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NSS Mathematics in Action Module 2 Vol.1 Full Solutions

52

4 More about Trigonometric

Functions (I)

Classwork

Classwork (p. 4.3)(a) 2 rad (b) 4 rad

Quick Practice

Quick Practice 4.1 (p. 4.5)

(a)

rad5

3

rad180

108108

rad,180

1Since

=

=

=

(b)

rad3

rad180

540540

rad,180

1Since

=

=

=


(a)

=

=

=

40

9

1802rad

9

2

,180radSince

(b)

=

=

=

342

10

18019rad

10

19

,180radSince

(c)

fig.)sig.3to(cor.143

1805.2rad.52

,180

rad1Since

=

=

=


(a)

cm

2

5

cm

12

56

=

=)

AB

(b) Area of sectorOAB

2

22

cm2

15

cm12

56

2

1

=

=


(a) ROS

rad3

rad180

60

=

=

(b) Area of the shaded region

fig.)sig.3to(cor.cm181.0

cm3

sin)2(2

1

3)2(

2

1

2

222

=

=


(a) LetAB =x cm andBOC= rad.In sectorOAD, we have

2

3

6

=

=

In sectorOBC, we have

2

1062

3

102

3

10

=

=+

+=

+=

x

x

x

x

OB = (2 + 3) cm = cm5

rad2=BOC

(b) Area of the shaded region= area of sectorOBC area of sectorOAD

= 222 cm232

125

2

1

= 2cm16


(a) Let rbe the length ofOP. Then, 543 22 =+=r .

By definition, we have

4

3cot

3

5sec

4

5cosec

3

4tan

5

3cos

5

4sin

==

==

==

==

==

==

y

x

x

r

y

r

x

y

r

x

r

y

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4 More about Trigonometric Functions (I)

53

2

2

22

242

cos

sin

1sincos

cosec)coscos(

=

=

(b) Let rbe the length ofOP. Then, 133)2( 22 =+=r .


3

2

3

2cot

2

13

2

13sec

3

13cosec

2

3

2

3tan

13

2

13

2cos

13

3sin

=

==

=

==

==

=

==

=

==

==

y

x

x

r

y

r

x

y

r

x

r

y


2

3

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54

1

)cosec(sin

2

3secsin

2

3secsin

2

3sec

2

3cos

=

=

=

=

+

(b)

cot

sin

cossincossin

cossin

sin

1

sin

cos

cos

sinsin

1

cottan

cosec

2

22

2

22

=

=

+=

+

=+


L.H.S.

R.H.S.

1sin

cot

)cosec1(sin

cosec1

cosec1

cosec1

sin

cosec1

sin

cosec1

2

2

=

=

=

+=

+=

1sin

cot

sin

cosec1 2

=

+


(a)

1sin

cos

sin

1

cotcosec

)2(cot2

3sec

2

2

2

22

22

=

=

=

(b)


42

2

22

2

22

2

4

2

42

21

2

)1(

2

)sin1(

1sin1

cos

1cos

secsec

kk

k

k

k

+

=

=

+=

+=

+


(a)

32cotcosec

1)cotcosec(2

3

1)cot)(coseccotcosec(

1cotcosec

coseccot1

22

22

=

=

=+

=

=+

xx

xx

xxxx

xx

xx

(b)

=

=+

(2)3

2cotcosec

(1)2

3cotcosec

LL

LL

xx

xx

(1) + (2):

12

13cosec

6

13cosec2

=

=

x

x

(1) (2):

12

5cot

6

5cot2

=

=

x

x


(a)

The graph repeats itself at intervals of3

2.

y = cosec 3x is a periodic function with period3

2.

(b) From the graph,y = cosec 3x does not have maximum or

minimum values. Also, it cannot take any value in the

interval (1, 1). Thus, the range is < cosec 3x1 or1 cosec 3x < + .


(a) Draw the straight liney = 1 on the graph ofy = cosecx.

The two graphs intersect atx =2

3for 0 x 2.

The solution of cosecx = 1 for 0 x 2isx =2

3.

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55

(b)

xx

xx

x

cosecsin

sin

1sin

1sin2

=

=

=

The two graphs intersect atx =2

and

2

3

for 0 x 2.

The solutions of sin2x = 1 for 0 x 2arex =2

or

2

3.


