higher mathematics unit 2 topic 3.2 compound … 13 sina = 4 5 cosa = 3 5 sin(a + b) = sinacosb +...

HIGHER MATHEMATICS

Unit 2 – Topic 3.2

Compound Angle Formula

REMINDERS Let OP = r & POX = A

O x

y

P(x,y)

sinA =y

r

sin(90 - A) = xr = cosA

cos(90 - A) =yr

= sinA

P’(y ,x)

Now reflect OP in the line y = x

A

So OP = OP’ = r

& P’OY = A

A

P’OX = 90 - A

This gives the following

cosA =x

r

tanA =y

x

x

ry

So OP’ = r & P’OX = -A

sin(-A) =

cos(-A) =

-yr = -sinA

xr

= cosA

Ox

y

A

P(x,y)

r

Let OP = r & POX = A

P’(x,-y)

-A

r

Now reflect OP in the x-axis

So OP’ = r & P’OX = 180 - A

sin(180-A) =

cos(180-A) =

-yr = sinA

-xr

= -cosA

Let OP = r& POX = A

Now reflect OP in the y-axis

Ox

y

A

P(x,y)P’(-x,y)

A

Consider:

So Tan A =Sin ACos A

Ox

y

A

P(x,y)

r Sin ACos A

=

yrxr

= ÷yr

xr

= xyr

rx

= yrrx

= yx

= Tan A

Consider:

So Sin2A + Cos2A

Ox

y

A

P(x,y)

r= y2

r2

x2

r2+

= 1

So Sin2A + Cos2A = 1

= y2 + x2

r2

= r2

r2

SUMMARY

sin(90 - A) = cosA cos(90 - A) = sinA

sin(-A) = -sinA cos(-A) = cosA

sin(180 - A) = sinA cos(180 - A)= -cosA

sinAcosA

= tanA sin2A + cos2A = 1

Exercises from MIA book:

Page 153 Ex 1 All Qu.

Exercises from Heinemann book:

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Formulae for cos(A + B) and cos(A – B)

Let the circle’s radius be 1 and Q be the point (x , y)

OA

Q

cosA =xr

x = rcosAcosA

So Q is (cosA , sinA)

sinA =yr

y = rsinAsinA-B

P Consider the point P(x , y)

cos(-B) =xr

x = rcos(-B)cos(-B) So P is(cosB , -sinB)x = cosB

sin(-B) =yr

y = rsin(-B)sin(-B)x = -sinB

Use the distance formula to find PQ

P(cosB , -sinB) & Q(cosA , sinA)

PQ2 = (cosA – cosB)2 + (sinA + sinB)2

PQ2 = 1 + 1 – 2cosAcosB + 2sinAsinB

PQ2 = 2 – 2(cosAcosB - sinAsinB)

Now rotate ΔPOQ

anti-clockwise through angle B

PQ2 = cos2A – 2cosAcosB + cos2B + sin2A + 2sinAsinB + sin2B

PQ2 = cos2A + sin2A + sin2B + cos2B – 2cosAcosB + 2sinAsinB

O -B

P

A

Q

A+BP’

Q’cos(A+B) =

xr

x = rcos(A+B)cos(A+B)

sin(A+B) =yr

y = rsin(A+B)sin(A+B)

Q’ is (x , y) and P’ is (1 , 0)

Use the distance formula to find P’Q’

P’(1 , 0) & Q’(cos(A+B) , sin(A+B))

P’Q’2 = (1 – cos(A+B))2 + (0 – sin(A+B))2

P’Q’2 = 2 – 2cos(A+B)

P’Q’2 = 1 – 2cos(A+B) + cos2(A+B) + sin2(A+B)

P’Q’2 = 1 – 2cos(A+B) + 1

Remember that PQ = P’Q’ so PQ2 = P’Q’2

2 – 2cos(A+B) = 2 – 2(cosAcosB - sinAsinB)

We get another expansion by replacing B with -B

Cos(A + (-B)) = CosACos(-B) – SinASin(-B)

Cos(A + B) = CosACosB - SinASinBGiving:

Cos(A - B) = CosACosB + SinASinBGiving:

