜃 = −180°; 0°; 180°

2.8. 𝜃
tan 2 = 0,9 for −180° ≤ 𝑥 ≤ 180°

𝜃
ref ∠ for 2 = tan−2 0,9
𝜃
ref ∠ for 2 = 41,9872125°

1st Quadrant
𝜃
= 41,9872125° + 𝑛. 180° 𝑛∈ℤ
2
𝜃 = 83,97° + 𝑛. 360°

3rd Quadrant
𝜃
= 180° + 41,9872125° + 𝑛. 180°
2
𝜃
= 221,9872125° + 𝑛. 180°
2
𝜃 = 443,97° + 𝑛. 360°

∴ 𝜃 = 83,97° + 𝑛. 360° 𝑜𝑟 443,97° + 𝑛. 360°

𝜃 = 83,97° + 𝑛. 360° 𝜃 = 443,97° + 𝑛. 360°

𝑛 = −2 −636,03°  −276,03° 
𝑛 = −1 −276,03°  83,97° ✓
 𝑛=0 83,97° ✓ 443,97° 
𝑛=1 443,97°  803,97° 

∴ 𝜃 = 83,97°
2.9. 2 sin 2𝜃 = −
√3
for −180° ≤ 𝑥 ≤ 180°
2
√3
sin 2𝜃 = −4

√3
ref ∠ for 2𝜃 = sin−1 4
ref ∠ for 2𝜃 = 25,65890627°

3rd Quadrant

2𝜃 = 180° + 25,65890627° + 𝑛. 360° 𝑛∈ℤ

2𝜃 = 205,65890627° + 𝑛. 360°
𝜃 = 102,83° + 𝑛. 180°

4th Quadrant

2𝜃 = 360° − 25,65890627° + 𝑛. 360° 𝑛∈ℤ

2𝜃 = 334,3410937° + 𝑛. 360°
𝜃 = 167,17° + 𝑛. 180°

∴ 𝜃 = 102,83° + 𝑛. 180° 𝑜𝑟 167,17° + 𝑛. 180°

𝜃 = 102,83° + 𝑛. 180° 𝜃 = 167,17° + 𝑛. 180°

𝑛 = −2 −257,17°  −192,83° 
𝑛 = −1 −77,17° ✓ −12,83° ✓
𝑛=0 102,83° ✓ 167,17° ✓
𝑛=1 282,83°  347,17° 
∴ 𝜃 = −77,17°; −12,83°; 102,83°; 167,17°

Classwork/Homework
Activity 7A.7. P.38
1
1. Find 𝜃 if sin2 𝜃 + 2 sin 𝜃 = 0 for 𝜃 ∈ [0°; 360°].

2. Determine the general solution for each of the following:

2.1. 2 cos2 𝜃 − 3 cos 𝜃 = 2 2.5. sin(𝛼 + 15°) = 2 cos(𝛼 + 15°)
2
2.2. 3 tan 𝜃 + 2 tan 𝜃 = 0 2.6. sin2 𝜃 − 4 cos2 𝜃 = 0
2.3. cos(2𝜃+30°)
cos2 𝛼 = 0,64 2.7. + 0,38 = 0
2
2.4. sin(4𝛽 + 35°) = cos(10° − 𝛽)

1
3. Find 𝛽 if 3 tan 𝛽 = cos 200° for 𝛽 ∈ [−180°; 180°].
Answers:

