Introduction
• This chapter extends your knowledge of
Trigonometrical identities
• You will see how to solve equations involving
combinations of sin, cos and tan
• You will learn to express combinations of
these as a transformation of a single graph
Further Trigonometric Identities and their
Applications
You need to know and be able to
use the addition formulae
Q
By GCSE Trigonometry:
1
So the coordinates of P are:
B
P
1
A
O
M
N
So the coordinates of Q are:
𝑆𝑖𝑛𝐴 − 𝑆𝑖𝑛𝐵
Q
P
7A
Further Trigonometric Identities and their
Applications
You need to know and be able to
use the addition formulae
𝑃𝑄2 = (𝐶𝑜𝑠𝐴 − 𝐶𝑜𝑠𝐵)2 + (𝑆𝑖𝑛𝐴 − 𝑆𝑖𝑛𝐵)2
Multiply out
the brackets
𝑃𝑄2 = (𝐶𝑜𝑠 2 𝐴 − 2𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 + 𝐶𝑜𝑠 2 𝐵) + (𝑆𝑖𝑛2 𝐴 − 2𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵 + 𝑆𝑖𝑛2 𝐵)
Rearrange
𝑃𝑄2 = (𝐶𝑜𝑠 2 𝐴 + 𝑆𝑖𝑛2 𝐴) + (𝐶𝑜𝑠 2 𝐵 + 𝑆𝑖𝑛2 𝐵) − 2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 + 𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵)
𝐶𝑜𝑠 2 θ + 𝑆𝑖𝑛2 θ ≡ 1
𝑃𝑄2 = 2 − 2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 + 𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵)
7A
Further Trigonometric Identities and their
Applications
You need to know and be able to
use the addition formulae
Q
You can also work out PQ using the triangle OPQ:
Q
P
1
1
B-A
B
P
1
A
O
1
M
N
𝑎2 = 𝑏 2 + 𝑐 2 − 2bcCosA
Sub in the values
𝑃𝑄2 = 12 + 12 − 2Cos(B - A)
Group terms
𝑃𝑄2 = 2 − 2Cos(B - A)
𝑃𝑄2 = 2 − 2Cos(A - B)
Cos (B – A) = Cos (A – B)
eg) Cos(60) = Cos(-60)
7A
Further Trigonometric Identities and their
Applications
You need to know and be able to
use the addition formulae
𝑃𝑄2 = 2 − 2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 + 𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵)
𝑃𝑄2 = 2 − 2Cos(A - B)
2 − 2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 + 𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵) = 2 − 2Cos(A - B)
− 2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 + 𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵) = − 2Cos(A - B)
𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 + 𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵 = Cos(A - B)
Subtract 2 from both
sides
Divide by -2
Cos(A - B) = CosACosB + SinASinB
Cos(A + B) = CosACosB - SinASinB
7A
Further Trigonometric Identities and their
Applications
You need to know and be able to
use the addition formulae
Cos(A - B) ≡ CosACosB + SinASinB
Cos(A + B) ≡ CosACosB - SinASinB
Sin(A + B) ≡ SinACosB + CosASinB
Sin(A - B) ≡ SinACosB - CosASinB
7A
Further Trigonometric Identities and their
Applications
You need to know and be able to
use the addition formulae
Sin(A + B) = SinACosB + CosASinB
Sin(A - B) = SinACosB - CosASinB
Tan (A+B) ≡
Tan (A+B) ≡
Cos(A - B) = CosACosB + SinASinB
Cos(A + B) = CosACosB - SinASinB
Tan (A+B) ≡
Show that:
Tan (A + B) ≡
Tan θ ≡
𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵
1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵
Tan (A+B) ≡
𝑆𝑖𝑛(𝐴+𝐵)
𝐶𝑜𝑠(𝐴+𝐵)
𝑆𝑖𝑛𝐴𝐶𝑜𝑠𝐵+𝐶𝑜𝑠𝐴𝑆𝑖𝑛𝐵
𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵−𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵
𝑆𝑖𝑛𝐴𝐶𝑜𝑠𝐵 𝐶𝑜𝑠𝐴𝑆𝑖𝑛𝐵
+
𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵
𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵
−
𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵
TanA + TanB
1 - TanATanB
Rewrite
Divide top and
bottom by
CosACosB
Simplify each
Fraction
𝑆𝑖𝑛θ
𝐶𝑜𝑠θ
7A
Further Trigonometric Identities and their
Applications
You need to know and be able to
use the addition formulae
Cos(A - B) ≡ CosACosB + SinASinB
Cos(A + B) ≡ CosACosB - SinASinB
Sin(A + B) ≡ SinACosB + CosASinB
Sin(A - B) ≡ SinACosB - CosASinB
Tan (A + B) ≡
𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵
1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵
Tan (A - B) ≡
𝑇𝑎𝑛𝐴−𝑇𝑎𝑛𝐵
1+𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵
You may be asked to prove
either of the Tan
identities using the Sin and
Cos ones!
