π 2π - nulake.co.nz · IA 3.3 – ear 13 athematics and tatistics – Published by uake td ew...

57

IAS 3.3 – Year 13 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

IAS 3.3 – Trigonometry

Page 91.

2.

3.

4.

5. Period=2π

6.

Answers

f(x)

–1

π

1

2πx

–1

π

1

2π

f(x)

x

–1

π

1

2π

f(x)

x

–2

2

π 2πx

–1

1

f(x)2

–1

1f(x)

π 2πx

–1

1f(x)

πx

-π

07.

8.

9. Period4πamplitude3.

10.

11.

12. Frequency=4,period=π

2

andamplitude=1

2π–1

1f(x)

πx

–1

1f(x)

πx

-π

–2

2

–1

1

f(x)

πx

-π

2

1

f(x)

π

x

-πx

2πx

–1

1f(x)

π

1

f(x)

π-π

2

–2

–1

Page 13 (Other answers possible)

13. x =0.3514 (4sf)

14. x =2.3005 (4dp)

15. x =1.929 (4sf)

16. 4x = –0.22079 x = –0.0552 (4dp)

17. x =0.5773 (4dp)

18. x = –1.1398 or x =0.1990 (4dp)

19. x = –0.4343 or x =2.7073 (4dp)

20. x =0.6261 or x =3.1185 (4dp)

21. x =2.3562 or x = –2.3562 (4dp)

22. x =2.3562 or x = –0.7854 (4dp)

23. x = –2.8798 or x = –1.3090 (4dp)

24. x = 0

Page 15

25. x = π4

26. x = − π3

27. x2

= − π3

x = − 2π3

28. x + π6

= − π6

x = − π3

29. x = 5π6

30. x = π4

31. 3x=π3

x = π9

32. 4x = 5π6

x = 5π24


58 IAS 3.3 – Trigonometry

Page 19

33. 2x+1.426=nπ–1.1365

x=1.571nπ–1.281

34. x+1.571=2nπ±1.231

x=2nπ±1.231–1.571

35. 3x–π2=nπ+π

4 3x=nπ+3π

4 x = nπ3 + π

4

36. x – π4=nπ+(–1)n π

6⎛⎝⎜

⎞⎠⎟

x=nπ+(–1)n π6

⎛⎝⎜

⎞⎠⎟ + π4

=nπ+(–1)nx 0.5236 +0.7854

37. x=0.547,2.594,3.689,5.73638. x = –1.869,–1.272,0.225,0.822,

2.319,2.91739. x=3.4814,5.943340. x=1.112,2.029,4.254,5.171

0

41. x=0.262,1.309,3.403,4.451π12, 5π12,13π12,17π12

⎛⎝⎜

⎞⎠⎟

42. x=0.262,0.524,1.833,2.094, 3.404,3.665,4.974,5.236

π12, π6, 7π12, 2π3,13π12, 7π6,19π12, 5π3

⎛⎝⎜

⎞⎠⎟

43. x=0.654,1.702,3.796,4.843

5π24,13π24, 29π24

, 37π24

⎛⎝⎜

⎞⎠⎟

44. 2x–π=2nπ±0.5236 x=1.309,1.833,4.451,4.974

5π12, 7π12,17π12

,19π12

⎛⎝⎜

⎞⎠⎟

45. x=3.712,5.71346. x = –2.744,–0.398,0.398,2.744

Page 2147. x = –2.718,–1.995,0.424,1.14748. x=0.417,2.72449. x=0.424,1.147,1.995,2.718,