1. 1
Find 𝜃 if sin2 𝜃 + sin 𝜃 = 0 for 𝜃 ∈ [0°; 360°].
2

1
sin2 𝜃 + 2 sin 𝜃 = 0
1
sin 𝜃 (sin 𝜃 + 2) = 0

1
sin 𝜃 = 0 sin 𝜃 + 2 = 0
1
sin 𝜃 = − 2

1
ref ∠ for 𝜃 = sin−1 0 ref ∠ for 𝜃 = sin−1 2
ref ∠ for 𝜃 = 0° ref ∠ for 𝜃 = 30°

1st Quadrant 3rd Quadrant

𝜃 = 0° + 𝑛. 360° 𝑛∈ℤ 𝜃 = 180° + 0° + 𝑛. 360°

𝜃 = 180° + 𝑛. 360°

2nd Quadrant 4th Quadrant

𝜃 = 180° − 0° + 𝑛. 360° 𝜃 = 360° − 0° + 𝑛. 360°

𝜃 = 180° + 𝑛. 360° 𝜃 = 360° + 𝑛. 360°

∴ 𝜃 = 0° + 𝑛. 360° 𝑜𝑟 180° + 𝑛. 360° 𝑜𝑟 180° + 𝑛. 360° 𝑜𝑟 360° + 𝑛. 360°

𝜃 = 0° + 𝑛. 360° 𝜃 = 360° + 𝑛. 360°

𝑛 = −1 −360°  0° ✓
𝑛=0 0° ✓ 360° ✓
𝑛=1 360° ✓ 720° 
𝜃 = 180° + 𝑛. 360° 𝜃 = 360° + 𝑛. 360°
𝑛 = −1 −180° ✓ 0° ✓
𝑛=0 180° ✓ 360° ✓
𝑛=1 540°  720° 
∴ 𝜃 = 0°; 180°; 360°

2.1. 2 cos 2 𝜃 − 3 cos 𝜃 = 2

2 cos 2 𝜃 − 3 cos 𝜃 − 2 = 0
(2 cos 𝜃 + 1)(cos 𝜃 − 2) = 0

2 cos 𝜃 + 1 = 0 cos 𝜃 − 2 = 0
 1
cos 𝜃 = − 2 cos 𝜃 = 2
1
ref ∠ for 𝜃 = cos −1 2 N.A.
ref ∠ for 𝜃 = 60°

2nd Quadrant

𝜃 = 180° − 60° + 𝑛. 360°

𝑛∈ℤ
𝜃 = 120° + 𝑛. 360°

3rd Quadrant

𝜃 = 180° + 60° + 𝑛. 360°

𝜃 = 240° + 𝑛. 360°

∴ 𝜃 = 120° + 𝑛. 360° 𝑜𝑟 240° + 𝑛. 360°

2.2. 3 tan2 𝜃 + 2 tan 𝜃 = 0

tan 𝜃 (3 tan 𝜃 + 2) = 0

tan 𝜃 = 0 3 tan 𝜃 + 2 = 0
2
tan 𝜃 = −
3

2
ref ∠ for 𝜃 = tan−1 0 ref ∠ for 𝜃 = tan−1 3
ref ∠ for 𝜃 = 0° ref ∠ for 𝜃 = 33,69006753°

1st Quadrant 2nd Quadrant

 𝜃 = 0° + 𝑛. 180° 𝑛∈ℤ 𝜃 = 180° − 33,69006753° + 𝑛. 180°
𝜃 = 146,31 ° + 𝑛. 180°
3rd Quadrant 4th Quadrant

𝜃 = 180° + 0° + 𝑛. 180° 𝜃 = 360° − 33,69006753° + 𝑛. 180°

𝜃 = 180° + 𝑛. 180° 𝜃 = 326,31° + 𝑛. 180°

∴ 𝜃 = 0° + 𝑛. 180° 𝑜𝑟 180° + 𝑛. 180° 𝑜𝑟 146,31 ° + 𝑛. 180° 𝑜𝑟 326,31° + 𝑛. 180°

2.3. cos2 𝛼 = 0,64

√cos 2 𝛼 = ±√0,64
cos 𝛼 = ±0,8

cos 𝛼 = 0,8 cos 𝛼 = −0,8

ref ∠ for 𝛼 = cos−1 0,8 ref ∠ for 𝛼 = cos−1 0,8
ref ∠ for 𝛼 = 36,86989765° ref ∠ for 𝛼 = 36,86989765°

1st Quadrant 2nd Quadrant

𝛼 = 36,87° + 𝑛. 360°
𝛼 = 180° − 36,86989765° + 𝑛. 360°
𝑛∈ℤ
𝛼 = 143,13° + 𝑛. 360°
4th Quadrant 3rd Quadrant