EXAMPLE 1 Acute angles P and Q are such that

sinP = 1213

and cosQ = 35

Show that cos(P - Q) = 6365

P

Q

1213

3

5

5

4

sinP =1213

cosP =513

sinQ =45

cosQ =35

Cos(P - Q) = CosPCosQ + SinPSinQ

= +513

35

1213

45

x x

= +1565

4865

= as required6365

EXAMPLE 2 Acute angles X and Y are such that

sinX = 817

and tanY = 34

Show that cos(X + Y) = 3685

X

Y

817

4

5

15

3

sinX =817

cosX =1517

sinY =35

cosY =45

Cos(X + Y) = CosXCosY - SinXSinY

= -1517

45

817

35

x x

= -6085

2485

= as required3685

EXAMPLE 3 Use the formula for cos(A + B) to simplify cos(360 + y)

cos(A + B) = cosAcosB - sinAsinB

cos(360 + y) = cos360ocosy - sin360osiny

cos(360 + y) = (1)cosy - (0)siny

cos(360 + y) = cosy

EXAMPLE 4 Using the fact that 105 = 60 + 45

Show that the EXACT VALUE of

cos105o is 2 - 6 4

cos(105) = cos(60 + 45)

= cos60cos45 - sin60sin45

= x - x

= -

cos(A + B) = cosAcosB - sinAsinB

30º

60º 21

√3

45º

45º

1

1√2

12

√32

1√2

1√2

12√2

√32√2

= 1 - √32√2

Trying to get

2 - 6 4

x2

2

= (1 - √3) √2

2√2√2

= √2 - √6

4As required




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Formulae for sin(A + B) and sin(A – B)

Remember that sinX = cos(90 – X)

So sin(A + B) = cos(90 – (A + B))

= cos(90 – A - B)

= cos((90 – A) - B)

= cos(90–A)cosB + sin(90-A)sinB

= sinAcosB + cosAsinB

Sin(A + B) = SinACosB + CosASinBGiving:

Sin(A + (-B)) = SinACos(-B) + CosASin(-B)

Now what happens when we replace B with -B

Sin(A - B) = SinACosB - CosASinBGiving:

EXAMPLE 5 Acute angles X and Y are such that

sinX = 1√5

and tanY = 3

Find the exact value of sin(X - Y)

X

Y

1√5

1

√10

2

3

sinX =1√5

cosX =2√5

sinY =3

√10cosY =

1√10

sin(X - Y) = sinXcosY - cosXsinY

= -1√5

1√10

2√5

3√10

x x

= -1√50

6√50

= -5√50

= -55√2

= -1√2

EXAMPLE 6

For the diagram, express sin(ABC) as a fraction.

A B

C

D

4

3

13

aobo

5

12

sinABC = sin(a + b)

sinb =1213

cosb =513

sina =45

cosa =35

sin(a + b) = sinacosb + cosasinb

= +45

513

35

1213

x x

= +2065

3665

= 5665




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A Few More Formulae.

Replace B with A

sin(A + A) = sinAcosA + cosAsinA

sin2A = 2sinAcosA

So sin2A = 2sinAcosA

sin(A + B) = sinAcosB + cosAsinB

Replace B with A

cos(A + A) = cosAcosA - sinAsinA

cos2A = cos2A – sin2A

Remember that cos2A + sin2A = 1

So cos2A = 1 - sin2A

& sin2A = 1 - cos2A

Cos(A + B) = cosAcosB - sinAsinB


= cos2A – (1 - cos2A)

= 2cos2A – 1

Hence:

So cos2A = 2cos2A – 1


= 1 - sin2A – sin2A

= 1 – 2sin2A

And:

So cos2A = 1 – 2sin2A

These also give:

So cos2A = ½(1 + cos2A)

So sin2A = ½(1 - cos2A)

SUMMARY Cos(A + B) = CosACosB - SinASinB

Cos(A - B) = CosACosB + SinASinB

Sin(A + B) = SinACosB + CosASinB

Sin(A - B) = SinACosB - CosASinB

sin2A = 2sinAcosA

cos2A = 2cos2A – 1

cos2A = 1 – 2sin2A

cos2A = ½(1 + cos2A)

sin2A = ½(1 - cos2A)


