grade 12 chapter 5+6 trigonometry day 1 (revision for the ... · 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 2𝑥

1+cos 2𝑥= tan 𝑥

Solution

LK = sin 2𝑥

1+cos 2𝑥

= 2 sin 𝑥 cos 𝑥

1+(cos2 𝑥−sin2 𝑥)


1−sin2 𝑥+cos2 𝑥


2 cos2 𝑥

= sin 𝑥

cos 𝑥

= tan 𝑥 = RK

Prove that


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Example

Prove that

Homework Exercise 8 p 61 no a) 2 b) 1 c) 1 d) 1, 2


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Day2 - Trigonometrical graphs Sinus graph

Standard 𝒚 = 𝐬𝐢𝐧 𝒙 graph:

Period = 360𝑜 (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: (0𝑜; 0), (90𝑜; 1), (180𝑜 ; 0), (270𝑜; −1), (360𝑜 ; 0)

Maximum value = 1 (greatest y-value)

Minimum value = −1 (smallest y-value)

𝒚 = 𝟐𝐬𝐢𝐧 𝒙

If there is a number in front of the sin, the amplitude will be influenced. (In other words on the y-axis)

Period = 360𝑜 (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: (0𝑜; 0), (90𝑜; 2), (180𝑜 ; 0), (270𝑜; −2), (360𝑜 ; 0)

Maximum value = 2 (greatest y-value)

Minimum value = −2 (smallest y-value)


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𝒚 = 𝐬𝐢𝐧 𝟐𝒙

If there is a number in front of the 𝑥, the period will be influenced. (In other words on the 𝑥-as) It means that there will fit two sin graphs between 0𝑜 and 360𝑜.

Period = 360𝑜

2= 180𝑜 (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 = −1 (smallest y-value)

𝒚 = −𝐬𝐢𝐧 𝒙

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 = 360𝑜 (How many degrees do it take to complete one graph) Amplitude = 1 (From the middle to the max value or middle to min value)


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𝒚 = 𝐬𝐢𝐧 𝒙 + 1 graph:

The sin-graph will move one place up Period = 360𝑜 (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: (0𝑜; 1), (90𝑜; 2), (180𝑜 ; 1), (270𝑜; 0), (360𝑜; 1) Maximum value = 2 (greatest y-value) Minimum value = 0 (smallest y-value)

𝒚 = 𝐬𝐢𝐧(𝒙 − 𝟑𝟎𝒐) graph:

The sin-graph will move 30𝑜 right Period = 360𝑜 (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: (30𝑜; 0), (120𝑜 ; 1), (210𝑜; 0), (300𝑜 ; −1), (390𝑜; 0) Maximum value = 1 (greatest y-value) Minimum value = −1 (smallest y-value)


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Cosinus graph

Standard 𝒚 = 𝐜𝐨𝐬 𝒙 graph:

Period = 360𝑜 (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: (0𝑜; 1), (90𝑜; 0), (180𝑜 ; −1), (270𝑜; 0), (360𝑜 ; 1) Maximum value = 1 (greatest y-value) Minimum value = −1 (smallest y-value)

𝒚 = 𝟐𝐜𝐨𝐬 𝒙

If there is a number in front of the cos is, the amplitude will be influenced. (In other words on the y-axis) Period = 360𝑜 (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: (0𝑜; 2), (90𝑜; 0), (180𝑜 ; −2), (270𝑜; 0), (360𝑜 ; 2) Maximum value = 2 (greatest y-value) Minimum value = −2 (smallest y-value)


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𝒚 = 𝐜𝐨𝐬 𝟐𝒙

If there is a number in front of the 𝑥 is, the period will be influenced. (In other words on the 𝑥-as) It means that there will fit two cos graphs between 0𝑜 and 360𝑜.

