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HIGHER MATHEMATICS
Unit 2 – Topic 3.2
Compound Angle
Formula
REMINDERS
y
Let OP = r & POX = A
This gives the following
P’(y ,x)
A
O
r
A
x
sinA =
P(x,y)
y
x
Now reflect OP in the line y = x
x
sin(90 - A) = r
y
cos(90 - A) =
r
= cosA
= sinA
y
r
cosA =
x
r
y
tanA =
x
So OP = OP’ = r
& P’OY = A
P’OX = 90 - A
y
P(x,y)
r
O
A
-A
Let OP = r & POX = A
Now reflect OP in the x-axis
So OP’ = r & P’OX = -A
x
r
P’(x,-y)
-y
sin(-A) =
= -sinA
r
x
cos(-A) =
= cosA
r
Let OP = r
& POX = A
y
P’(-x,y)
A
P(x,y)
A
O
x
Now reflect OP
in the y-axis
So OP’ = r &
P’OX = 180 - A
-y
sin(180-A) =
= sinA
r
-x
cos(180-A) =
= -cosA
r
y
P(x,y)
r
O
A
x
Consider:
Sin A
Cos A
=
=
=
=
y
r = y ÷ x
r
r
x
r
y
r
x
r
x
yr
rx
y
= Tan A
x
Sin A
So Tan A =
Cos A
y
P(x,y)
So Sin2A + Cos2A
r
O
A
Consider:
x
=
=
=
=
y2
+
2
r
y2 + x2
r2
r2
r2
x2
r2
1
So Sin2A + Cos2A = 1
SUMMARY
sin(90 - A) = cosA
cos(90 - A) = sinA
sin(-A) = -sinA
cos(-A) = cosA
sin(180 - A) = sinA
cos(180 - A)= -cosA
sinA
cosA
=
tanA
sin2A + cos2A = 1
Exercises from MIA book:
Page 153 Ex 1 All Qu.
Exercises from Heinemann book:
Page ? Ex ? Qu ??
Formulae for cos(A + B) and cos(A – B)
Let the circle’s radius be 1
and Q be the point (x , y)
Q
x
cosA = r
x = cosA
rcosA
A
O -B
y
sinA = r
y = rsinA
sinA
So Q is (cosA , sinA)
P
Consider the point P(x , y)
y
x
cos(-B) = r sin(-B) = r
rsin(-B)
x = cos(-B)
rcos(-B) y = sin(-B)
x = cosB
x = -sinB
So P is
(cosB , -sinB)
P(cosB , -sinB) & Q(cosA , sinA)
Use the distance formula to find PQ
PQ2 = (cosA – cosB)2 + (sinA + sinB)2
PQ2
PQ2
PQ2
PQ2
=
=
=
=
Q’
cos2A – 2cosAcosB + cos2B + sin2A + 2sinAsinB + sin2B
cos2A + sin2A + sin2B + cos2B – 2cosAcosB + 2sinAsinB
1 + 1 – 2cosAcosB + 2sinAsinB
2 – 2(cosAcosB - sinAsinB)
Q
Now rotate
ΔPOQ
anti-clockwise
P’ through angle B
A+B
A
O -B
P
Q’ is (x , y)
and P’ is (1 , 0)
x
cos(A+B) = r
x = cos(A+B)
rcos(A+B)
y
sin(A+B) = r
y = sin(A+B)
rsin(A+B)
P’(1 , 0) & Q’(cos(A+B) , sin(A+B))
Use the distance formula to find P’Q’
P’Q’2 = (1 – cos(A+B))2 + (0 – sin(A+B))2
P’Q’2 = 1 – 2cos(A+B) + cos2(A+B) + sin2(A+B)
P’Q’2 = 1 – 2cos(A+B) + 1
P’Q’2 = 2 – 2cos(A+B)
Remember that PQ = P’Q’ so PQ2 = P’Q’2
2 – 2cos(A+B) = 2 – 2(cosAcosB - sinAsinB)
Giving:
Cos(A + B) = CosACosB - SinASinB
We get another expansion by replacing B with -B
Cos(A + (-B)) = CosACos(-B) – SinASin(-B)
Giving:
Cos(A - B) = CosACosB + SinASinB
EXAMPLE 1
Acute angles P and Q are such that
and cosQ = 3
5
sinP = 12
13
Show that cos(P - Q) = 63
65
13
12
P
5
4
5
3
Q
sinP =
12
13
cosP =
5
13
sinQ =
4
5
cosQ =
3
5
Cos(P - Q) = CosPCosQ + SinPSinQ
= 5 x 3 + 12 x 4
5
13
5
13
= 15 + 48
65
65
= 63 as required
65
EXAMPLE 2
Acute angles X and Y are such that
and tanY = 3
4
sinX = 8
17
Show that cos(X + Y) = 36
85
8
17
15
3
X
5
4
Y
sinX =
8
17
cosX =
15
17
sinY =
3
5
cosY =
4
5
Cos(X + Y) = CosXCosY - SinXSinY
