further trigonometric identities
DESCRIPTION
Further Trigonometric Identities. and their Applications. Introduction. This chapter extends your knowledge of Trigonometrical identities You will see how to solve equations involving combinations of sin, cos and tan - PowerPoint PPT PresentationTRANSCRIPT
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
7A
Q
P
N
11
AB
By GCSE Trigonometry:
O
So the coordinates of P are:
M So the coordinates of Q
are:
Q
P
𝑆𝑖𝑛𝐴−𝑆
𝑖𝑛𝐵
Further Trigonometric Identities and their Applications
You need to know and be able to use the addition
formulae
7A
𝑃𝑄2=¿¿
𝑃𝑄2=¿(𝐶𝑜 𝑠2 𝐴−2𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵+𝐶𝑜𝑠2𝐵)+(𝑆𝑖𝑛2 𝐴−2𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵+𝑆𝑖𝑛2𝐵)
𝑃𝑄2=¿(𝐶𝑜 𝑠2 𝐴+𝑆𝑖𝑛2 𝐴) −2 (𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵+𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵)+(𝐶𝑜𝑠2𝐵+𝑆𝑖𝑛2𝐵)
𝑃𝑄2=¿2−2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵+𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵)
Multiply out the brackets
Rearrange
≡ 1
Further Trigonometric Identities and their Applications
You need to know and be able to use the addition
formulae
7A
Q
P
N
11
A
B
O M
You can also work out PQ using the triangle
OPQ:
B - A1
1
2bcCosA 2Cos(B - A)
Q
P
2Cos(B - A) 2Cos(A - B)
Sub in the values
Group terms
Cos (B – A) = Cos (A – B) eg) Cos(60) = Cos(-60)
Further Trigonometric Identities and their Applications
You need to know and be able to use the addition
formulae
7A
2Cos(A - B)𝑃𝑄2=¿2−2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵+𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵)
2−2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵+𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵) 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 + SinASinBCos(A + B) = CosACosB - SinASinB
Further Trigonometric Identities and their Applications
You need to know and be able to use the addition
formulae
7A
Cos(A - B) ≡ CosACosB + SinASinBCos(A + B) ≡ CosACosB - SinASinBSin(A + B) ≡ SinACosB + CosASinBSin(A - B) ≡ SinACosB - CosASinB
Further Trigonometric Identities and their Applications
You need to know and be able to use the addition
formulae
Show that:
7A
Cos(A - B) = CosACosB + SinASinBCos(A + B) = CosACosB - SinASinB
Sin(A + B) = SinACosB + CosASinBSin(A - B) = SinACosB - CosASinB
Tan (A + B) Tan θ
Tan (A+B) Tan (A+B)
Tan (A+B) Tan (A+B) TanA+ TanB1 - TanATanB
Rewrite
Divide top and bottom by CosACosB
Simplify each Fraction
Further Trigonometric Identities and their Applications
You need to know and be able to use the addition
formulae
7A
Cos(A - B) ≡ CosACosB + SinASinBCos(A + B) ≡ CosACosB - SinASinBSin(A + B) ≡ SinACosB + CosASinBSin(A - B) ≡ SinACosB - CosASinBTan (A + B)
Tan (A - B) You may be asked to
prove either of the Tan identities using the Sin
and Cos ones!
