NELSON SENIOR MATHS SPECIALIST 11mathsbooks.net/Nelson Specialist 11/Nelson...16 Use the graph of y...
Transcript of NELSON SENIOR MATHS SPECIALIST 11mathsbooks.net/Nelson Specialist 11/Nelson...16 Use the graph of y...
NELSON SENIOR MATHS SPECIALIST 11 FULLY WORKED SOLUTIONS
CHAPTER 9: Trigonometric identities Exercise 9.01: Reciprocal trigonometric functions
Concepts and techniques
1 cot (θ) is the reciprocal of tan (θ) so cot3π
= 13
A
2 cosec (θ) is the reciprocal of sin (θ) so cosec4π
= 2 D
3 sec (θ) is the reciprocal of cos (θ) so sec6π
= 23
D
4 ( )cosec 90° = 1 D
5 ( )cot 45° = 1 C
all
costan
sin all
seccot
cosec
Positve in quadrants.....
6 a cot( ) cot4 4π π π − = −
= –1
b sec sec 24 4π π π − = − = −
c 2cosec sin3 3 3π π π + = − = −
d sec 2 sec 23 3π π π − = − =
e cot cot 36 6π π π + = =
f sec sec 24 4π π π + = − = −
g sec 2 sec 24 4π π π + = =
h cot 2 cot 36 6π π π + = =
7 a ( )sec(2 ) secπ + θ = θ
b ( )sec( ) secπ − θ = − θ
c ( ) ( )cosec(2 ) cos ec cos ecπ − θ = −θ = − θ
d ( )cot( ) cotπ + θ = θ
e ( )cosec( ) cosecπ + θ = − θ
f ( ) ( )cot(2 ) cot cotπ − θ = −θ = − θ
© Cengage Learning Australia 2014 ISBN 9780170251501 2
8 For 0 2x π≤ ≤ sketch the graph of ( )3cosec 1y x= −
9 For 0 2x≤ ≤ π sketch the graph of sec 12
y x π = − − +
10 For 0 2x≤ ≤ π sketch the graph of ( )4cot 2 1y x= +
© Cengage Learning Australia 2014 ISBN 9780170251501 3
11 For x−π ≤ ≤ π sketch the graph of 2sec 22
y x π = − +
Reasoning and communication
12 sec 24π =
and cosec 2
4π =
Equal values.
NB 4π +
4 2π π=
NB cos4π =
sin
4π
13 sec3π
= 2 and cosec 26π =
NB. 3π +
6 2π π=
NB. cos3π =
sin
6π
NB. cos (θ) = sin 2π − θ
so sec (θ) = cosec
2π − θ
© Cengage Learning Australia 2014 ISBN 9780170251501 4
14 a
π 2sec6 3
=
b πcot 14
=
c π 2cosec3 3
− = −
d 2π π 2cosec cos ec3 3 3
= =
e sec (0) = 1
f cot (0) is not defined as tan (0) = 0
g 4π π 1cot cot3 3 3
− = − = −
h 7π πsec sec 24 4
− = =
i 3πcosec 12
= −
j πsec3
= 2
15 Use the graph of y = tan (x) to sketch y = cot (x) from − π2
to π2
.
© Cengage Learning Australia 2014 ISBN 9780170251501 5
16 Use the graph of y = cos (3x) to sketch the graph of y = sec (3x) from − π2
to π2
.
17 Use the graph of y = sin (4x) to sketch the graph of y = cosec (4x) from 0 to π.
18 Use the graph of y = tan (2x) to sketch the graph of y = cot (2x) from −π to π.
19 Use the graph of πcos4
y x = −
to sketch the graph of πsec4
y x = −
from −π to π.
© Cengage Learning Australia 2014 ISBN 9780170251501 6
20 Use the graph of πsin6
y x = +
to sketch the graph of πcosec6
y x = +
from
0 to 2π.
21 Use the graph of πtan4
y x = +
to sketch the graph of πcot4
y x = +
from
− π2
to π2
.
22 Sketch the graph of πcosec4
y x = +
from 0 to 2π.
© Cengage Learning Australia 2014 ISBN 9780170251501 7
23 Sketch the graph of πsec 34
y x = +
from 0 to π.
24 Sketch the graph of πcot 24
y x = −
from 0 to π.
25 Sketch the graph of πsec 23
y x = −
from 0 to π.
© Cengage Learning Australia 2014 ISBN 9780170251501 8
Exercise 9.02: The Pythagorean identities
Concepts and techniques
1 Prove that sin2 (A) + cos2 (A) 1≡ .
Using the equation 2 2 2c a b= + we get the relationship 2 2 2r x y= +
We know that ( )cos xr
θ = and ( )sin yr
θ =
( ) ( )cos cosx x rr
θ = ⇒ = θ and ( ) ( )sin siny y rr
θ = ⇒ = θ
giving ( ) ( )2 2 2 2 2cos sinr r r= θ + θ
So for any r , ( ) ( )2 21 cos sin= θ + θ
2 Prove that ( ) ( )( )
sintan
cosA
AA
≡ .
Consider the diagram showing angle θ, general point P(x, y) and radius of circle, r.
Using ( )tan yx
θ = then we have ( ) ( )( )
sintan
cosθ
θ ≡θ
i.e. ( ) ( )( )
sintan
cosA
AA
≡
© Cengage Learning Australia 2014 ISBN 9780170251501 9
3 Prove that 2 21 cot ( ) cosec ( )A A+ ≡ .
( ) ( )
( )( )
( )( )
( ) ( )( )
( )( )
22
2
2
2
2
2 2
2
2
2
11 cot 1tan
11sin
cos
cos1
sin
sin +cossin
1sin
= cosec A
AA
AA
AA
A AA
A
+ ≡ +
= +
= +
=
=
4 Prove that ( ) ( )( )
coscot
sinA
AA
≡
.
( ) ( )
( )( )
( )( )
1cottan
1sin
cos
cossin
AA
AA
AA
=
=
=
5 Show that
2sin2 3tan
23 cos3
π π = π
.
Given ( ) ( )( )
sintan
cosA
AA
≡ then if A = 23π then
2sin2 3tan
23 cos3
π π = π
© Cengage Learning Australia 2014 ISBN 9780170251501 10
Reasoning and communication
6 Prove that [1 − sin (θ)][1 + sin (θ)] = cos2 (θ).
Given sin2 (A) + cos2 (A) 1≡ then
cos2 (θ) = 1 – sin2 (θ)
= [1 − sin (θ)][1 + sin (θ)]
7 Prove that sin4 (θ) − cos4 (θ) = 1 − 2 cos2 (θ).
sin4 (θ) − cos4 (θ) = [sin2 (θ) +cos2 (θ)][sin2 (θ) – cos2 (θ)]
= 1 × {[1 – cos2 (θ)] – cos2 (θ)}
= 1 − 2 cos2 (θ).
8 Prove that ( ) ( ) ( )( )
2 1 coscot cosec .
1 cosx
x xx
++ = −
( ) ( ) ( )( ) ( )
( )( )
( )( )( ) ( )( ) ( )( )( )
22
2
2
2
cos 1cot cosecsin sin
cos 1sin
cos 11 cos
1 cos 1 cos
1 cos 1 cos
1 cos.
1 cos
xx x
x x
xx
xx
x xx x
xx
+ = +
+=
+ =−
+ + =− + +
=−
9 Prove that ( ) ( ) ( )( )
1 sinsec tan .
cosy
y yy
++ =
( ) ( ) ( )
( )( )
( )( )
sin1sec tancos cos
1 sincos
yy y
y y
yy
+ = +
+=
© Cengage Learning Australia 2014 ISBN 9780170251501 11
10 Prove that ( ) ( ) ( )2 1 12 cosec .
1 cos 1 cosa
a a= +
− +
( ) ( )( ) ( )( ) ( )
( ) ( )
( )
( )( )
2
2
2
1 cos 1 cos1 11 cos 1 cos 1 cos 1 cos
21 cos 1 cos
21 cos
2sin
2cos ec
a aa a a a
a a
a
a
a
+ + −+ =
− + − +
=− +
=−
=
=
11 Prove that ( ) ( )( ) ( ) ( ) ( ) ( ) ( )
sin cos βsin β 1 tan β cos β 1 cot β .
