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P2 Chapter 7 :: Trigonometry And Modelling
jfrost@tiffin.kingston.sch.ukwww.drfrostmaths.com
@DrFrostMaths
Last modified: 23rd November 2017
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Chapter Overview
sin π + π = sin π cos π + cosπ sin π
1a:: Addition Formulae
βSolve, for 0 β€ π₯ < 2π, the equation2 tan 2π¦ tan π¦ = 3
giving your solutions to 3sf.β
1b:: Double Angle Formulae
βFind the maximum value of 2 sin π₯ + cos π₯ and the value of π₯for which this maximum occurs.β
3:: Simplifying π cos π₯ Β± π sin π₯
This chapter is very similar to the trigonometry chapters in Year 1. The only difference is that new trig functions: π ππ, πππ ππ and πππ‘, are introduced.
βThe sea depth of the tide at a beach can be
modelled by π₯ = π π ππ2ππ‘
5+ πΌ , where π‘ is
the hours after midnight...β
4:: Modelling
Teacher Note: There is again no change in this chapter relative to the old pre-2017 syllabus.
Addition Formulae
Addition Formulae allow us to deal with a sum or difference of angles.
π¬π’π§ π¨ + π© = π¬π’π§π¨ ππ¨π¬π© + ππ¨π¬π¨ π¬π’π§π©π¬π’π§ π¨ β π© = ππππ¨ππππ© β ππππ¨ππππ©
ππ¨π¬ π¨ + π© = ππ¨π¬π¨ ππ¨π¬π© β π¬π’π§π¨ π¬π’π§π©ππ¨π¬ π¨ β π© = ππ¨π¬π¨ ππ¨π¬π© + π¬π’π§π¨ π¬π’π§π©
πππ§ π¨ + π© =πππ§π¨ + πππ§π©
π β πππ§π¨ πππ§π©
πππ π¨ β π© =ππππ¨ β ππππ©
π + ππππ¨ ππππ©
Do I need to memorise these?Theyβre all technically in the formula booklet, but you REALLY want to eventually memorise these (particularly the π ππ and πππ ones).
How to memorise:
First notice that for all of these the first thing on the RHS is the same as the first thing on the LHS!
β’ For sin, the operator in the middle is the same as on the LHS.
β’ For cos, itβs the opposite.β’ For tan, itβs the same in
the numerator, opposite in the denominator.
β’ For sin, we mix sin and cos.
β’ For cos, we keep the cosβs and sinβs together.
Common Schoolboy/Schoolgirl Error
Why is sin(π΄ + π΅) not just sin π΄ + sin(π΅)?Because πππ is a function, not a quantity that can be expanded out like this. Itβs a bit like how π + π π β’ ππ + ππ.We can easily disprove it with a counterexample.
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Addition Formulae
Now can you reproduce them without peeking at your notes?
π¬π’π§ π¨ + π© β‘ π¬π’π§π¨ ππ¨π¬π© + ππ¨π¬π¨ π¬π’π§π©π¬π’π§ π¨ β π© β‘ππππ¨ππππ© β ππππ¨ππππ©
ππ¨π¬ π¨ + π© β‘ππ¨π¬π¨ ππ¨π¬π© β π¬π’π§π¨ π¬π’π§π©ππ¨π¬ π¨ β π© β‘ ππ¨π¬π¨ ππ¨π¬π© + π¬π’π§π¨ π¬π’π§π©
πππ§ π¨ + π© β‘πππ§π¨ + πππ§π©
π β πππ§π¨ πππ§π©
πππ π¨ β π© β‘ππππ¨ β ππππ©
π + ππππ¨ ππππ©
How to memorise:
First notice that for all of these the first thing on the RHS is the same as the first thing on the LHS!
β’ For sin, the operator in the middle is the same as on the LHS.
β’ For cos, itβs the opposite.β’ For tan, itβs the same in
the numerator, opposite in the denominator.
β’ For sin, we mix sin and cos.
β’ For cos, we keep the cosβs and sinβs together.
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Proof of sin π΄ + π΅ β‘ sin π΄ cosπ΅ + cosπ΄ sin π΅
π΅
π΄
1
(Not needed for exam)
1: Suppose we had a line of length 1 projected an angle of π΄ + π΅ above the horizontal.Then the length of ππ =sin π΄ + π΅It would seem sensible to try and find this same length in terms of π΄ and π΅ individually.