EXAMPLE 7 Given that 0 < A < 90 and that

sinA = , obtain exact values for

sin2A and cos2A

A

817

15

cosA =1517

sin2A = 2sinAcosA

= 2 x x817

1517

= 240289

817

cos2A = 2cos2A - 1

= 2 x - 1152

172

= 2 x - 1225289

= -450289

289289

= 161289

When finding cos2A you can choose any of the 3 formulae that you want

EXAMPLE 8 Given that 0 < A < 90 and that

tanX = , obtain exact values for

sin2X, cos2X and hence sin4X

X

1

√15

4

sinX =14

sin2X = 2sinXcosX

= 2 x x14

√154

= 2√1516

1√15

cos2X = 2cos2X - 1

= 2 x - 1152

42

= 2 x - 11516

= -3016

1616

= 1416

When finding cos2A you can choose any of the 3 formulae that you want

cosX =√154

= √158

= 78

sin4X = 2sin2Xcos2X

= 2 x x 78

√158

= 14√1564

= 7√1532


Page 158 Ex 4A/B All Qu.


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Solving Trig Equations

EXAMPLE 9 Solve the equation 2sin2A = 1, 0<A<2

2sin2A = 1

sin2A = ½

sinA = √½

sinA = ± 12

C

AS

T

451

1√2

A = 45,

A = 45, 135, 225, 315

180 – 45, 180 + 45, 360 – 45

A = π4

3π4

5π4

7π4

EXAMPLE 10 Solve the equation 4cos2A = 3,0 < A < 2

4cos2A = 3

cos2A = ¾

cosA = √¾

cosA = ±32

C

AS

T

A = 30,

A = 30, 150, 210, 330

180 – 30, 180 + 30, 360 – 30

A = π6

5π6

7π6

11π6

30º

60º 21

√3

EXAMPLE 11 Solve the equation sin2x - cosx = 0,0 ≤ x ≤ 2

sin2x - cosx = 0

2sinxcosx - cosx = 0

cosx(2sinx – 1) = 0

cosx = 0 C

AS

T

x = 90,

x = 30, 90, 150, 270

270, 30, 180 – 30

x = π6

π2

5π6

3π2

30º

60º 21

√32sinx – 1 = 0

sinx = ½

x =

EXAMPLE 12 Solve the equationcos2x + 7cosx + 4 = 0, 0 ≤ x ≤ 2

cos2x + 7cosx + 4 = 0

2cos2x - 1 + 7cosx + 4 = 0

2cos2x + 7cosx + 3 = 0

(2cosx + 1)(cosx + 3) = 0

cosx = -½

C

AS

T

x = 180 - 60,

x = 120, 240

180 + 60

x = 2π3

4π3

30º

60º 21

√3

cosx = -3

Not possible

EXAMPLE 13 Solve the equation3cos2x + sinx - 1 = 0, 0 ≤ x ≤ 360

3cos2x + sinx - 1 = 0

3(1 - 2sin2x) + sinx - 1 = 0

-6sin2x + sinx + 2 = 0

-(6sin2x - sinx - 2) = 0

(3sinx + 2)(2sinx - 1) = 0

sinx = -

C

AS

T

x = 180 + 41.81,

x = 30, 150, 221.81, 318.59

360 – 41.81

sinx = ½23

x = 30, 180 – 30

30º

60º 21

√3

EXAMPLE 14 Solve the equation3sin(2x + 10) = 2, 0 ≤ x ≤ 180

3sin(2x + 10) = 2

sin(2x + 10) =

C

AS

T

2x + 10 = 41.81,

2x + 10 = 41.81, 138.19, 401.81, 498.19

180 – 41.81

23

2x = 31.81, 128.19, 391.81, 488.19

x = 15.9, 64.1, 195.9, 244.1

x = 15.9, 64.1

EXAMPLE 15 Solve the equationcosxcos50 –sinxsin50 = 0.45,0 ≤ x ≤ 180

cosxcos50 –sinxsin50 = 0.45

cos(x + 50) = 0.45

C

AS

T

x + 50 = 63.26,

x + 50 = 63.26, 296.74

360 – 63.26

x = 13.26, 246.74




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higher mathematics unit 2 topic 3.2 compound … 13 sina = 4 5 cosa = 3 5 sin(a + b) = sinacosb +...

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