Further Trigonometric Identities and their Applications
You need to know and be able to use the addition
formulae
Show, using the formula for Sin(A – B), that:
7A
Cos(A - B) ≡ CosACosB + SinASinBCos(A + B) ≡ CosACosB - SinASinBSin(A + B) ≡ SinACosB + CosASinBSin(A - B) ≡ SinACosB - CosASinBTan (A + B)
Tan (A - B)
𝑆𝑖𝑛15=√6−√24
𝑆𝑖𝑛15=𝑆𝑖𝑛(45−30)
Sin(A - B) ≡ SinACosB - CosASinBSin(45 - 30) ≡ Sin45Cos30 – Cos45Sin30
Sin(45 - 30) ≡√22
√32
√2212× ×−
Sin(45 - 30) ≡√64
√24−
Sin(15) ≡ √6−√24
A=45, B=30
These can be written as surds
Multiply each pair
Group the fractions up
Further Trigonometric Identities and their Applications
You need to know and be able to use the addition
formulae
Given that:
Find the value of:
7A
< A < 270˚CosTan(A+B)
Tan (A + B)
A
B
35
4
12
13
𝑆𝑖𝑛𝐴=𝑂𝑝𝑝𝐻𝑦𝑝
𝑆𝑖𝑛𝐴=− 35
𝐶𝑜𝑠𝐵=𝐴𝑑𝑗𝐻𝑦𝑝 5
𝑇𝑎𝑛𝐴=𝑂𝑝𝑝𝐴𝑑𝑗
𝑇𝑎𝑛𝐴=34
𝑇𝑎𝑛𝐵=𝑂𝑝𝑝𝐴𝑑𝑗
𝑇𝑎𝑛𝐵=512𝐶𝑜𝑠𝐵=− 1213
𝑇𝑎𝑛𝐵=− 512
Use Pythagoras’ to find the missing side (ignore negatives)Tan is positive in the range 180˚ -
270˚
Use Pythagoras’ to find the missing side (ignore negatives)Tan is negative in the range 90˚ -
180˚
90 180
270 360y = Tanθ
Further Trigonometric Identities and their Applications
You need to know and be able to use the addition
formulae
Given that:
Find the value of:
7A
< A < 270˚CosTan(A+B)
Tan (A + B)
Tan (A + B)
𝑇𝑎𝑛𝐴=34 𝑇𝑎𝑛𝐵=− 512
Tan (A + B) Tan (A + B) Tan (A + B)
Tan (A + B)
Substitute in TanA and TanB
Work out the Numerator and Denominator
Leave, Change and Flip
Simplify
Although you could just type the whole thing into your calculator, you still need to show the stages for
the workings marks…
Further Trigonometric Identities and their Applications
You need to know and be able to use the addition formulaeGiven that:
Express Tanx in terms of Tany…
7A
2𝑠𝑖𝑛 (𝑥+ 𝑦 )=3𝑐𝑜𝑠 (𝑥− 𝑦 )
2𝑠𝑖𝑛 (𝑥+ 𝑦 )=3𝑐𝑜𝑠 (𝑥− 𝑦 )
2(𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦+𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦 )¿3 (𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦+𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦 )
2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦+2𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦¿3𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦+3 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦+2𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦¿3𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦+3 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦
2 𝑡𝑎𝑛𝑥+2 𝑡𝑎𝑛𝑦¿3+3 𝑡𝑎𝑛𝑥𝑡𝑎𝑛𝑦2 𝑡𝑎𝑛𝑥−3 𝑡𝑎𝑛𝑥𝑡𝑎𝑛𝑦¿3−2𝑡𝑎𝑛𝑦𝑡𝑎𝑛𝑥 (2−3 𝑡𝑎𝑛𝑦 )¿3−2𝑡𝑎𝑛𝑦