π π
π
sin(π΄
+π΅)
2: We can achieve this by forming two right-angled triangles.
π΄
3: Then weβre looking for the combined length of these two lines.
4: We can get the lengths of the top triangleβ¦
sinπ΄cosπ΅
cosπ΄sinπ΅
5: Which in turn allows us to find the green and blue lengths.
6: Hence sin π΄ + π΅ =sinπ΄ cosπ΅ + cosπ΄ sinπ΅
β‘
Proof of other identities
Can you think how to use our geometrically proven result sin π΄ + π΅ = sinπ΄ cosπ΅ + cos π΄ sinπ΅ to prove the identity for sin(π΄ β π΅)?
π¬π’π§ π¨ β π© = π¬π’π§ π¨ + βπ©
= π¬π’π§ π¨ ππ¨π¬ βπ© + ππ¨π¬ π¨ π¬π’π§ βπ©= π¬π’π§π¨ ππ¨π¬π© β ππ¨π¬π¨ π¬π’π§π©
What about cos(π΄ + π΅)? (Hint: what links π ππ and πππ ?)
ππ¨π¬ π¨ + π© = π¬π’π§π
πβ π¨ + π© = π¬π’π§
π
πβ π¨ + βπ©
= π¬π’π§π
πβ π¨ ππ¨π¬ βπ© + ππ¨π¬
π
πβ π¨ π¬π’π§ βπ©
= ππ¨π¬π¨ ππ¨π¬π© β π¬π’π§π¨ π¬π’π§π©
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Proof of other identities
Edexcel C3 Jan 2012 Q8
The important thing is that you explicit show the division of each term by πππ π΄ cosπ΅.
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Examples
Given that 2 sin(π₯ + π¦) = 3 cos π₯ β π¦ express tan π₯ in terms of tan π¦.
Using your formulae:π π¬π’π§π ππ¨π¬ π + πππ¨π¬ π π¬π’π§π = πππ¨π¬ π ππ¨π¬ π + π π¬π’π§π π¬π’π§π
We need to get πππ§ π and πππ§π in there. Dividing by ππ¨π¬ π ππ¨π¬ π would seem like a sensible step:
π πππ§ π + π πππ§ π = π + π πππ§ π πππ§πRearranging:
πππ§ π =π β π πππ§π
π β π πππ§π
Q
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Exercises 7A
Pearson Pure Mathematics Year 2/ASPage 144
(Classes in a rush may wish to skip this exercise)
Uses of Addition Formulae
[Textbook] Using a suitable angle formulae, show that sin 15Β° =6β 2
4.Q
π¬π’π§ππ = π¬π’π§ ππ β ππ = π¬π’π§ππ ππ¨π¬ππ β ππ¨π¬ ππ π¬π’π§ππ
=π
πΓ
π
πβ
π
πΓπ
π
=π β π
π π
=π β π
π
[Textbook] Given that sin π΄ = β3
5and 180Β° < π΄ < 270Β°, and that cosπ΅ = β
12
13,
and B is obtuse, find the value of: (a) cos π΄ β π΅ (b) tan π΄ + π΅
πππ π¨ β π© β‘ πππ π¨ πππ π© + πππ π¨ πππ π©
πππ π¨ β‘ π β πππππ¨ = Β±π
π. But we know from the graph of πππ that it is negative
between πππΒ° < π¨ < πππΒ°, so πππ π¨ = βπ
π.
Similarly πππ π© = Β±π
ππ, and similarly considering the graph for πππ, πππ π© = +
π
ππ.
β΄ πππ π¨ β π© β‘ βπ
πΓ β
ππ
ππ+ β
π
πΓ
π
ππ=ππ
ππ
Continuedβ¦
Q
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Fro Tip: You can get πππ in terms of π ππ and vice versa by using a rearrangement of sin2 π₯ + cos2 π₯ β‘ 1.
