Euclidean Geometry and Acceptable proofs

1. A joint initiative between the Western Cape Education Department and Stellenbosch
University.

2. TRIGONOMETRIC
EQUATIONS
Grade 11
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3. The standard forms of angles in the 4 quadrants
𝟎 ≤ 𝛉 ≤ 𝟗𝟎°
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4. 4

5. Graphs of the Trigonometric functions
No maximum or minimum value
Range (−∞; ∞)
The period is 𝟏𝟖𝟎°.
Asymptotes at 𝒙 = 𝟗𝟎° and 𝒙 = 𝟐𝟕𝟎°
𝒚 = tan 𝜽
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6. 𝒚 = 𝐬𝐢𝐧 𝜽 𝒚 = 𝐜𝐨𝐬 𝜽
For graphs, the
Maximum value is +1,
Minimum value is -1
Period is 𝟑𝟔𝟎°.
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7. Consider the sine ratio o𝐟 𝟑𝟎° in the different quadrants.
𝒚 = 𝐬𝐢𝐧 𝟑𝟎°
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8. Consider the sine ratio o𝐟 𝟑𝟎° in the different quadrants
𝐬𝐢𝐧 𝒙 =
𝟏
𝟐
(positive), then
𝒙 = 𝟑𝟎° (Quadrant I)
Or 𝒙 = 𝟏𝟖𝟎° − 𝟑𝟎° (Quadrant II)
𝐬𝐢𝐧 𝒙 = −
𝟏
𝟐
(negative), then
𝒙 = 𝟏𝟖𝟎° + 𝟑𝟎° (Quadrant III)
𝒙 = 𝟐𝟏𝟎°
Or 𝒙 = 𝟑𝟔𝟎° − 𝟑𝟎° (Quadrant IV)
𝒙 = 𝟑𝟑𝟎°
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9. The sine ratio of 𝟑𝟎° has infinitely many solutions
𝒔𝒊𝒏 𝜽 =
𝟏
𝟐
, will have infinitely many solutions.
The graph repeats itself every 𝟑𝟔𝟎°.
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10. Steps to solve an equation
1. Isolate the trig ratio: Trig ratio (angle) = Number.
2. Determine the reference angle – ignore the sign of the ratio.
3. Now use the sign of the ratio, together with the given trig ratio, to determine
in which two quadrants the solutions lie.
4. Add 𝒌. 𝟑𝟔𝟎° to get the general solution of sin and cos and 𝒌. 𝟏𝟖𝟎° for tan.
5. Simplify
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11. Determine the general solution of sin 2𝜃 =
1
4
• sin 2𝜃 =
1
4
• Reference ∠ = sin−1 1
4
= 14,48°
• I : 2𝜃 = 14,48° + 𝑘. 360°; 𝑘 ∈ 𝑍
• 𝜃 = 7,24° + 𝑘. 180°; 𝑘 ∈ 𝑍
• II : 2𝜃 = 180° − 14,48° + 𝑘. 360; 𝑘 ∈ 𝑍
• 2𝜃 = 165,52° + 𝑘. 360°; 𝑘 ∈ 𝑍
• 𝜃 = 82, 76° + 𝑘. 180°; 𝑘 ∈ 𝑍
• 𝜃 = 7,24° + 𝑘. 180° or 𝜃 = 82, 76° + 𝑘. 180°; 𝑘 ∈ 𝑍
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12. Determine the general solution of 3 tan 𝜃 + 1 = 0
• 3 tan 𝜃 + 1 = 0
• 3 tan 𝜃 = −1
• tan 𝜃 = −
1
3
• Reference ∠ = tan−1 1
3
= 18,43°
• Quadrant II and IV: 𝜃 = 180° − 18,43° + 𝑘. 180°; 𝑘 ∈ 𝑍
• ∴ 𝜃 = 161,57° + 𝑘. 180°; 𝑘 ∈ 𝑍
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13. Determine the general solution of 2 cos 𝜃 − 20° = − 3
2 cos 𝜃 − 20° = − 3
cos 𝜃 − 20° = −
3
2
Reference ∠ = cos−1 3
2
= 30°
II : 𝜃 − 20° = 180° − 30° + 𝑘. 360; 𝑘 ∈ 𝑍
𝜃 − 20° = 150° + 𝑘. 360; 𝑘 ∈ 𝑍
𝜃 = 170° + 𝑘. 360; 𝑘 ∈ 𝑍
III : 𝜃 − 20° = 180° + 30° + 𝑘. 360; 𝑘 ∈ 𝑍
𝜃 − 20° = 210° + 𝑘. 360; 𝑘 ∈ 𝑍
𝜃 = 230° + 𝑘. 360; 𝑘 ∈ 𝑍