cos β sin ββ +
= + + +
( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )
( )
( ) ( ) ( )( ) ( ) ( ) ( )
( )
( ) ( ) ( )( )
( )( )
( ) ( ) ( ) ( )( ) ( )
2
sin β cos βsin β 1 tan β cos β 1 cot β sin β 1 cos β 1
cos sin
cos sin sin cossin β cos
cos sin
sin cossin β cos
cos sin
sin cos 2βsin β cos
cos sin
si
+ + + = + + + β β
β + β β + β= + β β β
β β= + β + β β
β += + β β β
=( ) ( )( ) ( )
n cos βcos β sin β
β +
© Cengage Learning Australia 2014 ISBN 9780170251501 12
12 Prove that ( ) ( )( ) ( ) ( ) ( ) ( ) ( )
( ) ( )sin θ cos θ cos θ cot θ sin θ tan θ
1 sin θ cos θ .cos θ sin θ
− + =−
( ) ( ) ( ) ( ) ( ) ( )( ) ( )
( ) ( )( ) ( ) ( ) ( )
( ) ( ) ( )( )
( ) ( )( ) ( )
( ) ( )( ) ( )
( ) ( )( ) ( )
( ) ( ) ( ) ( ) ( )
3 3
3 3
2
sin θ cos θ cos θ cot θ sin θ tan θcos θ sin θ
sin θ cos θ cos θ sin θcos θ sin θ
cos θ sin θ sin θ cos θ
sin θ cos θ cos θ sin θcos θ sin θ sin θ cos θ
cos θ sin θcos θ sin θ
cos θ sin θ cos θ +cos θ sin θ sin
− −
= − − −
= − −
=−
− + ( )( ) ( )
( ) ( ) ( ) ( )( ) ( )
2
2 2
θ
cos θ sin θ
cos θ +cos θ sin θ sin θ
1 sin θ cos θ .
−
= +
= +
© Cengage Learning Australia 2014 ISBN 9780170251501 13
Exercise 9.03: Angle sum and difference identities
Concepts and techniques
1 a sin (2x + y) = sin (2x) cos (y) + sin (y) cos (2x)
b cos (4x − 3y) = cos (4x) cos (3y) + sin (4x) sin (3y)
c sin (3p − 5q) = sin (3p) cos (5y) – cos (3p) sin (5y)
d tan (2e − f) = ( ) ( )( ) ( )
tan 2 tan1 tan 2 tan
e fe f−
+
e cos (3α + 2β) = ( ) ( ) ( ) ( )cos 3 cos 2 sin 3 sin 2α β − α β
f sin (4m + n) = ( ) ( ) ( ) ( )sin 4 cos cos 4 sinm n m n+
g tan (w + 3x) = ( ) ( )( ) ( )
tan tan 31 tan tan 3
w xw x+
−
h sin (2D − 5E) = ( ) ( ) ( ) ( )sin 2 cos 5 cos 2 sin 5D E D E−
i cos (P − 4Q) = cos (P) cos (4Q )+ sin (P) sin (4Q)
2 a sin (π − x) = sin (π) cos (x) − sin (x) cos (π) = 0 – sin (x) × (–1) = sin (x)
b cos (π + x) = cos (π) cos (x) − sin (x) sin (π) = –1 × cos (x) – 0 = – cos (x)
c ( )πtan cot2
x x − =
d
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )π π π2 2 2
π 1 1 1 1cosec sec2 sin sin cos cos sin 0 cos 1 cos
x xx x x x x
+ = = = = = + + + ×
© Cengage Learning Australia 2014 ISBN 9780170251501 14
e
3π πcot cot2 2
x x − = π + −
Consider tan π2
x π + −
( ) ( )( ) ( )( )
( )( )( )( )( )( )
π2π2
π2
π2
π2π2
tan tanπtan2 1 tan tan
0 tan1 0
tan
sincos
cossin
cot
xx
x
x
x
xx
xx
x
π + − π + − = − π −
+ −=
−= −
−=
−
=
=
( )
( )
3π 1cot2 cot
3πcot tan2
xx
x x
− =
− =
f πsec2
x +
( ) ( )
( )( )
( )
( )
π2
π π πcos co
Cons
s
ider c
sin sin2 2 2
0 sin 1
sin
π 1sec2 sin
πsec
os
c
cos ec2
os
x
x x x
x
x
xx
x x
+
+ = −
= − ×
= −
∴ + = −
∴ + = −
g sin (2π − x) = sin(–x) = –sin (x)
© Cengage Learning Australia 2014 ISBN 9780170251501 15
h ( ) ( )3π 3π 3πcos cos cos sin sin2 2 2
x x x − = +
( )( )
0 ( 1) sin
sin
x
x
= + − ×
= −
i tan (x − π) ( ) ( )( ) ( )
( ) ( )tan tan tan 0
tan1 tan tan 1 0
x xx
x− π −
= = =+ π −
3 Using ( ) ( ) ( ) ( )cos( ) cos cos sin sinx y x y x y− = + and putting 2
x π=
( ) ( )
( ) ( )
( )
cos cos cos sin sin2 2 2
0 cos 1 sin
cos sin2
x x x
x x
x x
π π π − = +
= × + ×
π − =
Using ( ) ( ) ( ) ( )sin( ) sin cos cos sinx y x y x y− = − and putting 2
x π=
( ) ( )
( ) ( )
( )
sin sin cos cos sin2 2 2
1 cos 0 sin
sin cos2
x x x
x x
x x
π π π − = −
= × − ×
π − =
4
( )( )
( )( )
( )( ) ( )
π π2 2π π2 2
sin sin costan cot
2 cos cos sinx x x
x xx x x
+ −π + = = = = − + − − −
© Cengage Learning Australia 2014 ISBN 9780170251501 16
Reasoning and communication
5 Prove that sin (x + y) = sin (x) cos (y) + cos (x) sin (y). Given ( ) ( ) ( ) ( )cos( ) cos cos sin sinx y x y x y− = + found using cosine laws and distance formula
Putting y = 2
yπ− we have ( ) ( ) ( ) ( )cos( ) cos cos sin sinx y x y x y− = +
becomes
cos 2
x y π − − = cos (x) cos
2yπ −
+ sin (x) sin
2yπ −
i.e. cos2
x y π − − − = cos (x) sin (y) + sin (x) cos (y) but cos(–θ) = cos (θ)
cos2
x yπ − −
= cos (x) sin (y) + sin (x) cos (y)
∴ sin (x + y) = sin (x) cos (y) + cos (x) sin (y).
6 Prove that ( ) ( ) ( )( ) ( )
tan tantan .
1 tan tanx y
x yx y−
− =+
( )
( ) ( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( )
( ) ( )( ) ( ) ( ) ( )
( ) ( )( ) ( ) ( ) ( )
( ) ( )( )( )
( )( )
( )( )
( )( )
( ) ( )( ) ( )
sin( )tancos( )
1sin cos sin cos cos cos
1cos cos sin sincos cos
sin cos sin coscos cos
cos cos sin sincos cos
sin sincos cos
sin sin1
cos cos
tan tan1 tan tan
x yx yx y
x y y x x yx y x y
x y
x y y xx y
x y x yx y
x yx y
x yx y
x yx y
−− =
−
−= ×
+
−
=−
−=
+
−=
+
© Cengage Learning Australia 2014 ISBN 9780170251501 17
7 Prove that ( ) ( ) ( )( ) ( )
cot cot 1cot .
cot cotx y
x yy x
−+ =
+
Consider tan(x+y)
( ) ( ) ( )( ) ( )
( ) ( ) ( )( ) ( )
( ) ( )
( ) ( )( ) ( )
( ) ( )( ) ( )( ) ( )
( ) ( )( )
( ) ( )( )
( ) ( )
( ) ( )
( ) ( )( ) ( )( ) ( )
tan tan1 tan tan
11 tan tan tan tan
cot1tan tan
tan tan
1 tan tantan tan
tan tantan tan
1 1tan tan
tan tantan tan tan tan
1 1tan tan
1 1tan tan
cot cot 1
ta
t t
n
co co
x yx y
x y x yx y
x y
x yx y
x yx y
x yx y
x y x y
x y
x y
y
x
y x
x
y
x y
+=
−
−∴ = ×
+
−
=+
−=
+
−
+
+
=+
+
=
−
8 Prove that ( ) ( ) ( ) ( )( )
sec sectan tan .
cosecx y
x yx y
+ =+
( ) ( )( ) ( ) ( )
( ) ( ) ( ) ( )( ) ( )
( )( )
( )( )
( ) ( )
sec sec 1sin( )cosec cos cos
sin cos sin coscos cos
sin sincos cos
tan tan
x yx y
x y x y
x y y xx y
x yx y
x y
= + ×+
+=
= +
= +
© Cengage Learning Australia 2014 ISBN 9780170251501 18
9 Prove that sin (x + y) + cos (x − y) = (sin (x) + cos (x)(sin (y) + cos (y)).
( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )
(
sin cos sin cos cos cos sin sin
sin
s
cos sin cos sin cos
cos sin sin cos
si
in cos
n cos cos s
)
in
x y x
x y y x x y x y
x y y x y y
y y x x
x x y
y
y
= + + +
= + + + = − + = + −
+ +
−
10 Prove that sin (x + y) sin (x − y) = sin2 x − sin2 y.
( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )
2 2 2 2
2 2 2 2
2 2 2 2
2 2 2 2 2
sin cos cos sin sin cos cos sin
sin cos sin cos cos sin cos sin sin cos cos s
sin si
in
sin cos cos sin
sin 1 sin 1 sin sin
sin sin sin sin si s n
n
n i
x y x y x y x y
x y x y x y x y x y x y
x y x y
x y x y
x x y y
x x y
x
y
= + − = − + −
= −
= − − − +
−
= −
+
− ( )( ) ( )
2
2 2sin sin
y
x y= −
11 Prove that ( ) ( )( ) ( )
( )( )
cot cot sin.
