p1 chapter 9 :: trigonometric ratios
TRANSCRIPT
P1 Chapter 9 :: Trigonometric Ratios
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Last modified: 18th August 2017
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Chapter Overview
1:: Sine/Cosine Rule
Sketch π¦ = sin 2π₯ for 0 β€ π₯ β€ 360Β°
3:: Graphs of Sine/Cosine/Tangent
In Ξπ΄π΅πΆ, π΄π΅ = 5, π΅πΆ = 6 and β π΄π΅πΆ = π₯. Given that the area of β π΄π΅πΆ is 12cm2 and that π΄πΆ is the longest side, find the value of π₯.
2:: Areas of Triangles
60Β°
5π₯
π₯ + 2
Determine π₯.
There is technically no new content in this chapter since GCSE.However, the problems might be more involved than at GCSE level.
RECAP :: Right-Angled Trigonometry
hyp
adj
opp
π
You are probably familiar with the formula: sin π =πππ
βπ¦π
But what is the conceptual definition of π ππ ?sin is a function which inputs an angle and gives the ratio between the opposite and hypotenuse.
Remember that a ratio just means the βrelative sizeβ between quantities (in this case lengths). For this reason, sin/cos/tan are known as βtrigonometric ratiosβ.
20Β°
4
π₯
Find π₯.
sin 20 =4
π₯
π₯ =4
sin 20= 11.7 (3π π)
Fro Tip: You can swap the thing youβre dividing
by and the result. e.g. 8
2= 4 β
8
4= 2. I call this
the βswapsie trickβ.
Find π.
π3
5
tan π =5
3
π = tanβ15
3= 59.0Β°
Froflections: You may have been taught βuse π‘ππβ1 whenever youβre finding an angleβ, and therefore write the second line directly. This is fine, but I prefer to always write the first line, then see the problem as a βchanging the subjectβ one. We need to remove the tan on front of the π, so apply π‘ππβ1 to each side of the equation to βcancel outβ the tan on the LHS.
Just for your interestβ¦
Have you ever wondered why βcosineβ contains the word βsineβ?
π
π
Supplementary Anglesadd to 180Β°
Complementary Anglesadd to 90Β°
ππ
Therefore these angles are complementary.
π
π
50Β°
40Β°
π¦
π₯
π§cos 50 =
π
π
sin 40 =π
π
β΄ ππ¨π¬ ππ = π¬π’π§ ππ
i.e. The cosine of an angle is the sine of the complementary angle.Hence cosine = COMPLEMENTARY SINE
ME-WOW!
OVERVIEW: Finding missing sides and angles
You have You want Use
#1: Two angle-side
opposite pairs
Missing
angle or side
in one pair
Sine rule
#2 Two sides known and a
missing side opposite a
known angle
Remaining
side
Cosine rule
#3 All three sides An angle Cosine rule
#4 Two sides known and a
missing side not opposite
known angle
Remaining
side
Sine rule
twice
When triangles are not right-angled, we can no longer use simple trigonometric ratios, and must use the cosine and sine rules.
Cosine Rule
We use the cosine rule whenever we have three sides (and an angle) involved.
15
12
115Β°
π₯
Cosine Rule:π2 = π2 + π2 β 2ππ cos π΄
π
π΄
π
π
The only angle in formula is π΄, so label angle in diagram π΄, label opposite side π, and so on (π and π can go either way).
π₯2 = 152 + 122 β 2 Γ 15 Γ 12 Γ cos 115π₯2 = 521.14257β¦π₯ = 22.83
Proof at end of PowerPoint.
Dealing with Missing Angles
7
49
πΌ
ππ = ππ + ππ β π Γ π Γ π Γ ππ¨π¬πΆππ = πππ β πππππ¨π¬πΆπππππ¨π¬πΆ = πππ β ππ
ππ¨π¬πΆ =πππ
πππ
πΆ = ππ¨π¬βππππ
πππ= ππ. πΒ°
Label sides then substitute into formula.
Simplify each bit of formula.