(a) cotx=2.5

tanx = 0.4

x 21.80 or 180 + 21.80

x = 8.21 (cor. to 3 sig. fig.) or 202 (cor. to 3 sig.

fig.)

(b) Let 2x+ 60 =.

Since 0 x 360, 60 780.

Now, we solve the equation 2 sin 1 = 0 for 60

780.

=

+

++=

=

=

750or510,390,150

30720

or)30180(360,30360,30180

2

1sin

01sin2

Then,

=

=+

345or225,165,45

750or510,390,150602

x

x


+

=

=

=+

=+

=+=+

46.348or54.191

54.11360or54.11180

(rejected)2or5

1sin

2

1or5cosec

0)1cosec2)(5cosec(

05cosec9cosec2

03cosec92cosec203cosec9cot2

2

2

2

x

x

x

x

xx

xx

xxxx

= 192x (cor. to 3 sig. fig.) or 348 (cor. to 3 sig. fig.)


0)1cossin2(cos

0coscossin2

sin

coscos2

cotcos2

2

2

2

=+

=+

=

=

xxx

xxx

x

xx

xx

When cosx = 0,

2

3or

2

=x

When 2sinx cosx + 1 = 0,

4

7or

4

3

1tan

0)cos(sin

0)cos(sincossin2

2

22

=

=

=+

=++

x

x

xx

xxxx

4

7or

2

3,

4

3,

2

=x

Exercise

Exercise 4A (p. 4.10)

Level 1

1. (a)

rad

5

rad180

3636

rad,180

1Since

=

=

=

(b)

rad5

4

rad180

144144

rad,180

1Since

=

=

=

(c)

rad4

9

rad180

405405

rad,180

1Since

=

=

=

2. (a)

fig.)sig.3to(cor.rad1.22

rad180

7070

rad,180

1Since

=

=

=

(b)


rad180

22.522.5

rad,180

1Since

=

=

=

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56

(c)


rad180

130130

rad,180

1Since

=

=

=

3. (a)

=

=

=

270

1805.1rad1.5,180radSince

(b)

=

=

=

288

5

1808rad

5

8

,180radSince

(c)

fig.)sig.3to(cor.19.8

180

7

1rad

7

1

,180

rad1Since

=

=

=

4.

5. Angle sum of triangle = rad

rad18

7

rad9

4

6

=

=C

6. Let A =x.Then B = 2x, C= 3x and D = 4x.

rad5

rad2432

=

=+++

x

xxxx

rad5

=A , rad

5

2=B , rad

5

3=C ,

rad5

4=D

7. (a)

cm2

cm12

6

=

=)CD

(b) Area of sectorOCD

2

22

cm2

3

cm12

62

1

=

=

8. (a) Let XOY= rad.

5

2

20102

1 2

=

=

rad5

2=XOY

(b) Perimeter of the sector

cm)4(20

cm5

210102

+=

+=

++=)

XYOYOX

9. (a) Area of OAB = 10 cm2


100.8sin2

1 2

=

=

r

r

(b) Area of sectorOAB

2

22

cm1514.11

cm8.028.52

1

=

=

Area of the shaded region

= area of sectorOAB area of OAB


cm)101514.11(

2

2

=

=

10. Let OM= rcm and MON= rad.

(2)82

1

(1)4

2KK

KK

=

=

r

r

4

22

1:

)1(

)2(

=

=

r

r

By substituting r= 4 into (1), we have

1

44

=

=

rad1=MON

11. (a) Let COD = rad. Area of the shaded region = 24 cm2

5.1

2422

16

2

1 22

=

=

rad5.1=COD

(b) Perimeter of the shaded region

cm20

cm]2)26(5.165.12[

=

++=

+++= BDACCDAB))

12. (a))2(

2+=

+=

r

rrP

(b) IfP=A,

2

4

42

2

1)2( 2

=

=+

=+

r

r

rr

Level 2

13. (a) (i) AC= rsin

Angle (in degree) 30 45 90 120 150

Angle (in radian)6

4

2

3

2

6

5

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57

cossin2

1

sincos2

1

2

1ofArea

2r

rr

ACOCOAC

=

=

=(ii)

(iii) Area of the shaded region

= area of sectorOAB area of OAC

)cossin(2

1

cossin2

1

2

1

2

22

=

=

r

rr

(b) Area of the shaded region > 0

cos

1sin

cossin

0)cossin(2

1 2

>r

14. (a) Let OA = rcm and COD = rad.

