Download pptx - Further Trigonometric Identities

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



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



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)



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



formulae

7A

Cos(A - B) ≡ CosACosB + SinASinBCos(A + B) ≡ CosACosB - SinASinBSin(A + B) ≡ SinACosB + CosASinBSin(A - B) ≡ SinACosB - CosASinB



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



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!



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



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



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…


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)


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




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)




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




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




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




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




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


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




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




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




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!


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° )




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



𝑠𝑖𝑛𝑥−√3𝑐𝑜𝑠𝑥¿2sin (𝑥− 𝜋3 )




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




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



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


acceptable range

2𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃¿√29cos (𝜃−68.2)




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




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



𝑅=√122+52𝑅=13

𝑅𝑐𝑜𝑠𝛼=1213𝑐𝑜𝑠𝛼=12𝑐𝑜𝑠𝛼=

1213

𝛼=𝑐𝑜𝑠−1 1213

𝛼=22.6

R = 13

Divide by 13

Inverse cos


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




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!


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!




7E





Using the formulae for Sin(A + B) and Sin (A – B), derive the result that:


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




7E





Show that:

𝑠𝑖𝑛105−𝑠𝑖𝑛15= 1√2


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




7E




𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛( 𝑃+𝑄2 )𝑠𝑖𝑛( 𝑃−𝑄2 )

Solve in the range indicated:𝑠𝑖𝑛4 𝜃−𝑠𝑖𝑛3 𝜃=0 0≤ 𝜃≤𝜋


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




7E





Solve in the range indicated:𝑠𝑖𝑛4 𝜃−𝑠𝑖𝑛3 𝜃=0 0≤ 𝜃≤𝜋


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




Prove that:

7E





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


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)




Prove that:

7E







𝑐𝑜𝑠 (𝑥+2 𝑦 )+𝑐𝑜𝑠 (𝑥+𝑦 )+𝑐𝑜𝑠𝑥In the

denominator:

𝑐𝑜𝑠 (𝑥+2 𝑦 )+𝑐𝑜𝑠𝑥


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


Denominator:

𝑐𝑜𝑠 (𝑥+ 𝑦 )(2𝑐𝑜𝑠𝑦+1)




Prove that:

7E








Denominator:

𝑐𝑜𝑠 (𝑥+ 𝑦 )(2𝑐𝑜𝑠𝑦+1)


¿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