3.566,4.28850. x = –0.937,0.152,0.633,1.723,

2.204,3.294,3.775,4.864

Page 23

51. a) h=2.25sin 2π11(t −1.75)⎛

⎝⎜⎞⎠⎟+4.35

b) t=12.236hours(12:14pm)

c) time=6.5hours

Page 24

52. a) s=salesin000 s=3.5sin 2π

26(t − 2.5)⎛

⎝⎜⎞⎠⎟+8.5

b) t=11100(3sf)

c) time=4.3weeks

53. a) c=4sin 2π8(t − 4.5)⎛

⎝⎜⎞⎠⎟+28

b) Exceeds30MWH8hoursaday

Page 24-2554. a) w=consumpt.in000L/min. w=9sin 2π

15(t − 3.25)⎛

⎝⎜⎞⎠⎟+36

b)w=27200litres/minute

c) time=8 x 4=32minutes

Page 25

55. a) Periodevery14days. Maximumeffect=1atday9and23.Thatis14thand28thJanuary. Minimumeffect=0atday2andday16.Thatis7thand21stofJanuary.

b) H(x)=1.2sin 2π0.52 (x−0.17)⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟+ 2

c) Springtideisonday9.Closestlunartidetoday9isday9.14.Thisis14thJanuaryat3:22am.

d)Mustrelatemodeltoreality.Otherfactorsthataffecttimingandheightoftidesarestormsurgesandatmosphericpressure.Alsothelunartideeffectisnotaconstantheightbutdependsuponwhetherthemoonisonthenearorfarsideoftheearth.

Page 27

56. 1tan1.257

=0.3245

57. 1sin0.2187

=4.6091

58. 1cos− 0.5478

=1.1714

59. 1

cos π3

⎛⎝⎜

⎞⎠⎟

= 2

60. 1

sin 4π5

⎛⎝⎜

⎞⎠⎟

=1.701

61. 1

tan 2π3

⎛⎝⎜

⎞⎠⎟

= –0.5774

Page 28

62. LHS =cotxsinx

= cosxsinx

sinx

=cosx =RHS

63. LHS =secxsinx

= 1cosx

xsinx

= sinxcosx

=tanx =RHS

64. LHS = secxcosec x

= 1

cos x ÷ 1sinx

= 1

cos x x sinx1

=tanx =RHS65. LHS =cos2x–sin2x =cos2x–(1–cos2x) =cos2x–1+cos2x =2cos2x – 1 =RHS

59



Page 2966. LHS =tanxcosx

= sinxcosx

cosx

=sinx =RHS

67. LHS = 1cosec x tan x

= 11sinx

xsinxcosx

= 11

cosx

=cosx =RHS68. LHS =cosxtanxcosecx

=cosxx sinxcosx

x 1sinx

= 1 =RHS

69. LHS =cosec2 x – 1

= 1sin2 x

– 1

= 1sin2 x

– sin2 x

sin2 x

= 1− sin2 xsin2 x

= cos2 x

sin2 x =cos2xcosec2 x =RHS

70. LHS= 1cosθ

xcosθsinθ

x1sinθ

= 1sin2 θ

=cosec2 θ =RHS

71. LHS=cosec2 x – 1 =cot2 x =RHS

72. LHS=1–sin2 θ =cos2 θ =RHS

73. LHS=sec2x–tan2 x = 1 =RHS

074. LHS =cos2x–sin2 x =(1–sin2x)–sin2 x =1–2sin2 x =RHS

75. LHS =cosxsinx

+ sinx1+ cosx

= cosx + cos2 x + sin2 x

sinx(1+ cosx)

= cosx +1sinx(1+ cosx)

= 1sinx

=RHS

76. LHS = cotxcot2 x

= 1cotx

=tanx

=RHS

77. LHS = 11+ sinx

+ 11− sinx

= 1+ sinx +1− sinx(1+ sinx)(1− sinx)

= 21− sin2 x

= 2cos2 x

=2sec2 x =RHS

78. RHS =cosec θ− sinθcosec θ

= 1 – sinθ1sinθ

=1–sin2 θ

=cos2 θ =LHS

79. LHS =secxsinx

– sinxcosx

= secxcosx − sin2 x

sinxcosx

= 1− sin2 x

sinxcosx

= cos2 xsinxcosx

= cosxsinx

=cotx =RHS

Page 3380. –cosx81. cosx82. –tanx83. –sinx

Page 3484. tanx85. sinx86. cosθ

87. 12(cosθ–sinθ)