Rearrange (I use βsubtraction swapsie trickβ to swap thing youβre subtracting and result)
ππ = ππ + ππ β πππ ππ¨π¬π¨
You have You want Use
All three sides An angle Cosine rule
Fro Tip: The brackets are not needed, but students who forget about BIDMAS see 42 = 72 + 92 β 2 Γ 7 Γ 9 cosπΌand hence incorrectly simplify to 16 = 4 cosπΌ
Textbook Note: The textbook presents the rearrangement of the cosine rule: cos π΄ =π2+π2βπ2
2ππto find missing angles.
Iβd personally advise against using this as: (a) Itβs another formula to remember. (b) Anything that gives you less practice of manipulating/rearranging equations is probably a bad thing. (c) You wonβt get to use the swapsie trick.
Harder Ones
60Β°
π₯ 2π₯ β 1
π₯ + 8
2π₯ β 1 2 = π₯2 + π₯ + 8 2 β 2π₯ π₯ + 8 cos 60Β°4π₯2 β 4π₯ + 1 = π₯2 + π₯2 + 16π₯ + 64 β π₯2 β 8π₯3π₯2 β 12π₯ β 63 = 0π₯2 β 4π₯ β 21 = 0π₯ + 3 π₯ β 7 = 0β΄ π₯ = 7
Determine the value of π₯.
π πππΆ
π΅
60Β°
π΄
π
18Β°4.8ππ
8 ππ
ππ = π. ππ + ππ β π Γ π. π Γ π Γ ππ¨π¬ ππΒ°π = π. ππ (πππ)
πͺ is 5.47 km from coastguard station π©.
[From textbook] Coastguard station π΅ is 8 km, on a bearing of 060Β°, from coastguard station π΄. A ship πΆ is 4.8 km on a bearing of 018Β°, away from π΄. Calculate how far πΆ is from π΅.
Test Your Understanding
70Β°
π₯
5
6
π₯ = 6.36Β°
1
3
60Β°
2π₯ β 3
π₯
2π₯ β 2
π₯ = 5
2
8
3
10
π = 124.2Β°
π
Fro Note: You will get an obtuse angle whenever you inverse cos a negative value.
Exercise 9A
Pearson Pure Mathematics Year 1/ASPages 177-179
Extension
[STEP I 2009 Q4i]The sides of a triangle have lengths π β π, π and π + π, where π > π > 0. The largest and smallest angles of the triangle are πΌ and π½ respectively. Show by means of the cosine rule that
4 1 β cosπΌ 1 β cos π½ = cosπΌ + cos π½
Note that the longest side of a triangle is opposite the largest angle, and the shortest opposite the smallest angle. Thus:
1
π
πΌπ½
π β π π = ππ + π + π π β ππ π + π ππ¨π¬π·π + π π = ππ + π β π π β ππ π β π ππ¨π¬πΆ
We can manipulate and combine these two equations to get the desired equation above.
2 [STEP I 2007 Q5] Note: a regular octahedron is a polyhedron with eight faces each of which is an equilateral triangle.(i) Show that the angle between any two
faces of a regular octahedron is
arccos β1
3
(ii) Find the ratio of the volume of a regular octahedron to the volume of the cube whose vertices are the centres of the faces of the octahedron.
Solutions for Q2 on next slide.
Solutions to Extension Question 2
(Official solutions) Big, clear diagram essential!
(i) Let the side length of the octahedron be 2π. Then sloping βheightβ of a triangular face is π 3.
Also, the vertical height of the whole octahedron is 2π 2.
Therefore, by the cosine rule, 8π2 = 3π2 + 3π2 β 2 π 3 π 3 cos π΄
Hence π΄ = πππππ β2π2
6π2= arccos β
1
3
(ii) The centre of each face is on any median of the equilateral triangle that is the face, and the centre is two-thirds of the way along the median from any vertex.This is a quotable fact, but can be worked out from a diagram, using the fact that the centre of an equilateral triangle is equidistant from the three vertices: the centre divides the median in the ratio 1: cos 60Β°.
The feet of the two medians from the apex of the octahedron in two adjacent triangles are π 2 apart.
