Post on 17-May-2018
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 209
Exercise 7A — Circular and reciprocal functions
1 a 2π θ+ is in quadrant 2 and sin is positive in that quadrant.
The complementary function is cos(θ).
sin2π θ +
= cos(θ)
b 32π θ− is in quadrant 3 and sin is negative in that
quadrant. The complementary function is cos(θ).
3sin2π θ −
= −cos(θ)
c 32π θ+ is in quadrant 4 and cos is positive in that
quadrant. The complementary function is sin(θ).
3cos2π θ +
= sin(θ)
d 2π θ+ is in quadrant 2 and tan is negative in that
quadrant. The complementary function is cot(θ).
tan2π θ +
= −cot(θ)
e 32π θ+ is in quadrant 4 and cot(tan) is negative in that
quadrant. The complementary function is tan(θ).
3cot2π θ +
= −tan(θ)
2 a sec2π θ −
= cosec(θ)
LHS = sec2π θ −
= 1
cos2π θ −
= 1sin( )θ
= cosec(θ) = RHS
b 3cosec2π θ +
= sec(θ)
LHS = 3cosec2π θ +
= 1
sin 22ππ θ − −
= 1
sin2π θ − −
= 1cos( )θ−
= −sec(θ) = RHS
c 3sec2π θ −
= −cosec(θ)
LHS = 3sec2π θ −
= 13cos2π θ −
= 1
cos2ππ θ + −
= 1
cos2π θ − −
= 1sin( )θ−
= −cosec(θ) = RHS
d 3tan2π θ −
= cot(θ)
LHS = 3tan2π θ −
=
3sin2
3cos2
π θ
π θ
− −
= sin
2
cos2
ππ θ
ππ θ
+ − + −
= sin
2
cos2
π θ
π θ
− − − −
= cos( )sin( )
θθ
= cot(θ) = RHS
e cos
2
sin2
π θ
πθ
+ −
= tan(θ)
LHS = cos
2
sin2
π θ
πθ
+ −
= cos
2
sin 02
ππ θ
π θ
− − − −
= cos
2
sin2
π θ
π θ
− − − −
Chapter 7 — Advanced periodic functions
M C 1 2 Q l d - 7 210 A d v a n c e d p e r i o d i c f u n c t i o n s
= sin( )cos( )
θθ
= tan(θ) = RHS
3 a cos sin( )2π θ θ − + −
= sin(θ) −sin(θ)
= 0
b sin sec( )2π θ π θ + × +
= 1sin2 cos( )ππ θ
π θ − − × +
= 1sin2 cos( )π θ
θ − × −
= 1cos( )cos( )
θθ
×−
= −1
c 3cosec cot(2 )2π θ π θ − × −
= 1 cos(2 )
3 sin(2 )sin2
π θπ π θθ
−×− −
= 1 cos(2 )sin(2 )sin
2
π θπ θππ θ
−×− + −
= 1 cos( )sin( )sin
2
θπ θθ
×− − −
= 1 cos( )cos( ) sin( )
θθ θ
×− −
= 1sin( )θ
= cosec(θ)
d cosec
sin( ) 2cos(2 ) sec
2
π θπ θ
ππ θ θ
− + ++ +
= cos
sin( ) 2cos(2 ) sin
2
π θπ θ
ππ θ θ
+ + ++ −
= cos
2sin( )cos(2 ) sin
2
ππ θπ θ
ππ θ θ
− − + ++ −
= cos
sin( ) 2cos( ) sin
2
π θθ
πθ θ
− − − + −
= sin( ) sin( )cos( ) cos( )
θ θθ θ
− −+
= −2 tan(θ)
e
3cot sec2 2
cosec2
π πθ θ
π θ
+ × −
+
=
cos12
3sin cos2 2
1
sin2
π θ
π πθ θ
π θ
+ × + −
+
= cos
12 sin3 2sin cos
2 2
π θπ θ
π πθ θ
+ × × + + −
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 211
= cos
2 1 sin2sin cos
2 2
ππ θππ θ
π ππ θ π θ
− − × × − − − − + −
= cos
12 sin2sin cos
2 2
π θπ θ
π πθ θ
− − × × − − − −
= sin( ) cos( )cos( ) sin( )
θ θθ θ
− ×−
= 1
f sin cos
2 2sec( )
π πθ θ
π θ
+ − +
− =
sin cos2 2
1cos( )
π ππ θ π θ
π θ
− − − − −
−
= sin cos cos( )2 2π πθ θ π θ − + − × −
= (cos(θ) + sin(θ)) × −cos(θ) = −cos(θ) (cos(θ) + sin(θ))
Exercise 7B — Exact values and the Pythagorean relationships
1 a 5sin6π
= sin6ππ −
= sin6π
= 12
b 5cosec6π
= 15sin6π
= 112
(from part a)
= 2
c 3cos4π
= cos4ππ −
= cos4π −
= 12
−
= 22
−
d 3sec4π
= 13cos4π
= 112
− (from part c)
= 2−
e 4tan3π
= tan3ππ +
= tan3π
= 3
M C 1 2 Q l d - 7 212 A d v a n c e d p e r i o d i c f u n c t i o n s
f 4cot3π
= 14tan3π
= 13
(from part e)
g cosec3π −
= 1
sin3π −
= 1
sin3π −
= 13
2−
= 23
−
= 2 33
−
h 3cot sec4 6π π ×
= 1 1
3tan cos4 6π π
×
= 1 1
tan cos4 6π ππ
× −
= 1 13tan
4 2π
× −
= 1 21 3
×−
= 2 33
−
i sec(315°) + cosec(−240°) = 1 1cos(315 ) sin( 240 )
+° − °
= 1 1cos(360 45 ) sin(240 )
+° − ° − °
= 1 1cos(45 ) sin(180 60 )
+° − ° + °
= 1 11 sin(60 )2
+°
= 123
2
+
= 223
+
= 3 2 2 33+
j cosec2(−330°) + cot2(−330°) + 1 = 2 2
1 1 1[sin( 330 )] [tan( 330 )]
+ +− ° − °
= 2 21 1 1
( sin(330 )) [ tan(330 )]+ +
− ° − °
= 2 21 1 1
[ sin(360 30 )] [ tan(360 30 )]+ +
− ° − ° − ° − °
= 2 21 1 1
sin (30 ) tan (30 )+ +
° °
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 213
= 2 2
1 1 11 12 3
+ +
= 1 1 11 14 3
+ +
= 4 + 3 + 1 = 8
2 sin(x) = 35
and cos(x) < 0 ⇒ Quad 2
a sin2(x) + cos2(x) = 1
2
23 cos ( )5
x +
= 1
cos2(x) = 9125
−
= 1625
cos(x) = 45
±
As x is in quad 2, cos(x) = 45
−
cot(x) = cos( )sin( )
xx
=
4535
−
= 43
−
b sec(x) = 1cos( )x
= 54
−
c (cosec(x) − sec(x))2 = 2
1 1sin( ) cos( )x x
−
=
2
1 13 45 5
− −
= 25 5
3 4 +
= 235
12
= 1225144
= 738144
d cot( ) sec( )cos( )x x
x+ =
4 53 4
45
− −+
−
= 16 15 512 12 4− − − + ×
= 31 512 4− −×
= 15548
= 11348
3 Quad 2, sin(θ) = 513
, sin2(θ) + cos2(θ) = 1
2
25 cos ( )13
θ +
= 1
cos2(θ) = 251169
−
= 144169
cos(θ) = 1213
±
As θ is in quad 2, cos(θ) = 1213
− , tan(θ) = 512
−
a cot(θ) − tan(θ) = 1 tan( )tan( )
θθ
−
= 12 55 12
− −−
= 144 2560
− +
= 11960
−
= 59160
−
b cosec(θ) − sec(θ) = 1 1sin( ) cos( )θ θ
−
= 13 135 12
−−
= 156 6560+
= 22160
= 41360
c cos( ) sin( )cos( ) sin( )
θ θθ θ
+−
=
12 513 1312 5
13 13
− +
− −
=
71317
13
−
−
= 7 1313 17− ×
−
= 717
4 sin2(θ ) + cos2(θ ) = 1
2
2 1sin ( )3
θ +
= 1
sin2(θ ) = 113
−
= 23
sin(θ ) = 23
±
M C 1 2 Q l d - 7 214 A d v a n c e d p e r i o d i c f u n c t i o n s
As θ is acute, sin(θ) = 23
, tan(θ) = 2
cot( )1 cosec( )
θθ−
=
1tan( )
11sin( )
θ
θ−
=
12
312
−
=
122
2 3 22
×−
= 1 2 32 3 2 3
+×− +
= ( 2 3)− +
5 sin(y) = 0.8 and y is in quad 1 sin2(y) + cos2(y) = 1 0.64 + cos2(y) = 1 cos2(y) = 0.36 cos(y) = ±0.6 As y is in quad 1, cos(y) = 0.6
a sec2
yπ +
= 1
cos2
yπ +
= 1
cos2
yππ − −
= 1
cos2
yπ − −
= 1sin( )y−
= 10.8−
= 54
−
b cosec2
yπ −
= 1
sin2
yπ −
= 1cos( )y
= 10.6
= 53
c 3sin2
yπ +
= sin 22
yππ − −
= sin2
yπ − −
= −cos(y) = −0.6
= 35
−
d cot2
yπ −
= cos
2
sin2
y
y
π
π
− −
= sin( )cos( )
yy
= 0.80.6
= 43
e sec(2π − y) = 1cos(2 )yπ −
= 1cos( )y
= 10.6
= 53
f 3cosec2
yπ −
= 13sin2
yπ −
= 1
sin2
yππ + −
= 1
sin2
yπ − −
= 1cos( )y−
= 10.6−
= 53
−
6 a cot2(x) − cosec2(x) = 2
2cos( ) 1sin( ) sin ( )
xx x
−
= 2
2 2cos ( ) 1sin ( ) sin ( )
xx x
−
= 2
2cos ( ) 1
sin ( )x
x−
= 2
2sin ( )
sin ( )x
x−
= −1
b (1 − sin2(x)) sec2(x) = 221cos ( )
cos ( )x
x×
= 1
c 1 1sec( ) 1 sec( ) 1θ θ
+− +
= (sec( ) 1) (sec( ) 1)(sec( ) 1)(sec( ) 1)
θ θθ θ
+ + −− +
= 2
2sec( )sec ( ) 1
θθ −
= 22sec( )tan ( )
θθ
= 2
22 cos ( )
cos( ) sin ( )θ
θ θ×
= 2cos( ) 1sin( ) sin( )
θθ θ
×
= 2cot(θ) cosec(θ) or 2cos(θ) cosec2(θ)
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 215
d cot( ) cot( )cosec( ) 1 cosec( ) 1
θ θθ θ
+− +
= cot( )(cosec( ) 1) cot( )(cosec( ) 1)(cosec( ) 1)(cosec( ) 1)
θ θ θ θθ θ
+ + −− +
= 2cot( )(2cosec( ))
cosec ( ) 1θ θ
θ −
= 2cot( )(2cosec( ))
cot ( )θ θ
θ
= 2cosec( )cot( )
θθ
= 2 sin( )sin( ) cos( )
θθ θ
×
= 2cos( )θ
= 2sec(θ)
e 2 2(1 cos ( ))(1 tan ( ))x x− + = 2 2sin ( ) sec ( )x x×
= 221sin ( )
cos ( )x
x×
= 2tan ( )x
= tan( )x
f tan( ) cot( )cosec( )sec( )
x xx x+ =
sin( ) cos( )cos( ) sin( )
1 1sin( ) cos( )
x xx x
x x
+
×
= sin( ) cos( ) sin( )cos( )cos( ) sin( )
x x x xx x
+ ×
= 2 2sin ( ) cos ( ) sin( )cos( )
sin( )cos( )x x x xx x
+ ×
= 1
g sin( ) tan
2
cot( )sin2
πθ θ
ππ θ θ
− − + +
=
sin2sin( )
cos2
cos( ) sinsin( ) 2
π θθ
π θ