1 cot cot cosx y x y
x y x y− −
=− +
( )( )
( ) ( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( )
( ) ( )( ) ( ) ( ) ( )
( ) ( )( ) ( ) ( ) ( )
( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )
1sin sin cos cos sin sin sin
1cos cos cos sin sinsin sin
sin cos cos sinsin sin
cos cos sin sinsin sin
cot cot 1cot cot 1 1
cot cot1 cot cot
x y x y x y x yx y x y x y
x y
x y x yx y
x y x yx y
y xx y
x yx y
− −= ×
+ −
−
=−
− −= ×
− −
−=
−
© Cengage Learning Australia 2014 ISBN 9780170251501 19
12 Prove that tan (x − y) + tan (y − z) + tan (z − x) = tan (x − y) tan (y − z) tan (z − x).
( ) ( ) ( )( )( )
( )( )
( )( )
( ) ( ) ( ){ } ( ) ( ) ( ){ } ( ) ( ) ( )( ) ( ) ( )
( ){ } ( ) ( ) ( ) ( ){ } ( ) ( ) ( )
sin sincos cos cos
cos cos sin cos cos sin cos coscos cos cos
cos c
tan tan tan
s
os sin cos sin co
in
sin
sin s cosco
x y y z z x
x y y z z xx y y z z x
x y y z z x y z x y z x z x x y y zx y y z z x
z x x y y z y z x y z x x y y z
− + − + −
− − −+ +
− − −
− − − − − − − − −
− − −
− − −
=
+ +=
+ +=
− − − − −
( ) ( ) ( )( ){ } ( ) ( ) ( ) ( ){ } ( ) ( ) ( )
( ) ( ) ( )( ){ }
s cos cos
cos cos cos sin sin cos coscos cos cos
co
sin
sis n
x y y z z x
z x x y y z x y y z z x x y y zx y y z z x
z x x y
− − −
− − − − − − − −+ +=
=
− − −
− − y+( ){ } ( ) ( ) ( )( ) ( ) ( )
( ){ } ( ){ } ( ) ( ) ( )( ) ( ) ( )
( )( ) ( ) ( ) ( ) ( )
( ) ( ) ( )( )( )
( ) ( ) ( )
sin cos cos
cos cos cos
cos sin cos coscos cos cos
and given that sin( ) sin
sin cos sin cos coscos cos cos
sin cos
s
cos coscos
in
z z x x y y z
x y y z z x
z x x z z x x y y zx y y z z x
z x z x z x x y y zx y y z z x
z x z x x y y zz x
− − − −
− − −
− − − − −
− − −
− − −
+
+=
− α = − α
+=
− +=
− −− − −
− − − −− ( ) ( )
( )( ) ( )( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )
( )( ) ( )
cos cos
cos cos costan
cos cos
cos cos costan
cos cos
costan
cos sin sin
cos
x y y z
z y y x x y y zz x
x y y z
z y y x z y y x x y y zz x
x y y z
z y y xz x
− +
= −
− −
− + − − −−
− −
− − + − +=
−
−=
− −−
− −
− −−
( ) ( ) ( ) ( )sin sin cos cosz y y x x y y z++ − − − −
( ) ( )
( ) ( ) ( )( ) ( )
( ) ( ) ( )( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )
cos cos
tancos cos
tancos cos
tan tan tan given sin = -sin and sin = -s
sin sin
sin sin
tan tan
in
tan tantan
x y y z
z y y xz x
x y y z
y x z yz x
x y y z
z x y x y z y x x y z y y z
x y y z z x x y y x
− −
− −−
− −
− −−
− −
− − − − − − −
− + −
=
=
=
−∴ =+ − − ( )tan z x−
© Cengage Learning Australia 2014 ISBN 9780170251501 20
Exercise 9.04: Double angle formulas
Concepts and techniques
1 Show that ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
2 2
2 2
2
cos 2 cos sin from the formula from cos ( ) = cos cos sin sin
= 1 sin sin
1 2sin
A A A A A A A A A
A A
A
= − + −
− − = −
2 Show that ( ) ( )2 22cos 1 1 2sinA A− = −
( ) ( )( )( )
2 2
2
2
2cos 1 2 1 sin 1
2 2sin 1
1 2sin
A A
A
A
− = − − = − −
= −
3 Show that ( ) ( ) ( ) ( )2 2 2 2cos sin 1 sin siny y A A − = − −
( )21 2sin A= −
4 Show that ( ) ( ) ( )sin 8 2sin 4 cos 4A A A= .
( )
( ) ( ) ( ) ( )( ) ( )
sin 8 sin(4 4 )
sin 4 cos 4 sin 4 cos 4
2sin 4 cos 4
A A A
A A A A
A A
= +
= +
=
5 Show that ( ) 2 5cos 5 1 2sin2AA = −
( )
2 2
2 2
2
5 52 2
5 5 5 5cos cos sin sin2 2 2 25 5cos
cos
sin2 2
5 51 sin sin2 2
51 2s
5 cos
2
in
A A
A A A A
A A
A A
A
A +
= − = −
= − −
=
= −
© Cengage Learning Australia 2014 ISBN 9780170251501 21
6 a sin (4x) = 2sin (2x) cos (2x)
b cos (6A) = 2cos2(3A) – 1
c ( ) ( )( )2
2 tan 2tan 4
1 tan 2y
yy
=−
d ( ) 2 2cos cos sin2 2y yy = −
e 2
2 tan(2 )tan(4 )1 tan (2 )
xxx
=−
7 a sin (4x) = 2 sin (2x) cos (2x)
= 2(2 sin (x)cos (x)[cos2 (x) – sin2 (x)]
= 4 sin (x) cos3(x) – 4 sin3(x) cos (x)
b cos (6A) = cos2(3A)– sin2(3A)
= (4cos3(A)–3cos (A))2 – (3sin (A) – 4cos3(A))2
= (16cos6(A) – 24 cos4(A) + 9cos2(A)) – (9sin2(A) – 24 sin4(A) +16sin6(A))
= 16cos6(A) – 16sin6(A)+ 9 cos2(A) – 9sin2(A) – 24(cos4(A)– sin4(A))
= 16(cos2(A) – sin2(A))(cos4(A)+ sin2(A)cos2(A) + sin4(A))+ 9(cos2(A)
– sin2(A)) – 24(cos2(A)– sin2(A)) (cos2(A)+ sin2(A))
= (cos2(A) – sin2(A))[16(cos4(A)+ sin2(A)cos2(A) + sin4(A)) + 9 – 24]
= (cos2(A) – sin2(A))[16(cos4(A)+ sin2(Acos2(A) + sin4(A)) – 15]
= (cos2(A) – sin2(A))[16(cos4(A)+2sin2(A)cos2(A) + sin4(A)– sin2(A)cos2(A)) – 15]
= (cos2(A )– sin2(A))[16(1– sin2(A)cos2(A) )– 15]
= (cos2(A) – sin2(A))[16 – 16sin2(A)cos2(A) – 15]
= (cos2(A) – sin2(A))[1 – 16sin2(A)cos2(A)]
= cos2(A) – sin2(A) – 16sin2(A)cos4(A) + 16sin4(Acos2(A)
NB. There are alternative answers. Answers may vary.
Sometimes answers are given in terms of cos (x) or sin (x) only.