Therefore, by similarity, adjacent centres of the triangular faces are 2
3Γ π 2 apart. Therefore, the volume
of the cube (whose vertices are the centres of the faces) is
2
3Γ π 2
3
=16π3 2
27
and the volume of the octahedron is 2 Γ4π2Γπ 2
3=
8π3 2
3.
Hence the ratio of the volume of the octahedron to the volume of the cube is 9: 2.
[STEP I 2007 Q5] Note: a regular octahedron is a polyhedron with eight faces each of which is an equilateral triangle.
(i) Show that the angle between any two faces of a regular octahedron is arccos β1
3
(ii) Find the ratio of the volume of a regular octahedron to the volume of the cube whose vertices are the centres of the faces of the octahedron.
The Sine Rule
65Β°
85Β°
30Β°
105.02
9.10
For this triangle, try calculating each side divided by the sin of its opposite angle. What do you notice in all three cases?
! Sine Rule:π
sin π΄=
π
sinπ΅=
π
sin πΆ
c
C
b
B
a
A
You have You want Use
#1: Two angle-side
opposite pairs
Missing angle or
side in one pair
Sine rule
Examples
45Β°
8
11.27
85Β°
Q1
You have You want Use
#1: Two angle-side
opposite pairs
Missing angle or
side in one pair
Sine rule
100Β°
8
15.7630Β°
Q2
50Β°
π
π¬π’π§ππ=
π
π¬π’π§ππ
π =ππ¬π’π§ππ
π¬π’π§ππ= ππ. ππ
π
π¬π’π§πππ=
π
π¬π’π§ππ
π =ππ¬π’π§πππ
π¬π’π§ππ= ππ. ππ
sin π
5=sin 85
6
sin π =5 sin 85
6
π = sinβ15 sin 85
6
= 56.11Β°
Examples
85Β°
6
5
56.11Β°
Q3 8
When you have a missing angle, itβs better to reciprocate to get:π¬π’π§π¨
π=π¬π’π§π©
πi.e. in general put the missing value in the numerator.
40.33Β°
10
126Β°
Q4
sin π
8=sin 126Β°
10
sin π =8 sin 126
10
π = sinβ18 sin 126
10
= 40.33Β°
Exercise 9B
Pearson Pure Mathematics Year 1/ASPages 181-183
Extension
[MAT 2011 1E]The circle in the diagram has centre πΆ. Three angles πΌ, π½, πΎ are also indicated.
The angles πΌ, π½, πΎ are related by the equation:A) πππ πΌ = π ππ(π½ + πΎ)B) π ππ π½ = π ππ πΌ π ππ πΎC) π ππ π½ 1 β πππ πΌ = π ππ πΎD) π ππ πΌ + π½ = πππ πΎ π ππ πΌ
1If we draw a vertical line down from πͺ, we have two triangles with a common length. This common lengths allows us to relate the two triangles. Let the radius be 1.
1
1π₯
Using bottom triangle:
π = π π¬π’π§πΆ β π =π
π¬π’π§πΆUsing sine rule on top:
π
π¬π’π§πΈ=
π
π¬π’π§π·Substituting in π from the first equation, and rearranging, we obtain (B).
The βAmbiguous Caseβ
π΅
π΄
4
44Β°
Suppose you are told that π΄π΅ = 4, π΄πΆ = 3and β π΄π΅πΆ = 44Β°. What are the possible values of β π΄πΆπ΅?
πΆ is somewhere on the horizontal line. Thereβs two ways in which the length could be 3. Using the sine rule:
π¬π’π§πͺ
π=π¬π’π§ππ
ππͺ = π¬π’π§βπ( π. ππππ)
πΆ1 πΆ2
Your calculator will give the acute angle of 67.9Β° (i.e. πΆ2). But if we look at a graph of sin, we can see thereβs actually a second value for sinβ1 0.9262 , corresponding to angle πΆ1.
3 3
180Β°67.9Β° 112.1Β°
0.9262
! The sine rule produces two possible solutions for a missing angle:
sin π = sin(180Β° β π)Whether we use the acute or obtuse angle depends on context.