π θ ππ θπ θ
− − × −
+ − − +
=
cos( )sin( )sin( )
cos( ) sinsin( ) 2
θθθ
θ π θθ
− ×
− × − −
= sin( )cot( )cot( )cos( )
θ θθ θ
−
= sin( )cos( )
θθ
−
= −tan(θ)
h
3cos( )cot2
tan( )cos2
πθ θ
ππ θ θ
− − − −
=
3cos2cos( )
3sin2
tan( )sin( )
π θθ
π θ
θ θ
− × −
−
=
cos2cos( )
sin2
tan( )sin( )
ππ θθ
ππ θ
θ θ
+ − × + −
−
M C 1 2 Q l d - 7 216 A d v a n c e d p e r i o d i c f u n c t i o n s
=
cos2cos( )
sin2
tan( )sin( )
π θθ
π θ
θ θ
− − × −
−
=
sin( )cos( )cos( )
tan( )sin( )
θθθ
θ θ
−×
−
= cos( ) tan( )tan( )sin( )
θ θθ θ
−
= cos( )sin( )
θθ
−
= −cot(θ)
7 a 1 11 sin( ) 1 sin( )x x
++ −
= 2sec2(x)
LHS = 1 11 sin( ) 1 sin( )x x
++ −
= (1 sin( )) (1 sin( ))(1 sin( ))(1 sin( ))
x xx x
− + ++ −
= 22
1 sin ( )x−
= 22
cos ( )x
= 2sec2(x) = RHS
b 1 cos( )1 cos( )
xx
+−
= (cosec(x) + cot(x))2
RHS = (cosec(x) + cot(x))2
= 2
1 cos( )sin( ) sin( )
xx x
+
= 2
1 cos( )sin( )
xx
+
= 2
2(1 cos( ))
sin ( )xx
+
= 2
2(1 cos( ))1 cos ( )
xx
+−
= 2(1 cos( ))
(1 cos( ))(1 cos( ))x
x x+
− +
= 1 cos( )1 cos( )
xx
+−
= LHS
c tan( ) cot( )cot( ) tan( )
x yx y
++
= tan( )tan( )
xy
LHS = tan( ) cot( )cot( ) tan( )
x yx y
++
=
sin( ) cos( )cos( ) sin( )cos( ) sin( )sin( ) cos( )
x yx yx yx y
+
+
=
sin( )sin( ) cos( )cos( )cos( )sin( )
cos( )cos( ) sin( )sin( )sin( )cos( )
x y x yx y
x y x yx y
+
+
= sin( )sin( ) cos( )cos( )cos( )sin( )
x y x yx y+
sin( )cos( )cos( )cos( ) sin( )sin( )
x yx y x y
×+
= sin( )cos( )cos( )sin( )
x yx y
= sin( ) cos( )cos( ) sin( )
x yx y
×
= tan( )tan( )
xy
= RHS
d 1sec( ) tan( )y y−
= sec(y) + tan(y)
LHS = 1sec( ) tan( )y y−
= 11 sin( )
cos( ) cos( )y
y y−
= 11 sin( )
cos( )y
y−
= cos( )1 sin( )
yy−
= cos( ) 1 sin( )1 sin( ) 1 sin( )
y yy y
+×− +
= 2cos( ) sin( )cos( )
1 sin ( )y y y
y+−
= 2cos( ) sin( )cos( )
cos ( )y y y
y+
= 2 2cos( ) sin( )cos( )cos ( ) cos ( )
y y yy y
+
= 1 sin( )cos( ) cos( )
yy y
+
= sec(y) + tan(y) = RHS
8 sin( ) sin( )1 cos( ) 1 cos( )
x xx x
−− +
= y2 cot(x)
sin( )(1 cos( )) sin( )(1 cos( ))(1 cos( ))(1 cos( ))
x x x xx x
+ − −− +
= y2 cot(x)
2sin( )(1 cos( ) 1 cos( ))
1 cos ( )x x x
x+ − +
− = y2 cot(x)
2sin( )(2cos( ))
sin ( )x x
x = y2 cot(x)
2 cot(x) = y2 cot(x) y2 = 2 y = 2± 9 a i As ∆OAB is an equilateral triangle, AB = 6 cm.
Therefore, DA = 3 cm. ii As ∆OAB is an equilateral triangle, OC = 6 cm
Consider ∆ADO. 2
AO =2
AD +2
OD
62 = 32 +2
OD
2
OD = 36 − 9 = 27 OD = 27 = 3 3
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 217
DC = OC OD− = 6 3 3− = 3(2 3)− b Consider ∆OAC, OA = OC , therefore ∆OAC = ∆OCA As ∆AOC = 30°, ∆OAC = 1
2(180° − 30°) = 75°
As ∆OAD = 60°, ∆DAC = 15° Consider ∆ADC,
tan(15)° = DCAD
= 3(2 3)3−
= 2 3− c tan(165)° = tan(180° − 15°) = −tan(15)° = (2 3)− −
= 3 2−
Exercise 7C — Modelling soundwaves
1 a 1f
= 2πω
1100
= 2πω
ω = 200π b y = 0.2 sin(200π t) c
2 a Fundamental wave: 0.1 = 2πω
ω = 20π
Overtone: 0.2 = 2πω
ω = 10π b Fundamental: y1 = 0.2 sin(20π t) Overtone: y2 = 0.2 sin(10π t) Combined: y = 0.2 sin(20π t) + 0.2 sin(10π t) c
d The amplitude of the combined waveforms is almost 0.4.
This means that the loudness is almost doubled. 3 a y = 0.005 sin(150π t) amplitude 0.005 ω = 150π
period = 2πω
= 2150
ππ
= 1 s75
frequency = 75 s−1
b y = 0.04 sin(400π t) amplitude 0.04 ω = 400π
period = 2πω
= 2400
ππ
= 1 s200
frequency = 200 s−1
c y = 0.2 sin(100π t) + 0.2 sin(300π t)
From the graph, amplitude = 0.31 period = 0.02
= 1 s50
frequency = 50 s−1
d y = ( )0.4sin 200 0.25sin 2002
t tππ π + +
From the graph, amplitude = 0.47 period = 0.01
= 1 s100
frequency = 100 s−1 e y = 0.5 sin(200π t) + 0.25 sin(100π t)
From the graph, amplitude = 0.68 period = 0.02
= 1 s50
frequency = 50 s−1
M C 1 2 Q l d - 7 218 A d v a n c e d p e r i o d i c f u n c t i o n s
4 a
b The composite wave and the partial waves all have a
frequency of 20 Hz. c The amplitude of the composite waveform is almost
double that of the partials, this means that the loudness is almost doubled.
5 a Fundamental period: 0.01 = 2πω
ω = 200π
Overtone period: 0.03 = 2πω
ω = 2003
π
Combined waveform:
y = 2000.5sin(200 ) 0.5sin3
tt ππ +
b
c Amplitude of composite is 0.77; that is, about 1 times the
partials. The frequency of the composite is the same as the
overtone; that is, 1003
Hz.
6 a For y1, 1f
= 2πω
= 2400
πω
= 1200
Frequency = 200 Hz The frequency of y2 will also be 200 Hz. This means that
the frequency of the composite function will be 200 Hz. b
c The sound will be inaudible as the soundwaves will cancel
each other out. 7 a If the frequency of the fundamental is x, the frequency of the 2nd harmonic is 2x, the frequency of the 3rd harmonic is 3x.
Fundamental frequency = 0.005, 10.005
= 2πω
ω = 100π
frequency 2nd harmonic = 0.01, 10.01
= 2πω
ω = 50π
frequency 3rd harmonic = 0.015, 10.015
= 2πω
ω = 3100
π
y = 32sin 2sin 2sin100 50 100
t t tπ π π + +
b
Frequency of the composite is the same as the
fundamental; that is,
1200 Hz. The amplitude of the
composite is 212 times the fundamental.
8
Amplitude 1.5 Period = 2π This is a ‘saw tooth’ waveform. The waveform in
example 10 is a square waveform.
Exercise 7D — Graphing the reciprocal trigonometric functions
1 a y = sec4
x π +
Period 2π
Asymptotes occur at 4
x π+ = 22
n ππ ±
x = 2 or4
n ππ +
x = 324
n ππ −
The turning points occur at y = ±1
b y = cosec3
x π −
Period 2π
Asymptotes occur at 3
x π− = nπ
x = 3
n ππ +
The turning points occur at y = ±1
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 219
c y = cot2
x π +
Period π
Asymptotes occur at 2
x π+ = nπ
x = 2
n ππ −
d y = 2sec24
x π −
Period 22π = π
Asymptotes occur at 24
x π −
= 22
n ππ ±
4
x π− = 4
n ππ ±
x = or2
n ππ +
x = nπ The turning points occur at y = ±2
e y = 1 cosec 32 2
x π +
Period 23π
Asymptotes occur at 32
x π+ = nπ
3x = 2
n ππ −
x = 13 6
n ππ −
The turning points occur at y = ± 12
f y = −sec(4x + π)
Period 24π =
2π
Asymptotes occur at 4x + π = 22
n ππ ±
4x = 22
n ππ ±
x = 12 8
n ππ +
The turning points occur at y = ±1
g y = cot(2x) Period π Asymptotes occur at 2x = nπ
x = 2
nπ
h y = cosec(π x) + 2 Period 2π Asymptotes occur at π x = nπ x = n The turning points occur at y = 2 ± 1
i y = 11 cosec2
x π − +
Period 212
π = 4π
Asymptotes occur at 12
x π+ = nπ
12
x = nπ
x = 2nπ The turning points occur at y = 1 ± 1
j y = 12sec 12
xπ −
Period 212
π = 4π
Asymptotes occur at 12
xπ = 22
n ππ ±
πx = 4nπ ± π x = 4n ± 1 The turning points occur at y = −1 ± 2
M C 1 2 Q l d - 7 220 A d v a n c e d p e r i o d i c f u n c t i o n s
2 a sec(3x) = 2, 0 ≤ x ≤ π Using the graphics calculator results in:
The smallest solution is x = 0.349 There are 3 possible solutions in the domain
b 1 cosec(2 )2
x = x, −π ≤ x ≤ π
Using the graphics calculator to graph
y = 1 cosec(2 )2
x x− results in:
The smallest solution is x = −1.386 There are 4 roots in the domain.