88. sinπ3cosπ

4–cosπ

3sinπ

4

= 6− 24

89. cosπ4cosπ

3–sinπ

4sinπ

3

= 2− 64

90.

tan π3 + tanπ4

1− tan π3 tanπ4

=

−2− 3

91. sinπ3 cos

π3 + sin

π3 cos

π3

= 32

92. cos(45°–60°)= 2+ 64

93. tan(60°–45°)= 2− 3

Page 35

94. LHS=cosAcosB – sinAsinBcosAcosB

=1–tanAtanB =RHS

95. LHS=cosxcoszsinzcosz

– sinxsinzsinzcosz

= cosxcosz− sinxsinzsinzcosz

= cos(x + z)0.5(sinzcosz+ sinzcosz)

= 2cos(x + z)sin(z+ z)

=RHS96. LHS=sin(x+y)+cos(x–y) =sinx.cosy+siny.cosx+cosx.cosy+sinx.siny =sinx.siny+sinx.cosy+cosx.siny+cosx.cosy =(sinx+cosx)(siny+cosy) =RHS



Page 37 cont...

103. RHS= 2cos2xsin 2x

= 2(cos2 x – sin2 x)

2sinxcosx

= cos2 x – sin2 xsinxcosx

= cosxsinx

– sinxcosx

=cotx–tanx

=LHS

104. sin3A=sin(A+2A) =sinAcos2A+cosAsin2A =sinA(1–2sin2A)+2sinAcos2A =sinA–2sin3A+2sinA(1–sin2A) =3sinA–4sin3A

105. cos3A=cos(A+2A) =cosAcos2A–sinAsin2A =cos3A–cosAsin2A–2sin2AcosA =cos3A–3cosAsin2A =4cos3A–3cosA

Page 39

106. x =2nπ±2.0944 or x =nπ+(–1)n0.3398 x = 0.3398,2.0944,2.8018,4.1888

107. x =nπ+(–1)n0.5236 or x =nπ–(–1)n0.7297 x = 0.5236,2.6180,3.8713,5.5535

108. x =2nπ±0.8411 or x =2nπ±2.4189

x = 0.8411,2.4189,3.8643,5.4421

109. x =nπ+1.2490 or x =nπ–1.1071

x = 1.2490,2.0344,4.3906,5.1760

110. (2sinx–1)(sinx+2)=0 x=nπ+(–1)n0.5236only

x = 0.5236,2.6180

111. x=2nπ±0.8411 or x=2nπ±3.1416 x = 0.8411,3.1416,5.4421

112. tanx=0 x = 0,π,2π

113. x=nπ+(–1)n0.3398 or x=nπ–(–1)n0.2526 x = 0.3398,2.8018,3.3943,6.0305

1

114. P=4sin9θ+4sinθ

115. Q = 12(cos4A+cos(2A+2B))

116. R=5cos10θ+5cos2θ

117. S=2(cos(3A+2B)–cos5A)

118. T=3(cos2x–cos4x)

119. U=cos 2x + π2

⎛⎝⎜

⎞⎠⎟+cosπ

2

=cos 2x + π2

⎛⎝⎜

⎞⎠⎟

120. V=sin4x+sinπ2

=sin4x+1121. W=0.5cosπ–0.5cos2x = –0.5–0.5cos2x122. X=3(cos2π+cos2x) =3+3cos2x

123.Y=5(cos–2x–cosπ4+ π4

⎛⎝⎜

⎞⎠⎟ )