Answer: 𝜃 = 170° + 𝑘. 360° or 𝜃 = 230° + 𝑘. 360°; 𝑘 ∈ 𝑍
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14. 4 Types of Trigonometric Equations
• A) a Ratio = a Value
• Eg 𝟑 sin 𝜽 = tan 𝟕𝟎°
• 𝟑 sin 𝜽 = 𝟐, 𝟕𝟒𝟕𝟒𝟖
sin 𝜽 = 𝟎, 𝟗𝟏𝟓𝟖
• Reference ∠ = sin−1
0,9158 = 66,32°
• Quadrant I : 𝜃 = 66,32° + 𝑘. 360°; 𝑘 ∈ 𝑍
• Quadrant II : 𝜃 = 180° − 66,32° + 𝑘. 360; 𝑘 ∈ 𝑍
• 𝜃 = 113,68° + 𝑘. 360°; 𝑘 ∈ 𝑍
• 𝜃 = 66,32° + 𝑘. 360°; of 𝜃 = 113,68° + 𝑘. 360°; 𝑘 ∈ 𝑍
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15. B) Ratio = co-ratio
• sin 𝑥 = cos 3𝑥
• sin 𝑥 = sin(90° − 3𝑥)
• ∴ 𝑥 = 90° − 3𝑥 is the reference angle
• I 𝑥 = 90° − 3𝑥 + 𝑘. 360° ; 𝑘 ∈ 𝑍
• 4𝑥 = 90° + 𝑘. 360°
• 𝑥 = 22,5° + 𝑘. 90° ; 𝑘 ∈ 𝑍
• II or 𝑥 = 180° − 90° − 3𝑥 + 𝑘. 360° ; 𝑘 ∈ 𝑍
• 𝑥 = 180° − 90° + 3𝑥 + 𝑘. 360°
• −2𝑥 = 90° + 𝑘. 360°
• 𝑥 = −45° + 𝑘. 180° ; 𝑘 ∈ 𝑍
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16. C) Ratio = Ratio
• sin 𝑥 = 3 cos 𝑥
•
sin 𝑥
cos 𝑥
=
3 cos 𝑥
cos 𝑥
• tan 𝑥 = 3
• Reference ∠ = tan−1 3 = 71,57°
• ∴ 𝑥 = 71,57° + 𝑘. 180° ; 𝑘 ∈ 𝑍
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17. D) Factorising and Identities
Factorising is required to solve certain trigonometric equations.
Look out for the following types of Factorising
 Common factor: sin 𝑥 + 3sin 𝑥. cos 𝑥 = 0
 Difference of Two Squares: sin2𝑥 − 9cos2𝑥 = 0
 Trinomial: sin2𝑥 − 2 cos 𝑥 . sin 𝑥 + cos2𝑥 = 0
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18. Determine the general solution of 2sin2
𝜃 = sin 𝜃
• 2sin2
𝜃 = sin 𝜃
• 2sin2𝜃 − sin 𝜃 = 0
• sin 𝜃 2 sin 𝜃 − 1 = 0
• ∴ sin 𝜃 = 0 or 2 sin 𝜃 − 1 = 0
• ∴ sin 𝜃 = 0 or sin 𝜃 =
1
2
• ∴ sin 𝜃 = 0 : 𝜃 = 0° + 𝑘. 360° or 𝜃 = 180° + 𝑘. 360° ; 𝑘 ∈ Z
• ∴ sin 𝜃 =
1
2
: 𝜃 = 30° + 𝑘. 360° or 𝜃 = 150° + 𝑘. 360° ; 𝑘 ∈ Z
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19. Determine the general solution of 2cos2
𝜃 = sin𝜃 + 1
• 2cos2
𝜃 − sin 𝜃 − 1 = 0
• 2 1 − sin2𝜃 − sin 𝜃 − 1 = 0
• 2 − 2sin2𝜃 − sin 𝜃 − 1 = 0
• 2sin2
𝜃 + sin 𝜃 − 1 = 0
• 2 sin 𝜃 − 1 sin 𝜃 + 1 = 0
• sin 𝜃 =
1
2
of sin 𝜃 = −1
• 𝜃 = 30° + 𝑘. 360° or 𝜃 = 150° + 𝑘. 360°
or 𝜃 = 270° + 𝑘. 360°; 𝑘 ∈ Z
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20. Determining the solutions in a given interval
• Solve for θ if sin 𝜃 = −
1
5
and θ ∈ −360°; 360°
• sin 𝜃 = −
1
5
• Reference ∠ = sin−1 1
5
= 11,54°
• III: 𝜃 = 180° + 11,54° + 𝑘. 360°; 𝑘 ∈ 𝑍
• 𝜽 = 𝟏𝟗𝟏, 𝟓𝟒° + 𝒌. 𝟑𝟔𝟎°; 𝒌 ∈ 𝒁
• IV: 𝜃 = 360° − 11,54° + 𝑘. 360°; 𝑘 ∈ 𝑍
• 𝜽 = 𝟑𝟒𝟖, 𝟒𝟔° + 𝒌. 𝟑𝟔𝟎°; 𝒌 ∈ 𝒁
• ∴ 𝜃 ∈ 191,54°; 348,46°; −168,46°; −11,54°
 For a given interval,
substitute the integer values
of 𝑘 in the obtained general
solution.
 Start with 0, 1, 2 and then
− 1, −2, etc as values for 𝑘
until the answer lies outside
the given interval.
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