3

4)1(

=+r (1)

(rejected)7

3or3

0)37)(3(

09187

169189

9

16)1(

9

16

2

1

)1(2

1

2

22

2

2

2

2

=

=+

=

=++

=+

=+

r

rr

rr

rrr

r

r

r

r

By substituting r= 3 into (1), we have

rad3

3

3

4)13(

=

=

=+

COD

(b) Area of the shaded region

2

222

cm6

7

cm3

321

3)13(

21

sectorofareasectorofarea

=

+=

= OABOCD

15.

cm3

25

cm6

510

=

=)

AB

cm6

25

cm3

25)(2

=

=

AN

AN

In OAN,

cm6

2510

2

2

22

=

= ANOAON

(Pyth. theorem)

Volume of the cone

fig.)sig.3to(cor.cm165

cm

6

2510

6

25

3

1

3

3

2

2

2

=

=

16. (a) BC=BEandBC=EC

EBCis an equilateral triangle.

EBC= rad3

(b) Area of the shaded region= area of sectorBEC+ area of sectorCBE

area of EBC


cm3

sin12

1

31

2

1

31

2

1

2

2222

=

+=

17. (a)

rad6

2

1sin

4

2)cos(90

=

=

=

(b) Area of the shaded region= area of rectanglePQRS area of sectorPST

area of PQT


cm3

tan2221

64

2142

2

22

=

=

18.

3

2

3

2

32

3

12

6cos

=

==

=

=

AOB

COD

OOD

OOD

Length of the thread))

DYCAXBBCAD +++=


cm3

49

3

236122 22

=

++=

Exercise 4B (p. 4.30)

Level 1

1. Let rbe the length ofOA. Then, r 2952 22 =+= .


sin

29

5==

r

y

cos 29

2==

r

x

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58

tan 2

5==

x

y

cosec 5

29==

y

r

sec

2

29==

x

r

cot 5

2==

y

x

2. Let rbe the length ofOB. Then, r 587)3( 22 =+= .


sin 58

7==

r

y

cos 58

3==

r

x

tan 37==

xy

cosec 7

58==

y

r

sec 3

58==

x

r

cot 7

3==

y

x

3. Let rbe the length ofOC. Then,

r 3.1)2.1()5.0(22

=+= .By definition, we have

sin 13

12

3.1

2.1=

==

r

y

cos 13

5

3.1

5.0=

==

r

x

tan 5

12

5.0

2.1=

==

x

y

cosec 12

13

2.1

3.1=

==

y

r

sec 5

13

5.0

3.1

=== x

r

cot 12

5

2.1

5.0=

==

y

x

4. Let rbe the length ofOD. Then, r 11)3()2( 22 =+= .


sin 11

3==

r

y

cos 11

2==

r

x

tan 2

3==

x

y

cosec 3

11==

y

r

sec

2

11==

x

r

cot 3

2==

y

x

5. (a) 7

9lies in quadrant III.

7

9sec

is negative.

(b)

=

6

11tan

6

11tan

6

11lies in quadrant IV.

6

11tan

is negative.

i.e.

6

11tan

is positive.

(c) 36

29lies in quadrant II.

36

29sin

is positive.

(d)3

5cot

3

5cot

=

3

5 lies in quadrant IV.

3

5cot

is negative.

i.e.

3

5cot

is positive.

6. sin is positive.

lies in quadrant I or quadrant II.

cot is negative.

must lie in quadrant II.

7.

2

3

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59

In particular, we may take r= 7 andy = 3.

Since lies in quadrant III,x < 0.

102

)3(7 22

22

=

=

= yrx

102

7sec

102

3tan

=

=

8.

22

3 0.

62

)1(5 22

22

=

=

= xry

62tan = ,

5

1cos =

3

65

6

10

5

6

62

5

11

62

cos1

tan

=

=

=

=

11. 05

3tan

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60

lies in quadrant II

5

3tan =

In particular, we may takex = 5 andy = 3.

34

3)5( 22

22

=

+=

+= yxr

34

3sin = ,

5

34sec =

25

343

5

34

34

3secsin

2

2

=

=

Case 2:

lies in quadrant IV

5

3tan =


34

)3(5 22

22

=

+=

+= yxr

34

3sin = ,

5

34sec =

25

343

5

34

34

3secsin

2

2

=

=

12.