So cosπ΄ = 1 β sin2 π΄
Continuedβ¦
Given that sinπ΄ = β3
5and 180Β° < π΄ < 270Β°, and that cosπ΅ = β
12
13, find the
value of: (b) tan π΄ + π΅
Q
πππ§ π¨ + π© =πππ§ π¨ + πππ§(π©)
π β πππ§π¨ πππ§π©
We earlier found π¬π’π§π¨ = βπ
π, π¬π’π§π© =
π
ππ, ππ¨π¬π¨ = β
π
π, ππ¨π¬π© = β
ππ
ππ
β΄ πππ§π¨ = βπ
πΓ· β
π
π=
π
πand πππ§π© =
π
ππΓ· β
ππ
ππ= β
π
ππ
By substitution:
πππ§ π¨ + π© =ππ
ππ
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Test Your Understanding
Without using a calculator, determine the exact value of:a) cos 75Β°b) tan 75Β°
ππ¨π¬ ππΒ° = ππ¨π¬ ππΒ° + ππΒ° = ππ¨π¬ ππΒ° ππ¨π¬ ππΒ° β π¬π’π§ ππΒ° π¬π’π§ ππΒ°
=π
πΓ
π
πβ
π
πΓπ
π
=π β π
π π=
π β π
π
πππ§ ππΒ° =π¬π’π§ ππΒ°
ππ¨π¬ ππΒ°π¬π’π§ ππΒ° = π¬π’π§ ππΒ° + ππΒ° = π¬π’π§ ππΒ° ππ¨π¬ ππΒ° + ππ¨π¬ ππΒ° π¬π’π§ ππΒ°
=π
πΓ
π
π+
π
πΓπ
π=
π + π
π
β΄ πππ§ ππΒ° =π + π
πΓ·
π β π
π=
π + π
π β π= π + π
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Challenging question
Edexcel June 2013 Q3
βExpandingβ both sides:2 cos π₯ cos 50 β 2 sin π₯ sin 50= sin π₯ cos 40 + cos π₯ sin 40
Since thing to prove only has 40 in it, use cos 50 = sin 40 and sin 50 =cos 40 .
2 cos π₯ sin 40 β 2 sin π₯ cos 40= sin π₯ cos 40 + cos π₯ sin 40
As per usual, when we want tans, divide by cos π₯ cos 40:
2 tan 40 β 2 tan π₯ = tan π₯ + tan 40tan 40 = 3 tan π₯1
3tan 40 = tan π₯
a
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Exercise 7B
Pearson Pure Mathematics Year 2/ASPages 172-173
[STEP I 2010 Q3] Show thatsin π₯ + π¦ β sin π₯ β π¦ = 2 cos π₯ sin π¦
and deduce that
sin π΄ β sin π΅ = 2 cos1
2π΄ + π΅ sin
1
2π΄ β π΅
Show also that
cosπ΄ β cosπ΅ = β2 sin1
2π΄ + π΅ sin
1
2π΄ β π΅
The points π, π, π and π have coordinates π cos π , π sin π and π cos π , π sin π respectively,
where 0 β€ π < π < π < π < 2π, and π and π are positive.Given that neither of the lines ππ and ππ is vertical, show that these lines are parallel if and only if
π + π β π β π = 2π
Extension
Double Angle Formulae
sin 2π΄ β‘ 2 sinπ΄ cosπ΄cos 2π΄ β‘ cos2 π΄ β sin2 π΄
β‘ 2 cos2 π΄ β 1β‘ 1 β 2 sin2 π΄
tan 2π΄ =2 tanπ΄
1 β tan2 π΄
Tip: The way I remember what way round these go is that the cos on the RHS is βattractedβ to the cos on the LHS, whereas the sin is pushed away.
These are all easily derivable by just setting π΄ = π΅ in the compound angle formulae. e.g.
sin 2π΄ = sin π΄ + π΄= sinπ΄ cosπ΄ + cosπ΄ sin π΄= 2 sinπ΄ cosπ΄
!This first form is relatively rare.
Double-angle formula allow you to halve the angle within a trig function.
