| MATHEMATICS |
WISKUNDE |
| GRADE 11 |
GRAAD 11 |
| MORE EXERCISES |
NOG OEFENINGE |
| Solving trigonometric equations. |
Oplossing van trigonometriese |
| : : answers. |
vergelykings : antwoorde. |
| |
Use the positive value of the function to |
|
Gebruik die positiewe funksie-waarde om die |
| |
calculate the reference angle ("ref. ∠"). |
|
verwysingshoek ("verw. ∠") te bereken. |
| |
Now use this angle to calculate the size |
|
Gebruik nou hierdie hoek om die groottes |
| |
of the required angles. |
|
van die gevraagde hoeke te bereken. |
|
1.1
sin θ = 0,819
1.2
cos α = 0,337
ref. ∠ / verw. ∠ = 54,98°
ref. ∠ / verw. ∠ = 70,31°
┃ ∴ θ in quadrants 1 and 2
┃ ∴ α in quadrants 1 and 4
sin θ > 0 ┃
cos α > 0 ┃
┃ ∴ θ in kwadrante 1 en 2
┃ ∴ α in kwadrante 1 en 4
θ = 54,98° + k .360° ; k ∈ Z
α = 70,31° + k .360° ; k ∈ Z
OR / OF
OR / OF
θ = (180° − 54,98°) + k .360° ; k ∈ Z
α = (360° − 70,31°) + k .360° ; k ∈ Z
= 125,02° + k .360° ; k ∈ Z
= 289,69° + k .360° ; k ∈ Z
1.3
tan θ = 2,605
1.4
sin α = − 0,468
ref. ∠ / verw. ∠ = 68,999..°
ref. ∠ / verw. ∠ = 27,9045..°
┃ ∴ θ in quadrants 1 and 3
┃ ∴ α in quadrants 3 and 4
tan θ > 0 ┃
sin α < 0 ┃
┃ ∴ θ in kwadrante 1 en 3
┃ ∴ α in kwadrante 3 en 4
θ = 68,999..° + k .180°
α = (180° + 27,9045..°) + k .360°
θ = 69,00° + k .180° ; k ∈ Z
α = 152,10° + k .360° ; k ∈ Z
OR / OF
α = (360° − 27,9045..°) + k .360° ; k ∈ Z
= 332,10° + k .360° ; k ∈ Z
1.5
cos Φ = − 0,888
1.6
tan α = − 0,43
ref. ∠ / verw. ∠ = 27,3770..°
ref. ∠ / verw. ∠ = 23,2677...°
┃ ∴ θ in quadrants 2 and 3
┃ ∴ α in quadrants 1 and 3
cos Φ < 0 ┃
tan α > 0 ┃
┃ ∴ θ in kwadrante 2 en 3
┃ ∴ α in kwadrante 1 en 3
Φ = (180° − 27,3770..°) + k .360°
α = 23,2677..° + k .180°
Φ = 152,62° + k .360° ; k ∈ Z
α = 23,27° + k .180° ; k ∈ Z
OR / OF
Φ = (180° + 27,3770..°) + k .360° ; k ∈ Z
Φ = 207,38° + k .360° ; k ∈ Z
1
1
1.7
cosec α = 2,103 ∴ sin α = ———
1.8
sec Φ = 1,257 ∴ cos Φ = ———
2,103
1,257
= 0,47551...
= 0,79554..
ref. ∠ / verw. ∠ = 28,3925..°
ref. ∠ / verw. ∠ = 37,2937..°
α = 28,3925..° + k .360°
Φ = 37,2937..° + k .360°
α = 28,39° + k .360° ; k ∈ Z
Φ = 37,29° + k .360° ; k ∈ Z
OR / OF
OR / OF
α = (180° − 28,3925..°) + k .360°
Φ = (360° − 37,2937..°) + k .360°
α = 151,61° + k .360° ; k ∈ Z
Φ = 322,71° + k .360° ; k ∈ Z
1
3
3
1.9
cot θ = − 0,315 ∴ tan α = ————
1.10
— tan α = 1,077 ∴ tan α = 1,077 x ——
−0,315
2
2
= − 3,17460...