1

cossin

cossin

)cos)(sincos(sin

cossin

cossin

22

22

2222

22

44

=

+=

+=

13. cosec2+ sec2 (tan + cot )2= (cosec2 cot2) + (sec2 tan2) 2= 1 + 1 2= 0

14.

2

2

cot

1cosec

sin

sin11cosec

sin

)1cosec)(sin1(

=

=

+=

+

15.

xx

xx

x

x

x

xxxx

xxxxx

seccosec

cossin

sin1

cos

sin

cos

1

cos

1

cossin

1

)tan(seccos

1

cossin

1

3

2

3333

22

=

=

+=

+

16. R.H.S.

L.H.S.

sec

1tan

tantan)1(tan

tantansec

2

2

2422

244

=

=

+=

+=

=

sec2 = sec4 tan4 tan2

17. (a)4

3cot

6

5sin

+

2

1

12

1

4cot

6sin

4cot

6sin

=

=

=

+

=

(b)

1

12

13

sec

13

sec4

tan3

2sec

=

+=

+=

+

=+

(c)

2

1

22

14

cosec4

cos

4cosec

42cos

4

5cosec

4

7cos

=

=

=

++

=+

18. (a) sin)sin( =

(b)

sin2

3cos =

+

(c) sec)sec( =

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61

(d)

cot

2tan

22tan

2

5tan

=

+=

++=

+

(e)

cot

)cot(

)2cot()3cot(

=+=

++=+

(f)

sec

2cosec

24cosec

2

7cosec

2

7cosec

2

7cosec

=

=

=

+=

+=

19.

sec

sin

tan

2

5cos

)2(tan

=

=

+

20.

2

5tan)(5cosec 22

1

cotcosec22

=

=

21.

The graph repeats itself at intervals of 2.

2

tanx

y = is a periodic function with period 2.

22.

The graph repeats itself at intervals of.

xy 2cosec2

1= is a periodic function with period .

23. Draw the straight liney = 1 on the graph of2

cotx

y = .

The two graphs intersect at2

=x for 0 x 2.

The solution of cot 12

=x

for 0 x 2is2

=x .

Level 2

24. cot = 2.5 < 0

lies in quadrant II or quadrant IV.

LetP(x,y) be a point on the terminal side ofand OP= r.

Case 1:

lies in quadrant II

255.2cot ==


29

2)5( 22

22

=

+=

+= yxr

29

2sin = ,

29

5cos =

292942

129

52

29

2)1cos2(sin

2

2

=

=

Case 2:

lies in quadrant IV

2

5cot =

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62


29

)2(5 22

22

=

+=

+= yxr

29

2sin = ,

29

5cos =

2929

42

129

52

29

2

)1cos2(sin

2

2

=

=

25. cosec = 2 > 0



cosec = 2


Case 1:

lies in quadrant I

Since lies in quadrant I,x > 0.

3

12 22

22

=

=

= yrx

2

1sin = ,

2

3cos = ,

3

1tan =

8

3

3

3

2

1

2

3

3

1sincostan

2

2

=

=

Case 2:

lies in quadrant II

Since lies in quadrant II,x < 0.

3

)1(2 22

22

=

=

= yrx

2

1sin = ,

2

3cos = ,

3

1tan =

sincostan 2

83

33

2

1

2

3

3

12

=

=

26.

m=

=

=

=

=

cot

sin

cos

cossin

sin1

cos

sin

cossin

1tan

cossin

1

2

27.

)2(

1

)1(1

1

)sin1(1

1

cos1

1

1sec

sec

22

22

22

44

4

=

=

=

=

28.

=

=+

)2(2

1cossin

)1(1cossin

22

22

KK

KK

xx

xx

(1) + (2):

3

2cosec

2

3sin

2

3sin2 2

=

=

=

x

x

x

(1) (2):

2sec

2

1cos

2

1cos2 2

=

=

=

x

x

x

29.