𝑡𝑎𝑛𝑥¿3−2𝑡𝑎𝑛𝑦2−3 𝑡𝑎𝑛𝑦
Rewrite the sin and cos parts
Multiply out the brackets
Divide all by cosxcosy
Simplify
Subtract 3tanxtanySubtract 2tany
Factorise the left side
Divide by (2 – 3tany)
Further Trigonometric Identities and their Applications
You can express sin2A, cos 2A and tan2A in terms of angle A,
using the double angle formulae
7B
Sin(A + B) ≡ SinACosB + CosASinBSin(A + A) ≡ SinACosA + CosASinASin2A ≡ 2SinACosA
Replace B with A
Simplify
Sin2A ≡ 2SinACosASin4A ≡ 2Sin2ACos2ASin2A ≡ SinACosA
Sin60 ≡ 2Sin30Cos303Sin2A ≡ 6SinACosA
÷ 2
2A 4A
x 3 2A = 60
Further Trigonometric Identities and their Applications
You can express sin2A, cos 2A and tan2A in terms of angle A,
using the double angle formulae
7B
Cos(A + B) ≡ CosACosB - SinASinBCos(A + A) ≡ CosACosA - SinASinA
Cos2A ≡ CoReplace B with A
Simplify
Cos2A ≡ CoCos2A ≡ (1 Cos2A ≡ Co1 - Co
Cos2A ≡ 1 Cos2A ≡ 2CoReplace Sin2A with (1 – Cos2A)Replace Cos2A with (1 – Sin2A)
Further Trigonometric Identities and their Applications
You can express sin2A, cos 2A and tan2A in terms of angle A,
using the double angle formulae
7B
Replace B with A
Simplify
Tan (A + B) Tan (A + A)
Tan 2A
Tan 2A Tan 60 Tan 2A
2Tan 2A Tan A ÷ 2 2A = 60
x 22A = A
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:
7B
2𝑠𝑖𝑛 𝜃2 𝑐𝑜𝑠
𝜃2 𝑐𝑜𝑠𝜃
𝑆𝑖𝑛 2𝜃≡2 𝑠𝑖𝑛𝜃𝑐𝑜𝑠 𝜃𝑆𝑖𝑛𝜃≡2 𝑠𝑖𝑛 𝜃2 𝑐𝑜𝑠
𝜃2
2𝑠𝑖𝑛 𝜃2 𝑐𝑜𝑠
𝜃2 𝑐𝑜𝑠𝜃
¿ 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
¿12 𝑠𝑖𝑛2𝜃
2θ θ
Replace the first part
Rewrite
Further Trigonometric Identities and their Applications
You can express sin2A, cos 2A and tan2A in terms of angle A,
using the double angle formulae
Show that:
Can be written as:
7B
1+𝑐𝑜𝑠 4𝜃
𝐶𝑜𝑠2 𝜃≡2𝑐𝑜𝑠2𝜃−1 Double the angle parts
2𝑐𝑜𝑠22𝜃
𝐶𝑜𝑠 4 𝜃≡2𝑐𝑜𝑠22𝜃−1
1+𝑐𝑜𝑠 4𝜃¿1+(2𝑐𝑜𝑠22𝜃−1)
¿2𝑐𝑜𝑠22 𝜃
Replace cos4θ
The 1s cancel out
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:
Find the exact value of:
7B
𝑐𝑜𝑠𝑥=34 180 ˚<𝑥<360 ˚
𝑠𝑖𝑛2𝑥
x
𝐶𝑜𝑠𝑥=𝐴𝑑𝑗𝐻𝑦𝑝
𝐶𝑜𝑠𝑥=34
𝑆𝑖𝑛𝑥=𝑂𝑝𝑝𝐻𝑦𝑝
𝑆𝑖𝑛𝑥=√74
Use Pythagoras’ to find the missing side (ignore negatives)
Cosx is positive so in the range 270 - 360
3
√74
Therefore, Sinx is negative
90 180
270 360
y = Sinθ
y = Cosθ
𝑆𝑖𝑛𝑥=− √74
Sin2x ≡ 2SinxCosxSin2x = 2Sin2x =
Sub in Sinx and Cosx
Work out and leave in surd form
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:
Find the exact value of:
7B
𝑐𝑜𝑠𝑥=34 180 ˚<𝑥<360 ˚
𝑡𝑎𝑛2𝑥
x
𝐶𝑜𝑠𝑥=𝐴𝑑𝑗𝐻𝑦𝑝
𝐶𝑜𝑠𝑥=34
𝑇𝑎𝑛𝑥=𝑂𝑝𝑝𝐴𝑑𝑗
𝑇𝑎𝑛𝑥=√73
Use Pythagoras’ to find the missing side (ignore negatives)
Cosx is positive so in the range 270 - 360
3
√74
Therefore, Tanx is negative