Examples
[Textbook] Use the double-angle formulae to write each of the following as a single trigonometric ratio.a) cos2 50Β° β sin2 50Β°
b)2 tan
π
6
1βtan2π
6
c)4 sin 70Β°
sec 70Β°
This matches ππ¨π¬ ππ¨ = ππ¨π¬π π¨ β π¬π’π§ππ¨:ππππ ππΒ° β ππππ ππΒ° = ππ¨π¬ πππΒ°
This matches πππ§ ππ¨ =π πππ§ π¨
πβπππ§π π¨
π ππππ π
π β πππππ π
= πππ§π
π
π π¬π’π§ππΒ°
π¬ππ ππΒ°=
ππ¬π’π§ππΒ°
π Γ· ππ¨π¬ππΒ°= ππ¬π’π§ππΒ° ππ¨π¬ ππΒ°
This matches π¬π’π§ ππ¨ = ππ¬π’π§π¨ ππ¨π¬π¨π π¬π’π§ππΒ° πππππΒ° = π π¬π’π§πππΒ°
sin 2π₯ = 2 sin π₯ cos π₯cos 2π₯ = cos2 π₯ β sin2 π₯
= 2 cos2 π₯ β 1= 1 β 2 sin2 π₯
tan 2π₯ =2 tan π₯
1 β tan2 π₯
a
b
c
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Examples
[Textbook] Given that π₯ = 3 sin π and π¦ = 3 β 4πππ 2π, eliminate π and express π¦ in terms of π₯.
sin 2π₯ = 2 sin π₯ cos π₯cos 2π₯ = cos2 π₯ β sin2 π₯
= 2 cos2 π₯ β 1= 1 β 2 sin2 π₯
tan 2π₯ =2 tan π₯
1 β tan2 π₯
Fro Note: This question is an example of turning a set of parametric equations into a single Cartesian one. You will cover this in the next chapter.
π = π β π π β ππ¬π’π§π π½
= ππ¬π’π§π π½ β π
π¬π’π§π½ =π
π
β΄ π = ππ
π
π
β π
Given that cos π₯ =3
4and π₯ is acute, find the exact value of
(a) sin 2π₯ (b) tan 2π₯
Since πππππ + ππ¨π¬ππ = π, πππ π = Β± π β πππππ
πππ π = π βπ
π
π=
π
π(and is positive given π < π± < ππΒ°)
β΄ πππ ππ = ππππ π πππ π = π Γπ
πΓπ
π=π π
π
πππ π =πππ π
πππ=π π/π
π/π=
π
πβ΄ πππππ =
π Γππ
π βππ
π = π π
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Solving Trigonometric Equations
This is effectively the same type of question you encountered in Chapter 6 and in Year 1, except you may need to use either the addition formulae or double angle formulae.
[Textbook] Solve 3 cos 2π₯ β cos π₯ + 2 = 0 for 0 β€ π₯ β€ 360Β°.
3 2 cos2 π₯ β 1 β cos π₯ + 2 = 06 cos2 π₯ β 3 β cos π₯ + 2 = 06 cos2 π₯ β cos π₯ β 1 = 03 cos π₯ + 1 2 cos π₯ β 1 = 0
cos π₯ = β1
3ππ cos π₯ =
1
2
π₯ = 109.5Β°, 250.5Β° π₯ = 60Β°, 300Β°
Recall that we have a choice of double angle identities for cos 2π₯. This is clearly the most sensible choice as we end up with an equation just in terms of πππ .?
Further Examples
[Textbook] By noting that 3π΄ = 2π΄ + π΄, :a) Show that sin 3π΄ = 3 sinπ΄ β 4 sin3π΄.b) Hence or otherwise, solve, for 0 < π < 2π,
the equation 16 sin3 π β 12 sinπ β 2 3 = 0
π¬π’π§ ππ¨ + π¨= π¬π’π§ ππ¨ ππ¨π¬π¨ + ππ¨π¬ ππ¨ π¬π’π§π¨
= π π¬π’π§π¨ ππ¨π¬ π¨ ππ¨π¬π¨ + π β π π¬π’π§π π¨ π¬π’π§π¨
= π π¬π’π§π¨ π β π¬π’π§π π¨ + π¬π’π§ π¨ β ππ¬π’π§π π¨
= π π¬π’π§π¨ β π π¬π’π§π π¨ + π¬π’π§π¨ β π π¬π’π§π π¨= π π¬π’π§π¨ β π π¬π’π§π π¨
Note that ππππππ π½ β ππ ππππ½
= βπ ππ¬π’π§ π¨ β π π¬π’π§π π¨ = βπ π¬π’π§ ππ½.
β΄ βπ π¬π’π§ ππ½ = π π
π¬π’π§ ππ½ = βπ
π
ππ½ =ππ
π,ππ
π,πππ
π,πππ
π,πππ
π,πππ
π
π½ =ππ
π,ππ
π,πππ
π,πππ
π,πππ
π,πππ
π
Fro Exam Note: A question pretty much just like this came up in an exam once.