)sin1(tan

)1(coscos

)sin1)(1)(cos1cos(

)1cos(

)sin1(cos

1cos

2sin

)sin1)(seccos(

2sin

)sin1)(sin1)(sec1(

2sin

)sin1)(sec1(

22

22

222

2

2

2

4

2

222

2

224

2

44

+=

+

++=

+

+

=

+=

+=

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63

30. sinx cosx (sinx + cosx)(secx + cosecx)

xx

xxxxxx

xxxxxx

cossin21

coscossincossinsin

)cosec)(seccossincos(sin

22

22

+=

+++=

++=

xx

xxxxxx

cosecsec

cossin21)cos(sincossin

+

+=+

31. (1 + cosx + cotx + cosecx)(1 + sinx cosx)

)sin1)(cossin1(

cossincossinsin21

sincosec

1eccoscossincossinsin

cotcos1eccoscossincossinsin

cot1seccosin

cos

coscotcoscossincoscossin1

2

2

2

2

2

xxx

xxxxx

xx

xxxxxx

xxxxxxxx

xxx

x

xxxxxxxx

+++=

++++=

+

+++++=

+++++=

++

+++++=

x

xxxxxxx

sin1coseccotcos1

cossin1cossin1

+ +++=+ ++

32. L.H.S.

R.H.S.

1

sin

sin

)1(cossin

1

sin

1

sin

cos

sin

1

sin

cos

)](cosec)()[cotcosec(cot

2

2

2

2

=

=

=

=

+

=

++=

x

x

xx

xx

x

xx

x

xxxx

(cotx cosecx)[cot (+x) cosec (+x)] = 1

33. L.H.S.

R.H.S.

tan

cos1

)1(costan

cos1

tansin

2

3sin1

2cotsin

=

=

=

=

+

++

=

x

x

xx

x

xx

x

xx

x

x

xx

tan

2

3sin1

2cotsin

=

+

++

34. (a)

The graph repeats itself at intervals of 2. y = cosec (x ) is a periodic function with

period 2.

(b) From the graph,y = cosec (x ) does not have

maximum or minimum values. Also, it cannot take

any value in the interval (1, 1). Thus, the range is

< cosec (x ) 1 or 1 cosec (x ) < + .

35. (a)


2.

xy 3sec2

1= is a periodic function with period

3

2.

(b) From the graph, xy 3sec21= does not have


any value in the interval

2

1,

2

1. Thus, the range is

2

13sec

2

1

8/2/2019 1254108407_M2V104_FS_01e

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64

2

6sin

2

3

6cos

rAD

r

AD

rOD

r

OD

=

=

=

=

0secsin

1secand1sin0

I,quadrantinliesWhen

>+

>

8/2/2019 1254108407_M2V104_FS_01e

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65

fig.)sig.3to(cor.rad915

orfig.)sig.3to(cor.rad52.3

3747.02or3747.0

(rejected)2

31or

2

31sin

01sin2sin22

.

x

x

x

xx

=

+

+=

=



82or

8

8coscos

=

=

=

Exercise 4C (p. 4.37)

Level 1

1.

2.

rad)3218.0(orrad3218.0

3

1tan

3cot

+

=

=



.

=

3.

3794.0or3794.0

7.2

1sin

7.2cosec

=

=



.

=

4.

2780.12or2780.132

1cos

32sec

=

=


orfig.)sig.3to(cor.rad28.1=

5.

18

cos)2(cosor

18

coscos

)182

(sin)2(cosor)182

(sincos

9

4sincos

==

==

=

xx

xx

x

18

35or

18

=x

6.

611or

65

62or

6

3

1tan

3cot

3tancot

=

=

=

=

=

x

x

x

x

7.

0)1cosec4)(9cosec(

09cosec35cosec4 2

=+

=+

xx

xx

111.02or111.0

9

1

sin

(rejected)4

1cosecor9cosec

01cosec4or09cosec

+=

==

==+

x

x

xx

xx


orfig.)sig.3to(cor.rad25.3=x

8.

9.

0)3sin2)(1sin2(

03sin4sin4

01sin4sin44

01sin4cos4

2

2

2

=+

=+

=

=

xx

xx

xx

xx

6

5or

6

6or

6

(rejected)2

3sinor

2

1sin

03sin2or01sin2

=

=

==

=+=

x

xx

xx


orfig.)sig.3to(cor.rad524.0=x

10.