= 1,6155
ref. ∠ / verw. ∠ = 72,5155..°
ref. ∠ / verw. ∠ = 58,24235..°
θ = (180° − 72,5155..°) + k .180° ; k ∈ Z
α = 58,24235..° + k .180° ; k ∈ Z
θ = 107,48° + k .180° ; k ∈ Z
α = 58,24° + k .180° ; k ∈ Z
1
3
2
1.11
— cos θ = − 0,468 ∴ cos θ = 2 x (− 0,4682)
1.12
— sin θ = 0,544 ∴ sin θ = 0,544 x ——
2
2
3
= − 0,9364
= 0,816
ref. ∠ / verw. ∠ = 20,5444..°
ref. ∠ / verw. ∠ = 54,68635..°
θ = (180° − 20,5444..°) + k .360°
θ = 54,68635..° + k .360°
θ = 159,46° + k .360° ; k ∈ Z
θ = 54,69° + k .360° ; k ∈ Z
OR / OF
OR / OF
θ = (180° + 20,5444..°) + k .360°
θ = (180° − 54,68635..°) + k .360°
θ = 200,54° + k .360° ; k ∈ Z
θ = 125,31° + k .360° ; k ∈ Z
1.13
sin (α + 12,5°) = 0,819
1.14
cos (θ − 22,3°) = − 0,273
ref. ∠ / verw. ∠ = 54,9848..°
ref. ∠ / verw. ∠ = 74,15713..°
α + 12,5° = 54,9848..° + k .360°
θ − 22,3° = (180° − 74,15713..°) + k .360°
α = 54,9848..° − 12,5° + k .360°
θ = 105,84287° + 22,3° + k .360°
α = 42,48° + k .360° ; k ∈ Z
θ = 128,14° + k .360° ; k ∈ Z
OR / OF
OR / OF
α + 12,5° = (180° − 54,9848..°) + k .360°
θ − 22,3° = (180° + 74,15713..°) + k .360°
α = 125,0152° − 12,5° + k .360°
θ = 254,15713° + 22,3° + k .360°
α = 112,52° + k .360° ; k ∈ Z
θ = 276,46° + k .360° ; k ∈ Z
1.15
tan (θ − 32,6°) = 2,836
1.16
cot (α + 34,5°) + 1,664 = 0
1,664
1
tan (θ − 32,6°) = − ———
1,664
ref. ∠ / verw. ∠ = 70,57687..°
ref. ∠ / verw. ∠ = 31,0042..°
θ − 32,6° = 70,57687..° + k .180° ; k ∈ Z
α + 34,5° = (180° − 31,0042..°) + k .180° ; k ∈ Z
θ = 103,18° + k .180° ; k ∈ Z
α= 114,50° + k .180° ; k ∈ Z
1.17
2 cosec (θ − 18,3°) = 4,822
1.18
tan (2θ − 28,4°) = − 0,430
1
sin (θ − 18,3°) = ———
2,411
ref. ∠ / verw. ∠ = 24,50455..°
ref. ∠ / verw. ∠ = 23,2677..°
θ − 18,3° = 24,50455..° + k .360°
2θ − 28,4° = (180° − 23,2677..°) + k .180°
θ = 42,80° + k .360° ; k ∈ Z
2θ = 185,1322952...° + k .180° ; k ∈ Z
OR / OF
θ = 92,57° + k .90° ; k ∈ Z
θ − 18,3° = (180° − 24,50455..°) + k .360°
θ = 173,80° + k .360° ; k ∈ Z
| 2. |
Find the general solution and then use it to |
2. |
Bepaal die algemene oplossing en gebruik dit |
| |
calculate the required sizes of the angles. |
|
om die gevraagde hoekegroottes te bereken. |
|
2.1
1,5 tan (θ − 10°) + 1,077 = 0
2.2
cos 2α = sin 12°
tan (θ − 10°) = − 0,718
cos 2α = 0,20791...