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[Textbook] Solve 4 cos π β 30Β° = 8 2sin π in the range 0 β€ π < 360Β°.
Your instinct should be to use the addition formulae first to break up the ππππ π½ β ππΒ° .
π ππ¨π¬π½ ππ¨π¬ππΒ° + π π¬π’π§π½ π¬π’π§ππΒ° = π π π¬π’π§π½
π π ππ¨π¬π½ + ππ¬π’π§π½ = π ππ¬π’π§π½
π πππ¨π¬π½ = π ππ¬π’π§π½ β π π¬π’π§π½
π π ππ¨π¬π½ = π π β π π¬π’π§π½
π π
π π β π=π¬π’π§π½
ππ¨π¬π½= πππ§π½
π½ = ππ. πΒ°, πππ. πΒ°
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Test Your Understanding
Edexcel C3 Jan 2013 Q6
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If you finish that quicklyβ¦
[Textbook] Solve 2 tan2π¦ tanπ¦ = 3 for 0 β€ π¦ < 2π, giving your answer to 2dp.
π πππ§ ππ πππ§ π = π
ππ πππ§ π
π β πππ§π ππππ§ π = π
β¦π πππ§π π = π
πππ§ π = Β±π
π
π = π. ππ, π. ππ, π. ππ, π. ππ
Forgetting the Β± is an incredibly common source of lost marks.
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Exercise 7D
Pearson Pure Mathematics Year 2/ASPages 180-181
Extension
[AEA 2013 Q2]
1
2 [AEA 2009 Q3]
(Solution on next slide)
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π sin π + π cos π
Hereβs a sketch of π¦ = 3 sin π₯ + 4 cos π₯.What do you notice?
Itβs a sin graph that seems to be translated on the π-axis and stretched on the π axis. This suggests we can represent it as π = πΉ π¬π’π§(π + πΆ), where πΆ is the horizontal translation and πΉ the stretch on the π-axis.
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π sin π + π sin π
Put 3 sin π₯ + 4 cos π₯ in the form π sin π₯ + πΌ giving πΌ in degrees to 1dp.Q
Fro Tip: I recommend you follow this procedure every time β Iβve tutored students whoβve been taught a βshortcutβ (usually skipping Step 1), and they more often than not make a mistake.
STEP 1: Expanding:π sin π₯ + πΌ = π sin π₯ cosπΌ + π cos π₯ sin πΌ
STEP 2: Comparing coefficients:π cosπΌ = 3 π sin πΌ = 4
STEP 3: Using the fact that π 2 sin2 πΌ + π 2 cos2 πΌ = π 2:
π = 32 + 42 = 5
STEP 4: Using the fact that π sin πΌ
π cos πΌ= tanπΌ:
tanπΌ =4
3πΌ = 53.1Β°
STEP 5: Put values back into original expression.3 sin π₯ + 4 cos π₯ β‘ 5 sin π₯ + 53.1Β°
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If π cos πΌ = 3 and π sinπΌ = 4then π 2 cos2 πΌ = 32 and π 2 sin2 πΌ = 42.π 2 sin2 πΌ + π 2 cos2 πΌ = 32 + 42
π 2 sin2πΌ + cos2 πΌ = 32 + 42
π 2 = 32 + 42
π = 32 + 42
(You can write just the last line in exams)
Test Your Understanding
Put sinπ₯ + cosπ₯ in the form π sin π₯ + πΌ giving πΌ in terms of π.Q
π sin π₯ + πΌ β‘ π sin π₯ cos πΌ + π cos π₯ sin πΌπ cosπΌ = 1 π sinπΌ = 1
π = 12 + 12 = 2tanπΌ = 1
πΌ =π
4
sin π₯ + cos π₯ β‘ 2 sin π₯ +π
4
Put sinπ₯ β 3 cos π₯ in the form π sin π₯ β πΌ giving πΌ in terms of π.Q
π sin π₯ + πΌ β‘ π sin π₯ cos πΌ β π cos π₯ sin πΌ
π cosπΌ = 1 π sin πΌ = 3π = 2
tanπΌ = 3 π π πΌ =π
3
sin π₯ + cos π₯ β‘ 2 sin π₯ βπ
3
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Further Examples
Put 2 cosπ + 5 sin π in the form π cos π β πΌ where 0 < πΌ < 90Β°Hence solve, for 0 < π < 360, the equation 2 cosπ + 5 sin π = 3
Q
2 cos π + 5 sin π β‘ 29 cos π β 68.2Β°Therefore:
29 cos π β 68.2Β° = 3
cos π β 68.2Β° =3
29π β 68.2Β° = β56.1β¦ Β°, 56.1β¦ Β°π = 12.1Β°, 124.3Β°
(Without using calculus), find the maximum value of 12 cosπ + 5 sinπ, and give the smallest positive value of π at which it arises.