7A
Further Trigonometric Identities and their
Applications
You need to know and be able to
use the addition formulae
Cos(A + B) ≡ CosACosB - SinASinB
𝑆𝑖𝑛15 = 𝑆𝑖𝑛(45 − 30)
Sin(A - B) ≡ SinACosB - CosASinB
A=45,
B=30
Cos(A - B) ≡ CosACosB + SinASinB
Sin(A + B) ≡ SinACosB + CosASinB
Sin(45 - 30) ≡ Sin45Cos30 – Cos45Sin30
Sin(A - B) ≡ SinACosB - CosASinB
𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵
Tan (A + B) ≡ 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵
𝑇𝑎𝑛𝐴−𝑇𝑎𝑛𝐵
Tan (A - B) ≡ 1+𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵
Show, using the formula for Sin(A – B),
that:
𝑆𝑖𝑛15 =
6− 2
4
Sin(45 - 30) ≡
Sin(45 - 30) ≡
Sin(15) ≡
2
3
×
−
2
2
6
−
4
6− 2
4
2
4
1
2
×
2
2
These can
be written
as surds
Multiply
each pair
Group the
fractions up
7A
Further Trigonometric Identities and their
Applications
You need to know and be able to
use the addition formulae
Given that:
SinA = −
3
5
CosB = −
𝑂𝑝𝑝
𝐻𝑦𝑝
3
𝑆𝑖𝑛𝐴 = −
5
180˚ < A < 270˚
12
13
𝑆𝑖𝑛𝐴 =
5
3
A
𝑂𝑝𝑝
𝐴𝑑𝑗
𝑇𝑎𝑛𝐴 =
3
4
4
B = Obtuse
Use Pythagoras’ to find the missing side (ignore negatives)
Find the value of:
Tan is positive in the range 180˚ - 270˚
Tan(A+B)
Tan (A + B) ≡
𝑇𝑎𝑛𝐴 =
𝐶𝑜𝑠𝐵 =
𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵
1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵
𝐴𝑑𝑗
𝐻𝑦𝑝
𝐶𝑜𝑠𝐵 = −
12
13
13
𝑇𝑎𝑛𝐵 =
𝑂𝑝𝑝
𝐴𝑑𝑗
𝑇𝑎𝑛𝐵 =
5
12
5
B
12
5
12
Use Pythagoras’ to find the missing side (ignore negatives)
𝑇𝑎𝑛𝐵 = −
90
180
270
360
y = Tanθ
Tan is negative in the range 90˚ - 180˚
7A
Further Trigonometric Identities and their
Applications
You need to know and be able to
use the addition formulae
𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵
Given that:
SinA =
3
−
5
CosB = −
12
13
Tan (A + B) ≡ 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵
180˚ < A < 270˚
B = Obtuse
Tan (A + B) ≡
3
5
+
−
4
12
3
5
1− 4×−12
Tan (A + B) ≡
1
3
63
48
Find the value of:
Tan(A+B)
3
𝑇𝑎𝑛𝐴 =
4
5
𝑇𝑎𝑛𝐵 = −
12
1
Work out the
Numerator and
Denominator
Leave, Change and
Flip
𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵
Tan (A + B) ≡ 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵
Substitute in TanA
and TanB
48
Tan (A + B) ≡ 3 × 63
Simplify
16
Tan (A + B) ≡ 63
Although you could just type the whole thing into your
calculator, you still need to show the stages for the
workings marks…
7A
Further Trigonometric Identities and their
Applications
You need to know and be able to use the addition formulae
Given that:
2 𝑠𝑖𝑛 𝑥 + 𝑦 = 3𝑐𝑜𝑠(𝑥 − 𝑦)
Express Tanx in terms of Tany…
2 𝑠𝑖𝑛 𝑥 + 𝑦 = 3𝑐𝑜𝑠(𝑥 − 𝑦)
2(𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦 + 𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦) = 3(𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 + 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦)
Rewrite the sin and cos parts
Multiply out the brackets
2 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦 + 2𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦 = 3 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 + 3𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦
2 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦 + 2𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦 = 3 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 + 3𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦
𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦
𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦
𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦
𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦
2 𝑡𝑎𝑛𝑥 + 2𝑡𝑎𝑛𝑦 = 3 +3𝑡𝑎𝑛𝑥𝑡𝑎𝑛𝑦
2 𝑡𝑎𝑛𝑥 − 3𝑡𝑎𝑛𝑥𝑡𝑎𝑛𝑦 = 3 −2𝑡𝑎𝑛𝑦