c cot(2x) = tan(x), 2 2
xπ π− ≤ ≤
Using the graphics calculator to graph y = cot(2x) and y = tan(x) results in:
The solutions are x = −0.524 and x = 0.524 d cosec(3x°) = sec(x°), 0° ≤ x ≤ 90° Using the graphics calculator to graph y = cosec(3x°) and
y = sec(x°)
There are 2 solutions in the domain, largest solution is x = 45°
cosec(3x°) < sec(x°) 22°30′ ≤ x ≤ 45° 3 a
Area of sector AOB = 2
2rθ π
π×
= 2
2aθ
Area of ∆AOB = 21 sin( )2
a θ
Using area of the segment, 214
aπ = 2 21 sin( )2 2
a aθ θ−
= 2 ( sin( ))
2a θ θ−
2π = sin( )θ θ−
sin(θ) = 2πθ −
b Graphing y = sin(θ) and y = ,2πθ − 0 < x < π results in
The solution is 2.310 4 a
h = x sin(θ ) y = x cos(θ )
Area of trapezium = 1 ( ( 2 ))2
h x x y+ +
= h(x + y) = x sin(θ )(x + x cos(θ )) = x2 sin(θ )(1 + cos(θ )) b The area will be a maximum when sin(θ )(1 + cos(θ )) is a
maximum. Using a graphics calculator, this occurs when θ = 60°
Exercise 7E — Addition identities for sin(x ± y) and cos(x ± y)
1 a sin4πθ +
= sin( )cos cos( )sin
4 4π πθ θ +
= 1 1sin( ) cos( )2 2
θ θ+
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 221
= 1 (sin( ) cos( ))2
θ θ+
= 2 (sin( ) cos( ))2
θ θ+
b cos6
x π −
= cos( )cos sin( )sin6 6
x xπ π +
= 3 1cos( ) sin( )2 2
x x+
= 1 ( 3 cos( ) sin( ))2
x x+
c 2sin 23
x π −
= 2 sin(2 )cos cos(2 )sin3 3
x xπ π −
= 1 32 sin(2 ) cos(2 )2 2
x x
−
= sin(2 ) 3 cos(2 )x x−
d cos2
xπ +
= cos cos sin sin2 2 2 2
x xπ π −
= 0 sin2x −
= sin2x −
2 a sin(θ) cos(2θ) − sin(2θ)cos(θ) = sin(θ − 2θ) b cos(25)° cos(30)° + sin(25)° sin(30)° = cos(25° − 30°) = cos(−5)° = cos(5)°
c cos cos( ) sin sin( )4 4π πθ θ θ θ + − +
= cos4π θ θ + +
= cos 24πθ +
d sin
6
cos6
πθ
πθ
+ −
= sin( )cos cos( )sin
6 6
cos( )cos sin( )sin6 6
π πθ θ
π πθ θ
+ +
=
3 1sin( ) cos( )2 23 1cos( ) sin( )
2 2
θ θ
θ θ
+
+
= 3 sin( ) cos( )3 cos( ) sin( )
θ θθ θ
++
3 a cos(2x) = cos(x) cos(x) − sin(x) sin(x) = cos2(x) − sin2(x) cos2(x) = 1 − sin2(x) cos(2x) = 1 − sin2(x) − sin2(x) = 1 − 2sin2(x) b (sin(A) + cos(A))(sin(B) + cos(B)) = sin(A + B)
+ cos(A − B) RHS = sin(A + B) + cos(A − B) = sin(A) cos(B) + cos(A) sin(B)
+ cos(A) cos(B) + sin(A) sin(B) = sin(A) sin(B) + sin(A) cos(B)
+ cos(A) sin(B) + cos(A) cos(B) = sin(A)(sin(B) + cos(B)) + cos(A)(sin(B) + cos(B)) = (sin(A) + cos(A))(sin(B) + cos(B)) = LHS
c cot(A) + cot(B) = sin( )sin( )sin( )
A BA B
+
RHS = sin( )sin( )sin( )
A BA B
+
= sin( )cos( ) cos( )sin( )sin( )sin( )
A B A BA B
+
= sin( )cos( ) cos( )sin( )sin( )sin( ) sin( )sin( )
A B A BA B A B
+
= cos( ) cos( )sin( ) sin( )
B AB A
+
= cot(B) + cot(A) = LHS
d 2sin sin4 4
x xπ π + −
= sin2(x) − cos2(x)
LHS = 2sin sin4 4
x xπ π + −
= 2 sin( )cos cos( )sin4 4
x xπ π +
sin( )cos cos( )sin4 4
x xπ π −
= 1 12 sin( ) cos( )2 2
x x +
1 1sin( ) cos( )2 2
x x −
= 2 21 12 sin ( ) cos ( )2 2
x x −
= sin2(x) − cos2(x) = RHS
4 a 7sin12π
= 3 4sin12 12π π +
= sin4 3π π +
7sin12π
= sin cos cos sin4 3 4 3π π π π +
= 1 1 1 32 22 2
× + ×
= 1 32 2+
= 2 64+
b 5cos12π
= 3 2cos12 12π π +
= cos4 6π π +
5cos12π
= cos cos sin sin4 6 4 6π π π π −
= 1 3 1 12 22 2
× − ×
= 3 12 2
−
= 6 24−
5 a To find cos8π
, consider cos 28π ×
= cos
4π
Using cos(2θ) = 2cos2(θ) − 1
cos 28π ×
= 22cos 1
8π −
M C 1 2 Q l d - 7 222 A d v a n c e d p e r i o d i c f u n c t i o n s
cos4π
= 22cos 18π −
12
= 22cos 18π −
22cos8π
= 112
+
2cos8π
= 1 112 2 +
= 1 22 2+
= 2 24+
cos8π
= 2 24+±
= 2 22
+±
As 8π is in quadrant 1, cos
8π
is positive, therefore
cos8π
= 2 22
+
b To find 3sin8π
, consider 3cos 28π ×
= 3cos
4π
Using cos(2θ) = 1 −2sin2(θ)
3cos 28π ×
= 2 31 2sin
8π −
3cos4π
= 2 31 2sin8π −
12
− = 2 31 2sin8π −
2 32sin8π
= 112
+
2 3sin8π
= 1 2 12 2 +
= 2 12 2
+
= 2 24
+
3sin8π
= 2 24
+±
= 2 22+±
As 38π is in quadrant 1, 3sin
8π
is positive, therefore
3sin8π
= 2 22+
6 a cos sin3 6
x xπ π − + −
= cos(x)
LHS = cos sin3 6
x xπ π − + −
= cos cos( ) sin sin( )3 3
x xπ π +
sin cos( ) cos sin( )6 6
x xπ π + −
= 1 3 1 3cos( ) sin( ) cos( ) sin( )2 2 2 2
x x x x+ + −
= cos(x) = RHS b sin(3x) = 3 sin(x) − 4 sin3(x) LHS = sin(3x) = sin(x + 2x) = sin(x) cos(2x) + cos(x) sin(2x) = sin(x)(1 − 2 sin2(x)) + cos(x)(2sin(x) cos(x)) = sin(x) − 2 sin3(x) + 2 sin(x) cos2(x) = sin(x) − 2 sin3(x) + 2 sin(x)(1 − sin2(x)) = sin(x) − 2 sin3(x) + 2 sin(x) − 2 sin3(x) = 3 sin(x) − 4 sin3(x) 7 a A + B + C = 180° sin(A) = sin(180° − (B + C)) = sin(180)° cos(B + C) − cos(180)° sin(B + C) = 0 − (−sin(B + C)) = sin(B) cos(C) + cos(B) sin(B) b cos(B) = cos(180° − (A + C)) = cos(180)° cos(A + C) + sin(180)° sin(A + C) = −cos(A + C) + 0 = −(cos(A) cos(C) − sin(A) sin(C)) = sin(C) sin(A) − cos(C) cos(A)
8 a sin4
a x π +
= cos4
b x π +
Dividing by cos4
x π +
gives
sin
4
cos4
a x
x
π
π
+ +
= cos
4
cos4
b x
x
π
π
+ +
tan4
a x π +
= b
sin
4
cos4
x
x
π
π
+ +
= ba
sin( )cos cos( )sin
4 4
cos( )cos sin( )sin4 4
x x
x x
π π
π π
+ −
= ba
1 1sin( ) cos( )2 2
1 1cos( ) sin( )2 2
x x
x x
+
− = b
a
1 (sin( ) cos( ))2
1 (cos( ) sin( ))2
x x
x x
+
− = b
a
a(sin(x) + cos(x)) = b(cos(x) − sin(x)) asin(x) + acos(x) = bcos(x) − bsin(x) asin(x) + bsin(x) = bcos(x) − acos(x) sin(x)(a + b) = cos(x)(b − a)
sin( )cos( )
xx
= b aa b
−+
tan(x) = b aa b
−+
b i tan(x) = 1n
, n > 0
1n
= b aa b
−+
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 223
a + b = n(b − a) = nb − na a + na = nb − b a(1 + n) = b(n − 1)
ab
= 11
nn
−+
= 1 21
nn+ −
+
= 211n
−+
ii As n → ∞, 2 01n
→−
. Therefore, as n → ∞,
1ab
→ meaning a → b
Exercise 7F — Addition identities for tan(x ± y)
1 a tan(x + a) = tan( ) tan( )1 tan( ) tan( )
x ax a+
−
b tan(45° − x) = tan(45 ) tan( )1 tan(45 ) tan( )
xx
° −+ °
= 1 tan( )1 tan( )
xx
−+
2 a tan(27 ) tan(18 )1 tan(27 ) tan(18 )
° + °− ° °
= tan(27° + 18°)
= tan(45°) = 1
b tan( 3 ) tan( )1 tan( 3 ) tan( )
A B B AA B B A
+ + −− + −
= tan((A + 3B) + (B − A))
= tan(4B) 3 a x = 45°, y = 60°
tan(x + y) = tan(45° + 60°)
= tan(45 ) tan(60 )1 tan(45 ) tan(60 )
° + °− ° °
= 1 31 3
+−
= 1 3 1 31 3 1 3
+ +×− +
= 1 2 3 31 3
+ +−
= 2(2 3)2
+−
= (2 3)− +
b tan(α) = 1 ,4
tan(β) = 35
tan( )α β− = tan( ) tan( )1 tan( ) tan( )
α βα β
−+ −
=
1 34 5
1 314 5
−
+ ×
=
5 122020
3 20120
−
×+
= 720 3
−+
= 723−
4 a 512π = 3 2
12 12π π+
= 4 6π π+
5tan12π
= tan4 6π π +
= tan tan
4 6
1 tan tan4 6
π π
π π
+ −
=
1133
1 313
+×
−
= 3 1 3 13 1 3 1
+ +×− +
= 3 2 3 13 1
+ +−
= 2(2 3)2+
= 2 3+
b 4π =
8 8π π+
tan4π
= tan8 8π π +
= tan tan
8 8
1 tan tan8 8
π π
π π
+ −
1 = 2
2 tan8
1 tan8
π
π
−
21 tan8π −
= 2 tan8π
0 = 2tan 2 tan 18 8π π + −
Using the quadratic formula,
tan8π
= 2 4 4 1 12
− ± − × × −
= 2 82
− ±
= 2 2 22
− ±
= 1 2− ±
As 8π is in quadrant 1, tan 0,
8π >
therefore tan8π
= 1 2− +
5 a 1 tan( )1 tan( )
yy
−+
= cos( ) sin( )cos( ) sin( )
y yy y
−+
LHS = 1 tan( )1 tan( )
yy
−+
=
sin( )1cos( )cos( )
sin( ) cos( )1cos( )
yyy
y yy
−×
+
M C 1 2 Q l d - 7 224 A d v a n c e d p e r i o d i c f u n c t i o n s
= cos( ) sin( )cos( ) sin( )
y yy y
−+
= RHS
b sin( )cos( )
x yx y
+−
= tan( ) tan( )1 tan( ) tan( )
x yx y+
+
LHS = sin( )cos( )
x yx y
+−
=
1sin( )cos( ) cos( )sin( ) cos( )cos( )
1cos( )cos( ) sin( )sin( )cos( )cos( )
x y x y x yx y x y
x y
+ ×+
=