90 180
270 360y = Cosθ
𝑇𝑎𝑛𝑥=− √73
Sub in Tanx
Work out and leave in surd form90 18
0270 360
y = Tanθ
Tan 2x Tan 2x
𝑇𝑎𝑛2 𝑥=−3√7
Further Trigonometric Identities and their Applications
The double angle formulae allow you to solve more
equations and prove more identities
Prove the identity:
7C
𝑡𝑎𝑛2𝜃≡ 2𝑐𝑜𝑡 𝜃− 𝑡𝑎𝑛𝜃
𝑡𝑎𝑛2𝜃≡ 2𝑡𝑎𝑛 𝜃1−𝑡𝑎𝑛2𝜃
𝑡𝑎𝑛2𝜃≡
2 𝑡𝑎𝑛𝜃𝑡𝑎𝑛𝜃
1𝑡𝑎𝑛𝜃 −
𝑡𝑎𝑛2𝜃𝑡𝑎𝑛𝜃
𝑡𝑎𝑛2𝜃≡ 2𝑐𝑜𝑡 𝜃−𝑡𝑎𝑛𝜃
Divide each part by tanθ
Rewrite each part
Further Trigonometric Identities and their Applications
The double angle formulae allow you to solve more
equations and prove more identities
By expanding:
Show that:
7C
𝑠𝑖𝑛 (2𝐴+𝐴)
Replace A and B
𝑠𝑖𝑛 (3 𝐴)≡3 𝑠𝑖𝑛𝐴−4 𝑠𝑖𝑛3 𝐴
𝑠𝑖𝑛 (𝐴+𝐵 )≡𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵+𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵
𝑠𝑖𝑛 (2𝐴+𝐴 )≡𝑠𝑖𝑛 2𝐴𝑐𝑜𝑠𝐴+𝑐𝑜𝑠2 𝐴𝑠𝑖𝑛𝐴
𝑠𝑖𝑛 (3 𝐴)≡(2 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴)𝑐𝑜𝑠𝐴+(1−2𝑠𝑖𝑛2 𝐴) 𝑠𝑖𝑛𝐴
𝑠𝑖𝑛 (3 𝐴)≡2𝑠𝑖𝑛𝐴𝑐𝑜𝑠2 𝐴+𝑠𝑖𝑛𝐴−2𝑠𝑖𝑛3𝐴
𝑠𝑖𝑛 (3 𝐴)≡2𝑠𝑖𝑛𝐴 (1−𝑠𝑖𝑛2 𝐴)+𝑠𝑖𝑛𝐴−2𝑠𝑖𝑛3𝐴
𝑠𝑖𝑛 (3 𝐴)≡2𝑠𝑖𝑛𝐴−2𝑠𝑖𝑛3𝐴+𝑠𝑖𝑛𝐴−2𝑠𝑖𝑛3𝐴
𝑠𝑖𝑛 (3 𝐴)≡3 𝑠𝑖𝑛𝐴−4𝑠𝑖𝑛3 𝐴
Replace Sin2A and
Cos 2AMultiply
outReplace cos2A
Multiply out
Group like terms
Further Trigonometric Identities and their Applications
The double angle formulae allow you to solve more
equations and prove more identities
Given that:
Eliminate θ and express y in terms of x…
7C
𝑥=3 𝑠𝑖𝑛𝜃 𝑦=3−4 𝑐𝑜𝑠2 𝜃and
𝑥=3 𝑠𝑖𝑛𝜃𝑥3=𝑠𝑖𝑛𝜃
𝑦=3−4 𝑐𝑜𝑠2 𝜃3−𝑦4 =𝑐𝑜𝑠 2𝜃
Divide by 3
Subtract 3, divide by 4Multiply by -1
𝑐𝑜𝑠2𝜃=1−2𝑠𝑖𝑛2𝜃
3−𝑦4 =¿1−2( 𝑥3 )
2
3− 𝑦=¿4−8( 𝑥3 )2
− 𝑦=¿1−8 (𝑥3 )2
𝑦=¿8 (𝑥3 )2
−1
Replace Cos2θ and Sinθ
Multiply by 4
Subtract 3
Multiply by -1
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:
(All trigonometrical parts must be in terms x, rather than 2x)
7C
3𝑐𝑜𝑠 2𝑥−𝑐𝑜𝑠𝑥+2=00 ° ≤ 𝑥≤360 °
3𝑐𝑜𝑠2𝑥−𝑐𝑜𝑠𝑥+2=0
3 (2𝑐𝑜𝑠2𝑥−1)−𝑐𝑜𝑠𝑥+2=0
6𝑐𝑜𝑠2𝑥−3−𝑐𝑜𝑠𝑥+2=0
6𝑐𝑜𝑠2𝑥−𝑐𝑜𝑠𝑥−1=0
Replace cos2x
Multiply out the bracket
Group terms
(3𝑐𝑜𝑠𝑥+1)(2𝑐𝑜𝑠𝑥−1)=0
𝑐𝑜𝑠𝑥=− 13 𝑐𝑜𝑠𝑥=12or
Factorise
90 180
270 360
y = Cosθ1
2− 13
𝑥=𝑐𝑜𝑠−1(− 13 ) 𝑥=𝑐𝑜𝑠−1( 12 )𝑥=109.5 °,250.5 °𝑥=60 °,300 °
𝑥=60 ° ,109.5 ° ,250.5 ° ,300 °
Solve both pairs
Remember to find additional answers!
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:
Can be expressed in the form:
So:
7D
3 𝑠𝑖𝑛𝑥+4𝑐𝑜𝑠𝑥
𝑅𝑠𝑖𝑛(𝑥+α )
𝑅𝑠𝑖𝑛(𝑥+α )¿𝑅𝑠𝑖𝑛𝑥𝑐𝑜𝑠 α+𝑅𝑐𝑜𝑠𝑥𝑠𝑖𝑛 α3 𝑠𝑖𝑛𝑥+4𝑐𝑜𝑠𝑥¿𝑅𝑠𝑖𝑛𝑥𝑐𝑜𝑠 α+𝑅𝑐𝑜𝑠𝑥𝑠𝑖𝑛 α𝑅𝑐𝑜𝑠α=3 𝑅𝑠𝑖𝑛α=4
𝑐𝑜𝑠α= 3𝑅 𝑠𝑖𝑛α= 4
𝑅 α
𝑅
3
4(𝑂𝐻 )( 𝐴𝐻 )So in the triangle, the Hypotenuse is
R…𝑅=√32+42 𝑅=5
𝑐𝑜𝑠α= 3𝑅
𝑐𝑜𝑠α=35
α=𝑐𝑜𝑠− 1 35
α=53.1°
Replace with the expression
Compare each term – they must be equal!