Test Your Understanding
20Β°
π₯
10
5π
Given that the angle π is obtuse, determine π and hence determine the length of π₯.
sin π
10=sin 20Β°
5
sinβ110 sin 20Β°
5= 43.1602Β°
β΄ π = 180Β° β 43.1602Β° = 136.8398Β°The other angle is:180Β° β 136.8398Β° β 20Β° = 23.1602Β°
Using sine rule again:π₯
sin 23.1602Β°=
5
sin 20Β°π₯ = 5.75 (3π π)
Exercise 9C
Pearson Pure Mathematics Year 1/ASPages 184-185
Area of Non Right-Angled Triangles
Area = 1
2π π sin πΆ
where πΆ is the angle between two sides π and π.
59Β°
3cm
7cm
Area = 0.5 x 3 x 7 x sin(59)= 9.00cm2
!
Fro Tip: You shouldnβt have to label sides/angles before using the formula. Just remember that the angle is between the two sides.
Test Your Understanding
π₯
π₯ + 3
30Β°
The area of this triangle is 10.Determine π₯.
1
2π₯ π₯ + 3 sin 30Β° = 10
1
4π₯ π₯ + 3 = 10
π₯ π₯ + 3 = 40π₯2 + 3π₯ β 40 = 0π₯ + 8 π₯ β 5 = 0
As π₯ > 0, π₯ = 5
The area of this triangle is also 10.If π is obtuse, determine π.
56
π
1
2Γ 5 Γ 6 Γ sin π = 10
sin π =10
15π = 180Β° β 41.8Β° = 138.2Β°
Exercise 9D
Pearson Pure Mathematics Year 1/ASPages 186-187
Sin or cosine rule?
100Β° π
5
3
40Β°45Β°
π₯
3
π
5
2
3
40Β°
5
π₯
3
Sine
Recall that whenever we have two βside-angle pairsβ involved, use sine rule. If thereβs 3 sides involved, we can use cosine rule. Sine rule is generally easier to use than cosine rule.Cosine
Sine Cosine
CosineSine
CosineSine
Using sine rule twiceYou have You want Use
#4 Two sides known
and a missing side not
opposite known angle
Remaining side Sine rule
twice
4
π₯
3 32Β°
Given there is just one angle involved, you might attempt to use the cosine rule:
ππ = ππ + ππ β π Γ π Γ π Γ ππ¨π¬πππ = ππ + ππ β ππππ¨π¬ ππ
This is a quadratic equation!Itβs possible to solve this using the quadratic formula (using π = π, π = βπππ¨π¬ππ ,π = π). However, this is a bit fiddly and not the primary method expected in the examβ¦
Using sine rule twiceYou have You want Use
#4 Two sides known
and a missing side not
opposite known angle
Remaining side Sine rule
twice
4
π₯
3 32Β°
πππ β ππ β ππ. ππππ = πππ. ππππ
1: We could use the sine rule to find this angle.
2: Which means we would then know this angle.
sin π΄
4=sin 32
3π΄ = 44.9556Β°
3: Using the sine rule a second time allows us to find π₯
π₯
sin 103.0444=
3
sin 32π₯ = 5.52 π‘π 3π π
!
Test Your Understanding
61Β°
53Β°
10
9
π¦
34
π¦ = 6.97
π΄πππ = 6.00
Problem Solving With Sine/Cosine Rule
[From Textbook] The diagram shows the locations of four mobile phone masts in a field, π΅πΆ = 75 π.πΆπ· = 80π, angle π΅πΆπ· = 55Β° and angle π΄π·πΆ = 140Β°.In order that the masts do not interfere with each other, they must be at least 70m apart.Given that π΄ is the minimum distance from π·, find:a) The distance π΄ is from π΅b) The angle π΅π΄π·c) The area enclosed by the four masts.