© Cengage Learning Australia 2014 ISBN 9780170251501 22
c
( ) ( )( )( )( )( )( )
( )( )( )( )( )( )
( ) ( )( )
( )( )
( )( )
( ) ( ) ( )( )
2
2
2
2
2
2
22
2
22 2
22
2
2 2
2 2 2
22
2 tan 2tan 4
1 tan 2
2[2 tan1 tan
2 tan1
1 tan
4 tan1 tan
4 tan1
1 tan
4 tan1 tan
1 tan 4 tan
1 tan
4 tan 1 tan1 tan 1 tan
1 2 tan tan 4 4 tan
1 tan
4 t
yy
y
yy
yy
yyy
y
yy
y y
y
y yy y
y y y
y
=−
−=
−
−
−=
− −
−= − −
− −
×− −
=− + −
−
=( ) ( )
( ) ( )
2
2 2
an 1 tan
1 6 tan tan 4
y y
y y
− − +
© Cengage Learning Australia 2014 ISBN 9780170251501 23
8 Prove that ( ) ( ) ( ) ( ) ( ) ( )1 1 2cos sec 2
cos sin cos sin+ = θ θ
θ + θ θ − θ
( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )
( )( ) ( ) ( ) ( )
( )( ) ( )( )
( )( ) ( )
2 2
1 1cos sin cos sin
cos sin cos sin
cos sin cos sin
2coscos sin cos sin
2coscos sin
2coscos 2
2cos sec 2
+θ + θ θ − θ
θ − θ + θ + θ =θ + θ θ − θ
θ=
θ + θ θ − θ θ
= θ − θ
θ=
θ
= θ θ
9 Prove that ( ) ( ) ( ) ( )( )
sec 4 sectan 4 tan .
cosec 3x x
x xx
− =
( ) ( ) ( )( )
( )( )
( ) ( ) ( ) ( )( ) ( )
( ) ( ) [ ]
( ) ( ) ( )
sin 4 sintan 4 tan
cos 4 cos
sin 4 cos cos 4 sincos 4 cos
1 sin(4 )cos 4 cos
sec 4 sec sin 3
x xx x
x x
x x x xx x
x xx x
x x x
− = −
−=
= −
=
10 Show that sin (4x) = 4sin (x) cos (x) – 8sin3(x)cos (x)
( ) ( ) ( )( ) ( ) ( )
( ) ( ) ( ) ( )
2
3
sin 4 2sin 2 cos 2
2 2sin cos 1 2sin
4sin cos 8sin cos
x x x
x x x
x x x x
=
= − = −
© Cengage Learning Australia 2014 ISBN 9780170251501 24
11 Show that ( ) ( )( )
2cot 1cot 2 .
2cotx
xx−
=
( ) ( )( )
( ) ( ) ( )
( ) ( )( )( )
( )( )
2
2
2
1 tan1 1 1cot 2 tantan 2 2 tan 2 tan
cot 11 1 1cot2 cot 2 cot
cot 1
2cot
xx x
x x x
xx
x x
xx
−= = = −
− = − =
− =
12 Prove that cot (θ) − tan (θ) = 2 cot (2θ).
2 cot (2θ) = 2 × ( )( )
2cot 12cot
θ −θ
= ( )( )
( )( ) ( ) ( ) ( )
2 2cot 1 cot 1 cot tancot cot cot
θ − θ= − = θ − θ
θ θ θ
13 Prove that ( ) ( ) ( )1 1tan 2 .
1 tan 1 tanx
x x= −
− +
( ) ( )( ) ( )( ) ( )( )( )
( )
2
1 tan 1 tan1 11 tan 1 tan 1 tan 1 tan
2 tan1 tan
tan 2
x xx x x x
xx
x
+ − +− =
− + − +
=−
=
14 Prove that tan (x) = cot (x) − 2 cot (2x).
( ) ( ) ( )( )
( )( )
( )( )
2
2
1 tan1cot 2 cot 2 2tan 2 tan
1 1 tantan
tan
xx x
x x
xx
x
−− = −
− +
=
=
© Cengage Learning Australia 2014 ISBN 9780170251501 25
Exercise 9.05: Finding and using exact values
Concepts and techniques
1 a tan (15°) = tan (45° − 30°)
( ) ( )( ) ( )
( )( )
13
13
113
1 13
2 1 4 23 33 3
1 23 3
1tan 45 tan 301 tan 45 tan 45 1 1
1
1
4 2 3 33 23 3
4 2 3 3 4 2 33 3 2 2 2
2 3
−
−
−° − °= = ×
+ ° ° + ×
− + −= =
−
= − × ×
= − × = −
= −
b ( )cos 75 cos (45 30 )° = ° + °
( ) ( ) ( ) ( )
( )
cos 45 cos 30 sin 45 sin 30
1 3 1 12 22 2
3 1 22 2 2
2 3 1
46 2
4
= ° ° − ° °
= × − ×
−= ×
−=
−=
© Cengage Learning Australia 2014 ISBN 9780170251501 26
Conversion between answers in textbook.
2 2
2
4(2 3)2 32 4
8 4 34
6 2 4 3 24
( 6) 2 6 2 ( 2)4
( 6 2)4
6 24
−−=
−=
− × +=
− × +=
−=
−=
c 7 4 3tan tan12 12 12π π π = +
( )
tan tan3 4
1 tan tan3 4
3 1 1 31 3 1 31 2 3 3
1 32 2 3
22 3
π π + =
π π −
+ += ×
− +
+ +=
−
+=
−= − −
© Cengage Learning Australia 2014 ISBN 9780170251501 27
d
5π π πsin sin12 6 4 = +
( )
π πsin cos sin cos6 4 4 6
1 1 1 32 22 21 3 22 2 2
2 1 3
4
π π = +
= × + ×
+= ×
+=
e
11π 3π 8π π 2πcot cot cot12 12 12 4 3
= + = +
( )( )
π 2πConsider tan4 3
2tan tanπ 2π 4 3tan
24 3 1 tan tan4 3
1 31 31 1 3 1 3
21 2 3 3
12 3
π 2πcot 3 24 3
+
π π + + = π π −
+−= ×
+ × +
−=
+ +−
=+
∴ + = − −
© Cengage Learning Australia 2014 ISBN 9780170251501 28
f sin (105°) = sin (60° + 45°)
= sin (60°) cos (45°) + sin (45°) cos (60°)
( )
3 1 1 12 22 23 1 2
2 2 2
2 3 1
4
= × + ×
+= ×
+=
g cos (195°) = cos (150° + 45°)
= cos (150°) cos (45°) – sin (45°) sin (150°)
( )
3 1 1 12 22 2
3 1 22 2 2
2 3 1
4
= − × − ×
+= − ×
+= −
h
7πsin12
= sin 4π 3π π πsin
12 12 3 4 + = +
= sin π3
cos π4
+ sin π4
cos π3
( )
3 1 1 12 22 23 1 2
2 2 2
2 3 1
4
= × + ×
+= ×
+=
© Cengage Learning Australia 2014 ISBN 9780170251501 29
i sin (195°) = sin (150° + 45°)
= sin (150°) cos (45°) + sin (45°) cos (150°)
( )
1 1 1 32 22 21 3 22 2 2
2 1 3
4
−= × + ×
−= ×
−=
© Cengage Learning Australia 2014 ISBN 9780170251501 30
2 ( )tan 75° = tan (x + y) where x = 45° and y = 30°
( ) ( )( ) ( )( ) ( )( ) ( )
13
13
13
13
3 13
3 13
tan tantan( )
1 tan tan
tan 45 tan 30tan(45 30 )
1 tan 45 tan 301
1 1
1
1
3 1 33 3 1
3 13 13 1 3 13 1 3 1
3 2 3 13 1
4 2 32
2 3
x yx y
x y
+
−
++ =
−
° + °° + ° =
− ° °
+=
− ×
+=
−
=
+= ×
−
+=
−
+ += ×
− +
+ +=
−+
=
= +
© Cengage Learning Australia 2014 ISBN 9780170251501 31
3 a sin ?12π =
1
12 2 6π π= ×
Use cos (2x) = 1 – 2 sin2(x)
( ) ( )
( )
2 2
2
2 3 2 2 32 4
8 4 34
6 2 12 24
6 2 6 2 2
4
6 2
46 2
4
− −=
−=
− +=
− +=
−=
−=
b 1cos ?8 8 2 4π π π = = ×
2
2
2
cos 2cos 14 8
2cos cos 18 4
1 1cos 18 2 2
1 2 2cos8 2 2 2
but 0 cos8
2 2so cos8 4
2 22
π π = −
π π = + π = +
π + = ± ×
π ≤
π + =
+=
© Cengage Learning Australia 2014 ISBN 9780170251501 32
c 5 5 3tan ?12 12 4 3π π π π = = −
( ) ( )( ) ( )
tan tanUse tan( )
1 tan tan
3tan tan5 4 3tan
312 1 tan tan4 3
1 31 ( 1) 3
1 3 1 31 3 1 31 2 3 3
1 34 2 3
22 3
x yx y
x y−
− =+
π π − π ∴ = π π +
− −=
+ − ×
− − += ×
− +
− − −=
−− −
=−
= +
d 1sin ?8 8 2 4π π π = = ×
2
2
2
cos 1 2sin4 8