𝑡𝑎𝑛𝑥(2 − 3𝑡𝑎𝑛𝑦) = 3 −2𝑡𝑎𝑛𝑦
𝑡𝑎𝑛𝑥 = 3 −2𝑡𝑎𝑛𝑦
2 −3𝑡𝑎𝑛𝑦
Divide all by cosxcosy
Simplify
Subtract 3tanxtany
Subtract 2tany
Factorise the left side
Divide by (2 – 3tany)
7A
Further Trigonometric Identities and their
Applications
You can express sin2A, cos 2A and
tan2A in terms of angle A, using
the double angle formulae
Sin(A + B) ≡ SinACosB + CosASinB
Replace B with A
Sin(A + A) ≡ SinACosA + CosASinA
Simplify
Sin2A ≡ 2SinACosA
1
Sin2A
2
≡ SinACosA
Sin4A ≡ 2Sin2ACos2A
÷ 2
2A  4A
Sin2A ≡ 2SinACosA
x 3
3Sin2A ≡ 6SinACosA
2A = 60
Sin60 ≡ 2Sin30Cos30
7B
Further Trigonometric Identities and their
Applications
You can express sin2A, cos 2A and
tan2A in terms of angle A, using
the double angle formulae
Cos(A + B) ≡ CosACosB - SinASinB
Cos(A + A) ≡ CosACosA - SinASinA
Replace B with A
Simplify
Cos2A ≡ Co𝑠 2 𝐴 − 𝑆𝑖𝑛2 𝐴
Cos2A ≡ Co𝑠 2 𝐴 − 𝑆𝑖𝑛2 𝐴
Replace Cos2A with (1 – Sin2A)
Replace Sin2A with (1 – Cos2A)
Cos2A ≡ (1−𝑆𝑖𝑛2 𝐴) − 𝑆𝑖𝑛2 𝐴
Cos2A ≡ Co𝑠 2 𝐴 − (1 - Co𝑠 2 𝐴)
Cos2A ≡ 1 − 2𝑆𝑖𝑛2 𝐴
Cos2A ≡ 2Co𝑠 2 𝐴 − 1
7B
Further Trigonometric Identities and their
Applications
You can express sin2A, cos 2A and
tan2A in terms of angle A, using
the double angle formulae
𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵
Tan (A + B) ≡ 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵
Replace B with A
𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐴
Tan (A + A) ≡ 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐴
Simplify
2𝑇𝑎𝑛𝐴
Tan 2A ≡ 1−𝑇𝑎𝑛2𝐴
1
Tan 2A
2
≡
2𝑇𝑎𝑛30
𝑇𝑎𝑛𝐴
1−𝑇𝑎𝑛2 𝐴
÷ 2
2A = 60
Tan 60 ≡ 1−𝑇𝑎𝑛2 30
2𝑇𝑎𝑛𝐴
x 2
2Tan 2A ≡
4𝑇𝑎𝑛𝐴
1−𝑇𝑎𝑛2 𝐴
Tan 2A ≡ 1−𝑇𝑎𝑛2𝐴
2A = A
𝐴
Tan A ≡
2𝑇𝑎𝑛 2
𝐴
1−𝑇𝑎𝑛2 2
7B
Further Trigonometric Identities and their
Applications
You can express sin2A, cos 2A and
tan2A in terms of angle A, using
the double angle formulae
Rewrite the following as a single
Trigonometric function:
𝜃
𝜃
2𝑠𝑖𝑛 𝑐𝑜𝑠 𝑐𝑜𝑠𝜃
2
2
𝑆𝑖𝑛2𝜃 ≡ 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
𝜃
𝜃
𝑆𝑖𝑛𝜃 ≡ 2𝑠𝑖𝑛 𝑐𝑜𝑠
2
2
𝜃
𝜃
2𝑠𝑖𝑛 𝑐𝑜𝑠 𝑐𝑜𝑠𝜃
2
2
2θ  θ
Replace the first
part
= 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
Rewrite
1
= 𝑠𝑖𝑛2𝜃
2
7B
Further Trigonometric Identities and their
Applications
You can express sin2A, cos 2A and
tan2A in terms of angle A, using
the double angle formulae
𝐶𝑜𝑠2𝜃 ≡ 2𝑐𝑜𝑠 2 𝜃 − 1
𝐶𝑜𝑠4𝜃 ≡ 2𝑐𝑜𝑠 2 2𝜃 − 1
Double the
angle parts
Show that:
1 + 𝑐𝑜𝑠4𝜃
Can be written as:
2𝑐𝑜𝑠 2 2𝜃
1 + 𝑐𝑜𝑠4𝜃
= 1 + (2𝑐𝑜𝑠 2 2𝜃 − 1)
= 2𝑐𝑜𝑠 2 2𝜃
Replace
cos4θ
The 1s
cancel out
7B
Further Trigonometric Identities and their
Applications
You can express sin2A, cos 2A and
tan2A in terms of angle A, using
the double angle formulae
Given that:
3
𝑐𝑜𝑠𝑥 =
4
𝐶𝑜𝑠𝑥 =
𝐴𝑑𝑗
𝐻𝑦𝑝
4
180˚ < 𝑥 < 360˚
𝑆𝑖𝑛𝑥 =
7
4
7
3
𝐶𝑜𝑠𝑥 =
4
x
Use Pythagoras’ to find the missing side (ignore negatives)
Cosx is positive so in the range 270 - 360
𝑠𝑖𝑛2𝑥
Therefore, Sinx is negative
180
𝑂𝑝𝑝
𝐻𝑦𝑝
3
Find the exact value of:
90
𝑆𝑖𝑛𝑥 =
270
360
y = Cosθ
7
4
Sin2x ≡ 2SinxCosx
Sin2x = 2 ×
y = Sinθ
𝑆𝑖𝑛𝑥 = −
Sin2x = −
3
×
4
3 7
8
−
7
4
Sub in Sinx and Cosx
Work out and leave in
surd form
7B
Further Trigonometric Identities and their
Applications
You can express sin2A, cos 2A and
tan2A in terms of angle A, using
the double angle formulae
Given that:
3
𝑐𝑜𝑠𝑥 =
4
𝐶𝑜𝑠𝑥 =
𝐴𝑑𝑗
𝐻𝑦𝑝
4
180˚ < 𝑥 < 360˚
x
270
360
y = Cosθ
270
360
y = Tanθ
𝑇𝑎𝑛𝑥 = −
7
3
2𝑇𝑎𝑛𝑥
Tan 2x ≡ 1−𝑇𝑎𝑛2𝑥
Tan 2x =
180
7
3
Use Pythagoras’ to find the missing side (ignore negatives)
Therefore, Tanx is negative
90
𝑇𝑎𝑛𝑥 =
7
3
𝐶𝑜𝑠𝑥 =
4
Cosx is positive so in the range 270 - 360
𝑡𝑎𝑛2𝑥
180
𝑂𝑝𝑝
𝐴𝑑𝑗
3
Find the exact value of:
90
𝑇𝑎𝑛𝑥 =
2×−
7
Sub in Tanx
7
3
7
1− − 3 ×− 3
𝑇𝑎𝑛2𝑥 = −3 7
Work out and leave in
surd form