R = 5
Inverse Cos
Find the smallest value in the acceptable range given
3 𝑠𝑖𝑛𝑥+4𝑐𝑜𝑠𝑥¿5sin (𝑥+53.1° )
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 you can express:
In the form:
So:
7D
𝑠𝑖𝑛𝑥−√3𝑐𝑜𝑠𝑥
𝑅𝑠𝑖𝑛(𝑥−α)
𝑅𝑠𝑖𝑛(𝑥−α)¿𝑅𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝛼−𝑅𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝑥−√3𝑐𝑜𝑠𝑥¿𝑅𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝛼−𝑅𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝛼𝑅𝑐𝑜𝑠𝛼=1 𝑅𝑠𝑖𝑛𝛼=√3
𝑅=√12+(√3 )2 𝑅=2
𝑅𝑐𝑜𝑠𝛼=12𝑐𝑜𝑠𝛼=1𝑐𝑜𝑠𝛼=
12
𝛼=𝑐𝑜𝑠−1 12
𝛼=𝜋3
R = 2
Divide by 2
Inverse cos
Find the smallest value in the
acceptable range
Replace with the expression
Compare each term – they must be equal!
𝑠𝑖𝑛𝑥−√3𝑐𝑜𝑠𝑥¿2sin (𝑥− 𝜋3 )
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 you can express:
In the form:
So:
7D
𝑠𝑖𝑛𝑥−√3𝑐𝑜𝑠𝑥
𝑅𝑠𝑖𝑛(𝑥−α)
𝑠𝑖𝑛𝑥−√3𝑐𝑜𝑠𝑥¿2sin (𝑥− 𝜋3 )
Sketch the graph of:𝑠𝑖𝑛𝑥−√3𝑐𝑜𝑠𝑥= Sketch the graph of:2sin (𝑥− 𝜋3 )
π/2 π 3π/2 2π
1
-1
π/2 π 3π/2 2π
1
y=sin 𝑥
y=sin (𝑥− 𝜋3 )
π/2 π 3π/2 2π
1
-1
y=2sin (𝑥− 𝜋3 )
π/34π/3-1
2
-2
Start out with sinx
Translate π/3 units
right
Vertical stretch, scale
factor 2π/34π/3
2sin (− 𝜋3 )¿−√3 At the y-
intercept, x = 0
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:
in the form:
So:
7D
2𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃
𝑅𝑐𝑜𝑠(𝜃−𝛼)
𝑅𝑐𝑜𝑠(𝜃−𝛼)¿𝑅𝑐𝑜𝑠 𝜃𝑐𝑜𝑠𝛼+𝑅𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛼2𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃¿𝑅𝑐𝑜𝑠 𝜃𝑐𝑜𝑠𝛼+𝑅𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛼𝑅𝑐𝑜𝑠𝛼=2 𝑅𝑠𝑖𝑛𝛼=5
Replace with the expression
Compare each term – they must be equal!
𝑅=√22+52 𝑅=√29
𝑅𝑐𝑜𝑠𝛼=2√29𝑐𝑜𝑠𝛼=2𝑐𝑜𝑠𝛼=
2√29
𝛼=𝑐𝑜𝑠−1 2√29
𝛼=68.2
R = √29
Divide by √29
Inverse cos
Find the smallest value in the
acceptable range
2𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃¿√29cos (𝜃−68.2)
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
Solve in the given range, the following equation:
7D
2𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃=30 °<𝜃<360 °
2𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃=√29cos (𝜃−68.2)
√29cos (𝜃−68.2)=30 °<𝜃<360 °
We just showed that the original equation can be rewritten…
Hence, we can solve this equation instead!
−68.2 °<𝜃−68.2<291.2°Remember to
adjust the range for (θ –
68.2)
√29cos (𝜃−68.2)=3
cos (𝜃−68.2)= 3√29
𝜃−68.2=𝑐𝑜𝑠− 1 3√29
𝜃−68.2=56.1,−56.1,303.9
𝜃=12.1,124.3
Divide by √29
Inverse Cos
Remember to work out other values in the
adjusted range
Add 68.2 (and put in order!)