𝛼 = 360° −
𝛼 = 180° + 36,86989765° + 𝑛. 360°
36,86989765° + 𝑛. 360°
𝛼 = 323,13° + 𝑛. 360° 𝛼 = 216,87° + 𝑛. 360°

∴ 𝛼 = 36,87° + 𝑛. 360° 𝑜𝑟 323,13° + 𝑛. 360° 𝑜𝑟 143,13° + 𝑛. 360° 𝑜𝑟 216,87° + 𝑛. 360°

2.4. sin(4𝛽 + 35°) = cos(10° − 𝛽)

sin(4𝛽 + 35°) = sin(90° −(10° −
𝛽))
sin(4𝛽 + 35°) = sin(80° + 𝛽)

ref ∠ for 4𝛽 + 35° = 80° + 𝛽

1st Quadrant

4𝛽 + 35° = 80° + 𝛽 + 𝑛. 360° 𝑛∈ℤ

3𝛽 = 45° + 𝑛. 360°
𝛽 = 15° + 𝑛. 120°

2nd Quadrant
 4𝛽 + 35° = 180° − (80° + 𝛽) +
𝑛. 360°
4𝛽 + 35° = 180° − 80° − 𝛽 + 𝑛. 360°
5𝛽 = 65° + 𝑛. 360°
𝛽 = 13° + 𝑛. 72°

∴ 𝛽 = 15° + 𝑛. 120° 𝑜𝑟 13° + 𝑛. 72°

2.5. sin(𝛼 + 15°) = 2 cos(𝛼 + 15°)

sin(𝛼+15°)
cos(𝛼+15°)
=2
tan(𝛼 + 15°) = 2

ref ∠ for 𝛼 + 15° = tan−1 2

ref ∠ for 𝛼 + 15° = 63,43494882°

1st Quadrant

𝛼 + 15° = 63,43494882° − 15° + 𝑛. 180° 𝑛∈ℤ

𝛼 = 63,43494882° − 15° + 𝑛. 180°
𝛼 = 48,43° − 15° + 𝑛. 180°
3rd Quadrant

𝛼 + 15° = 180° + 63,43494882° + 𝑛. 180°

𝛼 = 243,43494882° − 15° + 𝑛. 180°
𝛼 = 228,43° + 𝑛. 180°
∴ 𝛼 = 48,43° + 𝑛. 180° 𝑜𝑟 228,43° + 𝑛. 180°

2.6. sin2 𝜃 − 4 cos2 𝜃 = 0

sin2 𝜃 = 4 cos 2 𝜃
sin2 𝜃
=4
cos2 𝜃
tan2 𝜃 = 4
√tan2 𝜃 = ±√4
tan 𝜃 = ±2

tan 𝜃 = 2 tan 𝜃 = −2

ref ∠ for 𝜃 = tan−1 2 ref ∠ for 𝜃 = tan−1 2

ref ∠ for 𝜃 = 63,43494882° ref ∠ for 𝜃 = 63,43494882°

1st Quadrant 2nd Quadrant

𝜃 = 63,43° + 𝑛. 180° 𝑛∈ℤ 𝜃 = 180° − 63,43494882° + 𝑛. 180°

 𝜃 = 116,57° + 𝑛. 180°

3rd Quadrant 4th Quadrant

𝜃 = 180° + 63,43494882° + 𝑛. 180° 𝜃 = 360° − 63,43494882° + 𝑛. 180°

𝜃 = 243,43° + 𝑛. 180° 𝜃 = 296,57° + 𝑛. 180°

∴ 𝜃 = 63,43° + 𝑛. 180° 𝑜𝑟 243,43° + 𝑛. 180° 𝑜𝑟 116,57° + 𝑛. 180° or 296,57° + 𝑛. 180°

2.7. cos(2𝜃+30°)
+ 0,38 = 0
2
cos(2𝜃+30°)
= −0,38
2
cos(2𝜃 + 30°) = −0,76

ref ∠ for 2𝜃 + 30° = cos−1 0,76

ref ∠ for 2𝜃 + 30° = 40,53580211°

2nd Quadrant

2𝜃 + 30° = 180° − 40,53580211° + 𝑛. 360°

2𝜃 = 139,4641979° − 30° + 𝑛. 360°
2𝜃 = 109,4641979° + 𝑛. 360°
𝜃 = 54,73° + 𝑛. 180°