= 15 x 4 - 8 x 3
5
17
5
17
= 60 - 24
85
85
= 36 as required
85
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(A + B)
= cosAcosB - sinAsinB
cos(105) = cos(60 + 45)
= cos60cos45 - sin60sin45
1
1
=
x
- √3 x 1
2
√2
2
√2
= 1 - √3
2√2
2√2
= 1 - √3 x 2
2√2
2
(1 - √3) √2
=
2√2√2
√2 - √6
As required
=
4
1
2
60º
√3
45º
√2
45º
30º
1
1
Trying to get
2 - 6
4
Exercises from MIA book:
Page 154 Ex 2 All Qu.
Exercises from Heinemann book:
Page ? Ex ? Qu ??
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
Giving:
Sin(A + B) = SinACosB + CosASinB
Now what happens when we replace B with -B
Sin(A + (-B)) = SinACos(-B) + CosASin(-B)
Giving:
Sin(A - B) = SinACosB - CosASinB
EXAMPLE 5
Acute angles X and Y are such that
sinX = 1
√5
and tanY = 3
Find the exact value of sin(X - Y)
√5
1
X
2
3
√10
1
Y
sinX =
1
√5
cosX =
sinY =
3
√10
cosY =
2
√5
1
√10
sin(X - Y) = sinXcosY - cosXsinY
= 1 x 1 - 2 x 3
√5
√10
√10 √5
= 1 - 6
√50 √50
= -5
= -1
= -5
5√2
√2
√50
EXAMPLE 6
For the diagram, express
sin(ABC) as a fraction.
C
sina =
4
5
cosa =
3
5
sinb =
12
13
cosb =
5
13
sinABC = sin(a + b)
sin(a + b) = sinacosb + cosasinb
12
13
D
4
A
5
3
ao
bo
B
= 4 x 5 + 3 x 12
13
5
13
5
= 20 + 36
65
65
= 56
65
Exercises from MIA book:
Page 156 Ex 3 All Qu.
Exercises from Heinemann book:
Page ? Ex ? Qu ??
A Few More Formulae.
sin(A + B) = sinAcosB + cosAsinB
Replace B with A
sin(A + A) = sinAcosA + cosAsinA
sin2A
= 2sinAcosA
So sin2A = 2sinAcosA
Cos(A + B) = cosAcosB - sinAsinB
cos(A + A) = cosAcosA - sinAsinA
cos2A
= cos2A – sin2A
Remember that cos2A + sin2A = 1
So
cos2A = 1 - sin2A
&
sin2A = 1 - cos2A
Replace B with A
Hence:
cos2A
= cos2A – sin2A
= cos2A – (1 - cos2A)
= 2cos2A – 1
So cos2A = 2cos2A – 1
And:
cos2A
= cos2A – sin2A
= 1 - sin2A – sin2A
= 1 – 2sin2A
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 = cos2A – sin2A
cos2A = 2cos2A – 1
cos2A = 1 – 2sin2A
cos2A = ½(1 + cos2A)
sin2A = ½(1 - cos2A)
EXAMPLE 7
8
17
15
15
cosA =
17
Given that 0 < A < 90 and that
sinA = 8 , obtain exact values for
17
sin2A and cos2A
A
When finding cos2A you can choose
any of the 3 formulae that you want
sin2A = 2sinAcosA
cos2A = 2cos2A - 1
= 2 x 8 x 15
17
17
=
= 240
289
=
=
=
2
15
2 x
- 1
2
17
2 x 225 - 1
289
450 - 289
289
289
161
289
EXAMPLE 8
1
4
√15
sinX =
cosX =
1
4
√15
4
Given that 0 < A < 90 and that
tanX = 1
√15
, obtain exact values for
sin2X, cos2X and hence sin4X
X
When finding cos2A you can choose
any of the 3 formulae that you want
sin2X = 2sinXcosX
= 2 x
= 2√15
16
= √15
8
1 x √15
4
4
cos2X = 2cos2X - 1
= 2 x
= 2 x
= 30
16
= 14
16
152 - 1
42
15 - 1
16
- 16
16
= 7
8
sin4X = 2sin2Xcos2X
= 2 x √15 x
8
= 14√15
64
= 7√15
32
7
8
Exercises from MIA book:
Page 158 Ex 4A/B All Qu.
Exercises from Heinemann book:
Page ? Ex ? Qu ??
Solving Trig Equations
EXAMPLE 9
Solve the equation 2sin2A = 1, 0<A<2
2sin2A = 1
sin2A
= ½
sinA = √½
sinA = ± 1
2
A = 45,
A = 45,
A = π
4

