30: Trig addition formulae © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

30: Trig addition 30: Trig addition formulaeformulae

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules

Trig Addition Formulae

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Module C3

Edexcel

Module C4

AQA

MEI/OCROCR


1

Does ? 60sin30sin)6030sin(

and

371

So, 60sin30sin)6030sin(

We cannot simplify the brackets as we do in algebra because they don’t mean multiply.

90sin)6030sin(l.h.s. =

2

3

2

160sin30sin

r.h.s. =


BBAA cossin,cos,sin and

The result, however, is true for any size of angles.

We’ll find the formula for where A and B are in degrees and where

)sin( BA

90 BA

The proof is complicated but you are not expected to remember it !

However, can be written in terms of )sin( BA


Consider this rectangle

Tilt the rectangle through an angle A.

Let PR = 1

We can now find

using a right angled triangle

)sin( BA

1

Ba

b

R

Q

S

P

R

Q

P

S1

a

b

BA


Q

P

S1

BA

N

a

b

R

h

1

h )sin( BA

h

A

)90( A


Q

P

S1

BA

N

a

b

R

h

1

h

h

)sin( BA

h =But

A


Q

P

S1

BA

N

a

b

R

h

1

h

h

)sin( BA M

NM + MRh =But

A


Q

P

S1

BA

N

a

b

R

1

h

h

)sin( BA

NM + MR = TQ +T

h =But MR

A

M


Q

P

S1

BA

N

a

b

R

= TQ + MR

But, TQ = Aa sin

1

h

h

)sin( BA

h =But NM + MR

M

T

A


Q

P

S1

BA

N

a

b

R

T

M

= TQ + MR

But, TQ = Aa sin and MR =

1

h

h

)sin( BA

h =But NM + MR

A

Abcos


Q

P

S1

BA

N

a

b

R

T

M

= TQ + MR

But, TQ = Aa sin Abcosand MR =

1

h

h

)sin( BA

AbAah cossin

h =But NM + MR

A

h


AbAah cossin

Q

P

S1

BA

N

a

b

R

M

T

1

h

h

)sin( BA

Also Bcos1a

a


Q

P

S1

BA

N

a

b

R

M

T

h

)sin( BA

Also Bcos1a

a and

Bsin b1b

AbAah cossin

1

h


Q

P

S1

BA

N

a

b

R

M

h

)sin( BA

ABABh cossinsincos So,

Also Bcos1a

a and

Bsin b

hBA )sin( BABABA sincoscossin)sin(

1b

AbAah cossin

1

h

BABAh sincoscossin

T


xy sinxy cos

BABABA sincoscossin)sin(

Before we find the other addition formulae we need to notice 4 relationships between some of the trig ratios.

)1(

)cos( B

B B

)sin( B

BB

Bcos Bsin


BABABA sincoscossin)sin( )1(

xy cos

Acos

A

xy sin

)90sin( A

90 - A

)90sin(cos AA



xy cosxy sin


)90cos( A

90 - A

)90cos(sin AA

Asin

A



Now we can easily find 5 more addition formulae Replace B by (–B) in (1) : )sin(cos)cos(sin))(sin( BABABA

We now have

BB cos)cos( BB sin)sin( )90sin(cos AA AA sin)90cos(

)sin(coscossin)sin( BABABA BABABA sincoscossin)sin( )2(

BABABA sincoscossin)sin( and )1(

Trig Addition Formulae We now

haveBB cos)cos( BB sin)sin(

)90sin(cos AA AA sin)90cos(

BABABA sincoscossin)sin( )1(and


Replace A by ( 90 A ) in (2) :

BABABA sin)90cos(cos)90sin()90sin(

)3(

BABABA sinsincoscos))(90sin( BABABA sinsincoscos)cos(






)3(BABABA sinsincoscos)cos(

Exercise: Use (3) to find a formula for )cos( BA






Replace B by ( B ) in (3) :

)sin(sin)cos(cos)cos( BABABA

)sin(sincoscos)cos( BABABA BABABA sinsincoscos)cos( )4(

BABABA sinsincoscos)cos( )3(

)2(






BABABA sinsincoscos)cos(


)3(

These formulae are true for all values of A and B so they are identities. They should be written with identity signs.








