grade 12 chapter 5+6 trigonometry day 1 (revision for the ...Β Β· day 1 (revision for the test)...
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Grade 12 Chapter 5+6 Trigonometry
Day 1 (revision for the test)
Grade 12 Chapter 5 + 6 Trig
Grey College 2
Example
Prove that sin 2π₯
1+cos 2π₯= tan π₯
Solution
LK = sin 2π₯
1+cos 2π₯
= 2 sin π₯ cos π₯
1+(cos2 π₯βsin2 π₯)
= 2 sin π₯ cos π₯
1βsin2 π₯+cos2 π₯
= 2 sin π₯ cos π₯
2 cos2 π₯
= sin π₯
cos π₯
= tan π₯ = RK
Prove that
Grade 12 Chapter 5 + 6 Trig
Grey College 3
Example
Prove that
Homework Exercise 8 p 61 no a) 2 b) 1 c) 1 d) 1, 2
Grade 12 Chapter 5 + 6 Trig
Grey College 4
Day2 - Trigonometrical graphs Sinus graph
Standard π = π¬π’π§ π graph:
Period = 360π (How many degrees do it take to complete one graph)
Amplitude = 1 (From the middle to the max value or middle to min value)
Turning points: (0π; 0), (90π; 1), (180π ; 0), (270π; β1), (360π ; 0)
Maximum value = 1 (greatest y-value)
Minimum value = β1 (smallest y-value)
π = ππ¬π’π§ π
If there is a number in front of the sin, the amplitude will be influenced. (In other words on the y-axis)
Period = 360π (How many degrees do it take to complete one graph)
Amplitude = 2 (From the middle to the max value or middle to min value)
Turning points: (0π; 0), (90π; 2), (180π ; 0), (270π; β2), (360π ; 0)
Maximum value = 2 (greatest y-value)
Minimum value = β2 (smallest y-value)
Grade 12 Chapter 5 + 6 Trig
Grey College 5
π = π¬π’π§ ππ
If there is a number in front of the π₯, the period will be influenced. (In other words on the π₯-as) It means that there will fit two sin graphs between 0π and 360π.
Period = 360π
2= 180π (How many degrees do it take to complete one graph)
Amplitude = 1 (From the middle to the max value or middle to min value) Maximum value = 1 (greatest y-value) Minimum value = β1 (smallest y-value)
π = βπ¬π’π§ π
If there is a number in front of the sin is, the amplitude will be influenced. (In other words on the y-axis) The negative in front of the sin means that the graph will reflect in the x-axis. Period = 360π (How many degrees do it take to complete one graph) Amplitude = 1 (From the middle to the max value or middle to min value)
Grade 12 Chapter 5 + 6 Trig
Grey College 6
π = π¬π’π§ π + 1 graph:
The sin-graph will move one place up Period = 360π (How many degrees do it take to complete one graph) Amplitude = 1 (From the middle to the max value or middle to min value) Turning points: (0π; 1), (90π; 2), (180π ; 1), (270π; 0), (360π; 1) Maximum value = 2 (greatest y-value) Minimum value = 0 (smallest y-value)
π = π¬π’π§(π β πππ) graph:
The sin-graph will move 30π right Period = 360π (How many degrees do it take to complete one graph) Amplitude = 1 (From the middle to the max value or middle to min value) Turning points: (30π; 0), (120π ; 1), (210π; 0), (300π ; β1), (390π; 0) Maximum value = 1 (greatest y-value) Minimum value = β1 (smallest y-value)
Grade 12 Chapter 5 + 6 Trig
Grey College 7
Cosinus graph
Standard π = ππ¨π¬ π graph:
Period = 360π (How many degrees do it take to complete one graph) Amplitude = 1 (From the middle to the max value or middle to min value) Turning points: (0π; 1), (90π; 0), (180π ; β1), (270π; 0), (360π ; 1) Maximum value = 1 (greatest y-value) Minimum value = β1 (smallest y-value)
π = πππ¨π¬ π
If there is a number in front of the cos is, the amplitude will be influenced. (In other words on the y-axis) Period = 360π (How many degrees do it take to complete one graph) Amplitude = 2 (From the middle to the max value or middle to min value) Turning points: (0π; 2), (90π; 0), (180π ; β2), (270π; 0), (360π ; 2) Maximum value = 2 (greatest y-value) Minimum value = β2 (smallest y-value)
Grade 12 Chapter 5 + 6 Trig
Grey College 8
π = ππ¨π¬ ππ
If there is a number in front of the π₯ is, the period will be influenced. (In other words on the π₯-as) It means that there will fit two cos graphs between 0π and 360π.