140Β°
55Β°
75 π
80 π70 π
π·
π΄
π΅
πΆ
Using triangle π΅πΆπ·:π΅π·2 = 752 + 802 β 2 Γ 75 Γ 80 Γ cos 55Β°
π΅π· = 71.708β¦Then use sine rule to find β π΅π·πΆ:sin β π΅π·πΆ
75=
sin 55
71.708β β π΅π·πΆ = 58.954
β΄ β π΅π·π΄ = 81.045β¦
We can then use cosine rule on Ξπ΄π΅π·:π΄π΅2 = 702 + 71.7082 β 2 Γ 70 Γ 71.708 Γ cos 81.045
π΄π΅ = 92.1 π 3π π
Using sine rule on Ξπ΄π΅π·, β π΅π΄π· = 50.3Β° (3π π)
By adding areas of Ξπ΄π΅π· and Ξπ΅π·πΆ:π΄πππ π΄π΅πΆπ· = 4940 π2 (3π π)
a
b
c
Exercise 9E
Pearson Pure Mathematics Year 1/ASPages 189-191
[AEA 2009 Q5a] The sides of the triangle π΄π΅πΆ have lengths π΅πΆ = π, π΄πΆ = π and π΄π΅ = π, where π < π < π. The sizes of the angles π΄, π΅ and πΆ form an arithmetic sequence.(i) Show that the area of triangle π΄π΅πΆ is
ππ3
4.
Given that π = 2 and sin π΄ =15
5, find
(ii) the value of π,(iii) the value of π.
[STEP I 2006 Q8] Note that the volume of a tetrahedron is equal to
1
3Γ ππππ ππ πππ π Γ βπππβπ‘
The points π, π΄, π΅, πΆ have coordinates 0,0,0 , π, 0,0 , 0, π, 0 and (0,0, π)
respectively, where π, π, π are positive.(i) Find, in terms of π, π, π the volume of
the tetrahedron ππ΄π΅πΆ.(ii) Let angle π΄πΆπ΅ = π. Show that
cos π =π2
π2 + π2 π2 + π2
and find, in terms of π, π and π, the area of triangle π΄π΅πΆ.Hence show that π, the perpendicular distance of the origin from the triangle
π΄π΅πΆ, satisfies 1
π2=
1
π2+
1
π2+
1
π2
Solutions to extension problems on next slides.
1 2
Solution to Extension Problem 1
[AEA 2009 Q5a] The sides of the triangle π΄π΅πΆ have lengths π΅πΆ = π, π΄πΆ = π and π΄π΅ = π, where π < π < π. The sizes of the angles π΄, π΅ and πΆ form an arithmetic sequence.(i) Show that the area of triangle π΄π΅πΆ is
ππ3
4.
Given that π = 2 and sin π΄ =15
5, find
(ii) the value of π,(iii) the value of π.
Solution to Extension Problem 2
[STEP I 2006 Q8] Note that the volume of a tetrahedron is equal to 1
3Γ ππππ ππ πππ π Γ βπππβπ‘
The points π, π΄, π΅, πΆ have coordinates 0,0,0 , π, 0,0 , 0, π, 0 and (0,0, π)
respectively, where π, π, π are positive.(i) Find, in terms of π, π, π the
volume of the tetrahedron ππ΄π΅πΆ.
(ii) Let angle π΄πΆπ΅ = π. Show that
cos π =π2
π2 + π2 π2 + π2
and find, in terms of π, π and π, the area of triangle π΄π΅πΆ.Hence show that π, the perpendicular distance of the origin from the triangle π΄π΅πΆ,
satisfies 1
π2=
1
π2+
1
π2+
1
π2
This is FM content, but see a few lines below.
Sin Graph
What does it look like?
90 180 270 360-90-180-270-360
Key Features:β’ Repeats every 360Β°.β’ Range: β1 < sin π₯ < 1
Sin Graph
What do the following graphs look like?
90 180 270 360-90-180-270-360
Suppose we know that sin(30) = 0.5. By thinking about symmetry in the graph, how could we work out:
sin(150) = 0.5 sin(-30) = -0.5 sin(210) = -0.5
Cos Graph
What do the following graphs look like?
90 180 270 360-90-180-270-360
Key Features:β’ Repeats every 360Β°.β’ Range: β1 < sin π₯ < 1
Cos Graph
What does it look like?