)3(


BABA

BABA

sinsincoscos

sincoscossin

)5(

)cos(

)sin()tan(

BA

BABA

Divide numerator and denominator by :

BAcoscos

BA

BA

tantan1

tantan

)tan( BA

)tan( BA Formula for :

1

BAcoscos

BAcoscos BAcoscos

BAcoscos BAcoscos

BABA

BABA

sinsincoscos

sincoscossin

BAcoscos BAcoscos

BAcoscos


BA

BA

tantan1

tantan

)5()tan( BA

Exercise: Using this formula, or otherwise, find a formula for )tan( BA Solution:

Replace B by ( B ) in (5) :

)tan(tan1

)tan(tan)tan(

BA

BABA

)6(BA

BABA

tantan1

tantan)tan(

By dividing by we get

)sin()sin( BB )cos()cos( BB BB tan)tan(

so,

OR: Use the method used to find formula (5)


SUMMARY

BB cos)cos( BB sin)sin(

BA

BABA

tantan1

tantan)tan(



You need to remember the following results.

Check whether the addition formulae are in your formulae booklets. If so, they may be written as

Notice that the cos formulae have opposite signs on the 2 sides.

Use both top signs in a formula or both bottom signs.


Using the Addition Formulae

Solution:

)4590sin(135sin

You will need your formulae booklets for the rest of this presentation and all the remaining

trig work.

BABABA sincoscossin)sin( Using

45sin90cos45cos90sin

02

11

2

1

We can rationalise the surd by multiplying numerator and denominator by2 2

2

e.g. 1 Find the exact value of simplifying the answer

135sin

)4590sin(


Using the Addition Formulae e.g. 2 Prove the following: xyyxyx cossin2)sin()sin( Proof:

l.h.s. )sin()sin( yxyx )sincoscos(sin)sincoscos(sin yxyxyxyx

yxyxyxyx sincoscossinsincoscossin

yx sincos2... shr

( formulae (1) and (2) )

Trig Addition Formulae Exercise

s

(a)

1. Simplifying the answers as much as possible, find exact values for:

75cos (b) 105sin (c) 15tan

2. Prove the following:yxyxyx sinsin2)cos()cos( (a)

(b) )sin()tan(tancoscos yxyxyx

(c) yxyx

yxtantan

coscos

)sin(

You can assume some, or all, of the following: ,45sin45cos

21

2360sin30cos

2130sin60cos and


1(a)

30sin45sin30cos45cos)3045cos(75cos

(b) 45sin60cos45cos60sin)4560sin(105sin

Solutions:

,45sin45cos2

1 2360sin30cos

2130sin60cos and

2

1

2

1

2

3

2

1

22

13

4

)13(2

We can multiply numerator and denominator by to rationalise the surds.

2

2

1

2

1

2

1

2

3

22

13

4

)13(2


2

)32(2

)13)(13(

)13)(13(15tan

Solutions: ,45sin45cos

21

2360sin30cos

2130sin60cos and

145cos

45sin45tan

212

1

313

1

1

115tan

13

1315tan

Multiply numerator and denominator by

3Rationalise the surds 13

1323

(c)30tan45tan1

30tan45tan)3045tan(

15tan

3

1

30cos

30sin30tan

23

21

an

d


yxyxyx sinsin2)cos()cos( 2(a) Prove

Solutions:

Proof: l.h.s. )cos()cos( yxyx

)sinsincos(cos)sinsincos(cos yxyxyxyx

yxyxyxyx sinsincoscossinsincoscos

yx sinsin2

... shr



Solutions:

Proof:

l.h.s. )tan(tancoscos yxyx

y

y

x

xyx

cos

sin

cos

sincoscos

... shr

(b) )sin()tan(tancoscos yxyxyx

y

yyx

x

xyx

cos

sincoscos

cos

sincoscos

yxxy sincossincos )sin( yx using formula

(2):

A

AA

cos

sintan


yx

yx

yx

yx

coscos

sincos

coscos

cossin

Solutions:

Proof:

yx

yxyx

coscos

sincoscossin

... shr

(c) yxyx

yxtantan

coscos

)sin(

using formula (1):

l.h.s. yx

yx

coscos

)sin(

yx tantan A

AA

cos

sintan



The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

Trig Addition Formulae SUMMAR

Y

BB cos)cos( BB sin)sin(

BA

BABA

tantan1

tantan)tan(



You need to remember the following results.

Check whether the addition formulae are in your formulae booklets. If so, they may be written as

Notice that the cos formulae have opposite signs on the 2 sides.


Using the Addition Formulae e.g. Prove the following:

xyyxyx cossin2)sin()sin( Proof:

l.h.s. )sin()sin( yxyx )sincoscos(sin)sincoscos(sin yxyxyxyx

yxyxyxyx sincoscossinsincoscossin

yx sincos2... shr


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30: Trig addition formulae © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules