DEPARTMENT OF MATHEMATICS
Core 3
Topic 7 – Trigonometry 2
1
Trigonometry 2
The Addition Formulae
The Double Angle Formulae
Recall the addition formulae in the
formula book
Rewrite an expression using the
addition formulae
Substitute values into an addition
formulae to derive an exact value
Derive the addition formula for tan
using the addition formulae for sin
and cos
Solving Trigonometric
Equations
Harmonic Form
Use sin(A + B) to
derive:
sin2A = 2sinAcosA
Solve an equation using the double
angle formulae
Rewrite an expression in harmonic
form, given an addition formula
(i.e. Find R and θ)
Use cos(A + B) to
derive:
cos2A = cos2A - sin2A
cos2A = 1 - 2 sin2A
cos2A = 2cos2A - 1
Recognise the link between an
identity and 'hence solve'
Finding maximum/minimum values
(and the corresponding angle)
having put an expression in
harmonic form
Use tan(A + B) to
derive:
tan2A = 2 tanA_
1 - tan2A
Prove an identity using the double
angle formulae
Solve an equation using the
harmonic form
Solve a functional (real-life)
problem having put an expression
in harmonic form
2
Trigonometry 2
Topic 7 now – we’ll extend our Trigonometry a little further, taking some new
ideas on board.
The Addition Formulae:
The addition formulae are given in the formula book. You don’t need to be able to
derive the the ones for 𝑠𝑖𝑛 or 𝑐𝑜𝑠, you simply need to be able to use them. The
one for 𝑡𝑎𝑛, however, you do need to be able to derive – can you think how we
could?
Worked Example:
Working without a calculator:
Using the formula sin(𝐴 − 𝐵) = 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵 − 𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵, show that sin 15𝑜 =
√6−√2
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3
Notes:
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Worked Examples:
a)
b)
Rewrite as a single expression and hence evaluate:
𝑡𝑎𝑛70+𝑡𝑎𝑛65
1−𝑡𝑎𝑛70𝑡𝑎𝑛65
sin 100 cos 10 − cos 100 sin 10
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And a tougher example:
c)
Given that:
[June 2013]
2 cos(𝑥 + 50) = sin(𝑥 + 40)
1
Without a calculator, show that: tan 𝑥 = tan 40
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Notes:
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The Double Angle Formulae:
We derive the double angle formulae from the addition formulae.
Double Angle Formula 1:
Using:
sin(𝐴 + 𝐵) = 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵 + 𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵
Double Angle Formula 2:
Using:
sin 2𝐴
cos 2𝐴
we let B = A
(There are 3 variations of this one!)
cos(𝐴 + 𝐵) = 𝑐𝑜𝑠𝐴𝑐𝑜𝑠𝐵 − 𝑠𝑖𝑛𝐴𝑠𝑖𝑛𝐵
we let B = A
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Double Angle Formula 3:
Using:
tan(𝐴 + 𝐵) =
tan 2𝐴
𝑡𝑎𝑛𝐴+𝑡𝑎𝑛𝐵
1−𝑡𝑎𝑛𝐴𝑡𝑎𝑛𝐵
we let B = A
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The Double Angle Formula:
sin 2𝐴 = 2 sin 𝐴 cos 𝐴
cos 2𝐴 = cos 2 𝐴 − sin2 𝐴
cos 2𝐴 = 2 cos 2 𝐴 − 1
cos 2𝐴 = 1 − 2 sin2 𝐴
tan 2𝐴 =
2 tan 𝐴
1 − tan2 𝐴
Notes:
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Proving an identity using the double angle formulae:
We can now use these formulae to prove an identity. It can be reasonably tricky to
spot what you need to do – often, you’ll have to establish whether you have to use
any of the identities covered in Trigonometry 1, or whether a double angle
formula will be appropriate.
Let’s have a look at a few examples to get our heads around this:
Worked Examples:
a)
tan 2𝑥 =
Show that:
2
𝑐𝑜𝑡𝑥−tan 𝑥
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b)
sin 2𝑥
1+cos 2𝑥
= tan 𝑥
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c)
1
𝑐𝑜𝑠𝑒𝑐 2𝑥 = 𝑐𝑜𝑠𝑒𝑐 𝑥 sec 𝑥
2
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And a slightly tougher one, with some guidance:
d)
By expanding sin(2𝐴 + 𝐴), show that sin 3𝐴 = 3 sin 𝐴 − 4 sin3 𝐴
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Notes:
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Solving Trigonometric Equations using Addition/Double Angle Formulae:
Next step – solving equations involving these formulae. These are often some of
the tougher questions within C3 – you need to make sure you know your identities
off by heart and be willing to have a stab at the question, but there are a few
important things to look for:
 An obvious double angle by itself within the question (e.g. a sin 2𝑥, a
cos 2𝐴, etc, alone with no other double angles in the equation)
 A simple factorisation that you could perform if you use a double angle
formula (example below!)