2sin 1 cos8 4
1 1sin 18 2 2
2 1 2sin8 2 2 2
but sin 08
2 2so sin8 4
2 22
π π = −
π π = − π = −
π − = ± ×
π >
π − =
−=
© Cengage Learning Australia 2014 ISBN 9780170251501 33
e 5cos cos12 6 4π π π = +
( )
cos cos sin sin6 4 6 4
3 1 1 12 22 23 1 2
2 2 2
2 3 1
4
π π π π = −
= × − ×
−= ×
−=
f 1tan ?8 8 2 4π π π = = ×
Use ( ) ( )( )2
2 tan1 a
t n 2t
an
AA
A =−
and tan4π
= 1
2
2
2
2
2 tan81
1 tan8
1 tan 2 tan8 8
tan 2 tan 1 08 8
2 2 4( 1)1tan
8 2
2 82
2( 1 2)tan8 2
tan 08
tan 2 18
π =
π −
π π − =
π π + − =
− ± − −π =
− ±=
π − ± = π >
π = −
© Cengage Learning Australia 2014 ISBN 9780170251501 34
Exercise 9.06: Products to sums
Concepts and techniques
1 a ( ) ( )cos 10 cos 50° ° = ? given ( )sin 40 0.6428° ≈ and ( )cos 40 0.7660° ≈
( ) ( ) [ ]( ) ( ) [ ]
( )
12
12
12
12
Use cos cos cos( ) cos( )
so cos 10 cos 50 cos( 40 ) cos(60 )
0.7660 0.51.2660
0.6330
A B A B A B= × − + +
° ° = × − ° + °
= × +
= ×
=
b ( ) ( )sin 10 sin 50° ° = ? given ( )sin 40 0.6428° ≈ and ( )cos 40 0.7660° ≈
( ) ( ) [ ]( ) ( ) [ ]
( )
12
12
12
12
Use sin sin cos( ) cos( )
so sin 10 sin 50 cos( 40 ) cos(60 )
0.7660 0.50.2660
0.1330
A B A B A B= × − − +
° ° = × − ° − °
= × −
= ×
=
c ( ) ( )sin 10 cos 50° ° = ? given ( )sin 40 0.6428° ≈ and ( )cos 40 0.7660° ≈
( ) ( ) [ ]( ) ( ) [ ]
( )
12
12
312 2
Use sin cos sin( ) sin( )
so sin 10 cos 50 sin(60 ) sin( 40 )
0.6428
0.112
A B A B A B= + + −
° ° = × ° + − °
= × −
=
d ( ) ( )cos 10 sin 50° ° = ? given ( )sin 40 0.6428° ≈ and ( )cos 40 0.7660° ≈
( ) ( ) [ ]( ) ( ) [ ]
( )
12
12
312 2
Use cos sin sin( ) sin( )
so cos 10 sin 50 sin(60 ) sin( 40 )
0.6428
0.754
A B A B A B= + − −
° ° = × ° − − °
= × +
=
© Cengage Learning Australia 2014 ISBN 9780170251501 35
2 a 2 sin (5x) cos (7x) = sin (12x) + sin (–2x) = sin (12x) – sin (2x)
b 8 sin (3x) sin (9x) = 4[cos(–6x) – cos(12x)] = 4 cos (6x) – 4 cos (12x)
c 6 cos (4x) cos (6x) = 3 cos (10x) + 3 cos(–2x) = 3 cos (10x) + 3 cos (2x)
d 4 cos (5x) sin (6x) = 2[sin(11x) – sin(–2x)] = 2 sin (11x) + 2 sin (2x)
e sin (6x) cos (10x) = ( ) ( ) ( ) ( )1 1sin 16 sin 4 sin 16 sin 4
2 2x x x x+ − = −
f 3 cos (10x) cos (14x) = ( ) ( ) ( ) ( )3 3cos 24 cos 4 cos 24 cos 4
2 2x x x x+ − = +
g 5 sin (2x) sin (3x) = ( ) ( ) ( ) ( )5 5cos cos 5 cos cos 5
2 2x x x x− − = −
h 7 cos (9x) cos (11x) = ( ) ( ) ( ) ( )7 7cos 20 cos 2 cos 20 cos 2
2 2x x x x+ − = +
i cos (8x) sin (x) = ( ) ( ) ( ) ( )1 1sin 9 sin 7 sin 9 sin 7
2 2x x x x− − = +
3 a cos cos4 12π π
= ?
( ) ( )2cos cos cos( ) cos( )
1cos cos cos cos4 12 2 4 12 4 12
1 2 4cos cos2 12 12
1 cos cos2 6 3
1 3 12 2 2
3 1
Use
4
x y x y x y= − + +
π π π π π π = × − + + π π = × + π π = × +
= × +
+=
© Cengage Learning Australia 2014 ISBN 9780170251501 36
b sin cos4 12π π
= ?
( ) ( )2sin cos sin( ) sin( )
1sin cos sin sin4 12 2 4 12 4 12
1 4 2sin sin2 12 12
1 sin sin2 3 6
1 3 12 2 2
3 1
Use
4
x y x y x y= + + −
π π π π π π = × + + − π π = × + π π = × +
= × +
+=
c sin sin12 4π π
= ?
( ) ( )2sin sin cos( ) cos( )
1sin sin cos cos12 4 2 12 4 12 4
1 2 4cos cos2 12 12
1 cos cos2 6 3
1 3 12 2 2
3 1
U
4
se x y x y x y= − − +
π π π π π π = × − − + − π π = × − −π π = × −
= × −
−=
© Cengage Learning Australia 2014 ISBN 9780170251501 37
d 2sin cos12 4π π
= ?
( ) ( )2sin cos sin( ) sin( )
2sin cos sin sin12 4 12 4 12 4
4 2sin sin12 12
sin sin3 6
3 12 23 1
U e
2
s x y x y x y= + + −
π π π π π π = + + −
π − π = + π π = −
= −
−=
Reasoning and communication
4 Show that ( ) ( ) [ ]12sin cos sin( ) sin( )A B A B A B= + + −
[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
1 1sin( ) sin( ) sin cos cos sin sin cos sin cos2 2
1 2sin cos2sin cos
A B A B A B A B A B B A
A B
A B
+ + − = + + −
=
=
5 Show that ( ) ( )2cos sin sin( ) sin( )A B A B A B= + − −
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ){ }( ) ( )
sin( ) sin( ) sin cos cos sin sin cos sin cos
2cos sin
A B A B A B A B A B B A
A B
+ − − = + − − =
© Cengage Learning Australia 2014 ISBN 9780170251501 38
6 Show that ( ) ( ) [ ]1sin sin cos( ) cos( )2
g f f g f g= − − +
[ ] ( ) ( )
( ) ( )
( ) ( )
1 1 1 1cos( ) cos( ) 2sin sin2 2 2 2
1 1sin 2 sin 22 2
sin sin
f g f g f g f g f g f g
f g
f g
− − + = − + + + − − =
=
7 Show that ( ) ( )sin( ) sin( ) 2sin cosc d c d c d+ = − −
Using sin (x) – sin (y) = 2 cos sin 2 2
x y x y+ −
( ) ( )sin( ) sin( ) 2sin cosc d c d c d+ − − =
∴ ( ) ( )sin( ) sin( ) 2sin cosc d c d c d+ = − +
8 Show that sin (11x) cos (7x) − sin (8x) cos (4x) = sin (3x) cos (15x).
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ){ }
( ) ( )
( ) ( )
( ) ( )
sin 11 cos 7 sin 8 cos 41 sin 18 sin 4 sin 12 sin 421 sin 18 sin 1221 2sin 15 cos 32sin 15 cos 3
x x x x
x x x x
x x
x x
x x
−
= + − +
= +
=
=
9 Show that 4 sin (3x) sin (5x) sin (8x) = sin (6x) + sin (10x) − sin (16x).
( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
sin 6 sin 10 sin 16
2sin 8 cos( 2 ) 2sin 8 cos 8
2sin 8 cos 2 cos 8
2sin 8 2sin 5 sin 3
4sin 3 sin 5 sin 8
x x x
x x x x
x x x
x x x
x x x
+ −
= − −
= − = =
© Cengage Learning Australia 2014 ISBN 9780170251501 39
10 If w(t) = sin (5t) + sin (6t) + sin (7t) + sin (8t), then it can be seen in the diagram
below that every sixth wave is larger.