S
T

180 – 45,
135, 225,
3π 5π
4
4

A
C
√2
45

1
1
180 + 45, 360 – 45
315
7π
4
EXAMPLE 10
Solve the equation 4cos2A = 3,
0 < A < 2
4cos2A = 3
cos2A
= ¾
cosA = √¾
cosA = ±3
2
A = 30,
A = 30,
A = π
6


S
A
T
C

180 – 30,
150, 210,
5π 7π
6
6

1
2
60º
√3
180 + 30, 360 – 30
330
11π
6
30º
EXAMPLE 11
Solve the equation sin2x - cosx = 0,
0 ≤ x ≤ 2
sin2x - cosx = 0
2sinxcosx - cosx = 0
cosx(2sinx – 1) = 0
cosx = 0
2sinx – 1 = 0


S
A
T
C
sinx = ½
x = 90, 270,
x = 30, 180 – 30
x = 30, 90, 150, 270
5π 3π
x = π π
6 2
6
2
1
2
60º
√3
30º
EXAMPLE 12
Solve the equation
cos2x + 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

S
A
T
C

cosx = -3
Not possible
x = 180 - 60, 180 + 60
x = 120, 240
x = 2π 4π
3
3
1
2
60º
√3
30º
EXAMPLE 13
Solve the equation
3cos2x + 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 = - 2
3
x = 180 + 41.81, 360 – 41.81
x = 30, 150, 221.81, 318.59
S

T

A
C


sinx = ½
x = 30, 180 – 30
1
2
60º
√3
30º
EXAMPLE 14
Solve the equation
3sin(2x + 10) = 2, 0 ≤ x ≤ 180
3sin(2x + 10) = 2
2
sin(2x + 10) =
3
2x + 10 = 41.81, 180 – 41.81
S

T
2x + 10 = 41.81, 138.19, 401.81, 498.19
2x
= 31.81, 128.19, 391.81, 488.19
x
= 15.9, 64.1, 195.9, 244.1
x
= 15.9, 64.1

A
C
EXAMPLE 15
Solve the equation
cosxcos50 –sinxsin50 = 0.45,
0 ≤ x ≤ 180
cosxcos50 –sinxsin50 = 0.45
cos(x + 50) = 0.45
x + 50 =
63.26, 360 – 63.26
x + 50 =
63.26, 296.74
x
13.26, 246.74
=

S
A
T
C
Exercises from MIA book:
Page 161 Ex 5 All Qu.
Exercises from Heinemann book:
Page ? Ex ? Qu ??