sin( )cos( ) cos( )sin( )cos( )cos( ) cos( )cos( )cos( )cos( ) sin( )sin( )cos( )cos( ) cos( )cos( )
x y x yx y x yx y x yx y x y
+
+
=
sin( ) sin( )cos( ) cos( )
sin( ) sin( )1cos( ) cos( )
x yx y
x yx y
+
+ ×
= tan( ) tan( )1 tan( ) tan( )
x yx y+
+
= RHS
6 Given ( )( )
coscos
A BA B
+−
= ( )( )
sinsin
C DC D
−+
cos( )cos( ) sin( )sin( )cos( )cos( ) sin( )sin( )
A B A BA B A B
−+
= sin( )cos( ) cos( )sin( )sin( )cos( ) cos( )sin( )
C D C DC D C D
−+
1cos( )cos( ) sin( )sin( ) sin( )sin( )
1cos( )cos( ) sin( )sin( )sin( )sin( )
A B A B A BA B A B
A B
− ×+
=
1sin( )cos( ) cos( )sin( ) sin( )cos( )
1sin( )cos( ) cos( )sin( )sin( )cos( )
C D C D C DC D C D
C D
− ×+
cos( )cos( ) sin( )sin( )sin( )sin( ) sin( )sin( )cos( )cos( ) sin( )sin( )sin( )sin( ) sin( )sin( )
A B A BA B A BA B A BA B A B
−
+ =
sin( )cos( ) cos( )sin( )sin( )cos( ) sin( )cos( )sin( )cos( ) cos( )sin( )sin( )cos( ) sin( )cos( )
C D C DC D C DC D C DC D C D
−
+
cot( )cot( ) 1cot( )cot( ) 1
A BA B
−+
= 1 cot( ) tan( )1 cot( ) tan( )
C DC D
−+
(cot(A) cot(B) − 1)(1 + cot(C) tan(D)) = (cot(A) cot(B) + 1)(1 − cot(C) tan(D)) cot(A) cot(B) + cot(A) cot(B) cot(C) tan(D) − 1 − cot(C) tan(D) = cot(A) cot(B) − cot(A) cot(B) cot(D) tan(D) + 1 − cot(C) tan(D) 2cot(A) cot(B) cot(C) tan(D) = 2
cot(A) cot(B) cot(C) = 1tan( )D
= cot(D)
7 cot( )θ α+ = cot( )cot( )cot( ) cot( )
θ αθ α
−1+
LHS = cot( )θ α+
= 1tan( )θ α+
= 1tan( ) tan( )
1 tan( ) tan( )θ α
θ α+
−
=
11 tan( ) tan( ) tan( ) tan( )
1tan( ) tan( )tan( ) tan( )
θ α θ αθ α
θ α
− ×+
=
1 tan( ) tan( )tan( ) tan( ) tan( ) tan( )
tan( ) tan( )tan( ) tan( ) tan( ) tan( )
θ αθ α θ α
θ αθ α θ α
−
+
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 225
= cot( )cot( ) 11 1
tan( ) tan( )
θ α
α θ
−
+
= cot( )cot( ) 1cot( ) cot( )
θ αα θ
−+
= RHS
8 α + β = 4π
cot(α + β) = cot( )cot( ) 1cot( ) cot( )
α βα β
−+
cot4π
= cot( )cot( ) 1cot( ) cot( )
α βα β
−+
1 = cot( )cot( ) 1cot( ) cot( )
α βα β
−+
cot(α) + cot(β) = cot(α) cot(β) − 1 cot(α) = cot(α) cot(β) − 1 − cot(β) (cot(α))tan(β) = (cot(α) cot(β) − 1 − cot(β)) tan(β) cot(α) tan(β) = cot(α) − tan(β) − 1 cot(α) = cot(α) tan(β) + tan(β) + 1 9 a A + B + C = π, A + B = π − C
tan(A + B) = tan(π − C)
tan( ) tan( )1 tan( ) tan( )
A BA B+
− = tan( ) tan( )
1 tan( ) tan( )CC
ππ−
+
= 0 tan( )1
C−
= −tan(C)
b −tan(C) = tan( ) tan( )1 tan( ) tan( )
A BA B+
−
−tan(C)(1 − tan(A) tan(B)) = tan(A) + tan(B) −tan(C) + tan(A) tan(B) tan(C) = tan(A) + tan(B) tan(A) tan(B) tan(C) = tan(A) + tan(B) + tan(C)
10 Using the sine rule, sin( )
aA
= sin( )
bB
or sin(A) = sin( )a Bb
1tan ( )2
A B −
=
1sin ( )21cos ( )2
A B
A B
− −
= sin cos cos sin
2 2 2 2
cos cos sin sin2 2 2 2
A B A B
A B A B
− +
1tan ( )2
A B +
=
1sin ( )21cos ( )2
A B
A B
+ +
= sin cos cos sin
2 2 2 2
cos cos sin sin2 2 2 2
A B A B
A B A B
+ −
1tan ( )21tan ( )2
A B
A B
− +
= sin cos cos sin cos cos sin sin
2 2 2 2 2 2 2 2
cos cos sin sin sin cos cos sin2 2 2 2 2 2 2 2
A B A B A B A B
A B A B A B A B
− − × + +
=
2 2 2 2
2 2
sin cos cos sin sin cos cos sin cos sin cos sin2 2 2 2 2 2 2 2 2 2 2 2
cos sin cos cos sin cos2 2 2 2 2
A A B A B B A B B A A B
A A B A B
− − + +
2 2sin sin cos sin cos sin2 2 2 2 2 2 2B A B B A A B + +
M C 1 2 Q l d - 7 226 A d v a n c e d p e r i o d i c f u n c t i o n s
=
2 2 2 2
2 2
sin cos cos sin cos sin sin sin cos cos sin cos2 2 2 2 2 2 2 2 2 2 2 2
sin cos cos sin cos sin2 2 2 2 2
A A B A A B A B B A B B
A A B A A
+ − − +
2 2cos sin cos sin sin cos2 2 2 2 2 2 2B A B B A B B + +
=
2 2 2 2
2 2 2 2
sin cos cos sin sin cos sin cos2 2 2 2 2 2 2 2
sin cos cos sin sin cos cos sin2 2 2 2 2 2 2
A A B B B B A A
A A B B B B A
+ − +
+ + + 2
A
= sin cos sin cos
2 2 2 2
sin cos sin cos2 2 2 2
A A B B
A A B B
− +
=
1 1sin( ) sin( )2 21 1sin( ) sin( )2 2
A B
A B
−
+
= sin( ) sin( )sin( ) sin( )
A BA B
−+
=
sin( ) sin( )
sin( ) sin( )
a B Bb
a B Bb
−
+
= sin( ) 1
sin( ) 1
aBbb
a bBb
− × +
= a ba b
−+
Exercise 7G — The Werner or factorisation identities 1 2 sin(3x) cos(x) = sin(3x + x) + sin(3x − x) = sin(4x) + sin(2x) 2 2 cos(4x) sin(x) = sin(4x + x) − sin(4x − x) = sin(5x) − sin(3x) 3 2 cos(2x) cos(x) = cos(2x + x) + cos(2x − x) = cos(3x) + cos(x) 4 −2 sin(3x) sin(2x) = cos(3x + 2x) − cos(3x − 2x) = cos(5x) − cos(x) 5 4 cos(4x) sin(3x) = 2 [sin(4x + 3x) − sin(4x − 3x)] = 2(sin(7x) − sin(x)
6 cos(x + 30°) cos(30°) = 1 (cos[( 30 ) 30 ] cos[( 30 ) 30 ])2
x x+ ° + ° + + ° − °
= [ ]1 cos( 60 ) cos( )2
x x+ ° +
7 sin(5 )sin(3 )x x− = 1 (cos(5 3 ) cos(5 3 ))2
x x x x+ − −
= 1 (cos(8 ) cos(2 ))2
x x−
8 2 sin(a + b) cos(a – b) = sin[(a + b) + (a – b)] + sin[(a + b) – (a – b)] = sin(2a) + sin(2b)
9 cos(3 )cos(3 )a b b = [ ]cos(3 3 ) cos(3 3 )2a b b b b+ + −
= (cos(6 ) cos(0))2a b +
= (cos(6 1))2a b +
10 5cos(2 )sin( 2 )x y x y− + = [ ]5 sin((2 ) ( 2 )) sin((2 ) ( 2 ))2
x y x y x y x y− + + − − + +
= [ ]5 sin(3 ) sin( 3 )2
x y x y+ − −
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 227
11 1 1sin ( )cos ( )2 2
x y x y + −
= 1 1 1 1 1sin ( ) ( ) sin ( ) ( )2 2 2 2 2
x y x y x y x y + + − + + − −
= 1 1 1sin (2 ) sin (2 )2 2 2
x y +
= 1 (sin( ) sin( ))2
x y+
12 sin( )sin( )x y x y− + = 1(cos[( ) ( )] cos[( ) ( )])2
x y x y x y x y− − + + − − − +
= [ ]1 cos(2 ) cos( 2 ) 2
x y− − −
= 1 (cos(2 ) cos(2 )) 2
x y− −
Exercise 7H — The Simpson or half-sum/half-difference identities
1 a sin(4x) + sin(2x) = 4 2 4 22sin cos2 2
x x x x+ −
= 2 sin(3x) cos(x)
b sin(4x) − sin(2x) = 4 2 4 22cos sin2 2
x x x x+ −
= 2 cos(3x) sin(x)
c cos(3y) + cos(y) = 3 32cos cos2 2
y y y y+ −
= 2 cos(2y) cos(y)
d cos(3θ) − cos(θ) = 3 32sin sin2 2
θ θ θ θ+ − −
= −2 sin(2θ) sin(θ)
e sin(5α) + sin(3α) = 5 3 5 32sin cos2 2
α α α α+ −
= 2 sin(4α) cos(α)
f sin(4β) − sin(β) = 4 42cos sin2 2
β β β β+ −
= 5 32cos sin2 2β β
g cos(3x) + cos(4x) = 3 4 3 42cos cos2 2
x x x x+ −
= 72cos cos2 2x x−
= 72cos cos2 2x x
h cos(3x) − cos(4x) = 3 4 3 42sin sin2 2
x x x x+ − −
= 72sin sin2 2x x− −
= 72sin sin2 2x x
i 1 (sin(3 ) sin(4 ))2
x x+ = 3 4 3 4sin cos2 2
x x x x+ −
= 7sin cos2 2x x−
= 7sin cos2 2x x
j 1 (sin(2 ) sin(6 ))2
x x− = 2 6 2 6cos sin2 2
x x x x+ −
= cos(4x) sin(−2x) = −cos(4x) sin(2x)
M C 1 2 Q l d - 7 228 A d v a n c e d p e r i o d i c f u n c t i o n s
k [ ]1 cos(2 ) cos(2 )2
θ α− − = 2 2 2 2sin sin2 2
θ α θ α+ −
= sin( )sin( )θ α θ α+ −
l [ ]4 cos( ) cos( )x y x y+ + − = ( ) ( ) ( ) ( )8cos cos2 2
x y x y x y x y+ + − + − −
= 2 28cos cos2 2x y
= 8 cos(x) cos(y) m sin(α) + sin(2α) + sin(3α) = (sin(α) + sin(3α)) + sin(2α)
= 3 32sin cos sin 22 2
α α α α α+ − +
= 2 sin(2α) cos(−α) + sin(2α) = 2 sin(2α) cos(α) + sin(2α) = sin(2α)(2cos(α) + 1) n cos(y) + cos(3y) + cos(5y) = (cos(y) + cos(5y)) + cos(3y)
= 5 52cos cos cos32 2
y y y y y+ − +
= 2 cos(3y)cos(−2y ) + cos(3y) = 2 cos(3y)cos(2y) + cos(3y) = cos(3y)(2 cos(2y) + 1)
2 a sin( ) sin( )cos( ) cos( )
θ αθ α
−−
= 2cos sin
2 2
2sin sin2 2
θ α θ α
θ α θ α
+ −
+ − −
= cos
2
sin2
θ α
θ α
+ −
+
= cot2
θ α+ −
b cos( ) cos( )sin( ) sin( )
xx
αα
++
= 2cos cos
2 2
2sin cos2 2
x x
x x
α α
α α
+ −
+ −
= cos
2
sin2
x
x
α
α
+
+
= cot2