© Cengage Learning Australia 2014 ISBN 9780170251501 40
Exercise 9.07: Sums to products
Concepts and techniques
1 a Express 6 sin (x) + 8 cos (x) in each of the following forms and sketch the
graphs for x = 0 to 2π.
i A sin (x + θ) = Asin (x) cos (θ) + A cos (x) sin (θ)
= 6 sin (x) + 8 cos (x)
So 6 = A cos (θ) and 8 = A sin (θ)
62 + 82 = A2[cos2 (θ)+ sin2 (θ)]
∴62 + 82 = A2
A > 0, A = 10
( )( )
( )
sin 8cos 6
8tan6
0.9273
AA
θ=
θ
θ =
θ =
∴6 sin (x) + 8 cos (x) = 10 sin (x + 0.9273)
© Cengage Learning Australia 2014 ISBN 9780170251501 41
ii A cos (x + θ) = Acos (x) cos (θ) – A sin (x) sin (θ)
= 8 cos (x) + 6 sin (x)
So 8 = A cos (θ) and 6 = –A sin (θ)
82 + 62 = A2[cos2 (θ)+ sin2 (θ)]
∴82 + 62 = A2
A > 0, A = 10
( )( )
( )
sin 6cos 8
6tan8
0.6435
AA
θ −=
θ
−θ =
θ = −
∴6 sin (x) + 8 cos (x) = 10 cos (x – 0.6435)
© Cengage Learning Australia 2014 ISBN 9780170251501 42
iii A sin (x − θ) = Asin (x) cos (θ) – A cos (x) sin (θ)
= 6 sin (x) + 8 cos (x)
So 6 = A cos (θ) and 8 = –A sin (θ)
62 + 82 = A2[cos2 (θ)+ sin2 (θ)]
∴62 + 82 = A2
A > 0, A = 10
( )( )
( )
sin 8cos 6
8tan6
0.9273
AA
θ −=
θ
−θ =
θ = −
∴6 sin (x) + 8 cos (x) = 10 sin [x –(–0.9273)]
= 10 sin (x + 0.9273)
© Cengage Learning Australia 2014 ISBN 9780170251501 43
iv A cos (x − θ) = Acos (x) cos (θ) + A sin (x) sin (θ)
= 8 cos (x) + 6 sin (x)
So 8 = A cos (θ) and 6 = A sin (θ)
82 + 62 = A2[cos2 (θ)+ sin2 (θ)]
∴82 + 62 = A2
A > 0, A = 10
( )( )
( )
sin 6cos 8
6tan8
0.6435
AA
θ=
θ
θ =
θ =
∴6 sin (x) + 8 cos (x) = 10 cos(x – 0.6435)
b The graphs are all the same because they all come from the same the original
function.
© Cengage Learning Australia 2014 ISBN 9780170251501 44
2 Express ( ) ( )sin 3 cosx x+ in the form A sin (x + θ) and sketch the graph for
x = 0 to 2π.
( ) ( )sin 3 cosx x+ = A sin (x) cos (θ)+ A cos (x) sin (θ)
So 1 = A cos (θ) and 3 = A sin (θ)
12 + 3 2 = A2[cos2 (θ)+ sin2 (θ)]
∴12 + 3 2 = A2
A > 0, A = 2
( )( )
( )
sin 3cos 1
tan 3
3
AA
θ=
θ
θ =
πθ =
∴ ( ) ( )sin 3 cosx x+ = 2 sin 3x π +
© Cengage Learning Australia 2014 ISBN 9780170251501 45
3 A cos (x − θ) = Acos (x) cos (θ) + A sin (x) sin (θ)
= sin (x) + cos (x)
So 1 = A cos (θ) and 1 = A sin (θ)
12 + 12 = A2[cos2 (θ)+ sin2 (θ)]
∴12 + 12 = A2
A > 0 A = 2
( )( )
( )
sin 1cos 1
tan 1
4
AA
θ=
θ
θ =
πθ =
∴sin (x) + cos (x) = 2 cos 4x π −
© Cengage Learning Australia 2014 ISBN 9780170251501 46
4 A cos (x + θ) = Acos (x) cos (θ) – A sin (x) sin (θ)
= cos 3 sinx x− +
So –1 = A cos (θ) and – 3 = A sin (θ)
(–1)2 + (– 3 )2 = A2[cos2 (θ)+ sin2 (θ)]
∴ (–1)2 + (– 3 )2 = A2
A > 0, A = 2
( )( )
( )
sin 3cos 1
tan 343
AA
θ −=
θ −
θ =
πθ =
∴ ( ) ( )3 sin cosx x− = 2 cos
43
x π +
© Cengage Learning Australia 2014 ISBN 9780170251501 47
5 12 sin (x) + 5 cos (x) = 13 sin (x + θ) and tan–1(θ) = 512
so θ = 0.3948
12 sin (x) + 5 cos (x) = 13 sin (x +0.3948)
6 8 sin (x) − 15 cos (x) = 17 sin (x − θ) and tan–1(θ) = 158
so θ = 1.08
8 sin (x) − 15 cos (x) = 17 sin (x −1.08)
© Cengage Learning Australia 2014 ISBN 9780170251501 48
7 a sin (3x) + sin (7x) = 2 sin (5x) cos(–2x) = 2 sin (5x) cos (2x)
b sin (8x) − sin (2x) = 2 cos (5x) sin (3x)
c cos (5x) + cos (9x) = 2 cos (7x) cos (2x)
d cos (11x) − cos (7x) = 2 sin (9x) sin (–2x) = –2 sin (9x) sin (2x)
e sin (4x) + sin (9x) = 2 sin (6.5x) cos (2.5x)
f sin (2x) − sin (5x) = 2 cos (3.5x) sin(–1.5x) = –2 cos (3.5x) sin (1.5x)
g cos (x) + cos (5x) = 2 cos (3x) cos (2x)
h cos (3x) − cos (11x) = 2 sin (7x) sin (4x)
i sin (x) + sin (2x) = 2 sin (3.5x) cos (0.5x)
8 Using ( ) ( ) ( ) ( )1 12 2cos cos 2cos cosA B A B A B+ = + −
( ) ( )cos 120 cos 60 2cos(90 )cos(30 )
32 02
0
° + ° = ° °
= × ×
=
.
9 Using ( ) ( ) [ ] [ ]1 12 2sin sin 2sin ( ) cos ( )A B A B A B+ = + −
( ) ( )sin 75 sin 15 2sin (45 )cos (30 )
1 3222
3 22 26
2
° + ° = ° °
= × ×
= ×
=
.
10 Using ( ) ( ) [ ] [ ]1 12 2sin sin 2cos ( ) sin ( )A B A B A B− = + −
( ) ( ) ( )sin 135 sin( 45 ) 2cos 45 sin 9012 12
1 222 2
2
° − − ° = ° °
= × ×
= × ×
=
© Cengage Learning Australia 2014 ISBN 9780170251501 49
Reasoning and communication
11 Show that 2 cos (x) cos (y) = cos (x + y) + cos (x − y).
Given
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
cos cos sin sinancos
cos d
cos cos sin sin
x y x y
x y x y
x y
x y
+
−
= −
= +
by adding we obtain cos (x + y) + cos (x − y). = 2 cos (x) cos (y)
12 Show ( ) ( ) ( ) ( )1 1sin sin 2 sin cos .2 2
A B A B A B + = + −
Given
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
sin cos cos sinands
sin
in
sin cos cos sin
x y
x y
x y x y
x y x y
+ =
−−
+
=
By adding we get
( )sin x y+ + ( )sin x y− = 2 ( ) ( )sin cosx y
Put A = ( )x y+ and B = ( )x y−
then A + B = 2x and A – B = 2y
so and = 2 2
A B A Bx y+ −=
∴ ( )sin x y+ + ( )sin x y− = 2 ( ) ( )sin cosx y becomes
( ) ( ) ( ) ( )1 1sin sin 2 sin cos .2 2
A B A B A B+ = + −
© Cengage Learning Australia 2014 ISBN 9780170251501 50
13 Show that ( ) ( ) ( ) ( )1 1cos cos 2 cos cos .2 2
A B A B A B + = + −
Given
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
cos cos sin sinancos
cos d
cos cos sin sin
x y x y
x y x y
x y
x y
+
−
= −
= +
By adding we get
( ) ( ) ( )coco s( ) 2cos cos sx yx y x y+ − =+
Put A = ( )x y+ and B = ( )x y−
then A + B = 2x and A – B = 2y
so and =2 2
A B A Bx y+ −=
∴ ( ) ( ) ( )coco s( ) 2cos cos sx yx y x y+ − =+ becomes
( ) ( ) ( ) ( )1 1cos cos 2 cos cos .2 2
A B A B A B + = + −
14 Both notes are sine waves. At the same intensity (volume) they will add together to
make the sound.