=5cos2x

Page 42

124. 4 cos π3

⎛⎝⎜

⎞⎠⎟– cos(2x)⎛

⎝⎜⎞⎠⎟ = 4

2x=2nπ±2.0944 x=nπ±1.0472 x=1.0472,2.0944,4.1888,5.2359

or x = π3, 2π3, 4π3, 5π3

125.cos2x=1 2x=2nπ x=nπ x=0,π,2π

126. sin(2x–0.5)+sin(0.5)=2 x 0.4567 2x=nπ+(–1)n0.4489+0.5 x=0.5nπ+(–1)n0.2244+0.25 x=0.4744,1.596,3.616,4.738

127. 2x=nπ+(–1)n0.4058–0.5 x=0.5nπ+(–1)n0.2029–0.25 x=1.118,3.095,4.259,6.236

128. 2(cos(4x+π)+cosπ))=–3 4x+π=2nπ±2.0944 x=0.5nπ±0.5236–0.7854 x = 0.2618,1.3090,1.8326,2.8798,

3.4034,4.4506,4.9742,6.0214

129. (2x–0.5267)=2nπ±1.8140 x=nπ±0.9070+0.2634 x=1.1703,2.4980,4.3120,5.6396

Page 35 cont...97. LHS=sin(x+y).sin(x–y) =(sinxcosy+sinycosx)(sinxcosy–sinycosx) =sin2x.cos2y–sin2y.cos2 x = sin2x.(1–sin2y)–sin2y.(1–sin2 x) = sin2x–sin2x.sin2y–sin2y+sin2y.sin2 x =sin2x–sin2y =RHS

98. LHS=tan A– π4

⎛⎝⎜

⎞⎠⎟tan A + π

4⎛⎝⎜

⎞⎠⎟

= tanA − tan π

4⎛⎝⎜

⎞⎠⎟tanA + tan π

4⎛⎝⎜

⎞⎠⎟

1+ tanAtan π4

⎛⎝⎜

⎞⎠⎟1− tanAtan π

4⎛⎝⎜

⎞⎠⎟

= tanA −1( ) tanA +1( )1+ tanA( ) 1− tanA( )

= − 1− tanA( )1− tanA( )

= –1 =RHS

99. LHS=cot(A+B)

= 1tan A + B( )

= 1− tanAtanBtanA + tanB

Dividetopandbottomby tanA.tanB

= 1

tanAtanB– tanAtanBtanAtanB

tanAtanAtanB

+ tanBtanAtanB

= cotAcotB−1cotB+ cotA

=RHS

Page 37

100. cos24x–0.5=0.5(2cos2 4x – 1) =0.5cos(8x)

101. sin x2

⎛⎝⎜

⎞⎠⎟cos x

2⎛⎝⎜

⎞⎠⎟=0.5sinx

102. LHS=(sinA–cosA)2 =sin2A–2sinAcosA+cos2A =1–2sinAcosA =1–sin2A =RHS

61



44°27°48°

4900 m

B

D

O

A

C

6800 m

Page 44

130. T=2sin4Acos2A

131. U = –2sin2xsinx

132. V=8cos4xsinx

133. W=–4sinAsin B

or=4sinAsin(– B)

134. 2sin3xcosx=0

x = nπ3

or

x=2nπ±π2

x=0,π3,π2,2π3,π,4π

3,3π2,

5π3,2π

135. –2sin2xsinx=0

x = nπ2

x=0,π2,π,3π

2,2π

136. 2cos2xcos x + π3

⎛⎝⎜

⎞⎠⎟ = 0

x=nπ±π4

x=2nπ±π2

– π3

x = π6, π4,3π4, 7π6, 5π4,7π4

x=0.5236,0.7854,2.356,3.665, 3.927,5.498

137. 2sinxcosπ4

= 2 sinx=1 x=nπ+(–1)n π

2 x = π

2

Page 48138. a) AB=101m b)Usethecosinerule AngleABC=48.4° c) AngleATC=43.6°

Page 49

139. Lookforanisoscelestriangleh=asin2x+1.8140. AY=5.23 AW=3.59 WY= 3.592 + 5.232 − 2 x 3.59 x 5.23cos33˚ WY=3.0km(2sf)

0

141. a)Height=7.50+1.65 =9.15m(3sf) b)Usingsinerule

h −1.65sinA

= xsinB

sin((90−A)+ (90− B))

h = xsinAsinB

sin(180− (A + B))+1.65

h = xsinAsinBsin(A + B)