3rd Quadrant

2𝜃 + 30° = 180° + 40,53580211° + 𝑛. 360°

2𝜃 = 220,53580211° − 30° + 𝑛. 360°
2𝜃 = 190,53580211° + 𝑛. 360°
𝜃 = 95,27° + 𝑛. 180°
∴ 𝜃 = 54,73° + 𝑛. 180° 𝑜𝑟 95,27° + 𝑛. 180°

3. 1
Find 𝛽 if 3 tan 𝛽 = cos 200° for 𝛽 ∈ [−180°; 180°].
1
tan 𝛽 = cos 200°
3
1
tan 𝛽 = −0,93969262
3
tan 𝛽 = −2,819077862

ref ∠ for 𝛽 = tan−1 2

ref ∠ for 𝛽 = 70,46908474°

2nd Quadrant

𝛽 = 180° − 70,46908474° + 𝑛. 180°

𝛽 = 109,53° + 𝑛. 180°

4th Quadrant
 𝛽 = 360° − 70,46908474° + 𝑛. 180°
𝛽 = 289,53° + 𝑛. 180°

∴ 𝛽 = 109,53° + 𝑛. 180° 𝑜𝑟 289,53° + 𝑛. 180°

𝛽 = 109,53° + 𝑛. 180° 𝛽 = 289,53° + 𝑛. 180°

𝑛 = −2 −250,47°  −70,47° ✓
𝑛 = −1 −70,47° ✓ 109,53° ✓
𝑛=0 109,53° ✓ 289,53° 
𝑛=1 289,53°  496,53° 

𝛽 = −70,47°; 109,53°

Classwork/Homework
Activity 7A.8. P.39
1. Prove the following identities:
1 −1
1.1. = tan2 𝑥 cos2 𝑥
(cos 𝑥−1)(cos 𝑥+1)
1.2. (1 − tan 𝛼) cos 𝛼 = sin(90° + 𝛼) + cos(90° + 𝛼)

1 1
2.1. Prove: tan 𝑦 + tan 𝑦 = cos2 𝑦 tan 𝑦
2.2. For which values of 𝑦 ∈ [0°; 360°] is the identity above undefined?

3.1. sin(180°+𝜃) tan(360°−𝜃)

Simplify: sin(−𝜃) tan(180°+𝜃)
3.2. sin(180°+𝜃) tan(360°−𝜃)
Hence, solve the equation = tan 𝜃 for 𝜃 ∈ [0°; 360°].
sin(−𝜃) tan(180°+𝜃)

4. Given 12 tan 𝜃 = 5 and 𝜃 > 90°.

4.1. Draw a sketch.
4.2. Determine without using a calculator sin 𝜃 and cos(180° + 𝜃).
4.3. Use a calculator to find 𝜃 (correct to two decimal places).

Answers:
1 −1
1.1. = tan2 𝑥 cos2 𝑥
(cos 𝑥−1)(cos 𝑥+1)
 1 1
𝐿𝐻𝑆 = (cos 𝑥+1)(cos 𝑥−1) 𝑅𝐻𝑆 = − tan2 𝑥.cos2 𝑥
1 1
𝐿𝐻𝑆 = cos2 𝑥−1 𝑅𝐻𝑆 = − sin2 𝑥
.cos2 𝑥
cos2 𝑥
1 1
𝐿𝐻𝑆 = − sin2 𝑥 𝑅𝐻𝑆 = − sin2 𝑥

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

1.2. (1 − tan 𝛼) cos 𝛼 = sin(90° + 𝛼) + cos(90° + 𝛼)