7B
Further Trigonometric Identities and their
Applications
The double angle formulae allow you
to solve more equations and prove
more identities
Prove the identity:
𝑡𝑎𝑛2𝜃 ≡
2
𝑐𝑜𝑡𝜃 − 𝑡𝑎𝑛𝜃
𝑡𝑎𝑛2𝜃 ≡
2𝑡𝑎𝑛𝜃
1 − 𝑡𝑎𝑛2 𝜃
2𝑡𝑎𝑛𝜃
𝑡𝑎𝑛𝜃
𝑡𝑎𝑛2𝜃 ≡
1
𝑡𝑎𝑛2 𝜃
−
𝑡𝑎𝑛𝜃 𝑡𝑎𝑛𝜃
𝑡𝑎𝑛2𝜃 ≡
Divide each part by
tanθ
Rewrite each part
2
𝑐𝑜𝑡𝜃 − 𝑡𝑎𝑛𝜃
7C
Further Trigonometric Identities and their
Applications
The double angle formulae allow you 𝑠𝑖 𝑛 𝐴 + 𝐵 ≡ 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵 + 𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵
to solve more equations and prove
more identities
Replace A and B
𝑠𝑖 𝑛 2𝐴 + 𝐴 ≡ 𝑠𝑖𝑛2𝐴𝑐𝑜𝑠𝐴 + 𝑐𝑜𝑠2𝐴𝑠𝑖𝑛𝐴
By expanding:
𝑠𝑖 𝑛 3𝐴 ≡ (2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴)𝑐𝑜𝑠𝐴 + (1 − 2𝑠𝑖𝑛2 𝐴)𝑠𝑖𝑛𝐴
𝑠𝑖𝑛(2𝐴 + 𝐴)
Show that:
𝑠𝑖 𝑛 3𝐴 ≡ 3𝑠𝑖𝑛𝐴 − 4𝑠𝑖𝑛3 𝐴
𝑠𝑖 𝑛 3𝐴 ≡ 2𝑠𝑖𝑛𝐴𝑐𝑜𝑠 2 𝐴 + 𝑠𝑖𝑛𝐴 − 2𝑠𝑖𝑛3 𝐴
𝑠𝑖 𝑛 3𝐴 ≡ 2𝑠𝑖𝑛𝐴(1 − 𝑠𝑖𝑛2 𝐴) + 𝑠𝑖𝑛𝐴 − 2𝑠𝑖𝑛3 𝐴
Replace
Sin2A and
Cos 2A
Multiply
out
Replace
cos2A
Multiply out
𝑠𝑖 𝑛 3𝐴 ≡ 2𝑠𝑖𝑛𝐴 − 2𝑠𝑖𝑛3 𝐴 + 𝑠𝑖𝑛𝐴 − 2𝑠𝑖𝑛3 𝐴
𝑠𝑖 𝑛 3𝐴 ≡ 3𝑠𝑖𝑛𝐴 − 4𝑠𝑖𝑛3 𝐴
Group like
terms
7C
Further Trigonometric Identities and their
Applications
The double angle formulae allow you
to solve more equations and prove
more identities
Given that:
𝑥 = 3𝑠𝑖𝑛𝜃 and 𝑦 = 3 − 4𝑐𝑜𝑠2𝜃
Eliminate θ and express y in terms of x…
𝑥 = 3𝑠𝑖𝑛𝜃
𝑥
= 𝑠𝑖𝑛𝜃
3
3−𝑦
= 𝑐𝑜𝑠2𝜃
4
3−𝑦
𝑥
= 1−2
4
3
2
Replace Cos2θ and
Sinθ
Multiply by 4
𝑥
3−𝑦 = 4−8
3
2
Subtract 3
−𝑦 = 1 − 8
Divide
by 3
𝑦 = 3 − 4𝑐𝑜𝑠2𝜃
𝑐𝑜𝑠2𝜃 = 1 − 2𝑠𝑖𝑛2 𝜃
𝑥
3
2
Multiply by -1
𝑦 = 8
𝑥
3
2
−1
Subtract 3, divide by 4
Multiply by -1
7C
Further Trigonometric Identities and their
Applications
The double angle formulae allow you
to solve more equations and prove
more identities
Solve the following equation in the range
stated:
3𝑐𝑜𝑠2𝑥 − 𝑐𝑜𝑠𝑥 + 2 = 0
3𝑐𝑜𝑠2𝑥 − 𝑐𝑜𝑠𝑥 + 2 = 0
3(2𝑐𝑜𝑠 2 𝑥 − 1) − 𝑐𝑜𝑠𝑥 + 2 = 0
y = Cosθ
2
3
90
180
270
Group terms
2
6𝑐𝑜𝑠 𝑥 − 𝑐𝑜𝑠𝑥 − 1 = 0
Factorise
(3𝑐𝑜𝑠𝑥 + 1)(2𝑐𝑜𝑠𝑥 − 1) = 0
(All trigonometrical parts must be in terms
x, rather than 2x)
−1
Multiply out the
bracket
6𝑐𝑜𝑠 2 𝑥 − 3 − 𝑐𝑜𝑠𝑥 + 2 = 0
0° ≤ 𝑥 ≤ 360°
1
Replace cos2x
360
𝑐𝑜𝑠𝑥 = −
1
1
or 𝑐𝑜𝑠𝑥 =
3
2
Solve both pairs
𝑥 = 𝑐𝑜𝑠 −1 −
1
3
𝑥 = 109.5° , 250.5°
𝑥 = 𝑐𝑜𝑠 −1
1
2
Remember to find
additional answers!
𝑥 = 60° , 300°
𝑥 = 60°, 109.5°, 250.5°, 300°
7C
Further Trigonometric Identities and their
Applications
You can write expressions of the
form acosθ + bsinθ, where a and
b are constants, as a sine or
cosine function only
Show that:
3𝑠𝑖𝑛𝑥 + 4𝑐𝑜𝑠𝑥
Can be expressed in the form:
𝑅𝑠𝑖𝑛(𝑥 + α)
𝑅>0
0° < α < 90°
So:
3𝑠𝑖𝑛𝑥 + 4𝑐𝑜𝑠𝑥
= 5sin(𝑥 + 53.1°)
𝑅𝑠𝑖𝑛(𝑥 + α) = 𝑅𝑠𝑖𝑛𝑥𝑐𝑜𝑠α + 𝑅𝑐𝑜𝑠𝑥𝑠𝑖𝑛α
3𝑠𝑖𝑛𝑥 + 4𝑐𝑜𝑠𝑥 = 𝑅𝑠𝑖𝑛𝑥𝑐𝑜𝑠α + 𝑅𝑐𝑜𝑠𝑥𝑠𝑖𝑛α
𝑅𝑐𝑜𝑠α = 3
𝑐𝑜𝑠α =
𝑅𝑠𝑖𝑛α = 4
𝐴
𝐻
3
𝑅
𝑠𝑖𝑛α =
4
𝑅
𝑂
𝐻
So in the triangle, the Hypotenuse is R…
𝑅=
32 + 42
𝑐𝑜𝑠α =
𝑐𝑜𝑠α =
3
𝑅
α = 53.1°
Compare each term –
they must be equal!
𝑅
4
α
3
𝑅=5
R=5
3
5
α = 𝑐𝑜𝑠 −1
Replace with the
expression
3
5
Inverse Cos
Find the smallest value in the
acceptable range given
7D
Further Trigonometric Identities and their
Applications
You can write expressions of the
form acosθ + bsinθ, where a and
b are constants, as a sine or
cosine function only
𝑅𝑠𝑖𝑛(𝑥 − α) = 𝑅𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝛼 − 𝑅𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝛼
𝑠𝑖𝑛𝑥 − 3𝑐𝑜𝑠𝑥 = 𝑅𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝛼 − 𝑅𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝛼
𝑅𝑠𝑖𝑛𝛼 = 3
𝑅𝑐𝑜𝑠𝛼 = 1
Show that you can express:
𝑠𝑖𝑛𝑥 − 3𝑐𝑜𝑠𝑥
In the form:
𝑅>0
𝑅𝑠𝑖𝑛(𝑥 − α)
0<α<
𝜋
2
𝑅=
12 +
3
𝑅𝑐𝑜𝑠𝛼 = 1
𝑐𝑜𝑠𝛼 =
𝑠𝑖𝑛𝑥 − 3𝑐𝑜𝑠𝑥
= 2sin 𝑥 −
𝜋
3
𝛼=
𝜋
3
𝑅=2
Divide
by 2
1
2
𝛼 = 𝑐𝑜𝑠 −1
Compare each term –
they must be equal!