90 180 270 360-90-180-270-360
Suppose we know that cos(60) = 0.5. By thinking about symmetry in the graph, how could we work out:
cos(120) = -0.5 cos(-60) = 0.5 cos(240) = -0.5
Tan Graph
What does it look like?
90 180 270 360-90-180-270-360
Key Features:β’ Repeats every πππΒ°.β’ Roots: 0,180Β°, β180Β°, β¦β’ Range: tan π₯ β β (i.e. no
min/max value!)β’ Asymptotes: π₯ = Β±90Β°, Β±270Β°, β¦
Tan Graph
What does it look like?
90 180 270 360-90-180-270-360
Suppose we know that tan 30Β° =1
3. By thinking about symmetry in the
graph, how could we work out:
tan β30Β° = β1
3tan 150Β° = β
1
3
Transforming Trigonometric Graphs
There is no new theory here: just use your knowledge of transforming graphs, i.e. whether the transformation occurs βinsideβ the function (i.e. input modified) or βoutsideβ the function (i.e. output modified).
Sketch π¦ = 4 sin π₯, 0 β€ π₯ β€ 360Β°
π₯
π¦
90Β° 180Β° 270Β° 360Β°
4
β4
Sketch π¦ = cos π₯ + 45Β° , 0 β€ π₯ β€ 360Β°
π₯
π¦
90Β° 180Β° 270Β° 360Β°
1
β1
β90Β°
45°à 4
Transforming Trigonometric Graphs
Sketch π¦ = β tan π₯, 0 β€ π₯ β€ 360Β°
π₯
π¦
90Β° 180Β° 270Β° 360Β°
Γ (β1)
Sketch π¦ = sinπ₯
2, 0 β€ π₯ β€ 360Β°
π₯
π¦
90Β° 180Β° 270Β° 360Β°
1
β1
Γ 2
Exercise 9F/9GPearson Pure Mathematics Year 1/ASPages 194, 197-198
[MAT 2013 1B] The graph of π¦ = sin π₯ is reflected first in the line π₯ = π and then in the line π¦ = 2. The resulting graph has equation:A) π¦ = cos π₯B) π¦ = 2 + sin π₯C) π¦ = 4 + sin π₯D) π¦ = 2 β cos π₯
Solution: C
[MAT 2011 1D] What fraction of the interval 0 β€ π₯ β€ 360Β° is one (or both) of the inequalities:
sin π₯ β₯1
2, sin 2π₯ β₯
1
2true?
Solution: ππ
ππ(this is clear if you draw the graphs
π = π¬π’π§ π and π = π¬π’π§ ππ on the same axes)
[MAT 2007 1G] On which of the axes is a sketch of the graph
π¦ = 2βπ₯ sin2 π₯2
πβπ and ππππ are always positive eliminating (b). When π = π, π = π eliminating (d). Multiplying by πβπ causes the amplitude of the peaks to go down as π increases. The ππ increases more rapidly as πincreases, hence reducing the wavelength (i.e. distance between peaks). The answer is therefore (a).
1
Extension
2
3
APPENDIX :: Proof of Cosine Rule
ππ
β
π₯ π β π₯
We want to use Pythagoras, so split π into two so that we get two right-angled triangles.
By Pythagoras: β2 + π₯2 = π2 and β2 + π β π₯ 2 = π2
Subtracting to eliminate β:π₯2 β π β π₯ 2 = π2 β π2
π2 = π2 + π2 β 2ππ₯But π₯ = π cosπ΄ β΄
π2 = π2 + π2 β 2ππ cosπ΄
π΄
APPENDIX :: Proof of Sine Rule
ππ
β
π
The idea is that we can use the common length of Ξπ΄πΆπ and β ππ΅πΆ, i.e. β, to connect the two triangles, and therefore connect their angles/length.
Using basic trigonometry:β = π sinπ΅ and β = π sin π΄
β΄ π sin π΅ = π sin π΄
β΄π
sin π΄=
π
sin π΅
π΄ π΅
πΆ
π