Period = 360π
2= 180π (How many degrees do it take to complete one graph)
Amplitude = 1 (From the middle to the max value or middle to min value) Maximum value = 1 (greatest y-value) Minimum value = β1 (smallest y-value)
π = βππ¨π¬ π
If there is a number in front of the cos is, the amplitude will be influenced. (In other words on the y-axis) The negative in front of the cos means that the graph will reflect in the x-axis. Period = 360π (How many degrees do it take to complete one graph) Amplitude = 1 (From the middle to the max value or middle to min value)
Grade 12 Chapter 5 + 6 Trig
Grey College 9
π = ππ¨π¬ π β 1
The cos-graph will move one place down Period = 360π (How many degrees do it take to complete one graph) Amplitude = 1 (From the middle to the max value or middle to min value) Turning points: (0π; 0), (90π; β1), (180π ; β2), (270π; β1), (360π ; 0) Maximum value = 0 (greatest y-value) Minimum value = β2 (smallest y-value)
π = ππ¨π¬(π β πππ)
The cos-graph will move 30π right Period = 360π (How many degrees do it take to complete one graph) Amplitude = 1 (From the middle to the max value or middle to min value) Turning points: (30π; 1), (12; 0), (210π ; β1), (300π; 0), (390π ; 1) Maximum value = 1 (greatest y-value) Minimum value = β1 (smallest y-value)
Grade 12 Chapter 5 + 6 Trig
Grey College 10
Tan graph
The tan graph looks different because it is a fraction function: tan π₯ =sin π₯
cos π₯
Remember that you canβt divide with 0, therefor cos π₯ β 0 It means that π₯ β 90π , 270π . If there is π₯-values which is undefined, you get an asymptote.
Standard π = πππ§ π graph:
Period = 180π (How many degrees do it take to complete one graph) Amplitude = none. None Maximum β of minimum β value. Point of interest : (45π; 1)
Asymptotes by π₯ = 90π and π₯ = 270π.
π = ππππ§ π
Period = 180π (How many degrees do it take to complete one graph) Amplitude = none. None Maximum β of minimum β value. Point of interest : (πππ;π)
Asymptotes by π₯ = 90π and π₯ = 270π.
Grade 12 Chapter 5 + 6 Trig
Grey College 11
π = πππ§ ππ
If there is a number in front of the π₯, the period will be influenced. (In other words on the π₯-axis) It means that there will fit two tan graphs between 0π and 360π .
Period = 180π
2= 90π (How many degrees do it take to complete one graph)
Amplitude = none. None Maximum β of minimum β value. Point of interest : (ππ, ππ; π)
Asymptotes by π₯ = 45π, π₯ = 135π, π₯ = 225π
and π₯ = 315π
.
π = βπππ§ π
If there is a number in front of the tan is, the amplitude will be influenced. (In other words on the y-axis) The negative in front of the tan means that the graph will reflect in the x-axis. Period = 180π (How many degrees do it take to complete one graph) Point of interest : (45π; β1)
Grade 12 Chapter 5 + 6 Trig
Grey College 12
π = πππ§ π + 2 graph:
The graph will move 2 places up Period = 180π (How many degrees do it take to complete one graph) Amplitude = none. None Maximum β of minimum β value. Point of interest : (45π; 3)
Asymptotes by π₯ = 90π and π₯ = 270π.
π = πππ§(π β πππ) graph:
Hele tan-graph will move 30π right Period = 180π (How many degrees do it take to complete one graph) Amplitude = none. None Maximum β of minimum β value. Point of interest : (75π; 1)
Asymptotes by π₯ = 120π and π₯ = 300π .
Grade 12 Chapter 5 + 6 Trig
Grey College 13
Example
Grade 12 Chapter 5 + 6 Trig
Grey College 14
Solution
Grade 12 Chapter 5 + 6 Trig
Grey College 15
Example Determine the Maximum and minimum values of:
a. 2 sin π + 1 = 0
b. 1
3 cos2π+2 sin2 π= 0
Solution
a. 2 sin π + 1 = 0
Maximum = 3 Minimum = -1
b. 1
3 cos2π+2 sin2 π= 0
1
cos2 π + 2 cos2 π + 2 sin2 π= 0
1
cos2 π + 2= 0
cos2 π se Maximum is 1 and minimum is 0 The graph is moved two units up, in other words cos2 π + 2 will have a maximum at 3 and minimum at 2
1
cos2π+2= 0 Maximum is
1
2 and the minimum is
1
3
Homework
Exercise 12 p 73 no b) 1, 2, 4, 5, 6, 7 ; c ; d) 2,4,6,8,10,12 e) 9,10
Grade 12 Chapter 5 + 6 Trig
Grey College 16
Grade 12 Chapter 5 + 6 Trig
Grey College 17
Grade 12 Chapter 5 + 6 Trig
Grey College 18
Grade 12 Chapter 5 + 6 Trig
Grey College 19
Day 3 - Trig 3D~problems
Problems in three dimensions (Two planes)
Example1 A tower AB stands in a horizontal plane BCD. From A the angle of depression of C is 30,70. If BD = 70m; CD = 52m andπ΅οΏ½ΜοΏ½πΆ = 44,80 , calculate the height of the tower.
Note The height of the tower, AB, lies in a triangle in which no side is given. We must first find the length of the side shared by βABC, which contains the required height, and βCBD, a horizontal triangle in which sufficient information is given for finding CB.