+1.65

or h = xsinAsinBsinAcosB+ cosAsinB

+1.65

h = xtanAtanBtanA + tanB

+1.65

142. h=54.9m(3sf)

Page 51143. a)

b)OC =6800cot44˚ OD =4900cot27˚ CD2 = OC2 + OD2 – 2OC x ODcos48 CD =7172metres(7170m3sf)

144. a) x+23 =hcotAandx=hcotB

hcotB+23 =hcotA

23 =hcotA–hcotB

23 =h(cotA–cotB)

h = 23cotA − cotB

b) h =12metres



PossiblestudentresponsesforMerit.

Studentshaveformedtwocorrectmodelsforbothcitiesandusedthemodelstoidentifyhowmanydaysthroughtheyearbothcitiesreceiveinexcessof13hoursofdaylighthoursandthenidentifywhichreceivesmostandbyhowmuch.

For city at latitude 37.0˚ SSolve2.5sin0.0172(t–264.75)+12=13 togetx=58.2and288.7

Numberofdayswith13hoursofdaylightormoreperyearis58.2+365–288.7=134.5

For City at latitude 49.0˚ NSolve4sin0.0172(t–79.75)+12=13 togetx=94.4and247.7

Numberofdayswith13hoursofdaylightormoreperyearis247.7–94.4=153.3

Cityat49.0˚Nreceives18moredaysof13ormorehoursofdaylightthancityat37.0˚S.

For merit the student needs to have applied trigonometric methods, using relational thinking, in solving problems. The student needs to have formed and used a model and related findings in context or communicated thinking using appropriate mathematical statements.

PossiblestudentresponsesforExcellence.

Effectoflatitudeonthetrigonometricgraphparameters.

Theamplitude(A)ofbothcurvesisdifferent.AincreasesthefurtherwemoveawayfromtheequatorAttheequatorAwouldequal0.

Bisconstantforbothgraphs(2π÷365)

Cremainsthesamefornorthernhemispherecities.

Cremainsthesameforsouthernhemispherecities.ThedifferencebetweenCfornorthernandsouthernhemispheresis6months(approximately185days).Disconstantforbothgraphs,i.e.12.

For excellence the student needs to have applied trigonometric methods, using extended abstract thinking, in solving problems. The student needs to have devised a strategy to investigate the situation and used correct mathematical statements or communicated mathematical insight.

Pages 52 – 55Practice Internal Assessment – Trigonometry

PossiblestudentresponsesforAchievement.

City at latitude 37.0˚ SAmplitude(A)=(14.5–9.5)÷2=2.5B=2π÷365=0.01721...Shift(C)=–(356–0.25(365))=–264.75Verticaltranslation(D)=(14.5+9.5)÷2=12Equation2.5sin0.0172(t–264.75)+12

City at latitude 49.0˚ NAmplitude(A)=(16.0–8.0)÷2=4B=2π÷365=0.01721...Shift(C)=–(171–0.25(365))=–79.75Verticaltranslation(D)=(16.0+8.0)÷2=12Equation4sin0.0172(t–79.75)+12

Graphsforbothlatitudes(seebelow).

For achievement the student needs to have applied trigonometric methods in solving problems. The student needs to have correctly selected and used methods, demonstrated knowledge of concepts and terms and communicated using appropriate representations.Evidence from both features of trigonometric functions and solving trigonometric equations is required.

49.0˚N

37.0˚S

13hoursdaylight

π 2π - nulake.co.nz · IA 3.3 – ear 13 athematics and tatistics – Published by uake td ew...

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Transcript of π 2π - nulake.co.nz · IA 3.3 – ear 13 athematics and tatistics – Published by uake td ew...