x α+
c sin(2 ) sin(2 )cos(2 ) cos(2 )
θ βθ β
−+
=
2 2 2 22cos sin2 2
2 2 2 22cos cos2 2
θ β θ β
θ β θ β
+ −
+ −
= sin( )cos( )
θ βθ β
−−
= tan(θ − β)
d sin(2 )cos( ) cos(5 )
xx x−
= sin(2 )5 52sin sin
2 2
xx x x x+ − −
= sin(2 )2sin(3 )sin( 2 )
xx x− −
= sin(2 )2sin(3 )sin(2 )
xx x
= 12sin(3 )x
= 1 cosec(3 )2
x
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 229
e cos(2 ) cos(4 )cos(3 )sin(2 )
x xx x
+ =
2 4 2 42cos cos2 2
cos(3 )sin(2 )
x x x x
x x
+ −
= 2cos(3 )cos( )cos(3 )sin(2 )
x xx x
−
= 2cos( )sin(2 )
xx
= 2cos( )2sin( )cos( )
xx x
= 1sin( )x
= cosec(x) f sin(y) + sin(2y) + sin(3y) = (sin(y) + sin(3y)) + sin(2y)
= 3 32sin cos sin(2 )2 2
y y y y y+ − +
= 2sin(2y) cos(−y) + sin(2y) = sin(2y)(2 cos(y) + 1) cos(y) + cos(2y) + cos(3y) = (cos(y) + cos(3y)) + cos(2y)
= 3 32cos cos cos(2 )2 2
y y y y y+ − +
= 2cos(2y) cos(−y) + cos(2y) = cos(2y)(2 cos(y) + 1)
sin( ) sin(2 ) sin(3 )cos( ) cos(2 ) cos(3 )
y y yy y y
+ ++ +
= sin(2 )(2cos( ) 1)cos(2 )(2cos( ) 1)
y yy y
++
= sin(2 )cos(2 )
yy
= tan(2y) 3 a sin(nx) = 2cos(x) sin[(n − 1)x] − sin[(n − 2)x]
sin(nx) + [(n −2)x] = ( 2) ( 2)2sin cos2 2
nx n x nx n x+ − − −
= 2 2 22sin cos2 2
nx x x−
= 2sin[(n − 1)x]cos(x) Therefore sin(nx) = 2sin[(n − 1)x]cos(x) − sin[(n − 2)x] b cos(nx) = 2cos(x) cos[(n − 1)x] − cos[(n − 2)x]
cos(nx) + cos[(n − 2)x]
= ( 2) ( 2)2cos cos2 2
nx n x nx n x+ − − −
= 2 2 22cos cos2 2
nx x x−
= 2cos[(n − 1)x]cos(x) Therefore
cos(nx) = 2cos[(n − 1)x] cos(x) − cos[(n − 2)x] 4 a sin(2x) + sin(x) = 0 2sin(x)cos(x) + sin(x) = 0 sin(x)(2cos(x) + 1) = 0 sin(x) = 0 or 2cos(x) = 1
x = 0, π, 2π cos(x) = 12
x = ,3 3π ππ π− +
= 2 4,3 3π π
Solution set 2 40, , , , 23 3π ππ π
b cos(2x) + cos(x) = 0 2cos2(x) − 1 + cos(x) = 0 2cos2(x) + cos(x) − 1 = 0 (2cos(x) − 1)(cos(x) + 1) = 0
M C 1 2 Q l d - 7 230 A d v a n c e d p e r i o d i c f u n c t i o n s
2cos(x) − 1= 0 or cos(x) + 1 = 0
cos(x) = 12
cos(x) = −1
x = ,23 3π ππ − cos(x) = π
= 5,3 3π π
Solution set 5, ,3 3π ππ
5 a sin(x) + sin(2x) + sin(3x) = 0 (sin(x) + sin(3x)) + sin(2x) = 0
3 32sin cos sin(2 )2 2
x x x x x+ − +
= 0
2sin(2x) cos(−x) + sin(2x) = 0 sin(2x)(2cos(x) + 1) = 0 sin(2x) = 0 or 2cos(x) +1 = 0
2x = ± nπ cos(x) = 12−
x = 2nπ± x = 22
3n ππ ±
Where n ∈ Z b cos(x) + cos(3x) + cos(5x) = 0 (cos(x) + cos(5x)) + cos(3x) = 0
5 52cos cos cos(3 )2 2
x x x x x+ − +
= 0
2cos(3x) cos(−2x) + cos(3x) = 0 cos(3x)(2cos(2x) + 1) = 0 cos(3x) = 0 or 2cos(2x) + 1 = 0
3x = 22
n ππ ± cos(2x) = 12−
x = 23 6nπ π± 2x = 22
3n ππ ±
x = 3
n ππ ±
Where n ∈ Z
Exercise 7I — Modelling and problem solving 1 a To write cos(x) − sin(x) in the form A cos(x − α):
A cos(x − α) = A cos(x) cos(α) + A sin(x) sin(α) Equating coefficients of sin(x) and cos(x) gives: A cos(α) = 1 and A sin(α) = −1 A2 cos2(α) + A2 sin2(α) = 1 + 1 A2(cos2(α) + sin2(α)) = 2 A = 2 ( is > 0)A As cos(α) is positive and sin(α) is negative, α is in
quadrant 4
sin( )cos( )
AA
αα
= tan(α) = −1
tan(45)° = 1, therefore α = 360° − 45° = 315° cos(x) − sin(x) = 2 cos( 315 )x − ° b To write sin(x) + cos(x) in the form A sin(x + α): A sin(x + α) = A sin(x) cos(α) + A cos(x) sin(α) Equating coefficients of sin(x) and cos(x) gives: A cos(α) = 1 and A sin(α) = 1 A2 cos2(α) + A2 sin2(α) = 1 + 1 A2(cos2(α) + sin2(α)) = 2 A = 2 ( is > 0)A
sin( )cos( )
AA
αα
= tan(α) = 1, is in quadrant 1
tan(45)° = 1 sin(x) + cos(x) = 2 sin( 45 )x + °
c To write 4sin(x) − 3cos(x) in the form A sin(x − α): A sin(x − α) = A sin(x) cos(α) − A cos(x) sin(α) Equating coefficients of sin(x) and cos(x) gives: A cos(α) = 4 and A sin(α) = 3 A2 cos2(α) + A2 sin2(α) = 42 + 32 A2(cos2(α) + sin2(α)) = 25 A = 5(A is > 0)
sin( )cos( )
AA
αα
= tan(α) = 34
, is in quadrant 1
α = 0.644 4sin(x) − 3cos(x) = 5sin(x − 0.644) d To write cos(x) − 2sin(x) in the form A cos(x + α): A cos(x + α) = A cos(x) cos(α) − A sin(x) sin(α) Equating coefficients of sin(x) and cos(x) gives: A cos(α) = 1 and A sin(α) = 2 A2 cos2(α) + A2 sin2(α) = 12 + 22 A2(cos2(α) + sin2(α)) = 5 A = 5( is 0)A >
sin( )cos( )
AA
αα
= tan(α) = 2, is in quadrant 1
α = 1.107 cos(x) − 2sin(x) = 5 cos( 1.107)x +
e To write cos 3 sinx x− in the form A sin(x − α): A sin(x − α) = A sin(x) cos(α) − A cos(x) sin(α) Equating coefficients of sin(x) and cos(x) gives: A cos(α) = 3− and −A sin(α) = 1
A2 cos2(α) + A2 sin2(α) = 2 23 1+
A2(cos2(α) + sin2(α)) = 4 A = 2(A is > 0)
sin( )cos( )
AA
αα
= tan(α) = 1 ,3
is in quadrant 3
tan(30)° = 13
α = 180° + 30° = 210° cos( ) 3sin( )x x− = 2 sin(x − 210°)
f To write sin(x) + 2 cos(x) in the form A cos(x − α): A sin(x − α) = A cos(x) cos(α) + A sin(x) sin(α) Equating coefficients of sin(x) and cos(x) gives: A cos(α) = 2 and A sin(α) = 1
A2 cos2(α) + A2 sin2(α) = 2 22 1+
A2(cos2(α) + sin2(α) = 3 A = 3 (A is > 0) As cos(α) is positive and sin(α) is positive, α is in
quadrant 1
sin( )cos( )
AA
αα
= tan(α) = 12
α = 35° 16′ sin( ) 2 cos( )x x− = 3 cos( 35 16 )x ′− °
2 a i To write 3 cos( ) 2sin( )x x− in the form A sin(x − α):
A sin(x − α) = A sin(x) cos(α) + A cos(x) sin(α) Equating coefficients of sin(x) and cos(x) gives: A cos(α) = −2 and −A sin(α)= 3 A sin(α) = 3−
A2 cos2(α) + A2 sin2(α) = 22( 2) 3− +
A2(cos2(α) + sin2(α)) = 7 A = 7 ( is > 0)A As cos(α) is negative and sin(α) is negative, α is in
quadrant 3
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 231
sin( )cos( )
AA
αα
= tan(α) = 32
tan(0.714) = 32
α = α + 0.714 = 3.855 3 cos( ) 2sin( )x x− = 7 sin( 3.855)x −
ii To write 3 cos(x) − 2 sin(x) in the form B sin(x + β):
B sin(x + β) = Bsin(x) cos(β) + B cos(x) sin(β) Equating coefficients of sin(x) and cos(x) gives: B cos(β) = −2 and B sin(β) = 3
B2 cos2(β) + A2 sin2(β) = 22( 2) 3− +
B2(cos2(β) + sin2(β)) = 7 B = 7( is 0)B > As cos(β) is negative and sin(β) is positive, β is in
quadrant 2
sin( )cos( )
BB
ββ
= tan(β) = 32
−
tan(0.714) = 32
= π − 0.714 = 2.428 3 cos( ) 2sin( )x x− = 7 sin( 2.428)x +
iii To write 3 cos( ) 2sin( )x x− in the form C cos(x − θ):
C cos(x − θ) = C cos(x) cos(θ)+ C sin(x) sin(θ) Equating coefficients of sin(x) and cos(x) gives: C cos(θ) = 3 and C sin(θ) = −2
C2 cos2(θ) + C2 sin2(θ) = 22( 2) 3− +
C2(cos2(θ) + sin2(θ) = 7 C = 7( is 0)C > As cos(θ) is positive and sin(θ) is negative, θ is in
quadrant 4
sin( )cos( )
CC
θθ
= tan(θ) = 23
−
tan(0.857) = 23
θ = 2θ − 0.857 = 5.426 3 cos( ) 2sin( )x x− = ( )7 cos 5.426x −
iv To write 3 cos(x) − 2 sin(x) in the form D cos(x + φ):
D cos(x + φ) = D cos(x)cos(φ) − D sin(x)sin(φ) Equating coefficients of sin(x) and cos(x) gives: D cos(φ) = 3 and D sin(φ) = 2
D2 cos2(φ) + C2 sin2(φ) = 22( 2) 3− +
D2(cos2(φ) + sin2(φ)) = 7 D = 7 ( is > 0)D As cos(φ) is positive and sin(φ) is positive, (φ) is in
quadrant 1
sin( )cos( )
DD
φφ
= tan(φ) = 23
= 0.857 3 cos( ) 2sin( )x x− − = 7 cos( 0.857)x +
b As A = B = C = D = 7, the maximum is 7 and the
minimum is 7− c All the equations will give the same graph (as they are
different forms of 3 cos( ) 2sin( ).x x− The first minimum in the domain 0 ≤ x ≤ 2 occurs when x = 2.285.
The first maximum in the domain 0 ≤ x ≤ 2π occurs when x = 5.426.