90 180
270 360
y = Cosθ
-90
56.1
-56.1
303.9
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
Find the maximum value of the following expression, and the smallest positive value of θ at
which it arises:
7D
12𝑐𝑜𝑠𝜃+5 𝑠𝑖𝑛𝜃
𝑅𝑐𝑜𝑠(𝜃−𝛼)¿𝑅𝑐𝑜𝑠 𝜃𝑐𝑜𝑠𝛼+𝑅𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛼12𝑐𝑜𝑠𝜃+5 𝑠𝑖𝑛𝜃¿𝑅𝑐𝑜𝑠 𝜃𝑐𝑜𝑠𝛼+𝑅𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛼
𝑅𝑐𝑜𝑠𝛼=12 𝑅𝑠𝑖𝑛𝛼=5
Replace with the expression
Compare each term – they must be equal!
𝑅=√122+52𝑅=13
𝑅𝑐𝑜𝑠𝛼=1213𝑐𝑜𝑠𝛼=12𝑐𝑜𝑠𝛼=
1213
𝛼=𝑐𝑜𝑠−1 1213
𝛼=22.6
R = 13
Divide by 13
Inverse cos
Find the smallest value in the
acceptable range
¿13cos (𝜃−22.6 )
13cos (𝜃−22.6)13(1)𝑀𝑎𝑥=13𝜃−22.6=0𝜃=22.6
Max value of cos(θ - 22.6) =
1Overall maximum
therefore = 13Cos peaks at 0
θ = 22.6 gives us 0
Rcos(θ – α) chosen as it gives us the same
form as the expression
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
7D
𝑎𝑠𝑖𝑛 𝜃±𝑏𝑐𝑜𝑠𝜃
𝑎𝑐𝑜𝑠 𝜃±𝑏𝑠𝑖𝑛𝜃
𝑅𝑠𝑖𝑛 (𝜃±𝛼 )
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!
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’
7E
𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2 𝑠𝑖𝑛( 𝑃+𝑄2 )𝑐𝑜𝑠( 𝑃−𝑄2 )
𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑐𝑜𝑠 (𝑃 −𝑄2 )
𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃 −𝑄2 )
You get given all these in the formula booklet!
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’
7E
𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2 𝑠𝑖𝑛( 𝑃+𝑄2 )𝑐𝑜𝑠( 𝑃−𝑄2 )
𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑐𝑜𝑠 (𝑃 −𝑄2 )
𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃 −𝑄2 )
Using the formulae for Sin(A + B) and Sin (A – B), derive the result that:
𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2 𝑠𝑖𝑛( 𝑃+𝑄2 )𝑐𝑜𝑠( 𝑃−𝑄2 )
𝑆𝑖𝑛 ( 𝐴+𝐵 )=𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵+𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵𝑆𝑖𝑛 ( 𝐴−𝐵 )=𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵−𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵𝑆𝑖𝑛 ( 𝐴+𝐵 )+𝑠𝑖𝑛 (𝐴−𝐵)=2 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵
𝐴+𝐵=𝑃𝐴−𝐵=𝑄2 𝐴=𝑃+𝑄𝐴=
𝑃+𝑄2
𝐴+𝐵=𝑃𝐴−𝐵=𝑄2𝐵=𝑃−𝑄𝐵=
𝑃 −𝑄2
Add both sides together (1 + 2)
1)
2)
Let (A+B) = P Let (A-B) = Q
1)
2)