Q
Use either π sin(π + πΌ) or π cos(π β πΌ) before that way the + sign in the middle matches up.
β‘ 13 cos π β 22.6Β°
πππ is at most 1,thus the expression has value at most 13.This occurs when π β 22.6 = 0 (as cos 0 = 1) thus π = 22.6
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Fro Tip: This is an exam favourite!
Quickfire Maxima
What is the maximum value of the expression and determine the smallest positive value of π (in degrees) at which it occurs.
Expression Maximum (Smallest) π½ at max
20 sin π 20 90Β°
5 β 10 sin π 15 270Β°
3 cos π + 20Β° 3 340Β°
2
10 + 3 sin π β 30
2
7
300Β°
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Proving Trigonometric Identities
Prove that tan 2π β‘2
cot πβtan π
Prove that 1βcos 2π
sin 2πβ‘ tan π
Just like Chapter 6 had βproveyβ and βsolveyβ questions, we also get the βproveyβ questions in Chapter 7. Just use the appropriate double angle or addition formula.
tan 2π β‘2 tan π
1 β tan2 πβ‘
2
1tan π
β tanπ
β‘2
cot π β tanπ
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1 β 1 β 2 sin2 π
2 sin π cos πβ‘
2 sin2 π
2 sin π cos πβ‘ tanπ
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Test Your Understanding
[OCR] Prove that cot 2π₯ + πππ ππ 2π₯ β‘ cot π₯
ππ¨π¬ ππ
π¬π’π§ππ+
π
π¬π’π§ππ
β‘ππ¨π¬ ππ + π
π¬π’π§ππβ‘πππ¨π¬π π β π + π
ππ¬π’π§π ππ¨π¬ π
β‘πππ¨π¬π π
π π¬π’π§π ππ¨π¬ πβ‘ππ¨π¬ π
π¬π’π§πβ‘ ππ¨π π
[OCR] By writing cos π₯ = cos 2 Γπ₯
2or otherwise,
prove the identity 1βcos π₯
1+cos π₯β‘ tan2
π₯
2
ππ¨π¬ π β‘ πππ π Γπ
πβ‘ πππ¨π¬π
π
πβ π
β΄π β πππ π
π + πππ πβ‘π β πππ¨π¬π
ππ
β π
π + πππππππ
β π
β‘π π β ππ¨π¬π
ππ
π ππ¨π¬πππ
β‘π¬π’π§π
ππ
ππ¨π¬πππ
β‘ πππ§ππ
π
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Exercise 7F
Pearson Pure Mathematics Year 2/ASPages 187-188
Extension:
[STEP I 2005 Q4]
(a) β117
125(b) π‘ππ π = 8 + 75? ?
Very Challenging Exam Example
Edexcel C3 June 2015 Q8
? b
We eventually want cos π΄ β sinπ΄so cos2 π΄ β sin2π΄ is best choice of double-angle formula because this can be factorise to give cos π΄ β sinπ΄ as a factor.
This is even less obvious. Knowing that weβll have cosπ΄ β sin π΄ cosπ΄ + sin π΄ in the denominator and that the cosπ΄ + sin π΄ will cancel, we might work backwards from the final result and multiply by cosπ΄ + sin π΄! Working backwards from the thing weβre trying to prove is occasionally a good strategy (provided the steps are reversible!)
? a
Modelling
? a
? b
? c
? d
When trigonometric equations are in the form π π ππ ππ₯ Β± π or π πππ ππ₯ Β± π , they can be used to model various things which have an oscillating behaviour, e.g. tides, the swing of a pendulum and sound waves.
Test Your Understanding
? a? bi? bii
? c
? d
Tip: Reflect carefully on the substitution you use to allow (bii) to match your identity in (a). π =?
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