3 a 8 cos(x) − 15 sin(x) = r cos(x + α) r cos(x + α) = r cos(x) cos(α) − r sin(x) sin(α) Equating coefficients of cos(x) and sin(x) gives: r cos(α) = 8 and r sin(α) = 15 r2 cos2(α) + r2 sin2(α) = 82 + 152 r2(cos2(α) + sin2(α)) = 289 r = 17 As cos(α) is positive and sin(α) is positive, α is in
quadrant 1
sin( )cos( )
rr
αα
= tan(α) = 158
α = 61°56′ b Using a graphics calculator, graph y = 8 cos(x) − 15 sin(x)
and y = 9 over the domain 0° < x < 360°. The first intercept occurs at x = 240°02′ (Remember to
set the angle to degrees)
4 a 2sin(2 )cos( )dx x x∫ = [sin(2 ) sin(2 )]dx x x x x+ + −∫
= (sin(3 ) sin( ))dx x x+∫
= 1cos(3 ) cos( )3
x x c− − +
= 1(cos(3 ) 3cos( ))3
x x c− − +
b 2cos(3 )sin( )dx x x∫ = [ ]sin(3 ) sin(3 ) dx x x x x+ − −∫
= (sin(4 ) sin(2 ))dx x x−∫
= 1 1cos(4 ) cos(2 )4 2
x x c− + +
= 1(cos(4 ) 2cos(2 ))4
x x c− − +
c 3 cos(4 )cos( )dx x x∫ = [ ]13 cos(4 ) cos(4 ) d2
x x x x x+ + −∫
= 3 (cos(5 ) cos(3 ))d2
x x x+∫
= 3 1 1sin(5 ) sin(3 )2 5 3
x x c + +
= 1 (3sin(5 ) 5sin(3 ))10
x x c+ +
d sin(3 )sin( )dx x x−∫ = [ ]1 cos(3 ) cos(3 ) d2
x x x x x+ − −∫
= 1 (cos(4 ) cos(2 ))d2
x x x−∫
M C 1 2 Q l d - 7 232 A d v a n c e d p e r i o d i c f u n c t i o n s
= 1 1 1sin(4 ) sin(2 )2 4 2
x x c − +
= 1 (sin(4 ) 2sin(2 ))8
x x c− +
e 4 cos( )sin( )dx x x∫ = [ ]2 sin( ) sin( ) dx x x x x+ − −∫
= 2 sin(2 )dx x∫
= 12 cos(2 )2
x c−× +
= cos(2 )x c− + f This can be solved by using sin(x + π) = −sin(x) and cos(2x) = 1 − 2sin2(x)
1 sin( )sin( )d2
x x xπ− +∫ = 1 sin( )( sin( ))d2
x x x− −∫
= 21 sin ( )d2
x x∫
= 1 (1 cos(2 ))d4
x x−∫
= 1 1 sin(2 )4 2
x x c − +
= 1 (2 sin(2 ))8
x x c− +
Alternatively, using the same method as earlier questions:
1 sin( )sin( )d2
x x xπ− +∫ = 1 1 [cos( ) cos( ( ))]d2 2
x x x x xπ π+ + − − +∫
= 1 [cos(2 ) cos( )]d4
x xπ π+ − −∫
= 1 ( cos(2 ) 1)d4
x x− +∫
= 1 1sin(2 )4 2
x x c− + +
= 1[sin(2 ) 2 ]8
x x c− − +
= 1[2 sin(2 )]8
x x c− +
g sin(3 )cos(2 )dx x x−∫ = 1 [sin(3 2 ) sin(3 2 )]d2
x x x x x− + + −∫
= 1 (sin(5 ) sin( ))d2
x x x− +∫
= 1 1cos(5 ) cos( )2 5
x x c− − − +
= 1 (cos(5 ) 5cos( ))10
x x c+ +
h Using cos cos( )2πθ θ − =
and cos sin( ):
2πθ θ + = −
3 1cos cos ( 1) d2 2
x x xπ π − ∫ = 1 3 1 3 1cos ( 1) cos ( 1) d
2 2 2 2 2x x x x xπ π π π + − + − −
∫
= 1 cos 2 cos d2 2 2
x x xπ ππ π − + + ∫
= 1 (sin(2 ) sin( ))d2
x x xπ π−∫
= 1 1 1cos(2 ) cos( )2 2
x x cπ ππ π
− + +
= 1 (cos(2 ) 2cos( ))4
x x cπ ππ
− − +
= 1 (2cos( ) cos(2 ))4
x x cπ ππ
− +
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 233
5 a i 3cos(3 )sin( )dx x x∫ = [sin(3 ) sin(3 )]d2
x x x x x3 + − −∫
= 3 (sin(4 ) sin(2 ))d2
x x x−∫
= 3 1 1cos(4 ) cos(2 )2 4 2
x x c− + +
2
4
3cos(3 )sin( )dx x xπ
π∫ = 2
4
3 1 1cos(4 ) cos(2 )2 4 2
x xπ
π
− +
= 3 1 1 1 1cos(2 ) cos( ) cos( ) cos2 4 2 4 2 2
ππ π π − − + − +
= 3 1 1 1 02 4 2 4 − − − +
= 32
−
ii sin(4 )cos( )dx x x∫ = 1 [sin(4 ) sin(4 )]d2
x x x x x+ + −∫
= 1 (sin(5 ) sin(3 ))d2
x x x+∫
= 1 1 1cos(5 ) sin(3 )2 5 3
x x c − − +
= 1[3cos(5 ) 5sin(3 )]30
x x c− + +
3
sin(4 )cos( )dx x xππ∫ = [ ]
3
1 3cos(5 ) 5sin(3 )30
x x ππ
− +
= 1 53cos(5 ) 5cos(3 ) 3cos 5sin( )30 3
ππ π π − + − +
= 1 13 5 3 530 2
− − − − × −
= ( )3 2 1130 2
− +− ×
= 320
iii 4 sin(4 )cos(2 )dx x x∫ = [ ]2 sin(4 2 ) sin(4 2 ) dx x x x x+ + −∫
= 2 (sin(6 ) sin(2 ))dx x x+∫
= 1 12 cos(6 ) cos(2 )6 2
x x c − − +
= 1(cos(6 ) 3cos(2 ))3
x x c− + +
2
4 sin(4 )cos(2 )dx x xπ
π∫ = [ ]21 cos(6 ) 3cos(2 )3
x x ππ
− +
= 1[cos(12 ) 3cos(4 ) (cos(6 ) 3cos(2 ))]3
π π π π− + − +
= 1[1 3 (1 3)]3− + − +
= 0
iv 1 sin(7 )sin( )d2
x x x−∫ = 1 1 [cos(7 ) cos(7 )]d
2 2x x x x x+ − −∫
= 1 (cos(8 ) cos(6 ))d4
x x x−∫
= 1 1 1sin(8 ) sin(6 )4 8 6
x x c − +
= 1 (3sin(8 ) 4sin(6 ))96
x x c− +
M C 1 2 Q l d - 7 234 A d v a n c e d p e r i o d i c f u n c t i o n s
2
2
1 sin(7 )sin( )d2
x x xπ
π−−∫ = [ ]2
2
1 3sin(8 ) 4sin(6 )96
x xπ
π−−
= [ ]1 3sin(4 ) 4sin(3 ) (3sin( 4 ) 4sin( 3 ))96
π π π π− − − − −
= 1 096
×
= 0 b i
Integral is −1.5 ii
Integral is 0.379 iii
Area is 0 iv
Area is 0
6 a cos(2x) = cos(x + x) = cos(x) cos(x) − sin(x) sin(x) = cos2(x) − sin2(x) = cos2(x) − (1 − cos2(x)) = 2 cos2(x) − 1 b cos(2x) = 2 cos2(x − 1)
= cos2(x) = 1 (cos(2 ) 1)2
x +
2
4cos ( )dx x
ππ∫ =
4
1 (cos(2 ) 1)d2
x xππ +∫
= 4
1 1 sin(2 )2 2
x xπ
π
+
= [ ]4
1 sin(2 ) 24
x x ππ+
= 1 sin(2 ) 2 sin4 2 2
π ππ π + − +
= 1 2 14 2
ππ − +
= 1 3 14 2
π −
= 1 (3 2)8
π −
= 0.928
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 235
7 a cos(3x) = cos(2x + x) = cos(2x) cos(x) − sin(2x) sin(x) = (cos2(x) − sin2(x)) cos(x) − 2sin(x) cos(x) sin(x) = cos3(x) − sin2(x) cos(x) − 2sin2(x) cos(x) = cos3(x) − 3sin2(x) cos(x) = cos3(x) − 3(1 − cos2(x)) cos(x) = cos3(x) − 3cos(x) + 3cos3(x) = 4cos3(x) − 3cos(x) b cos(3x) = 4cos3(x) − 3cos(x)
cos3(x) = 1 (cos(3 ) 3cos( ))4
x x+
32
2
cos ( )dx xπ
π−∫ = 2
2
1 (cos(3 ) 3cos( ))4
x xπ
π−+∫
= 2
2
1 1 sin(3 ) 3sin( )4 3
x xπ
π−
+
= [ ]2
2
1 sin(3 ) 9sin( )12
x xπ
π−+
= 1 3sin 9sin12 2 2
π π + 3sin 9sin
2 2π π − − − +
= 1 [ 1 9 (1 9)]12
− + − −
= 1 (8 8)12
− −
= 1612
= 43
= 113
8 a 3cos(3x) sin(x) = 3 [sin(3 ) sin(3 )]2
x x x x+ − −
= 3 (sin(4 ) sin(2 ))2
x x−
cos(3x) = 0 when x = 3 5, ,6 6 6π π π…
sin(x) = 0 when x = 0, ,2π π… Sketching gives:
The graph is below the x-axis for 6 2
xπ π< <
The graph is above the x-axis for 52 6
xπ π< <
The areas are equal in these regions
Area = 56
2
2 3cos(3 )sin( )dx x xπ
π∫
= 56
2
3 (sin(4 ) sin(2 ))dx x xπ
π −∫
= 56
2
1 13 cos(4 ) cos(2 )4 2
x xπ
π
− +
= [ ]56
2
3 cos(4 ) 2cos(2 )4
x xπ
π− −
M C 1 2 Q l d - 7 236 A d v a n c e d p e r i o d i c f u n c t i o n s
= 3 10 5cos 2cos4 3 3
π π− − (cos(2 ) 2cos( ))π π − −
= ( )3 1 12 1 2 14 2 2
− − − × − − × −
= 3 3 34 2
− − −
= 3 94 2
− −×
= 278
= 338
b Finding the area in 2 parts:
The areas are both 1.6875
The combined area is 3.375 = 338
9 Using sin3π
= 32
and cos3π
= 12
sin(3x) = 3sin(x) − 4sin3(x)
sin 39π ×
= 33sin 4sin
9 9π π −
sin3π
= 2sin 3 4sin9 9π π −
32
= 2sin 3 4 1 cos9 9π π − −
= 2sin 3 4 4cos9 9π π − +
= 2sin 4cos 19 9π π −
= sin 2cos 1 2cos 19 9 9π π π − +
= 1 1sin 2 cos 2 cos9 9 2 9 2π π π × − × +
= sin 2 cos cos 2 cos cos9 9 3 9 3π π π π π × − × +
=
3 3 3 39 9 9 9 9 9 9 9sin 4 sin sin 4 cos cos
9 2 2 2 2
π π π π π π π ππ
+ − + − × − ×
= 2 2sin 4sin sin 4cos cos9 9 9 9 9π π π π π− − × − ×
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 237
= 2 216sin sin sin cos cos9 9 9 9 9π π π π π
= 2 24sin 2sin cos 2sin cos9 9 9 9 9π π π π π × ×
= 2 44sin sin sin9 9 9π π π
Therefore 2 48sin sin sin9 9 9π π π
= 322
× = 3
10 a i cos(mx) cos(nx) = 1 [cos( ) cos( )]2
mx nx mx nx+ − −
= 1 [cos( ) cos( ) ]2
m n x m n x+ − −
cos( )cos( )dmx nx x∫ = [ ]1 cos( ) cos( ) d2
m n x m n x x+ − −∫
= 1 1 1sin( ) sin( )2
m n x m n x cm n m n + − − + + −
We know that 0 = sin(0) = sin(π) = sin(2π) = … = sin(kπ)
0
cos( )cos( )dmx nx xπ∫ =
0
1 1 1sin( ) sin( )2
m n x m n xm n m n
π + − − + −
= 1 1 1 1 1sin( ) sin( ) sin(0) sin(0)2
m n m nm n m n m n m n
π π + − − − − + − + −
= 12
(0 − 0) if m and n are integers
= 0
ii cos(mx)sin(nx) = [ ]1 sin( ) sin( )2
mx nx mx nx+ − −
= [ ]1 sin( ) sin( )2
m n x m n x+ − −
cos( )sin( )dmx nx x∫ = [ ]1 sin( ) sin( ) d2
m n x m n x x+ − −∫
= 1 1 1cos( ) cos( )2
m n x m n x cm n m n
− + + − + + −
= 1 1 1 [ ( )cos( ) ( )cos( ) ]2
m n m n x m n m n x cm n m n
× × − − + + + − ++ −
= 2 21 [( )cos( ) ( )cos( ) ]
2( )m n m n x m n m n x c
m n+ − − − + +
−
We know that 1 = cos(0) = cos(2π) = cos(4π) = … = cos(2kπ) If m is even and n is even, m + n is even and m − n is even. If m is odd and n is odd, m + n is even and m − n is odd.