1)
2)1 + 2
Divide by
2
1 - 2
Divide by
2
𝑆𝑖𝑛 ( 𝐴+𝐵 )+𝑠𝑖𝑛 (𝐴−𝐵)=2 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵
𝑆𝑖𝑛𝑃+𝑆𝑖𝑛𝑄¿2 𝑠𝑖𝑛( 𝑃+𝑄2 ) ( 𝑃−𝑄2 )𝑐𝑜𝑠
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’
7E
𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2 𝑠𝑖𝑛( 𝑃+𝑄2 )𝑐𝑜𝑠( 𝑃−𝑄2 )
𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑐𝑜𝑠 (𝑃 −𝑄2 )
𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃 −𝑄2 )
Show that:
𝑠𝑖𝑛105−𝑠𝑖𝑛15= 1√2
𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑠𝑖𝑛105−𝑠𝑖𝑛15=2𝑐𝑜𝑠( 105+152 )𝑠𝑖𝑛(105−152 )𝑠𝑖𝑛105−𝑠𝑖𝑛15=2𝑐𝑜𝑠60 𝑠𝑖𝑛 45
𝑠𝑖𝑛105−𝑠𝑖𝑛15=¿2×12
1√2×
𝑠𝑖𝑛105−𝑠𝑖𝑛15=¿1√2
P = 105 Q
= 15Work out the fraction parts
Sub in values for Cos60 and
Sin45Work out the
right hand side
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’
7E
𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2 𝑠𝑖𝑛( 𝑃+𝑄2 )𝑐𝑜𝑠( 𝑃−𝑄2 )
𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑐𝑜𝑠 (𝑃 −𝑄2 )
𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
Solve in the range indicated:𝑠𝑖𝑛4 𝜃−𝑠𝑖𝑛3 𝜃=0 0≤ 𝜃≤𝜋
𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑠𝑖𝑛4 𝜃−𝑠𝑖𝑛3 𝜃=2𝑐𝑜𝑠( 4 𝜃+3𝜃2 )𝑠𝑖𝑛( 4 𝜃−3𝜃2 )
𝑠𝑖𝑛4 𝜃−𝑠𝑖𝑛3 𝜃=2𝑐𝑜𝑠( 7𝜃2 )𝑠𝑖𝑛(𝜃2 )2𝑐𝑜𝑠( 7 𝜃2 )𝑠𝑖𝑛( 𝜃2 )=0𝑐𝑜𝑠( 7 𝜃2 )=07𝜃2 =𝑐𝑜𝑠− 100≤ 7𝜃2 ≤
7𝜋2
0≤ 𝜃≤𝜋
7𝜃2 =
𝜋2 ,3𝜋2 ,5𝜋2 ,7𝜋2
𝜃=𝜋7 ,3𝜋7 ,5𝜋7 ,𝜋
P = 4θ Q = 3θWork 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
Adjust the range
π/2 π 3π/2 2πy = Cosθ
0
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’
7E
𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2 𝑠𝑖𝑛( 𝑃+𝑄2 )𝑐𝑜𝑠( 𝑃−𝑄2 )
𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑐𝑜𝑠 (𝑃 −𝑄2 )
𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
Solve in the range indicated:𝑠𝑖𝑛4 𝜃−𝑠𝑖𝑛3 𝜃=0 0≤ 𝜃≤𝜋
𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑠𝑖𝑛4 𝜃−𝑠𝑖𝑛3 𝜃=2𝑐𝑜𝑠( 4 𝜃+3𝜃2 )𝑠𝑖𝑛( 4 𝜃−3𝜃2 )
𝑠𝑖𝑛4 𝜃−𝑠𝑖𝑛3 𝜃=2𝑐𝑜𝑠( 7𝜃2 )𝑠𝑖𝑛(𝜃2 )2𝑐𝑜𝑠( 7 𝜃2 )𝑠𝑖𝑛( 𝜃2 )=0𝑠𝑖𝑛( 𝜃2 )=0𝜃2 =𝑠𝑖𝑛− 100≤ 𝜃2 ≤
𝜋2
0≤ 𝜃≤𝜋
𝜃2 =0
𝜃=0
P = 4θ Q = 3θWork 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
Adjust the range
π/2 π 3π/2 2πy = Sinθ
0
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’
Prove that:
7E
𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2 𝑠𝑖𝑛( 𝑃+𝑄2 )𝑐𝑜𝑠( 𝑃−𝑄2 )
𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑐𝑜𝑠 (𝑃 −𝑄2 )
𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑠𝑖𝑛 (𝑥+2 𝑦 )+𝑠𝑖𝑛 (𝑥+𝑦 )+𝑠𝑖𝑛𝑥𝑐𝑜𝑠 (𝑥+2 𝑦 )+𝑐𝑜𝑠 (𝑥+𝑦 )+𝑐𝑜𝑠𝑥
=𝑡𝑎𝑛 (𝑥+ 𝑦 )
𝑠𝑖𝑛 (𝑥+2 𝑦 )+𝑠𝑖𝑛 (𝑥+𝑦 )+𝑠𝑖𝑛𝑥In the numerator:
𝑠𝑖𝑛 (𝑥+2 𝑦 )+𝑠𝑖𝑛𝑥
𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2 𝑠𝑖𝑛( 𝑃+𝑄2 )𝑐𝑜𝑠( 𝑃−𝑄2 )
𝑠𝑖𝑛 (𝑥+2 𝑦 )+𝑠𝑖𝑛𝑥=2𝑠𝑖𝑛( 𝑥+2 𝑦+𝑥2 )𝑐𝑜𝑠( 𝑥+2 𝑦−𝑥
2 )¿2 𝑠𝑖𝑛 (𝑥+ 𝑦 )𝑐𝑜𝑠𝑦
¿2 𝑠𝑖𝑛 (𝑥+𝑦 )𝑐𝑜𝑠𝑦+𝑠𝑖𝑛 (𝑥+𝑦 )
¿ 𝑠𝑖𝑛 (𝑥+ 𝑦 )(2𝑐𝑜𝑠𝑦+1)
Ignore sin(x + y) for now…
Use the identity for adding 2 sines
P = x + 2y Q = x
Simplify Fractions
Bring back the sin(x + y) we ignored earlier
Factorise
Numerator: 𝑠𝑖𝑛 (𝑥+ 𝑦 )(2𝑐𝑜𝑠𝑦+1)
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’
Prove that:
7E
𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2 𝑠𝑖𝑛( 𝑃+𝑄2 )𝑐𝑜𝑠( 𝑃−𝑄2 )
𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑐𝑜𝑠 (𝑃 −𝑄2 )
𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑠𝑖𝑛 (𝑥+2 𝑦 )+𝑠𝑖𝑛 (𝑥+𝑦 )+𝑠𝑖𝑛𝑥𝑐𝑜𝑠 (𝑥+2 𝑦 )+𝑐𝑜𝑠 (𝑥+𝑦 )+𝑐𝑜𝑠𝑥
=𝑡𝑎𝑛 (𝑥+ 𝑦 )
𝑐𝑜𝑠 (𝑥+2 𝑦 )+𝑐𝑜𝑠 (𝑥+𝑦 )+𝑐𝑜𝑠𝑥In the
denominator:
𝑐𝑜𝑠 (𝑥+2 𝑦 )+𝑐𝑜𝑠𝑥
𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑐𝑜𝑠 (𝑃 −𝑄2 )
𝑐𝑜𝑠 (𝑥+2 𝑦 )+𝑐𝑜𝑠𝑥=2𝑐𝑜𝑠( 𝑥+2 𝑦+𝑥2 )𝑐𝑜𝑠( 𝑥+2 𝑦−𝑥
2 )¿2𝑐𝑜𝑠 (𝑥+ 𝑦 )𝑐𝑜𝑠𝑦
¿2𝑐𝑜𝑠 (𝑥+𝑦 )𝑐𝑜𝑠𝑦+𝑐𝑜𝑠 (𝑥+𝑦 )
¿𝑐𝑜𝑠 (𝑥+ 𝑦 )(2𝑐𝑜𝑠𝑦+1)
Ignore cos(x + y) for now…
Use the identity for adding 2 cosines
P = x + 2y Q = x
Simplify Fractions
Bring back the cos(x + y) we ignored earlier
Factorise
Numerator: 𝑠𝑖𝑛 (𝑥+ 𝑦 )(2𝑐𝑜𝑠𝑦+1)
Denominator:
𝑐𝑜𝑠 (𝑥+ 𝑦 )(2𝑐𝑜𝑠𝑦+1)
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’
Prove that:
7E
𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2 𝑠𝑖𝑛( 𝑃+𝑄2 )𝑐𝑜𝑠( 𝑃−𝑄2 )
𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠( 𝑃+𝑄2 )𝑐𝑜𝑠 (𝑃 −𝑄2 )
𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )
𝑠𝑖𝑛 (𝑥+2 𝑦 )+𝑠𝑖𝑛 (𝑥+𝑦 )+𝑠𝑖𝑛𝑥𝑐𝑜𝑠 (𝑥+2 𝑦 )+𝑐𝑜𝑠 (𝑥+𝑦 )+𝑐𝑜𝑠𝑥
=𝑡𝑎𝑛 (𝑥+ 𝑦 )
Numerator: 𝑠𝑖𝑛 (𝑥+ 𝑦 )(2𝑐𝑜𝑠𝑦+1)
Denominator:
𝑐𝑜𝑠 (𝑥+ 𝑦 )(2𝑐𝑜𝑠𝑦+1)
𝑠𝑖𝑛 (𝑥+2 𝑦 )+𝑠𝑖𝑛 (𝑥+𝑦 )+𝑠𝑖𝑛𝑥𝑐𝑜𝑠 (𝑥+2 𝑦 )+𝑐𝑜𝑠 (𝑥+𝑦 )+𝑐𝑜𝑠𝑥
¿𝑠𝑖𝑛(𝑥+𝑦)(2𝑐𝑜𝑠𝑦+1)𝑐𝑜𝑠(𝑥+𝑦 )(2𝑐𝑜𝑠𝑦+1)
¿𝑠𝑖𝑛(𝑥+ 𝑦)𝑐𝑜𝑠(𝑥+𝑦 )
¿ 𝑡𝑎𝑛 (𝑥+ 𝑦 )
Replace the numerator and denominator
Cancel out the (2cosy + 1)
brackets
Use one of the identities from C2
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