Period = 360𝑜


Amplitude = 1 (From the middle to the max value or middle to min value) Maximum value = 1 (greatest y-value) Minimum value = −1 (smallest y-value)

𝒚 = −𝐜𝐨𝐬 𝒙

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 = 360𝑜 (How many degrees do it take to complete one graph) Amplitude = 1 (From the middle to the max value or middle to min value)


Grey College 9

𝒚 = 𝐜𝐨𝐬 𝒙 − 1

The cos-graph will move one place down Period = 360𝑜 (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: (0𝑜; 0), (90𝑜; −1), (180𝑜 ; −2), (270𝑜; −1), (360𝑜 ; 0) Maximum value = 0 (greatest y-value) Minimum value = −2 (smallest y-value)

𝒚 = 𝐜𝐨𝐬(𝒙 − 𝟑𝟎𝒐)

The cos-graph will move 30𝑜 right Period = 360𝑜 (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: (30𝑜; 1), (12; 0), (210𝑜 ; −1), (300𝑜; 0), (390𝑜 ; 1) Maximum value = 1 (greatest y-value) Minimum value = −1 (smallest y-value)


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Tan graph

The tan graph looks different because it is a fraction function: tan 𝑥 =sin 𝑥

cos 𝑥

Remember that you can’t divide with 0, therefor cos 𝑥 ≠ 0 It means that 𝑥 ≠ 90𝑜 , 270𝑜 . If there is 𝑥-values which is undefined, you get an asymptote.

Standard 𝒚 = 𝐭𝐚𝐧 𝒙 graph:

Period = 180𝑜 (How many degrees do it take to complete one graph) Amplitude = none. None Maximum – of minimum − value. Point of interest : (45𝑜; 1)

Asymptotes by 𝑥 = 90𝑜 and 𝑥 = 270𝑜.

𝒚 = 𝟐𝐭𝐚𝐧 𝒙

Period = 180𝑜 (How many degrees do it take to complete one graph) Amplitude = none. None Maximum – of minimum − value. Point of interest : (𝟒𝟓𝒐;𝟐)



Grey College 11

𝒚 = 𝐭𝐚𝐧 𝟐𝒙

If there is a number in front of the 𝑥, the period will be influenced. (In other words on the 𝑥-axis) It means that there will fit two tan graphs between 0𝑜 and 360𝑜 .

Period = 180𝑜


Amplitude = none. None Maximum – of minimum − value. Point of interest : (𝟐𝟐, 𝟓𝒐; 𝟏)

Asymptotes by 𝑥 = 45𝑜, 𝑥 = 135𝑜, 𝑥 = 225𝑜

and 𝑥 = 315𝑜

.

𝒚 = −𝐭𝐚𝐧 𝒙

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 = 180𝑜 (How many degrees do it take to complete one graph) Point of interest : (45𝑜; −1)


Grey College 12

𝒚 = 𝐭𝐚𝐧 𝒙 + 2 graph:

The graph will move 2 places up Period = 180𝑜 (How many degrees do it take to complete one graph) Amplitude = none. None Maximum – of minimum − value. Point of interest : (45𝑜; 3)


𝒚 = 𝐭𝐚𝐧(𝒙 − 𝟑𝟎𝒐) graph:

Hele tan-graph will move 30𝑜 right Period = 180𝑜 (How many degrees do it take to complete one graph) Amplitude = none. None Maximum – of minimum − value. Point of interest : (75𝑜; 1)

Asymptotes by 𝑥 = 120𝑜 and 𝑥 = 300𝑜 .


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Example


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Solution


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Example Determine the Maximum and minimum values of:

a. 2 sin 𝜃 + 1 = 0

b. 1

3 cos2𝜃+2 sin2 𝜃= 0

Solution

a. 2 sin 𝜃 + 1 = 0

Maximum = 3 Minimum = -1

b. 1

3 cos2𝜃+2 sin2 𝜃= 0

1

cos2 𝜃 + 2 cos2 𝜃 + 2 sin2 𝜃= 0

1

cos2 𝜃 + 2= 0

cos2 𝜃 se Maximum is 1 and minimum is 0 The graph is moved two units up, in other words cos2 𝜃 + 2 will have a maximum at 3 and minimum at 2

1

cos2𝜃+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


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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 and𝐵�̂�𝐶 = 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 ∆ABC, which contains the required height, and ∆CBD, a horizontal triangle in which sufficient information is given for finding CB.