ref. ∠ / verw. ∠ = 35,6789..°
ref. ∠ / verw. ∠ = 78,00°
θ − 10° = (180°− 35,6783..°) + k .180°
2α = 78° + k .360°
θ = 154,32° + k .180° ; k ∈ Z
α = 39° + k .180° ; k ∈ Z
∴ θ = 154,32°
∴ α = 39°
2.3
4 sin (θ − 15°) + 2 = 0
2.4
sec (2 Φ + 24,3°) − 2,5 = 0
1
sin (θ − 15°) = −0,5
cos (2 Φ + 24,3°) = ———
2,5
ref. ∠ / verw. ∠ = 30°
ref. ∠ / verw. ∠ = 66,42182..°
θ − 15° = (180°+ 30°) + k .360°
2 Φ + 24,3° = 66,42182..° + k .360°
θ = 225° + k .360°
Φ = 21,06° + k .180°
OR / OF
OR / OF
θ − 15° = (360°− 30°) + k .360°
2 Φ + 24,3° = (360°− 66,42182..°) + k .360°
θ = 345° + k .360°
Φ = 134,64° + k .180°
∴ θ = 225°
∴ Φ = 21,06°
2.5
tan θ − 1 = − 6 and // en θ ∈ [− 180° ; 180°]
2.6
cot (Φ − 11,8°) = − 1,664 and // en θ ∈ [− 90° ; 270°]
1
tan θ = − 5
tan (Φ − 11,8°) = − ————
1,664
ref. ∠ / verw. ∠ = 78,69006..°
ref. ∠ / verw. ∠ = 31,004248..°
θ = (180° − 78,69006..°) + k .180°
Φ − 11,8° = (180° − 31,004248..°) + k .180°
θ = 101,31° + k .180°
Φ = 160,80° + k .180°
∴ θ = − 78,69° or/of 101,31°
∴ Φ = −19,20° or/of 160,80°
3.1
sin (2θ − 40°) + 0,5 = 0
3.2
√2 cos (α + 34°) = 1
1
sin (2θ − 40°) = − 0,5
cos (α + 34°) = ——
√2
ref. ∠ / verw. ∠ = 30°
ref. ∠ / verw. ∠ = 45°
2θ − 40° = (180° + 30°)
α + 34° = 45°
θ = 125°
α = 11°
OR / OF
OR / OF
2θ − 40° = (360° − 30°)
α + 34° = (360° − 45°)
θ = 185°
α = 281°
θ = 125° OR/OF θ = 185°
α = 11° OR/OF α = 281°
√3
1
3.3
sin (θ − 14°) = ——
3.4
tan (2α − 18°) = ——
2
√3
ref. ∠ / verw. ∠ = 60°
ref. ∠ / verw. ∠ = 30°
θ − 14° = 60°
2α − 18° = 30°
θ = 74°
α = 24°
OR / OF
OR / OF
θ − 14° = 180° − 60°
2α − 18° = 180° + 30°
θ = 134°
α = 114°
θ = 74° OR/OF θ = 134°
α = 24° OR/OF α = 114°
3.5
2 sec (5 θ + 15°) + 1 = 5
3.6
2 cos (2α − 15°) = tan 45°
1
1
cos (5 θ + 15°) = —
cos (2α − 15°) = ——
2
2
ref. ∠ / verw. ∠ = 60°
ref. ∠ / verw. ∠ = 60°
5 θ + 15° = 60°
2α − 15° = 60°
θ = 9°
α = 37,5°
OR / OF
OR / OF
5 θ + 15° = 360° − 60°
2α − 15° = 360° − 60°
θ = 57°
α = 157,5°
4.1
sin x . (2 cos x − 1) = 0
sin x = 0
OR / OF
2 cos x − 1 = 0 i.e. cos x = 0,5
ref. ∠ / verw. ∠ = 0°
ref. ∠ / verw. ∠ = 60°
x = 0° , 180° , 360°
x = 60° OR / OF x = 360° − 60°
x = 60° OR / OF x = 300°
4.2
sin x = cos 48°
= cos (90° − 42°)
. . . change cos to sin /
verander cos na sin
= sin 42°
x = 42°
OR / OF
x = 180° − 42°
= 138°
x = 42°
OR / OF
x = 138°
4.3
cos2 x + sin x = 1
sin x = 1 − cos2 x
sin x = sin2 x
sin2 x − sin x = 0
sin x(sin x − 1) = 0
sin x = 0
OR / OF
sin x − 1 = 0 ⇒ sin x = 1
ref. ∠ / verw. ∠ = 0°
ref. ∠ / verw. ∠ = 90°
x = 0° , 180° , 360°
x = 90°
4.4
3 cos2 x − 5 sin x = 1
3 (1 − sin2 x )− 5 sin x = 1
. . . sin2 x + cos2 x = 1
3 sin2 x + 5 sin x − 2 = 0
(3 sin x − 1)(sin x + 2) = 0
3 sin x − 1 = 0
OR / OF
sin x + 2 = 0
1
sin x = —
sin x = − 2
3
ref. ∠ / verw. ∠ = 19,47122...°
no solution / geen oplossing − 1 ≤ sin x ≤ 1
x = 19,47° OR/OF x = 180° −19,47122...°
x = 19,47° OR/OF x = 160,53