𝐿𝐻𝑆 = (1 − tan 𝛼) cos 𝛼 𝑅𝐻𝑆 = sin(90° + 𝛼) + cos(90° + 𝛼)

sin 𝛼 𝑅𝐻𝑆 = cos 𝛼 − sin 𝛼
𝐿𝐻𝑆 = (1 − cos 𝛼) cos 𝛼
𝐿𝐻𝑆 = cos 𝛼 − sin 𝛼

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

1 1
2.1. tan 𝑦 + tan 𝑦 = cos2 𝑦 tan 𝑦

1 1
𝐿𝐻𝑆 = tan 𝑦 + tan 𝑦 𝑅𝐻𝑆 = cos2 𝑦 tan 𝑦
sin 𝑦 1 1
𝐿𝐻𝑆 = cos 𝑦 + sin 𝑦 𝑅𝐻𝑆 = sin 𝑦
cos2 𝑦×
cos 𝑦 cos 𝑦
sin 𝑦 cos 𝑦 1
𝐿𝐻𝑆 = + 𝑅𝐻𝑆 =
cos 𝑦 sin 𝑦 cos 𝑦.sin 𝑦
sin2 𝑦+cos2 𝑦
𝐿𝐻𝑆 = cos 𝑦.sin 𝑦
1
𝐿𝐻𝑆 = cos 𝑦.sin 𝑦

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

2.2. For which values of 𝑦 ∈ [0°; 360°] is the identity above undefined?

Denominator may not equal to zero

∴ cos 𝑦 . sin 𝑦 ≠ 0

cos 𝑦 = 0 sin 𝑦 = 0

ref ∠ for 𝑦 = cos−1 0 ref ∠ for 𝑦 = sin−1 0

ref ∠ for 𝑦 = 90° ref ∠ for 𝑦 = 0°
 1st Quadrant 1st Quadrant

𝑦 = 90° + 𝑛. 360° 𝑛∈ℤ 𝑦 = 0° + 𝑛. 360°

4th Quadrant 2nd Quadrant

𝑦 = 360° − 90° + 𝑛. 360° 𝑦 = 180° − 0° + 𝑛. 360°

𝑦 = 270° + 𝑛. 360° 𝑦 = 180° + 𝑛. 360°

𝑦 = 0°; 90°; 180°; 270°; 360°

3.1. sin(180°+𝜃) tan(360°−𝜃)

sin(−𝜃) tan(180°+𝜃)
(− sin 𝜃)(− tan 𝜃)
= (− sin 𝜃)(tan 𝜃)
= −1

3.2. sin(180°+𝜃) tan(360°−𝜃) for 𝜃 ∈ [0°; 360°]

= tan 𝜃
sin(−𝜃) tan(180°+𝜃)
tan 𝜃 = −1

ref ∠ for 𝜃 = tan−1 1

ref ∠ for 𝜃 = 45°

2nd Quadrant

𝜃 = 180° − 45° + 𝑛. 180°

𝜃 = 135° + 𝑛. 180°

4th Quadrant

𝜃 = 360° − 45° + 𝑛. 180°

𝜃 = 315° + 𝑛. 180°

∴ 𝜃 = 135° + 𝑛. 180° 𝑜𝑟 315° + 𝑛. 180°

∴ 𝜃 = 135° 𝑜𝑟 315°

4. Given 12 tan 𝜃 = 5 and 𝜃 > 90°.

4.1. Draw a sketch.

𝑟2 = 𝑥2 + 𝑦2
𝑟 = ±√(−5)2 + (−12)2
𝑟 = ±√25 + 144
𝑟 = ±√169
𝑟 = ±13

∴ 𝑟 = 13
4.2. 5
sin 𝜃 = − 13
12
cos(180° + 𝜃) = −
13

4.3. 12 tan 𝜃 = 5
5
tan 𝜃 = 12
5
ref ∠ for 𝜃 = tan−1 12
ref ∠ for 𝜃 = 22,61986495°

3rd Quadrant

𝜃 = 180° + 22,61986495°
𝜃 = 202,62°

∴ 𝜃 = 202,62°

Exam Questions P.39

If cos 23° = 𝑝, express, without the use of a calculator, the following in

1.
terms of 𝑝:
1.1. cos 203° (2)
1.2. sin 293° (3)

2. Simplify the following expression to a single trigonometric term:

sin(360°−𝑥).tan(−𝑥)
cos(180°+𝑥).(sin2 𝐴+cos2 𝐴) (6)