Say one is at f = 110 and the other is at 110 + a
Then sum will be
( ) ( )sin 110 2 sin 110 2 2sin 110 2 cos 22 2a aa t t t t + π + × π = + π × × π
As a decreases, the beats will decrease in frequency (will get further apart) and the
dominant note 110 hz remain essentially unchanged.
© Cengage Learning Australia 2014 ISBN 9780170251501 51
15 a sin (x + h) − sin (x) = ( ) ( )2cos 0.5 sin 0.5x h h+
b ( ) ( ) ( )2cossin ( ) sin si0 n.5 0.5x x h hh
hx
h++ −
= .
c Given that ( )
0
sinlim 1z
zz→
= , find ( )sind xdx
from the definition of the derivative.
( ) ( ) ( )
( ) ( )
( ) ( )
( )( )
0
0
0
2cos 0.5 0.5
cos 0.5 0.
sinsin lim
sinlim
sinlim
50.5
0.50.5cos
cos 1
c s
5
o
0.
h
h
h
x h hh
x h hh
hx h
d xdx
x
x
h
→
→
→
=
=
=
= ×
+
+
=
+
© Cengage Learning Australia 2014 ISBN 9780170251501 52
Exercise 9.08: Trigonometric identities
Reasoning and communication
1 Prove the identity 2 2 21 1sin (4 ) sin (2 )cos (2 )8 2
x x x=
( ) ( )
( ) ( )
( ) ( )
22
2 2
2 2
1 1sin (4 ) 2sin 2 cos 28 2
1 4sin 2 cos 222sin 2 cos 2
x x x
x x
x x
=
=
=
2 Show that 2 2 42sin (6 ) 8 sin (3 ) sin (3 )x x x = −
( ) ( )( ) ( )
( ) ( )
22
2 2
2 2
2sin (6 ) 2 2sin 3 cos 3
2 4sin 3 cos 3
8sin 3 cos 3
x x x
x x
x x
= =
=
3 Prove the following identities.
a ( ) ( ) ( ) ( ) 22 22sin 2sin cos sin 1x x x x− − = − −
( ) ( ) ( ) ( ) ( ) ( ){ }( ) ( )
( )
( ) ( )
( )
2 2 2 2
2
2
22
2
2sin 2sin cos 2sin 2sin 1 sin
sin 2sin 1
sin 1
1 1 sin
1 sin
x x x x x x
x x
x
x
x
− − = − − + −
= − − +
= − −
= − − −
= − −
© Cengage Learning Australia 2014 ISBN 9780170251501 53
b ( )1 cos cos cos cos2 3 2 6 2 6
x xx π π π + = + −
( )
( )
( )
1cos cos cos cos2 6 2 6 2 2 6 2 6 2 6 2 6
1 2cos cos2 6
1 cos cos2 3
1 cos cos2 3
x x x x x x
x
x
x
π π π π π π + − = + − − + + + − π = + π = + π = +
4 Prove the following identities.
a ( ) ( )
1cosec cotx x
=+
[ ]( ) ( )
12
12
1 cos (2 )tan cos sin (2 )
xx x x
−
+
Take RHS:
[ ]( ) ( )
( ){ }( ) ( )
( ) ( ) ( ){ }( )
( ) ( ) ( )( )
( )[ ]( )
( ) ( )
( )
( )( )
( )
( ) ( )
21122
1 12 2
212
2
sin coscos
1 1 2sin1 cos (2 )tan cos sin (2 ) 2sin cos
2sinsin sin cos
sinsin 1 cos ( )
sin1 cos ( )
1sin sin
11 cos ( )sin
1cos1
sin sin1
cos ec cot
x xx
xxx x x x x
xx x x
xx x
xx
x xx
x
xx x
x x
− −− =+ +
×=
+
=+
=+
= ×+
=+
=+
© Cengage Learning Australia 2014 ISBN 9780170251501 54
b [ ]21 sin ( ) sin ( )4
x y x y+ + − =
( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2sin sin sin 2sin sin 2sin sinx x y x y x y− − +
Take LHS:
[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )( ) ( )
22
2
2 2
1 1sin ( ) sin ( ) sin cos cos sin sin cos cos sin4 4
sin cos
sin cos
x y x y x y x y x y x y
x y
x y
+ + − = + + −
= =
Take RHS:
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ){ }( )
2 2 2 2
22 2
2
sin sin sin 2sin sin 2sin sin
sin 1 sin 2sin 2sin
sin 1 2sin
x x y x y x y
x y y y
x y
− − +
= − − +
= − ( ) ( ) ( )2 2sin 2sin 2siny y y+ − +
( ) ( )( ) ( )
2 2
2 2
sin 1 sin
sin cos
x y
x y
= −
=
∴ [ ]21 sin( ) sin( )4
x y x y+ + − =
( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2sin sin sin 2sin sin 2sin sinx x y x y x y− − +
© Cengage Learning Australia 2014 ISBN 9780170251501 55
c ( ) ( )
1sin cosecx x
=+
( )( ) ( ) ( )
12
2 2 2
sec sin (2 )1 cos cosec cot
x xx x x− + −
Take RHS:
( )( ) ( ) ( )
( ) ( ) ( )
( ) ( )( )( )
( )( ) ( )
( )
( )( ) ( )
( )( )
( ) ( )( )( )
( )( )
( )
( )
( ) ( )
( ) ( )
11 222 2 2
2
4 2
2
2 2
2
2
2
2 2
4 2
2
1cos
cos1sin sin
sin 1 cossin
2sin cossec sin (2 )1 cos cosec cot sin
sin
sinsin sin
sin
sinsin sin 1
sin1
sin sin1sin 1
sin1
1sin sin1
sin cosec
x
xx x
x xx
x xx xx x x x
x
xx x
x
xx x
x
x xx
x
x x
x x
+ −
× ×=
− + − + −
=
=+
=+
= ×+
=+
=+
© Cengage Learning Australia 2014 ISBN 9780170251501 56
5 1 sin(3 ) cos(2 )3
x x+ in terms of sin only
( )
( ) ( ) ( ) ( )( ) ( )
( ) ( ) ( ) ( ) ( ){ } ( )
( ) ( ) ( ) ( ){ } ( )
( ) ( ) ( ) ( )
( )
2
2
2 2
2 2 2
2 2
3
1 sin(2 ) 1 2sin31 sin 2 cos sin cos 2 1 2sin31 2sin cos cos sin 1 2sin 1 2sin31 2sin 1 sin sin 1 2sin 1 2sin3sin 21 1 2sin sin 1 sin
3 3
2 sin 2si3
x x x
x x x x x
x x x x x x
x x x x x
xx x x
x
= + + −
= + + −
= + − + −
= − + − + −
= + − + −
= − − ( ) ( ) ( ) ( )
( ) ( ) ( )
2 3
3 2
sin 2 2n 1 sin sin3 3 3
4 sin 2sin sin 13
xx x x
x x x
+ + + −
= − − + +
6 ( )22sin cos(2 )x x− in terms of ( )cos x only.
