31 double angle formulae

8/8/2019 31 Double Angle Formulae

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31: Double Angle31: Double Angle

FormulaeFormulae

Christine Crisp

Teach A Level MathsTeach A Level Maths

Vol. 2: A2 Core ModulesVol. 2: A2 Core Modules


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Double Angle Formulae

)5(BA

BA

tantan1

tantan

|)tan( BA

BABABAsincoscossin)sin(

|)1(

BABABA sinsinosos)os( | )3(

The double angle formulae are used to express an

angle such as 2A in terms of A.

We derive the formulae from 3 of the additionformulae.

What do we need to do to obtain formulae for

?tancos,sin AAA and

ANS: Replace B by A in(1),(3) and (5).


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)3(

B

B

tantan1

tantan

|)tan( B

Finally,

These formulae are probably NOT in your formulaebook but please check!

If they are not there, you need to remember them.

N.B. The formulae can be derived quite quickly fromthe addition formulae. However, you may notrecognise the need to use them unless you havememorised them.

AA

AAA

tantan1

tantan2tan

|

A

AA

2tan1

tan22tan

|


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N.B. The formulae link any angle with double the

angle.For example, they can be used for

x2 xand

x2

xand

y33 y

and

We use them to solve equations

to prove other identities

to integrate some functions

U4 and U2


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Ive then used the method I used in AS to find the

other solutions.

Some of you will use a different method and if you

are happy with it, dont change.Whichever method you use, make sure you get allthe required solutions!

In the following examples, once Ive reduced theequations to simple ones, Ive used a calculator tofind the principal values.

So, for the equations andI have sketched graphs.

c!sin c!cos

Solving Equations


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e.g. 1 Solve the following equation:

;0cossin ! xx Q3600 ee x

Solution:When solving trig equations we must aim for onlyone trig ratio of one angle.

Here we have two trig ratios ( cos and sin )

We can use AAA cinin | x2sin

0cos2sin ! xx

becomes 0coscossin2 ! xxx

We still have 2 trig ratios but is a commonfactor

xc

and two angles ( xand 2x).


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Solution:0coscossin2 ! xxx

0)1sin2(cos !xx

0cos !x or 01si !x

21sin !x

We now have 2 simple trig equations which we cansolve.


;0cossin ! xx Q3600 ee xDont cancel will give solutons. )

0cos !x

xcos


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xy cos!

!y

0cos !x Q

90!x ( principal value )

QQ270,90!x

Q90

Q270

e.g. 1 Solve the following equation:e.g. 1 Solve the following equation:



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xy sin!

Q30!x ( principal value )

QQ150,0!x

21!y

Q30

Q150

QQQQ270,150,90,30!xANS:


21si !x




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;3sin5cos ! xx T20 ee x

Solution:

Again we have 2 trig ratios and 2 angles.

AAA 22 sis2cos |

1cos|

A

A2

si21 |

By choosing the 3rd version . . .

We use the double angle formula for buhave 3 versions to choose from:

A2cos


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Solution:

Again we have 2 trig ratios and 2 angles.

AAA 22 sicos2cos |

1cos|

A

A2

si21 |

we will reduce the equation to 1 angle and 1 trig ratio.By choosing the 3rd version . . .

)2(

)2( a

)2( b

We use the double angle formula for buhave 3 versions to choose from:

A2cos


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Solution:

So, ;3s n52cos xxQuadraticequation3sin5sin2

2 xx

25202

ss

35212

ss )sin( xs !

)2)(12(0 ! ss

21i x 2si !xor

( no solutions )


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xy sin!


(b) ;3sin5cos ! xx T20 ee x

!21si x

21

y

67T

630(

TQx

( outside therequired range )

6

T

6

TTx ,

7T

611

62 TTT

6

11,

6

7 TTxANS:

6

11T


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e.g. 3 Solve the following equation giving the answer

correct to the nearest degree:;tan5tan UU ! QQ 180180 ee U

Solution: Use

A

AA

2

tan1

tan22tan

|

UU tan52tan2 !

U

U

Utan5

tan1

tan22

2!

tt

t5

1

42!

)1(54

2

ttt !


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;tan5tan UU ! QQ 180180 ee U


correct to the nearest degree:

)1(542ttt !

5

1

0 stt or

3554 ttt !

053 ! tt

0)5( 2 !tt


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;tan5tan UU ! QQ 180180 ee U

0tanU or


correct to the nearest degree:

447

5

1ta }

450 y

QQ180,0,180U

QQQ15618024

,24Q

!U

,24Q!UQQQ

15618024 !QQQQQQ

180156,4,0,4,156,180 !ANS:

Q24 Q156


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We can now use the double angle formulae to prove

other identities. The general method of proof isthe same as you have met before.

Proof:

U

UUU

sin

)1cos21(cossin22

|

l.h. . =

)coscos UU|

U3

cos4|

UU

UU 3

cos4si

)2cos1(2si

|

e.g. 1 Prove that

)0(sin {U

= r.h. .

( Double angle

formulae )

U

UU

sin

)2cos1(2sin


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Which double angle formula can we use to changethe function so that it can be integrated?

e.g. Find d2

si

ANS: AA2

sin212cos )2( b

Rearranging the formula:

AA2

si212cos |

)2cos1(si212 AA |

So, ! dxxdxx )2os1(sin 212

2

2sin x

!2

1x C


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The previous example is an important application of

a double angle formula.The next 2 are also important. Try them yourself.

Exercise

1. Find dxx2

cos

2. Find dcossin

( This one is a product. Why cant it beintegrated by parts ? )


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Double Angle FormulaeExercise

1. Find

dxx

2s

Solution: 1coscos | AA

AA2

21 co2co1 |

! dxxdxx )2cos1(cos 212So,

Cx

x

!

2

2sin

21


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2. Find dxxxcossin

AAA cossin22sin |Solution:

AAA cossin2sin2

1 |

! dxxdxxx 2sincossin 21

So,

Cx

!

2

2co

2

1

Cx

!4

cos


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SUMMARY

A2cos

The rearrangements of the double angle formulaefor are

)2cos1cos212

AA |

)2cos1(sin2

12 AA |

They are important in integration so you shouldeither memorise them or be able to obtain themvery quickly.

31 double angle formulae

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