3.1. cos 𝑥 1+sin 𝑥 2

Prove the identity: + = (5)
1+sin 𝑥 cos 𝑥 cos 𝑥
3.2. For which values of 𝑥 in the interval 0° ≤ 𝑥 ≤ 360° will the
identity in QUESTION 3.1. be undefined? (2)

4. In the diagram below 𝑃(𝑥; √3) is a point on the Cartesian plane

such that 𝑂𝑃 = 2. 𝑄 (𝑎; 𝑏) is a point such that 𝑇𝑂̂𝑄 = 𝛼 and 𝑂𝑄 =
20. 𝑃𝑂̂𝑄 = 90°.
4.1. Calculate the value of 𝑥. (2)
4.2. Hence, calculate the size of 𝛼. (3)
4.3. Determine the coordinates of 𝑄. (5)

5. In the diagram below, 𝑃(𝑥; 24) is a point such that 𝑂𝑃 = 25 and

𝑅𝑂̂𝑃 = 𝛽, where 𝛽 is an obtuse angle.

5.1. Calculate the value of 𝑥. (2)

5.2. Determine the value of each of the following WITHOUT using a
calculator:
5.2.1. sin 𝛽 (1)
5.2.2. cos(180° − 𝛽) (2)
5.2.3. tan(−𝛽) (2)
5.3. 𝑇 is a point on 𝑂𝑃 such that 𝑂𝑇 = 15. Determine the coordinates
of 𝑇 WITHOUT using a calculator. (4)

6. Determine the value of the following expression:

2 sin 𝑥.cos 𝑥(1+tan2 𝑥)
tan 𝑥 (4)
 7. In the diagram below, 𝑃(−8; 𝑡 ) is a point in the Cartesian plane
such that 𝑂𝑃 = 17 units and reflex X𝑂̂𝑃 = 𝜃.

7.1. Calculate the value of 𝑡. (2)

7.2. Determine the value of each of the following WITHOUT using a
calculator:
7.2.1. cos(−𝜃) (2)
7.2.2. 1 − sin 𝜃 (2)

8. If sin 17° = 𝑎. WITHOUT using a calculator, express the following

in terms of 𝑎.
8.1. tan 17° (3)
8.2. sin 107° (2)
8.3. cos2 253° + sin2 557° (4)

9. Simplify fully. WITHOUT the use of a calculator:

cos(−225°).sin 135°+sin 330°
tan 225° (6)

1 −1
10. Prove the identity: =
(cos 𝑥+1)(𝑐𝑜𝑠𝑥−1) tan2 𝑥.cos2 𝑥 (4)

11. Determine the general solution for 2 sin 𝑥 . cos 𝑥 = cos 𝑥 (6)

12. Determine the general solution of: sin 2𝑥 = 4 cos 2𝑥 (5)

13. 1−cos2 𝐴
Consider:
4 cos(90°+𝐴)
13.1. Simplify the expression to a single trigonometric term. (3)
13.2. 1−cos2 2𝑥
Hence, determine the general solution of = 0,21 (6)
4 cos(90°+2𝑥)

Answers:

1. cos 23° = 𝑝

1.1. cos 203°

 = cos(180° + 23°)
= − cos 23°
= −𝑝

1.2. sin 293°

= sin(360° − 67°)
= − sin 67°
= − sin (90° − 23°)
= − cos 23°
= −𝑝

2. sin(360°−𝑥).tan(−𝑥)
cos(180°+𝑥).(sin2 𝐴+cos2 𝐴)
− sin 𝑥.− tan 𝑥
=
− cos 𝑥.1
= − tan 𝑥 . tan 𝑥
= − tan2 𝑥
3.1. cos 𝑥 1+sin 𝑥 2
+ =
1+sin 𝑥 cos 𝑥 cos 𝑥
cos 𝑥 1+sin 𝑥
𝐿𝐻𝑆 = +
1+sin 𝑥 cos 𝑥
cos 𝑥(cos 𝑥)+(1+sin 𝑥)(1+sin 𝑥)
𝐿𝐻𝑆 =
(1+sin 𝑥)(cos 𝑥)
cos2 𝑥+1+2 sin 𝑥+sin2 𝑥
𝐿𝐻𝑆 =
(1+sin 𝑥)(cos 𝑥)
sin2 𝑥+cos2 𝑥+1+2 sin 𝑥
𝐿𝐻𝑆 =
(1+sin 𝑥)(cos 𝑥)
1+1+2 sin 𝑥
𝐿𝐻𝑆 =
(1+sin 𝑥)(cos 𝑥)
2+2 sin 𝑥
𝐿𝐻𝑆 =
(1+sin 𝑥)(cos 𝑥)
2(1+sin 𝑥)
𝐿𝐻𝑆 =
(1+sin 𝑥)(cos 𝑥)
2
𝐿𝐻𝑆 =
(cos 𝑥)
𝐿𝐻𝑆 = 𝑅𝐻𝑆

3.2. cos 𝑥 = 0
∴ 90°; 270°
4.

4.1. 𝑥2 = 𝑟2 − 𝑦2
2
𝑥 = ±√22 − (√3)
𝑥 = ±√4 − 3
𝑥 = ±√1
𝑥 = ±1
∴𝑥=1 𝑥 lies in Quadrant 1

4.2. sin 𝑃𝑂̂𝑇 =

√3
2
√3
𝑃𝑂̂𝑇 = sin−1
2
̂
𝑃𝑂𝑇 = 60°

𝑃𝑂̂𝑄 = 90°
𝛼 = 90° − 60°
𝛼 = 30°

4.3 𝑏
sin 𝛼 =
𝑟
𝑏
sin 30° =
20
𝑏 = 20. sin 30°
𝑏 = 10

𝑎
cos 𝛼 =
𝑟
𝑎
cos 30° =
20
𝑎 = 20. cos 30°
𝑎 = 10√3
5.

5.1. 𝑥2 = 𝑟2 − 𝑦2
𝑥 = ±√252 − 242
𝑥 = ±√625 − 576
𝑥 = ±√49
𝑥 = ±7
∴ 𝑥 = −7 Point P lies in the 2nd quadrant

5.2.1. sin 𝛽
24
=
25

5.2.2. cos(180° − 𝛽)
= − cos 𝛽
7
= − (− )
25
7
=
25

5.2.3. tan(−𝛽)
= − tan 𝛽
24
=−
−7
24
=
7

24 𝑦
5.3. =
25 15
24×15
𝑦=
25
42
𝑦=
5

−7 𝑥
=
25 15
−7×15
𝑥=
25
−21
𝑦=
5
6. 2 sin 𝑥.cos 𝑥(1+tan2 𝑥)
tan 𝑥
sin2 𝑥
2 sin 𝑥.cos 𝑥(1+ )
cos2 𝑥
= sin 𝑥
cos 𝑥
cos2 𝑥+sin2 𝑥 cos 𝑥
= 2 sin 𝑥. cos 𝑥 ( )×
cos2 𝑥 sin 𝑥
2 sin 𝑥.cos2 𝑥 1
= ( )
sin 𝑥 cos2 𝑥
=2

7.

7.1. 𝑡 2 = 𝑟2 − 𝑥2
𝑡 = ±√172 − (−82 )
𝑡 = ±√289 − 64
𝑡 = ±√225
𝑡 = ±15
∴ 𝑡 = −15 Point P lies in the 3rd quadrant

7.2.1. cos(−𝜃)
= cos 𝜃
−8
=
17

7.2.2. 1 − sin 𝜃
−15
=1−( )
17
15
=1+
17
32
=
17
8.