Solution

In ∆CBD, by the cosine rule: CB2 = 702 + 522 − 2(70)(52) cos 44,80 =2 438,3... ∴ 𝐶𝐵 = 49,3

In ∆ABC: 𝐴𝐵

𝐶𝐵= tan 30,7𝑜

∴ 𝐴𝐵 = (49,3 … )tan 30,7𝑜

∴ 𝐴𝐵 = 29,3 …. ∴ The height of the tower is 29,3m. (Correct to 1 decimal)

The area of ∆ABC = 1

2𝑎𝑏 sin 𝐶

(i) a2 = b2 + c2 – 2bc cos A (ii) b2 = a2 + c2 – 2ac cos B (iii) c2 = a2 + b2 – 2ab cos C

𝑎

sin 𝐴=

𝑏

sin 𝐵=

𝑐

sin 𝐶


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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 𝜃𝑜 W of his observation point. He then walks k metres due east and now finds the bearing

of the church to be N 𝛼𝑜 W. Show that the height of the church spire above the ground is cos∝ tan 𝑥

sin(𝛼−𝜃).

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:

�̂�1 = 90𝑜

∴ 𝐻�̂�𝐵 = 90𝑜 + 𝜃

�̂�1 = 90𝑜 − 𝛼 ∴ 𝐻 = 180𝑜 − (90𝑜 + 𝜃 + 90𝑜 − 𝛼)

= 180𝑜 − (180𝑜 + 𝜃 − 𝛼) = 180𝑜 − 180𝑜 − 𝜃 + 𝛼 = 𝛼 − 𝜃

In ∆ABH, by the sin rule: 𝐴𝐻

sin(90𝑜 − 𝛼)=

𝑘

sin(𝛼 − 𝜃)

∴ 𝐴𝐻 = 𝑘 cos 𝛼

sin(𝛼 − 𝜃)

The diagram with the church spire introduced looks like this:

In ∆AHC:

𝐶𝐻

𝐴𝐻= tan 𝑥

∴ 𝐶𝐻 = AH tan 𝑥

∴ 𝐶𝐻 = 𝑘 tan 𝑥 cos 𝛼

sin(𝛼−𝜃)


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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 – 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 – otherwise use the sine rule.

Homework

Exercise 1 p 86 no a, c, d


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Day 4 - Trig 3D~problems

Application of compound angle identities in one and two planes Example 1

Prove : Area of ∆ABC = 𝑎2 sin 𝐵 sin 𝐶

2 sin 𝐴

Solution

Area of ∆ABC = 1

2ab sin 𝐶 ........ (1)

But 𝑎

sin 𝐴 =

𝑏

sin 𝐵

∴ 𝑏 = 𝑎 sin 𝐵

sin 𝐴 ........ (2)

Substitute (2) in (1): Area of ∆ABC = 1

2a (

𝑎 sin 𝐵

sin𝐴 ) sin 𝐶

= 𝑎2 sin 𝐵 sin 𝐶

2 sin 𝐴

Example 2 B, C and D are three points in the same horizontal plane such that BD = CD = d and C �̂�D = 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 = √2 units , x = 75𝑜 and y = 30𝑜, calculate AB, without using a calculator.

Solution a) In ∆BCD: 𝐶1 = 𝑥

�̂�1 = 180𝑜 − 2𝑥

𝐵𝐶

sin(1800 − 2𝑥)=

𝑑

sin 𝑥

∴ 𝐵𝐶 = 𝑑 sin(1800 − 2𝑥)

sin 𝑥

∴ 𝐵𝐶 = 𝑑 sin 2𝑥

sin 𝑥

∴ 𝐵𝐶 = 𝑑 2sin 𝑥 . cos 𝑥

sin 𝑥

∴ 𝐵𝐶 = 2𝑑 cos 𝑥 ....... (1) Since ∆ABC is a right-angled triangle:

tan 𝑦 = 𝐴𝐵

𝐵𝐶

∴ 𝐴𝐵 = BC tan 𝑦 ....... (2) Substitute (1) into (2): AB = 2dcosx.tany


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b) AB = 2√2 cos 75𝑜. tan 300

= 2√2 cos(30𝑜 + 45𝑜 ). tan 300

= 2√2(tan 300)(cos300 cos 450 − sin 300 sin 45𝑜)

= 2√2 (1

√3) ⌈(

√3

2) (

√2

2) − (

1

2) (

√2

2)⌉

= 1 −1

√3

= √3−1

√3

Homework

Exercise 2 p 89 No. a, c, d, f


Grey College 25

grade 12 chapter 5+6 trigonometry day 1 (revision for the ... · day 1 (revision for the test)...

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