0
cos( )sin( )dmx nx xπ∫ = [ ]2 2 0
1 ( )cos( ) ( )cos( )2( )
m n m n x m n m n xm n
π+ − − − +−
= [2 21 ( )cos( ) ( )cos( )
2( )m n m n m n m n
m nπ π+ − − − +
−
][( )cos( )0 ( )cos( )0]m n m n m n m n− + − − − +
= [ ]2 21 ( ) 1 ( ) 1 [( ) 1 ( ) 1]
2( )m n m n m n m n
m n+ × − − × − + × − − ×
−
= 2 21 (2 2 )
2( )n n
m n−
−
= 0 iii We know that −1 = cos(π) = cos(3π) = cos(5π) = … = cos(2k + 1)π If m is even and n is odd, m + n is odd and m − n is odd. If m is odd and n is even, m + n is odd and m − n is odd.
0
cos( )sin( )dmx nx xπ∫ = [ ]2 2 0
1 ( )cos( ) ( )cos( )2( )
m n m n x m n m n xm n
π+ − − − +−
= [2 21 ( )cos( ) ( )cos( )
2( )m n m n m n m n
m nπ π+ − − − +
−
][( )cos( )0 ( )cos( )0]m n m n m n m n− + − − − +
M C 1 2 Q l d - 7 238 A d v a n c e d p e r i o d i c f u n c t i o n s
= [ ]2 21 ( ) 1 ( ) 1 [( ) 1 ( ) 1]
2( )m n m n m n m n
m n+ × − − − × − − + × − − ×
−
= 2 21 ( 2 2 )
2( )n n
m n− −
−
= 2 24
2( )n
m n−
−
= 2 22n
m n−
−
b If m = n,
i 0
cos( )cos( )dmx mx xπ∫ =
0
1 [cos(2 ) 1]d2
mx xπ
+∫
= 0
1 1 sin(2 )2 2
mx xm
π +
= 1 1 1sin(2 ) sin(0) 02 2 2
mm m
π π + − +
= 1 1 0 02 2m
π × + −
= 2π
ii 0
cos( )sin( )dmx mx xπ∫ =
0
1 sin(2 )d2
mx xπ∫
= 0
1 1 cos(2 )2 2
mxm
π−
= 1 [cos(2 ) cos(0)]4
mm
π− −
= 1 (1 1)4m− −
= 0
The results would not change for cos(mx) sin(nx), but 0
cos( )cos( )dmx mx xπ∫ =
2π
11 a y = 2 2 2
cos(3 ) cos(5 ) cos(7 )cos3 5 7
x x xx + + + +…
Graphing with 4 terms gives
The amplitude is 1.1715 The x-intercepts occur at 1.571, 4.712, 7.854, … The period is 7.854 − 1.571 = 6.283
The frequency = 16.283
= 0.159
b y = 3 3 3
sin(2 ) sin(3 ) sin(4 )sin( )2 3 4
x x xx + + + +…
Graphing with 4 terms gives
The amplitude is 0.9887 The x-intercepts occur at 0, 3.142, 6.283, ... The period is 6.283 − 0 = 6.283
The frequency = 16.283
= 0.159
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 239
Chapter review 1 sin(90° − x) is in quadrant 1 and sin is positive in this
quadrant. The complementary function is cos. sin(90° − x) = cos(x) The answer is B. 2 The reciprocal of cos(x) is sec(x). The answer is A.
3 cosec(90 )θ° − = 1sin(90 )θ° −
= 1cos( )θ
= sec(θ)
4
1cot( )cosec2
sin( )
π θ π θ
θ
− + = −cosec2(θ)
We know that π − θ is in quadrant 2.2π θ+ is also in
quadrant 2 (sin +ve, cos −ve)
sin(π − θ) = sin(θ), cos(π − θ) = −cos(θ), sin2π θ +
= cos(θ)
LHS =
1cot( )cosec2
sin( )
π θ π θ
θ
− +
= cos( ) 1 1sin( ) sin( )sin
2
π θππ θ θθ
− × ×
− +
= cos( ) 1 1sin( ) cos( ) sin( )
θθ θ θ
− × ×
= 21
sin ( )θ−
= −cosec2(θ) = RHS 5 cos2(x) + sin2(x) = 1 The answer is D.
6 sec(30)° = 1cos(30)°
= 13
2
= 23
The answer is C.
7 21 cos ( )x− = 2sin ( )x = |sin(x)| The answer is D.
8 a cos(θ) = 5 ,13− 0 < θ < 180°. As cos(θ) is negative, θ is in
quadrant 2. cos2(θ) + sin2(θ) = 1
sin2(θ) = 251
13− −
= 251169
−
= 144169
sin(θ) = 1213
±
As θ is in quadrant 2, sin(θ) = 1213
cos( ) sin( )cos( ) sin( )
θ θθ θ
−+
=
5 1213 13
5 1213 13
− −
− +
=
171313
7 1313
−
×
= 177
−
= 327
−
b cot(θ) + tan(θ) = cos( ) sin( )sin( ) cos( )
θ θθ θ
+
= 5 1212 5− +
−
= 25 14460
+−
= 16960
−
= 49260
−
9 cos( ) cos( )1 sin( ) 1 sin( )
x xx x
+− +
= 2sec(x)
LHS = cos( ) cos( )1 sin( ) 1 sin( )
x xx x
+− +
= cos( )(1 sin( )) cos( )(1 sin( ))(1 sin( ))(1 sin( )) (1 sin( ))(1 sin( ))
x x x xx x x x
+ −+− + − +
= 2cos( )(1 sin( ) 1 sin( ))
1 sin ( )x x x
x+ + −
−
= 22cos( )cos ( )
xx
= 2cos( )x
= 2sec(x) = RHS 10 Pitch is an attribute of frequency The answer is A.
11 a y1 = 0.6 sin(50πt), y2 = 10.6sin 503
tπ π +
Sketching y = y1 + y2:
The amplitude is 1.04 The first maximum occurs at 0.00667. The next
maximum occurs at 0.04667 The period is 0.04667 − 0.00667 = 0.04
The frequency is 10.04
= 25
b The partials both have a period of 2 0.0450
ππ
= and
therefore a frequency of 25 The amplitude of the partials is 0.6
M C 1 2 Q l d - 7 240 A d v a n c e d p e r i o d i c f u n c t i o n s
This means that the frequency is not altered, but the loudness is almost doubled.
12 cosec(x) = 1sin( )x
Therefore the graph is related to the graph of sin(x) The answer is B. 13 y = 2 sec( )tπ Consider y1 = 2 cos( ).tπ 12 2y− ≤ ≤
Therefore 2y ≤ − or 2y ≥ The answer is E (both B and D). 14 y = 2 sec2(x + 45°)
Period 2
2π = π
Asymptotes occur at 2(x + 45°) = 360n° ± 90° x + 45° = 180n° ± 45° x = 180n° or 180n° − 90° The turning points occur at y = ± 2 15 Sketch y1 = 2 sec2(x + 45°) and y2 = cot(x − 45°) for the
domain −45° < x < 45° Sketching and finding the intersection:
The intersection is at x = 24.38942° or x = 24°23′ 16 cos(α + β) = cos(α) cos(β) − sin(α) sin(β) The answer is A. 17 sin(2y) = sin(y + y) = sin(y)cos(y) + cos(y)sin(y) = 2 sin(y)cos(y) The answer is E. 18 If sin(x) = t then sin2(x) + cos2(x) = 1 cos2(x) = 1 − sin2(x) = 1 − t2 cos(x) = 21 t± −
As x is acute, cos(x) = 21 t− sin(2x) = 2 sin(x) cos(x)
= 22 1t t− The answer is B.
19 cot( 45 )θ − ° = cos( ) sin( )sin( ) cos( )
θ θθ θ
+−
LHS = cot(θ − 45°)
= cos( 45 )sin( 45 )
θθ
− °− °
= cos( )cos(45 ) sin( )sin(45 )sin( )cos(45 ) cos( )sin(45 )
θ θθ θ
° + °° − °
=
1 1cos( ) sin( )2 2
1 1sin( ) cos( )2 2
θ θ
θ θ
× + ×
× − ×
=
1 (cos( ) sin( ))2
1 (sin( ) cos( ))2
θ θ
θ θ
+
−
= cos( ) sin( )sin( ) cos( )
θ θθ θ
+−
= RHS
cot(θ − 45°) = cos( ) sin( )sin( ) cos( )
θ θθ θ
+−
= cos( ) sin( ) sin( ) cos( )sin( ) cos( ) sin( ) cos( )
θ θ θ θθ θ θ θ
+ +×− +
= 2 2
2 2cos ( ) 2sin( )cos( ) sin ( )
sin ( ) cos ( )θ θ θ θ
θ θ+ +
−
= 2 2
2 2(cos ( ) sin ( )) 2sin( )cos( )
(cos ( ) sin ( ))θ θ θ θ
θ θ+ +
− −
= 1 sin 2( )cos2( )
θθ
+−
= 1 sin(2 )cos(2 ) cos(2 )
θθ θ
− −
= −sec(2θ) − tan(2θ) 20 a cos(2x) = cos(x + x)
= cos(x)cos(x) − sin(x)sin(x) = cos2(x) − sin2(x) = cos2(x) − (1 − cos2(x)) = cos2(x) − 1 + cos2(x) = 2cos2(x) − 1
b i cos 212π ×
= cos
6π
cos6π
= 22cos 112π −
22cos12π
= cos 16π +
= 3 12
+
= 2 32
+
2cos12π
= 2 34
+
cos12π
= 2 34
+±
= 1 2 32
± +
As 12π is in quadrant 1, cos
12π
is positive.
cos12π
= 1 2 32
+
ii 13cos12
π
= cos12ππ +
= cos12π −
= 1 2 32
− +
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 241
21 tan(x + y) = tan( ) tan( )1 tan( ) tan( )
x yx y+
−
The answer is C. 22 tan(2x) = tan(x + x)
= tan( ) tan( )1 tan( ) tan( )
x xx x+
−
= 22 tan( )
1 tan ( )x
x−
a tan 212π ×
= tan
6π
= 13
tan6π
= 2
2 tan12
1 tan12
π
π
−
21 1 tan123π −
= 2 tan
12π
21 tan12π −
= 2 3 tan
12π
0 = 2tan 2 3 tan 112 12π π + −
Using the quadratic formula
tan12π
= 2 3 12 4 12
− ± − × −
= 2 3 162
− ±
= 2 3 42
− ±
= 3 2− ±
As 12π is in quadrant 1, tan
12π
is positive.
