Chapter 7 — Advanced periodic...

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

+ × × + + −


= 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


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 )+ +

° °


= 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

±


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)


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

+ − × + −

−


=

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


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


c y = 0.2 sin(100π t) + 0.2 sin(300π t)

From the graph, amplitude = 0.31 period = 0.02

= 1 s50


d y = ( )0.4sin 200 0.25sin 2002

t tππ π + +


= 1 s100

frequency = 100 s−1 e y = 0.5 sin(200π t) + 0.25 sin(100π t)


= 1 s50



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


x π− = nπ

x = 3

n ππ +



c y = cot2

x π +

Period π


x π+ = nπ

x = 2

n ππ −

d y = 2sec24

x π −

Period 22π = π


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


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 +


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


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


xπ = 22

n ππ ±

πx = 4nπ ± π x = 4n ± 1 The turning points occur at y = −1 ± 2


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+


= 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 −


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 + 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

−×

+


= 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

+

+ ×


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

−

+


= 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 + +


=

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+ − −


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)


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


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


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


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−∫


= 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

− +


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− +


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


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− −


= 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− − × − ×


= 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− + − − − +


= [ ]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


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


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

− +


21 tan(x + y) = tan( ) tan( )1 tan( ) tan( )

x yx y+

−

The answer is C. 22 tan(2x) = tan(x + x)


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


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


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


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:


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…


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 .

Chapter 7 — Advanced periodic...

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