 In particular, a 𝐜𝐨𝐬 𝟐𝒙 within a question. Remember that this can be
swapped for 2 cos 2 𝑥 − 1 or 1 − 2 sin2 𝑥 , so is incredibly useful for making a
hidden quadratic!
Worked Examples:
a)
Solve, for 0 ≤ 𝜃 < 2𝜋:
sin 2𝜃 = 𝑐𝑜𝑠𝜃
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Solve, for 0 < 𝑥 < 360:
b)
3 cos 2𝑥 − cos 𝑥 + 2 = 0
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Notes:
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Solve, for for 0 ≤ 𝑥 ≤ 360:
c)
3cos 2𝑥 − sin 𝑥 + 2 = 0
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A slightly tougher example – we can use the same method with 4𝜃:
Solve, for 0 ≤ 𝜃 ≤ 𝜋:
d)
𝑠𝑖𝑛4𝜃 = √3 cos 2𝜃
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Solve, for −180 < 𝜃 ≤ 180
e)
2 sin 2𝜃 = tan 𝜃
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Notes:
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Harmonic Form:
‘Harmonic form’ is where we take an expression of the form acos 𝜃 + 𝑏𝑠𝑖𝑛 𝜃 ,
and write it in terms of only sin, or only cos.
The benefits of changing the expression into this form are numerous; mainly,
though, it allows us to easily solve equations, and to work out the minimum and
maximum values of such functions.
Worked Examples:
1)
a)
Express 3𝑐𝑜𝑠𝜃 + 4𝑠𝑖𝑛𝜃 in the form 𝑅𝑐𝑜𝑠(𝜃 − 𝛼), where 𝑅 and 𝛼 are
constants, 𝑅 > 0 and 0 < 𝛼 < 90𝑜 . Give answers to 1.d.p
b)
Hence, find the maximum value of 3𝑐𝑜𝑠𝜃 + 4𝑠𝑖𝑛𝜃 and the smallest
possible positive value of 𝜃 for which this occurs
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Notes:
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2)
a)
Express 7𝑐𝑜𝑠𝑥 − 24𝑠𝑖𝑛𝑥 in the form 𝑅𝑐𝑜𝑠(𝑥 + 𝛼), where 𝑅 and 𝛼 are
constants, 𝑅 > 0 and 0 < 𝛼 < 90𝑜 . Give answers to 1.d.p.
b)
Hence, find the minimum value of 7𝑐𝑜𝑠𝑥 − 24𝑠𝑖𝑛𝑥 and the smallest
possible positive value of 𝑥 for which this occurs
c)
Solve, for 0 ≤ 𝑥 < 2𝜋, the equation 7𝑐𝑜𝑠𝑥 − 24𝑠𝑖𝑛𝑥 = 10
Give all solutions to 2 decimal places.
d)
State the values of 𝑘 for which the equation 7𝑐𝑜𝑠𝑥 − 24𝑠𝑖𝑛𝑥 = 𝑘
has only one solution in the interval 0 ≤ 𝑥 < 2𝜋
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Notes:
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Finally, we’ll look at solving a real-life problem involving Harmonic Functions.
Let’s go back to Example 1 that we looked at previously; we’ll look at an
extension to this problem –
Worked Example:
1)
a)
Express 3𝑐𝑜𝑠𝜃 + 4𝑠𝑖𝑛𝜃 in the form 𝑅𝑐𝑜𝑠(𝜃 − 𝛼), where 𝑅 and 𝛼 are
constants, 𝑅 > 0 and 0 < 𝛼 < 90𝑜 . Give answers to 1.d.p
b)
Hence, find the maximum value of 3𝑐𝑜𝑠𝜃 + 4𝑠𝑖𝑛𝜃 and the smallest
possible positive value of 𝜃 for which this occurs
The temperature, 𝑓(𝑡), of a warehouse is modelled using the
equation
𝑓(𝑡) = 10 + 3 cos(15𝑡)𝑜 + 4 sin(15𝑡)𝑜
where 𝑡 is the time in hours from midday, and 0 ≤ 𝑡 < 24.
c)
Calculate the minimum temperature of the warehouse, as given
by this model
d)
Find the value of 𝑡 when this minimum temperature occurs.
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Notes:
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Trigonometry 2 – Questions
Exercise 1:
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Exercise 2:
36
Exercise 3: (Give answers to 1.d.p unless otherwise stated)
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Exercise 4:
39
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Solutions to all exercises:
Exercise 1:
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Exercise 2:
43
Exercise 3:
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Exercise 4:
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