tan12π
= 2 3−
b 11tan12
π
= tan12ππ −
= tan12π −
= 3 2− 23 a 2 sin(4x)cos(2y) = sin(4x + 2y) + sin(4x − 2y)
b a sin2(π x)cos(π x) = 1 sin( )(2sin( )cos( ))2
a x x xπ π π
= 1 sin( )sin(2 )2
a x xπ π
= [ ]1 1 cos( 2 ) cos( 2 )2 2
a x x x xπ π π π−× + − −
= [cos(3 ) cos( )]4a x xπ π− − −
= [ ]cos(3 ) cos( )4a x xπ π− −
= [ ]cos( ) cos(3 )4a x xπ π−
24 a i sin(4 ) sin(2 )cos(4 ) cos(2 )
x xx x
++
=
4 2 4 22sin cos2 2
4 2 4 22cos cos2 2
x x x x
x x x x
+ −
+ −
= 2sin(3 )cos( )2cos(3 )cos( )
x xx x
= sin(3 )cos(3 )
xx
= tan(3x)
ii cos(5 ) cos( )sin(5 ) sin( )
x xx x
−+
=
5 52sin sin2 2
5 52sin cos2 2
x x x x
x x x x
+ − −
+ −
= 2sin(3 )sin(2 )2sin(3 )cos(2 )
x xx x
−
= sin(2 )cos(2 )
xx
−
= −tan(2x)
b cos(5 ) cos( )sin(5 ) sin( )
x xx x
−+
= 0
−tan(2x) = 0
sin(2 )cos(2 )
xx
− = 0
sin(2x) = 0 2x = ±nπ
x = 2nπ± where n is an integer
25 a i cos(3x) sin(x) = [ ]1 sin(3 ) sin(3 )2
x x x x+ − −
= 1 (sin(4 ) sin(2 ))2
x x−
cos(3 )sin( )dx x x∫ = 1 (sin(4 ) sin(2 ))d2
x x x−∫
= 1 1 1cos(4 ) cos(2 )2 4 2
x x − +
+ c
= 1 ( cos(4 ) 2cos(2 ))8
x x c− + +
= 1 (2cos(2 ) cos(4 ))8
x x c− +
ii sin(4x)sin(2x) = [ ]1 cos(4 2 ) cos(4 2 )2
x x x x− + − −
= 1 (cos(6 ) cos(2 ))2
x x− −
40
sin(4 )sin(2 )dx x xπ
∫ = 40
1 (cos(6 ) cos(2 ))d2
x x xπ
− −∫
= 4
0
1 1 1sin(6 ) sin(2 )2 6 2
x xπ
− −
= [ ]40
1 sin(6 ) 3sin(2 )12
x xπ− −
= 1 6 2sin 3sin (sin(0) 3sin(0))12 4 4
π π − − − −
= 1( 1 3 0)12− − − −
= 13
b The graph cuts the x-axis when sin(4x) = 0 (sin(2x) will also be 0 at these points)
4x = 0, π, 2π, 3π, …
x = 2 30, , , ,4 4 4π π π
= 30, , , ,
4 2 4π π π
M C 1 2 Q l d - 7 242 A d v a n c e d p e r i o d i c f u n c t i o n s
The area bounded by y = sin(4x) sin(2x), x = 0 and 2
x π= is
40
2 sin(4 )sin(2 )dx x xπ
∫ = 123
× = 23
c Sketching the curve y = sin(4x)sin(2x). The required area is shaded.
Finding the area between 0 and :4π
Finding the area between 4π and :
2π
Both areas have the same magnitude of 0.3333333 The combined area is 0.66666
Modelling and problem solving 1 a f(x) = a cot(bx), x0 ≤ x ≤ 3.5 Given that f(3.5) = 0, the end of the chute is at x = 3.5 The end of the conveyor belt is 3 m before the end of the chute, i.e. at x = 0.5 This means that x0 = 0.5 b f(3.5) = 0, therefore a cot(3.5b) = 0 cot(3.5b) = 0
cos(3.5 )sin(3.5 )
bb
= 0
cos(3.5b) = 0
3.5b = 7π±
b = 7π±
f(0.5) = 2, therefore a cot(0.5b) = 2
1cot2 7
a π × ±
= 2
cot14
a π ±
= 2
a = 2 tan14π ±
As the graph is positive in the region x0 ≤ x ≤ 3.5, a is positive. Therefore b = 7π and a = 2 tan
14π
c When the toys are 1 m from the sack, x = 3.5 − 1 = 2.5
f(2.5) = 2 tan cot 2.514 7π π ×
= 0.22 The toys are 0.22 m above the sack
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 243
d f(x) = 2 tan cot14 7
xπ π
= cos
72 tan14 sin
7
x
x
ππ
π
×
f ′(x) = 2
sin sin cos cos7 7 7 7 7 72 tan
14 sin7
x x x x
x
π π π π π ππ
π
× − − × ×
=
2 2
2
sin cos7 7 72 tan
14 sin7
x x
x
π π ππ
π
− + ×
= 2
72 tan14 sin
7x
ππ
π
− ×
= 2
2 tan14
7sin7
x
ππ
π
−
f ′(x0) = f ′(0.5)
= 2
2 tan14
7sin14
ππ
π
−
= −4.14
f ′(3.5) = 2
2 tan14
7sin2
ππ
π
−
= −0.20
2 a i As x is in degrees, we need to use that cos( )x°∫ = cos180
xπ ∫ = 180 sin
180x cπ
π +
= 180 sin( )x cπ
° +
3 cos(4 )cos(2 )dx x x° °∫ = [ ]13 cos(4 2 ) cos(4 2 ) d2
x x x x x° + ° + ° − °∫
= 3 (cos(6 ) cos(2 ))d2
x x x° + °∫
= 3 180 1 180 1sin(6 ) sin(2 )2 6 2
x x cπ π
× ° + × ° +
= 3 180 1 (sin(6 ) 3sin(2 ))2 6
x x cπ
× × ° + ° +
= 45 (sin(6 ) 3sin(2 ))x x cπ
° + ° +
90
453 cos(4 )cos(2 )dx x x
°
°° °∫ = [ ]90
4545 sin(6 ) 3sin(2 )x xπ
°°° + °
= [ ]45 sin(540 ) 3sin(180 ) (sin(270 ) 3sin(90 ))π
° + ° − ° + °
= 45[0 ( 1 3)]π
− − +
= 45 2π
× −
= 90π
−
= −28.647
M C 1 2 Q l d - 7 244 A d v a n c e d p e r i o d i c f u n c t i o n s
ii As x is in degrees, we need to use that cos( )x°∫ = cos180
xπ ∫ = 180 sin
180x cπ
π +
= 180 sin( )x cπ
° +
sin(5 )sin( )dx x x− ° °∫ = [ ]1 cos(5 ) cos(5 ) d2
x x x x x° + ° − ° − °∫
= 1 (cos(6 ) cos(4 ))d2
x x x° − °∫
= 1 180 1 180 1sin(6 ) sin(4 )2 6 4
x x cπ π
× ° − × ° +
= 1 180 1 (2sin(6 ) 3sin(4 ))2 12
x xπ
× × ° − °
= 15 (2sin(6 ) 3sin(4 ))2
x xπ
° − °
180
45sin(5 )sin( )dx x x
°
°− ° °∫ = [ ]180
4515 2sin(6 ) 3sin(4 )2
x xπ
°°° − °
= [ ]15 2sin(1080 ) 3sin(720 ) (2sin(270 ) 3sin(180 ))2π
° − ° − ° − °
= 15 [0 ( 2 0)]2π
− − −
= 15π
= 4.775 b i
Area = −28.648 ii
Area = 4.775
3 a sin(πx) sin(2πx) = 1 [cos( 2 ) cos( 2 )]2
x x x xπ π π π− + − −
= 1 [cos(3 ) cos( )]2
x xπ π− − −
= 1 [cos(3 ) cos( )]2
x xπ π− −
sin(πx) = 0 when πx = 0, π, 2π, 3π, 4π, 5π, 6π … x = 0, 1, 2, 3, 4, 5, 6, … sin(2πx) = 0 when 2πx = 0, π, 2π, 3π, 4π, 5π, 6π …
x = 1 3 50, , 1, , 2, , 32 2 2
…
In the domain 2 ,3 3
xπ π< < the graph touches/crosses the x-axis when x = 32
and x = 2.
sin(πx) is negative for 1 < x < 2 and positive for 2 < x < 3
sin(2πx) is positive for 31 ,2
x< < negative for 3 22
x< < and positive for 522
x< <
This means that sin(πx) sin(2πx) is negative for 31 ,2
x< < positive for 3 22
x< < and positive for 52 .2
x< <
Sketching produces:
A d v a n c e d p e r i o d i c f u n c t i o n s M C 1 2 Q l d - 7 245
Area is found by integrating between 3,3 2π
and 3 2,2 3
π
sin( )sin(2 )dx x xπ π∫ = 1 [cos(3 ) cos( )]d2
x x xπ π− −∫
= 1 1 1sin(3 ) sin( )2 3
x x cπ ππ π
− − +
= 1 [sin(3 ) 3sin( )]6
x x cπ ππ
− − +
= 1 [3sin( ) sin(3 )]6
x x cπ ππ
− +
32
3
sin( )sin(2 )dx x xπ π π∫ = [ ]32
3
1 3sin( ) sin(3 )6
x x ππ ππ
−
= 2 21 3 9 33sin sin 3sin sin
6 2 2 3 3π π π π
π − − −
= −0.2115
23
32
sin( )sin(2 )dx x xπ
π π∫ = [ ]23
32
1 3sin( ) sin(3 )6
x xπ
π ππ
−
= 2 21 2 6 3 93sin sin 3sin sin
6 3 3 2 2π π π π
π − − −
= 0.2175 Required area = 0.2115 + 0.2175 = 0.429
b Finding area for 33 2
xπ < <
Area = −0.2115
Finding area for 3 22 3
x π< <
Area = 0.2175 Combined area is 0.2115 + 0.2175 = 0.429 4 a y1 = a sin(kx − ω t) and y2 = a sin(kt + ω t) The standing wave is found by finding y = y1 + y2 y = y1 + y2 = a sin(kx − ω t) + a sin(kt + ω t) = a[sin(kx − ω t) + sin(kt + ω t)]
= ( ) ( ) ( ) ( )2 sin cos2 2
kx t kx t kx t kx ta ω ω ω ω− + + − − +
= 2a sin(kx) cos(−ω t) = 2a sin(kx) cos(ω t) b The amplitude of the wave is 2a sin(kx).
The wavelength is 2 ,kπλ = rearranging gives 2k π
λ=
The largest amplitude occurs when sin(kx) = ±1.
kx = 3 5 7, , ,2 2 2 2π π π π …
x = 3 5 7, , ,2 2 2 2k k k kπ π π π …
= 3 5 7, , ,2 2 2 2 2 2 2 2π λ π λ π λ π λ
π π π π× × × × …
= 3 5 7, , ,4 4 4 4λ λ λ λ…
M C 1 2 Q l d - 7 246 A d v a n c e d p e r i o d i c f u n c t i o n s
5 a Using stats mode, and setting t = 0 as representing 4 February, 2009 6 am, a sin regression gives
The mathematical model would be (correct to 3 decimal places) d = 0.981 sin(0.521t − 0.202) + 1.589 b As 4 February, 2009 6 am represents t = 0, 7 February, 2009 6 pm would be 3 × 24 + 12 = 84 Need to investigate the graph between t = 84 and t = 96 to find when the tide is over 1.8 m between 6 pm and 6 am
The graph shows that mooring can occur. The intersections occur when t = 85.294 and t = 90.458 85.294 is 1 hour 14 minutes after 6 pm 90.498 is 6 hours 24 minutes after 6 pm The boat can moor after 7:14 pm and before 12:24 am 6 sin(θ) + sin(θ − α) + sin(θ + α) = 0
( ) ( ) ( ) ( )sin( ) 2sin cos2 2
θ α θ α θ α θ αθ − + + − − + +
= 0
sin(θ) + 2 sin(θ) cos(−α) = 0 sin(θ) + 2 sin(θ) cos(α) = 0 sin(θ)(1 + 2 cos(α)) = 0 0 ≤ θ ≤ 2π sin(θ) = 0 θ = 0, π, 2π 0 ≤ α ≤ π
cos(α) = 12−
cos3π
= 12
α = 3ππ −
= 23π
The problem can be solved for both θ and α . There are 3 solutions for θ and 1 solution for α .