𝑥2 = 𝑟2 − 𝑦2
𝑥 = √12 − 𝑎2
𝑥 = √1 − 𝑎2

8.1. tan 17°

𝑎
= √1−𝑎2

8.2. sin 107°

= sin(90° + 17°)
= cos 17°
√1−𝑎2
=
1
= √1 − 𝑎2

or

= sin 107°
= sin(180° − 73°)
= sin 73°
√1−𝑎2
=
1
= √1 − 𝑎2

8.3. cos2 253° + sin2 557°

= cos2 (180° + 73°) + sin2 (360° + 180° + 17°)
= −cos2 73° +sin2 (180° + 17°)
= −cos2 73° − sin2 17°
= − (𝑎 )2 − (𝑎 )2
= −𝑎2 − 𝑎2
= −2𝑎2

9. cos(−225°).sin 135°+sin 330°

tan 225°
cos(−(180+45)).sin(180°−45°)+sin(360°−30°)
=
tan(180°+45°)
 − cos 45°.sin 45°+(− sin 30°)
=
tan 45°
−1 1 1
( )( )−( )
√2 √2 2
=
1
1 1
− −
2 2
=
1
= −1

1 −1
10. =
(cos 𝑥+1)(𝑐𝑜𝑠𝑥−1) tan2 𝑥.cos2 𝑥

1
𝐿𝐻𝑆 =
(cos 𝑥+1)(𝑐𝑜𝑠𝑥−1)
1
𝐿𝐻𝑆 =
cos2 𝑥−1
1
𝐿𝐻𝑆 =
− sin2 𝑥

−1
𝑅𝐻𝑆 =
tan2 𝑥.cos2 𝑥
−1
𝑅𝐻𝑆 = sin2 𝑥
.cos2 𝑥
cos2 𝑥
−1
𝑅𝐻𝑆 =
sin2 𝑥

𝐿𝐻𝑆 = 𝑅𝐻𝑆

11. 2 sin 𝑥 . cos 𝑥 = cos 𝑥

2 sin 𝑥. cos 𝑥 − cos 𝑥 = 0
cos 𝑥 (2 sin 𝑥 − 1) = 0

1
cos 𝑥 = 0 sin 𝑥 =
2

1
ref ∠ for 𝑥 = cos −1 0 ref ∠ for 𝑥 = sin−1
2
ref ∠ for 𝑥 = 90° ref ∠ for 𝑥 = 30°
 1st Quadrant 1st Quadrant

𝑥 = 90° + 𝑛. 360° 𝑛∈ℤ 𝑥 = 30° + 𝑛. 360°

4th Quadrant 2nd Quadrant

𝑥 = 360° − 90° + 𝑛. 360° 𝑥 = 180° − 30° + 𝑛. 360°

𝑥 = 270° + 𝑛. 360° 𝑥 = 150° + 𝑛. 360°

12. sin 2𝑥 = 4 cos 2𝑥

sin 2𝑥
=4
cos 2𝑥
tan 2𝑥 = 4

ref ∠ for 2𝑥 = tan−1 4

ref ∠ for 2𝑥 = 75,96375653°

1st Quadrant

2𝑥 = 75,96375653° + 𝑛. 180° 𝑛∈ℤ

𝑥 = 37,98° + 𝑛. 90°

3rd Quadrant

2𝑥 = 180° + 75,96375653° +
𝑛. 180°
2𝑥 = 255,9637565° + 𝑛. 180°
𝑥 = 127,98° + 𝑛. 90°
13. 1−cos2 𝐴
Consider:
4 cos(90°+𝐴)
13.1. 1−cos2 𝐴
4 cos(90°+𝐴)
sin2 𝐴
=
4(− sin 𝐴)
1
= − sin 𝐴
4

13.2. 1−cos2 2𝑥
= 0,21
4 cos(90°+2𝑥)
1−cos2 2𝑥 1
= − sin 2𝑥
4 cos(90°+2𝑥) 4

1
− sin 2𝑥 = 0,21
4
sin 2𝑥 = −0,84

ref ∠ for 2𝑥 = sin−1 0,84

ref ∠ for 2𝑥 = 57,14011962°

3rd Quadrant

2𝑥 = 180 + 57,14011962° + 𝑛. 360° 𝑛∈ℤ

2𝑥 = 237,14011962° + 𝑛. 360°
𝑥 = 118,57° + 𝑛. 180°
4th Quadrant

2𝑥 = 360° − 57,14011962° + 𝑛. 360°

2𝑥 = 302,8598804° + 𝑛. 360°
𝑥 = 151,43° + 𝑛. 180°