R=2
2𝑐𝑜𝑠𝛼 = 1
So:
2
Replace with the
expression
1
2
Inverse
cos
Find the smallest
value in the
acceptable range
7D
Further Trigonometric Identities and their
Applications
Sketch the graph of: 𝑠𝑖𝑛𝑥 − 3𝑐𝑜𝑠𝑥
You can write expressions of the
form acosθ + bsinθ, where a and
b are constants, as a sine or
cosine function only
Show that you can express:
𝑠𝑖𝑛𝑥 − 3𝑐𝑜𝑠𝑥
In the form:
𝑅𝑠𝑖𝑛(𝑥 − α)
𝑅>0
0<α<
𝜋
2
So:
𝑠𝑖𝑛𝑥 − 3𝑐𝑜𝑠𝑥
= 2sin 𝑥 −
𝜋
2sin −
3
=− 3
= Sketch the graph of: 2sin 𝑥 −
1
𝜋
3
π/
2
π
3π/
2
1
π/
3
π/
2
π
4π/
3
3π/
2
𝜋
3
2π
y = 2sin 𝑥 −
1
-1
-2
2π
y = sin 𝑥 −
2
At the yintercept,
x=0
Start out
with sinx
y = sin 𝑥
-1
-1
𝜋
3
π/
3
π/
2
π
4π/
3
3π/
2
2π
𝜋
3
Translate
π/3 units
right
Vertical
stretch, scale
factor 2
7D
Further Trigonometric Identities and their
Applications
You can write expressions of the
form acosθ + bsinθ, where a and
b are constants, as a sine or
cosine function only
Express:
2𝑐𝑜𝑠𝜃 + 5𝑠𝑖𝑛𝜃
in the form:
𝑅𝑐𝑜𝑠(𝜃 − 𝛼)
𝑅>0
0° < 𝛼 < 90°
So:
2𝑐𝑜𝑠𝜃 + 5𝑠𝑖𝑛𝜃
= 29cos(𝜃 − 68.2)
𝑅𝑐𝑜𝑠(𝜃 − 𝛼) = 𝑅𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝛼 + 𝑅𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛼
2𝑐𝑜𝑠𝜃 + 5𝑠𝑖𝑛𝜃 = 𝑅𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝛼 + 𝑅𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛼
𝑅𝑐𝑜𝑠𝛼 = 2
𝑅=
𝑅𝑠𝑖𝑛𝛼 = 5
22 +52
𝑅𝑐𝑜𝑠𝛼 = 2
29𝑐𝑜𝑠𝛼 = 2
𝑐𝑜𝑠𝛼 =
2
29
𝛼 = 𝑐𝑜𝑠 −1
𝛼 = 68.2
2
29
Replace with the
expression
Compare each term –
they must be equal!
𝑅 = 29
R = √29
Divide by
√29
Inverse
cos
Find the smallest
value in the
acceptable range
7D
Further Trigonometric Identities and their
Applications
You can write expressions of the
form acosθ + bsinθ, where a and
b are constants, as a sine or
cosine function only
29cos(𝜃 − 68.2) = 3
Divide by √29
cos(𝜃 − 68.2) =
3
29
Inverse Cos
Solve in the given range, the
following equation:
3
𝜃 − 68.2 = 𝑐𝑜𝑠 −1
29
2𝑐𝑜𝑠𝜃 + 5𝑠𝑖𝑛𝜃 = 3
𝜃 − 68.2 = 56.1, −56.1 , 303.9
0° < 𝜃 < 360°
We just showed that the original equation can
be rewritten…
2𝑐𝑜𝑠𝜃 + 5𝑠𝑖𝑛𝜃 = 29cos(𝜃 − 68.2)
Remember to work out
other values in the
adjusted range
Add 68.2 (and
put in order!)
𝜃 = 12.1 , 124.3
Hence, we can solve this equation instead!
29cos(𝜃 − 68.2) = 3
0° < 𝜃 < 360°
−68.2° < 𝜃 − 68.2 < 291.2°
-56.1
Remember to
adjust the
range for (θ –
68.2)
-90
56.1
90
303.9
180
270
y = Cosθ
360
7D
Further Trigonometric Identities and their
Applications
Rcos(θ – α) chosen as it
gives us the same form
as the expression
You can write expressions of the
form acosθ + bsinθ, where a and
b are constants, as a sine or
cosine function only
𝑅𝑐𝑜𝑠(𝜃 − 𝛼) = 𝑅𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝛼 + 𝑅𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛼
12𝑐𝑜𝑠𝜃 + 5𝑠𝑖𝑛𝜃 = 𝑅𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝛼 + 𝑅𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛼
𝑅𝑐𝑜𝑠𝛼 = 12
Find the maximum value of the
following expression, and the
smallest positive value of θ at which
it arises:
12𝑐𝑜𝑠𝜃 + 5𝑠𝑖𝑛𝜃
= 13cos(𝜃 − 22.6)
13cos(𝜃 − 22.6)
13(1)
𝑀𝑎𝑥 = 13
𝜃 − 22.6 = 0
𝜃 = 22.6
Max value of
cos(θ - 22.6) = 1
Overall maximum
therefore = 13
Cos peaks at 0
θ = 22.6 gives us 0
𝑅=
𝑅𝑠𝑖𝑛𝛼 = 5
122 +52
𝑅𝑐𝑜𝑠𝛼 = 12
13𝑐𝑜𝑠𝛼 = 12
12
𝑐𝑜𝑠𝛼 =
13
𝛼 = 𝑐𝑜𝑠 −1
𝛼 = 22.6
12
13
Replace with the
expression
Compare each term –
they must be equal!