[ ]( )
2
2 2
2 2
4 2 2
4 2
2sin ( ) cos (2 ) 1 cos (2 )
1 2cos ( ) 1 2cos ( ) 1
2cos ( ) 2 2cos ( ) 1
4cos ( ) 2cos ( ) 4cos ( ) 24cos ( ) 6cos ( ) 2
x x x
x x
x x
x x xx x
− = − +
= − + − −
= − − = − − +
= − +
7 Prove that sec2 (x) = sec (x) cosec (x) tan (x).
( ) ( ) ( ) ( ) ( )( )( )
( )( )
2
2
sin1 1sec cosec tancos sin cos
1cos
sec
xx x x
x x x
x
x
= × ×
=
=
© Cengage Learning Australia 2014 ISBN 9780170251501 57
8 Prove that ( ) ( ) ( ) ( )1 1 2 tan sec
cosec 1 cosec 1x x
x x+ =
+ −
( ) ( ) ( ) ( )
( )( )
( )( )
( )( )
( )( )
( ) ( ) ( )( ) ( )
( ) ( )( ) ( )
2
1 1sin sin
1 sin 1 sinsin sin
1 1 1 1cosec 1 cosec 1 1 1
1 1
sin sin1 sin 1 sin
1 sin 1 sinsin
1 sin 1 sin
2sincos
2 tan sec
x x
x x
x x
x x
x xx x
x xx
x x
xx
x x
+ −
+ = ++ − + −
= +
= ++ −
− + + = + −
=
=
.
9 Show that 1 + cos (2x) = 2 cos2 (x).
1 + cos (2x) = ( ) ( )2 21 2cos 1 2cosx x+ − =
10 Prove that ( ) ( ) ( ) ( )cosec coseccot cot
cosec ( )x y
x yx y
+ =+
.
( ) ( )( ) ( )( ) ( ) ( ) ( )
( ) ( )( )( )
( )( )
( ) ( )
cosec cosec sin( )cosec ( ) sin sin
sin cos cos sinsin sin
cos cossin sin
cot cot
x y x yx y x y
x y x yx y
y xy x
y x
+=
+
+=
= +
= +
© Cengage Learning Australia 2014 ISBN 9780170251501 58
Chapter 9 Review
Multiple choice
1 ( )sin 60° = 32
C
2 ( )cosec 60° =23
A
3 5 1sin sin6 6 2π π π − = =
E
4 ( ) ( ) ( )tan 360 30 tan 330 tan 30° − ° = ° = − ° C
5 cos(75°) =
( ) ( ) ( ) ( )
( )
75 cos 5 30 cos 5 cos 30 sin 5 sin 3
cos( ) (40
1 3 1 12 22
)4 4
23 1 2
2 2 2
2 3 1
4
° = ° + °
= ° ° − ° °
= × − ×
−= ×
−=
A
© Cengage Learning Australia 2014 ISBN 9780170251501 59
6 5sin8π
= ?
( ) ( )
( )
2
2 5 5cos8 4
5sin 08
1 cos 25s
1cos2
cos 2 1 2sin
12
2 12
2 1 22 2
2 2
in8 2
1
2
2
2
4
2 22
A
A
A
× π π
π >
−π = +
=
−
= −
= −
−=
+=
+=
+=
+=
×
E
7 E cos (5x) cos (9x)
[ ]
[ ]
[ ]
1 cos (5 9 ) cos (5 9 )21 cos (14 ) cos ( 4 )21 cos (14 ) cos (4 )2
x x x x
x x
x x
= + + −
= + −
= +
E
8 sin (12x) − sin (8x)
( ) ( )
12 8 12 82 cos sin2 2
2cos 10 sin 2
x x x x
x x
+ − = =
B
© Cengage Learning Australia 2014 ISBN 9780170251501 60
Short answer
9 a ( )sec(2 ) secπ − θ = θ
b ( )cosec( ) cos ecπ − θ = θ
c ( )cot(2 ) cotπ − θ = − θ
d ( )sec( ) secπ + θ = − θ
e ( )cot( ) cotπ + θ = θ
f ( )cosec(2 ) cos ecπ − θ = − θ
10 a ( )sec cosec2π − θ = θ
b ( )3cosec cosec cosec s ec2 2 2π π π − θ = π + − θ = − − θ = − θ
c ( )cot tan2π − θ = θ
d ( )3sec sec sec cosec2 2 2π π π + θ = π + + θ = − + θ = θ
e ( )3cot cot cot tan2 2 2π π π − θ = π + − θ = − θ = θ
f ( )cosec sec2π − θ = θ
© Cengage Learning Australia 2014 ISBN 9780170251501 61
11 ( )3cosecy x= for 0 2 ,x x< < π ≠ π
12 ( )2sec 1y x= − for 30 2 , ,2 2
x x π π≤ ≤ π ≠
13 Prove the ( ) ( ) ( )2 3sec 1 tan cotx x x− =
( ) ( )( )
( )( )( )
( ) ( )( )
( ) ( )
22
2
2
2
2
2
3
1sec 1 1cos
1 coscos
sincos
tantan
tan
tan cot
xx
xx
xx
xx
x
x x
− = −
−=
=
= ×
=
© Cengage Learning Australia 2014 ISBN 9780170251501 62
14 cos cos12 3 4π π π = −
cos cos sin sin3 4 3 4
1 1 3 12 22 21 3 22 2 22 6
4
π π π π = +
= × + ×
+= ×
+=
15 tan tan12 3 4π π π = −
( )
( )
2
tan tan3 4
1 tan tan3 4
3 1 1 31 3 1 3
1 3
1 31 2 3 3
24 2 3
22 3
π π − =
π π +
− −= ×
+ −
−= −
−
− += −
−−
=
= −
16 7sin sin12 3 4π π π = +
sin cos cos sin3 4 3 4
3 1 1 12 22 23 1 2
2 2 26 2
4
π π π π = +
= × + ×
+= ×
+=
© Cengage Learning Australia 2014 ISBN 9780170251501 63
Application
17 ( ) ( ) ( )sin 4 2sin 2 cos 2x x x=
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2 2
3 3
2 2sin cos cos sin
4sin cos 4sin cos
x x x x
x x x x
= − = −
18 Show that ( ) ( ) [ ]12sin sin cos( ) cos( )A B A B A B= − − + .
Given ( ) ( ) ( ) ( )cos( ) cos cos sin sinA B A B A B− = +
and ( ) ( ) ( ) ( )cos( ) cos cos sin sinA B A B A B+ = −
then cos (A –B) – cos (A + B) = ( ) ( ) ( ) ( )cos cos sin sinA B A B+
– ( ) ( ) ( ) ( )cos cos sin sinA B A B−
i.e. cos(A –B) – cos (A + B) = 2 ( ) ( )sin sinA B
∴ ( ) ( ) [ ]12sin sin cos( ) cos( )A B A B A B= − − +
19 Show that ( ) ( )
( )( ) ( )( ) ( )
2
2
2 tan cos 1cos seccot 2 cos 1 tan
x xx xx x x
++ = −
( ) ( )( ) ( )
( )( )
( )( )
( ) ( ) ( )( ) ( )
( )
2 2
22
2 tan cos 1 cos 12 tancos1 tancos 1 tan
tan 2 cos sec
cos seccot 2
x x xxxxx x
x x x
x xx
+ + = ×− −
= × + +
=
© Cengage Learning Australia 2014 ISBN 9780170251501 64
20 Show that ( ) ( ) ( ) ( ) ( ) 231 cos
4sin cos 2sin cos8
xx x x x
− = −
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( )
( ) ( ) ( )
( )
( )
2 223 2
2 2
2
2
2 2
4sin cos 2sin cos 2sin cos 2cos 1
sin 2 cos 2
2sin 2 cos 24
sin 44
1 2sin 4 but cos 2 =1 2sin81 1 cos 881 cos 8
8
x x x x x x x
x x
x x
x
x A A
x
x
− = −
=
=
=
= × −
= × −
−=
21 Show that ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2cos tan sin 3 sin cos 2 sin cos sin 2x x x x x x x x= +
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( )
( ) ( ) ( )( )
( ) ( ) ( )
2sin cos 2 sin cos sin 2 sin sin cos 2 cos sin 2
sin sin( 2 )
sin sin 3
cossin sin 3
cos
cos tan sin 3
x x x x x x x x x x
x x x
x x
xx x
x
x x x
+ = + = +
=
= ×
=
22 Rewrite the expression ( ) ( ) ( )3sin(2 )sin cos cos cos(2 )x x x x x+ in terms of
( )cos x only.
( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
2
2 2 2
3sin(2 )sin cos cos cos(2 )
3 2sin cos sin cos cos 2cos 1
6cos ( ) 1 cos cos 2cos 1
x x x x x
x x x x x x
x x x x
+
= + − = − + −
© Cengage Learning Australia 2014 ISBN 9780170251501 65