Solution
In βCBD, by the cosine rule: CB2 = 702 + 522 β 2(70)(52) cos 44,80 =2 438,3... β΄ πΆπ΅ = 49,3
In βABC: π΄π΅
πΆπ΅= tan 30,7π
β΄ π΄π΅ = (49,3 β¦ )tan 30,7π
β΄ π΄π΅ = 29,3 β¦. β΄ The height of the tower is 29,3m. (Correct to 1 decimal)
The area of βABC = 1
2ππ sin πΆ
(i) a2 = b2 + c2 β 2bc cos A (ii) b2 = a2 + c2 β 2ac cos B (iii) c2 = a2 + b2 β 2ab cos C
π
sin π΄=
π
sin π΅=
π
sin πΆ
Grade 12 Chapter 5 + 6 Trig
Grey College 20
Example2 A boy stands at a point A and observes that the angle of elevation of the top of a church spire is x and that
the church is N ππ W of his observation point. He then walks k metres due east and now finds the bearing
of the church to be N πΌπ W. Show that the height of the church spire above the ground is cosβ tan π₯
sin(πΌβπ).
Solution
The two observation points and the foundation of the church, H, are in the same horizontal plane. The ground plan looks like this diagram:
οΏ½ΜοΏ½1 = 90π
β΄ π»οΏ½ΜοΏ½π΅ = 90π + π
οΏ½ΜοΏ½1 = 90π β πΌ β΄ π» = 180π β (90π + π + 90π β πΌ)
= 180π β (180π + π β πΌ) = 180π β 180π β π + πΌ = πΌ β π
In βABH, by the sin rule: π΄π»
sin(90π β πΌ)=
π
sin(πΌ β π)
β΄ π΄π» = π cos πΌ
sin(πΌ β π)
The diagram with the church spire introduced looks like this:
In βAHC:
πΆπ»
π΄π»= tan π₯
β΄ πΆπ» = AH tan π₯
β΄ πΆπ» = π tan π₯ cos πΌ
sin(πΌβπ)
Grade 12 Chapter 5 + 6 Trig
Grey College 21
Hints on solving problems of triangles in two planes:
It often happens that the height or length to be found lies in a triangle in which insufficient information has been given. It is usually possible to find the length of a side which is common to this triangle and another triangle in which sufficient information has been given. We then set about finding the length of this shared side β which is often the dividing line between 2 different planes.
If compass bearings and angles of elevation or depression are given, first draw a ground plan. In right-angled triangles, trigonometric ratios (or the sine rule) should be used.
In a triangle which is not right-angled, use the cosine rule if two sides and the including angle are given or if three sides are given β otherwise use the sine rule.
Homework
Exercise 1 p 86 no a, c, d
Grade 12 Chapter 5 + 6 Trig
Grey College 22
Grade 12 Chapter 5 + 6 Trig
Grey College 23
Day 4 - Trig 3D~problems
Application of compound angle identities in one and two planes Example 1
Prove : Area of βABC = π2 sin π΅ sin πΆ
2 sin π΄
Solution
Area of βABC = 1
2ab sin πΆ ........ (1)
But π
sin π΄ =
π
sin π΅
β΄ π = π sin π΅
sin π΄ ........ (2)
Substitute (2) in (1): Area of βABC = 1
2a (
π sin π΅
sinπ΄ ) sin πΆ
= π2 sin π΅ sin πΆ
2 sin π΄
Example 2 B, C and D are three points in the same horizontal plane such that BD = CD = d and C οΏ½ΜοΏ½D = x. AB is perpendicular to the plane and the angle of elevation of A from C is y.
a) Prove: AB = 2dcosx.tany
b) Given that d = β2 units , x = 75π and y = 30π, calculate AB, without using a calculator.
Solution a) In βBCD: πΆ1 = π₯
οΏ½ΜοΏ½1 = 180π β 2π₯
π΅πΆ
sin(1800 β 2π₯)=
π
sin π₯
β΄ π΅πΆ = π sin(1800 β 2π₯)
sin π₯
β΄ π΅πΆ = π sin 2π₯
sin π₯
β΄ π΅πΆ = π 2sin π₯ . cos π₯
sin π₯
β΄ π΅πΆ = 2π cos π₯ ....... (1) Since βABC is a right-angled triangle:
tan π¦ = π΄π΅
π΅πΆ
β΄ π΄π΅ = BC tan π¦ ....... (2) Substitute (1) into (2): AB = 2dcosx.tany
Grade 12 Chapter 5 + 6 Trig
Grey College 24
b) AB = 2β2 cos 75π. tan 300
= 2β2 cos(30π + 45π ). tan 300
= 2β2(tan 300)(cos300 cos 450 β sin 300 sin 45π)
= 2β2 (1
β3) β(
β3
2) (
β2
2) β (
1
2) (
β2
2)β
= 1 β1
β3
= β3β1
β3
Homework
Exercise 2 p 89 No. a, c, d, f
Grade 12 Chapter 5 + 6 Trig
Grey College 25
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