𝑅 = 13
R = 13
Divide by 13
Inverse
cos
Find the smallest
value in the
acceptable range
7D
Further Trigonometric Identities and their
Applications
You can write expressions of the
form acosθ + bsinθ, where a and
b are constants, as a sine or
cosine function only
𝑎𝑠𝑖𝑛𝜃 ± 𝑏𝑐𝑜𝑠𝜃
𝑅𝑠𝑖𝑛 𝜃 ± 𝛼
𝑎𝑐𝑜𝑠𝜃 ± 𝑏𝑠𝑖𝑛𝜃
𝑅𝑐𝑜𝑠 𝜃 ∓ 𝛼
Whichever ratio is at the start, change the expression into
a function of that (This makes solving problems easier)
Remember to get the + or – signs the correct way round!
7D
Further Trigonometric Identities and their
Applications
You can express sums and differences of
sines and cosines as products of sines
and cosines by using the ‘factor
formulae’
𝑠𝑖𝑛𝑃 + 𝑠𝑖𝑛𝑄 = 2𝑠𝑖𝑛
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠
2
2
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛𝑃 − 𝑠𝑖𝑛𝑄 = 2𝑐𝑜𝑠
𝑠𝑖𝑛
2
2
𝑐𝑜𝑠𝑃 + 𝑐𝑜𝑠𝑄 = 2𝑐𝑜𝑠
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠
2
2
𝑐𝑜𝑠𝑃 − 𝑐𝑜𝑠𝑄 = −2𝑠𝑖𝑛
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛
2
2
You get given all these in the
formula booklet!
7E
Further Trigonometric Identities and their
Applications
Using the formulae for Sin(A + B) and Sin (A – B),
derive the result that:
You can express sums and differences of
sines and cosines as products of sines
and cosines by using the ‘factor
formulae’
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛𝑃 + 𝑠𝑖𝑛𝑄 = 2𝑠𝑖𝑛
𝑐𝑜𝑠
2
2
𝑠𝑖𝑛𝑃 − 𝑠𝑖𝑛𝑄 = 2𝑐𝑜𝑠
2) 𝑆𝑖𝑛 𝐴 − 𝐵 = 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵 − 𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵
𝑆𝑖𝑛 𝐴 + 𝐵 + 𝑠𝑖𝑛(𝐴 − 𝐵) = 2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵
𝑆𝑖𝑛 𝐴 + 𝐵 + 𝑠𝑖𝑛(𝐴 − 𝐵) = 2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵
𝑃+𝑄
𝑃−𝑄
𝑆𝑖𝑛𝑃 + 𝑆𝑖𝑛𝑄 = 2𝑠𝑖𝑛
𝑐𝑜𝑠
2
2
Add both sides
together (1 + 2)
Let (A+B) = P
Let (A-B) = Q
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠𝑃 + 𝑐𝑜𝑠𝑄 = 2𝑐𝑜𝑠
𝑐𝑜𝑠
2
2
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛
2
2
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠
2
2
1) 𝑆𝑖𝑛 𝐴 + 𝐵 = 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵 + 𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛
2
2
𝑐𝑜𝑠𝑃 − 𝑐𝑜𝑠𝑄 = −2𝑠𝑖𝑛
𝑠𝑖𝑛𝑃 + 𝑠𝑖𝑛𝑄 = 2𝑠𝑖𝑛
1)
𝐴+𝐵 =𝑃
2)
𝐴−𝐵 =𝑄
2𝐴 = 𝑃 + 𝑄
𝐴=
𝑃+𝑄
2
1+2
Divide
by 2
1)
𝐴+𝐵 =𝑃
2)
𝐴−𝐵 =𝑄
2𝐵 = 𝑃 − 𝑄
𝑃−𝑄
𝐵=
2
1-2
Divide
by 2
7E
Further Trigonometric Identities and their
Applications
Show that:
You can express sums and differences of
sines and cosines as products of sines
and cosines by using the ‘factor
formulae’
𝑠𝑖𝑛𝑃 + 𝑠𝑖𝑛𝑄 = 2𝑠𝑖𝑛
𝑠𝑖𝑛𝑃 − 𝑠𝑖𝑛𝑄 = 2𝑐𝑜𝑠
𝑐𝑜𝑠𝑃 + 𝑐𝑜𝑠𝑄 = 2𝑐𝑜𝑠
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠
2
2
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛
2
2
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠
2
2
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠𝑃 − 𝑐𝑜𝑠𝑄 = −2𝑠𝑖𝑛
𝑠𝑖𝑛
2
2
𝑠𝑖𝑛𝑃 − 𝑠𝑖𝑛𝑄 = 2𝑐𝑜𝑠
𝑠𝑖𝑛105 − 𝑠𝑖𝑛15 =
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛
2
2
𝑠𝑖𝑛105 − 𝑠𝑖𝑛15 = 2𝑐𝑜𝑠
105 + 15
105 − 15
𝑠𝑖𝑛
2
2
𝑠𝑖𝑛105 − 𝑠𝑖𝑛15 = 2𝑐𝑜𝑠60𝑠𝑖𝑛45
𝑠𝑖𝑛105 − 𝑠𝑖𝑛15 = 2 ×
𝑠𝑖𝑛105 − 𝑠𝑖𝑛15 =
1
1
1
1
×
2
2
2
P = 105
Q = 15
Work out the
fraction parts
Sub in values for
Cos60 and Sin45
Work out the
right hand side
2
7E
Further Trigonometric Identities and their
Applications
Solve in the range indicated:
You can express sums and differences of
𝑠𝑖𝑛4𝜃 − 𝑠𝑖𝑛3𝜃 = 0
0≤𝜃≤𝜋
sines and cosines as products of sines
and cosines by using the ‘factor
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛𝑃 − 𝑠𝑖𝑛𝑄 = 2𝑐𝑜𝑠
𝑠𝑖𝑛
formulae’
2
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛𝑃 + 𝑠𝑖𝑛𝑄 = 2𝑠𝑖𝑛
𝑐𝑜𝑠
2
2
𝑠𝑖𝑛4𝜃 − 𝑠𝑖𝑛3𝜃 = 2𝑐𝑜𝑠
4𝜃 + 3𝜃
4𝜃 − 3𝜃
𝑠𝑖𝑛
2
2
𝑠𝑖𝑛4𝜃 − 𝑠𝑖𝑛3𝜃 = 2𝑐𝑜𝑠
7𝜃
𝜃
𝑠𝑖𝑛
2
2
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛𝑃 − 𝑠𝑖𝑛𝑄 = 2𝑐𝑜𝑠
𝑠𝑖𝑛
2
2
𝑐𝑜𝑠𝑃 + 𝑐𝑜𝑠𝑄 = 2𝑐𝑜𝑠
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠
2
2
2𝑐𝑜𝑠
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠𝑃 − 𝑐𝑜𝑠𝑄 = −2𝑠𝑖𝑛
𝑠𝑖𝑛
2
2
0≤𝜃≤𝜋
𝑐𝑜𝑠
Adjust the
range
7𝜃 7𝜋
0≤
≤
2
2
π/
2
π
3π/
2
7𝜃
𝜃
𝑠𝑖𝑛
=0
2
2
7𝜃
=0
2
7𝜃
= 𝑐𝑜𝑠 −1 0
2
0
2π
y = Cosθ
2
7𝜃 𝜋 3𝜋 5𝜋 7𝜋
,
= ,
,
2
2 2 2 2
𝜋 3𝜋 5𝜋
𝜃= ,
,
,𝜋
7 7 7
P = 4θ
Q = 3θ
Work out
the
fractions
Set equal
to 0
Either the cos or sin
part must equal 0…
Inverse cos
Solve, remembering to take into account the
different range
Once you have all the values from 0-2π, add
2π to them to obtain equivalents…
Multiply by 2 and divide by 7
7E
Further Trigonometric Identities and their
Applications
Solve in the range indicated:
You can express sums and differences of
𝑠𝑖𝑛4𝜃 − 𝑠𝑖𝑛3𝜃 = 0
0≤𝜃≤𝜋
sines and cosines as products of sines
and cosines by using the ‘factor
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛𝑃 − 𝑠𝑖𝑛𝑄 = 2𝑐𝑜𝑠
𝑠𝑖𝑛
formulae’
2
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛𝑃 + 𝑠𝑖𝑛𝑄 = 2𝑠𝑖𝑛
𝑐𝑜𝑠
2
2
𝑠𝑖𝑛4𝜃 − 𝑠𝑖𝑛3𝜃 = 2𝑐𝑜𝑠
4𝜃 + 3𝜃
4𝜃 − 3𝜃
𝑠𝑖𝑛
2
2
𝑠𝑖𝑛4𝜃 − 𝑠𝑖𝑛3𝜃 = 2𝑐𝑜𝑠
7𝜃
𝜃
𝑠𝑖𝑛
2
2
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛𝑃 − 𝑠𝑖𝑛𝑄 = 2𝑐𝑜𝑠
𝑠𝑖𝑛
2
2
𝑐𝑜𝑠𝑃 + 𝑐𝑜𝑠𝑄 = 2𝑐𝑜𝑠
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠
2
2
0≤𝜃≤𝜋
0
π/
2
𝑠𝑖𝑛
Adjust the
range
𝜃 𝜋
0≤ ≤
2 2
π
3π/
2
7𝜃
𝜃
𝑠𝑖𝑛
=0
2
2
2𝑐𝑜𝑠
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠𝑃 − 𝑐𝑜𝑠𝑄 = −2𝑠𝑖𝑛
𝑠𝑖𝑛
2
2
𝜃
=0
2
𝜃
= 𝑠𝑖𝑛−1 0
2
2π
y = Sinθ
2
𝜃
=0
2
𝜃=0
P = 4θ
Q = 3θ
Work out
the
fractions
Set equal
to 0
Either the cos or sin
part must equal 0…
Inverse sin
Solve, remembering to take into account the
different range
Once you have all the values from 0-2π, add
2π to them to obtain equivalents
Multiply by 2
7E
Further Trigonometric Identities and their
Applications
You can express sums and differences of
sines and cosines as products of sines
and cosines by using the ‘factor
formulae’
𝑠𝑖𝑛𝑃 + 𝑠𝑖𝑛𝑄 = 2𝑠𝑖𝑛
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠
2
2
𝑠𝑖𝑛𝑃 − 𝑠𝑖𝑛𝑄 = 2𝑐𝑜𝑠
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛
2
2
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠𝑃 + 𝑐𝑜𝑠𝑄 = 2𝑐𝑜𝑠
𝑐𝑜𝑠
2
2
𝑐𝑜𝑠𝑃 − 𝑐𝑜𝑠𝑄 = −2𝑠𝑖𝑛
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛
2
2
Prove that:
𝑠𝑖𝑛 𝑥 + 2𝑦 + 𝑠𝑖𝑛 𝑥 + 𝑦 + 𝑠𝑖𝑛𝑥
= 𝑡𝑎𝑛(𝑥 + 𝑦)
𝑐𝑜𝑠 𝑥 + 2𝑦 + 𝑐𝑜𝑠 𝑥 + 𝑦 + 𝑐𝑜𝑠𝑥
Numerator:
In the numerator:
𝑠𝑖𝑛 𝑥 + 2𝑦 + 𝑠𝑖𝑛 𝑥 + 𝑦 + 𝑠𝑖𝑛𝑥
𝑠𝑖𝑛 𝑥 + 2𝑦 + 𝑠𝑖𝑛𝑥
Ignore sin(x + y)
for now…
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛𝑃 + 𝑠𝑖𝑛𝑄 = 2𝑠𝑖𝑛
𝑐𝑜𝑠
2
2
𝑠𝑖𝑛(𝑥 + 2𝑦) + 𝑠𝑖𝑛𝑥 = 2𝑠𝑖𝑛
Use the identity
for adding 2 sines
𝑥 + 2𝑦 + 𝑥
𝑥 + 2𝑦 − 𝑥
𝑐𝑜𝑠
2
2
= 2 𝑠𝑖𝑛 𝑥 + 𝑦 𝑐𝑜𝑠𝑦
= 2 𝑠𝑖𝑛 𝑥 + 𝑦 𝑐𝑜𝑠𝑦 + 𝑠𝑖𝑛(𝑥 + 𝑦)
P = x + 2y
Q=x
Simplify
Fractions
Bring back the sin(x + y)
we ignored earlier
Factorise
= 𝑠𝑖𝑛 𝑥 + 𝑦 (2𝑐𝑜𝑠𝑦 + 1)
𝑠𝑖𝑛 𝑥 + 𝑦 (2𝑐𝑜𝑠𝑦 + 1)
7E
Further Trigonometric Identities and their
Applications
You can express sums and differences of
sines and cosines as products of sines
and cosines by using the ‘factor
formulae’
𝑠𝑖𝑛𝑃 + 𝑠𝑖𝑛𝑄 = 2𝑠𝑖𝑛
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠
2
2
𝑠𝑖𝑛𝑃 − 𝑠𝑖𝑛𝑄 = 2𝑐𝑜𝑠
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛
2
2
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠𝑃 + 𝑐𝑜𝑠𝑄 = 2𝑐𝑜𝑠
𝑐𝑜𝑠
2
2
𝑐𝑜𝑠𝑃 − 𝑐𝑜𝑠𝑄 = −2𝑠𝑖𝑛
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛
2
2
Prove that:
𝑠𝑖𝑛 𝑥 + 2𝑦 + 𝑠𝑖𝑛 𝑥 + 𝑦 + 𝑠𝑖𝑛𝑥
= 𝑡𝑎𝑛(𝑥 + 𝑦)
𝑐𝑜𝑠 𝑥 + 2𝑦 + 𝑐𝑜𝑠 𝑥 + 𝑦 + 𝑐𝑜𝑠𝑥
Numerator:
In the denominator:
𝑐𝑜𝑠 𝑥 + 2𝑦 + 𝑐𝑜𝑠 𝑥 + 𝑦 + 𝑐𝑜𝑠𝑥
𝑐𝑜𝑠 𝑥 + 2𝑦 + 𝑐𝑜𝑠𝑥
Ignore cos(x + y)
for now…
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠𝑃 + 𝑐𝑜𝑠𝑄 = 2𝑐𝑜𝑠
𝑐𝑜𝑠
2
2
𝑐𝑜𝑠(𝑥 + 2𝑦) + 𝑐𝑜𝑠𝑥 = 2𝑐𝑜𝑠
Use the identity for
adding 2 cosines
𝑥 + 2𝑦 + 𝑥
𝑥 + 2𝑦 − 𝑥
𝑐𝑜𝑠
2
2
= 2 𝑐𝑜𝑠 𝑥 + 𝑦 𝑐𝑜𝑠𝑦
= 2 𝑐𝑜𝑠 𝑥 + 𝑦 𝑐𝑜𝑠𝑦 + 𝑐𝑜𝑠(𝑥 + 𝑦)
P = x + 2y
Q=x
Simplify
Fractions
Bring back the cos(x + y)
we ignored earlier
Factorise
= 𝑐𝑜𝑠 𝑥 + 𝑦 (2𝑐𝑜𝑠𝑦 + 1)
𝑠𝑖𝑛 𝑥 + 𝑦 (2𝑐𝑜𝑠𝑦 + 1)
Denominator: 𝑐𝑜𝑠 𝑥 + 𝑦 (2𝑐𝑜𝑠𝑦 + 1)
7E
Further Trigonometric Identities and their
Applications
You can express sums and differences of
sines and cosines as products of sines
and cosines by using the ‘factor
formulae’
𝑠𝑖𝑛𝑃 + 𝑠𝑖𝑛𝑄 = 2𝑠𝑖𝑛
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠
2
2
𝑠𝑖𝑛𝑃 − 𝑠𝑖𝑛𝑄 = 2𝑐𝑜𝑠
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛
2
2
𝑃+𝑄
𝑃−𝑄
𝑐𝑜𝑠𝑃 + 𝑐𝑜𝑠𝑄 = 2𝑐𝑜𝑠
𝑐𝑜𝑠
2
2
𝑐𝑜𝑠𝑃 − 𝑐𝑜𝑠𝑄 = −2𝑠𝑖𝑛
𝑠𝑖𝑛 𝑥 + 2𝑦 + 𝑠𝑖𝑛 𝑥 + 𝑦 + 𝑠𝑖𝑛𝑥
𝑐𝑜𝑠 𝑥 + 2𝑦 + 𝑐𝑜𝑠 𝑥 + 𝑦 + 𝑐𝑜𝑠𝑥
=
=
𝑠𝑖𝑛(𝑥 + 𝑦)(2𝑐𝑜𝑠𝑦 + 1)
𝑐𝑜𝑠(𝑥 + 𝑦)(2𝑐𝑜𝑠𝑦 + 1)
𝑠𝑖𝑛(𝑥 + 𝑦)
𝑐𝑜𝑠(𝑥 + 𝑦)
Replace the
numerator and
denominator
Cancel out the
(2cosy + 1) brackets
Use one of the
identities from C2
= 𝑡𝑎𝑛(𝑥 + 𝑦)
𝑃+𝑄
𝑃−𝑄
𝑠𝑖𝑛
2
2
Prove that:
𝑠𝑖𝑛 𝑥 + 2𝑦 + 𝑠𝑖𝑛 𝑥 + 𝑦 + 𝑠𝑖𝑛𝑥
= 𝑡𝑎𝑛(𝑥 + 𝑦)
𝑐𝑜𝑠 𝑥 + 2𝑦 + 𝑐𝑜𝑠 𝑥 + 𝑦 + 𝑐𝑜𝑠𝑥
Numerator:
𝑠𝑖𝑛 𝑥 + 𝑦 (2𝑐𝑜𝑠𝑦 + 1)
Denominator: 𝑐𝑜𝑠 𝑥 + 𝑦 (2𝑐𝑜𝑠𝑦 + 1)
7E
Summary
• We have extended the range of techniques
we have for solving trigonometrical equations
• We have seen how to combine functions
involving sine and cosine into a single
transformation of sine or cosine
• We have learnt several new identities