0)4)(sec3(sec

012sec7sec

013sec7tan

2

2

=

=+

=+

xx

xx

xx

4

1cosor

3

1cos

4secor3sec

04secor03sec

==

==

==

xx

xx

xx

When

052.5or231.1

1.2312or231.1

,3

1cos

=

x

x

x

When

.9654or318.1

1.3182or318.1

,4

1cos

=

x

x

x

fig.)sig.3to(cor.

rad05.5orfig.)sig.3to(cor.rad97.4

fig.),sig.3to(cor.

rad32.1fig.),sig.3to(cor.rad23.1=x

8/2/2019 1254108407_M2V104_FS_01e

15/21

8/2/2019 1254108407_M2V104_FS_01e

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67


cm41sin332

1ofArea

2

2

=

= .OAB

fig.)sig.3to(cor.rad.245

orfig.)sig.3to(cor.rad464

fig.),sig.3to(cor.rad.143

fig.),sig.3to(cor.rad821

fig.),sig.3to(cor.rad05.1

.

.

=

19. (a)

3322

3

cotcot3cot31

)cot1(

kkk

k

+++=

+

(b)

2

1cot

0)cot21(

01cot6cot12cot8

3

23

=

=+

=+++



.

=

Revision Exercise 4 (p. 4.41)

Level 1

1. (a)


rad180

4242

rad,180

1Since

.=

=

=

(b)


rad180

147147

rad,180

1Since

.=

=

=

(c)


rad180

345345

rad,180

1Since

.=

=

=

2. (a)


1803.3rad3.3

,180

rad1Since

=

=

=

(b)


21

18013

21

13

,180radSince

=

=

=

(c)


1802rad2

,180

rad1Since

=

=

=

3. (a) Let OA = rand AOB = rad.

The length of)

AB is twice that of the radius.

2

2

=

=

rr

rad2=AOB

(b)


10

10(2)2

1

cm10sectorofArea

2

2

2

.r

r

r

OAB

=

=

=

=

The radius of the sector is 5.60 cm.

4. (a)

(b)

Area of the shaded region


cm)435.44.13

2

1(

2

22

=

=

5. (a)

525 = 360 + 165

The angle 525 lies in quadrantII.

(b)

22

7

2

3

11

20 +=

The angle11

20lies in quadrantIV.

(c)

)12

5(

12

17

+=

The angle

12

17 lies in quadrantII.

6. (a)

Let rbe the length ofOP. Then,

.859)2( 22 =+=r


9

2cot

2

85sec

9

85cosec

2

9tan

85

2cos

85

9sin

==

==

==

==

==

==

y

x

x

r

y

r

x

y

r

x

r

y

(b)

Let rbe the length ofOP. Then,

.6)2(2 22 =+=r


8/2/2019 1254108407_M2V104_FS_01e

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68

R.H.S.

sinsin

sinsin

sin

sin

sin

sin

1sin

sin

sin

sin1

)sin(sinsin

1

sin

1

)sin)(sincoseccosec(L.H.S.

22

=

=

=

+=

+=

+=

yx

yx

x

y

y

x

y

x

x

y

yxyx

yxyx

22

2cot

2

3

2

6sec

32

6cosec

2

1

2

2tan

3

2

6

2cos

3

1

6

2sin

=

==

===

=

==

=

==

===

=

==

y

x

x

r

y

r

x

y

r

x

r

y

7. sec is negative.

lies in quadrant II or quadrant III.

cosec is negative.

lies in quadrant III or quadrant IV.

lies in quadrant III.

8.

2

3 0


sec < 0

must lie in quadrant II.


cosec = 6


Since lies in quadrant II,x < 0.

35

16 22

22

=

=

= yrx

cot = 35 ,6

1sin = ,

35

6sec =

35

34

)35

6(6

135secsincot

=

=

10. (a)

xxxx

cosec2sin

2

cosec)1(cos

22

==

(b)

sec

cossin

cos1

sin

cos

cossin

1cot)cosec

cossin

1(

2

2

222

=

=

=

11. (a)

yx

yx

yxyx

sinsin

sinsin

)sin)(sincoseccosec(

22 =

+

(b)

L.H.S.

sec)(

sec)1(tan

sectan

)sectan)(sectan(R.H.S.

222

2222

22222

2

=

=

+=

+=

++=

ba

ba

aba

ababa

2

222

)sectan)(sectan(

sec)(

ababa

ba

++=

8/2/2019 1254108407_M2V104_FS_01e

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69

2

)2()2(1)1(

4cosec4sec4cot4tan

4

32cosec

4

52sec

4

72cot

4tan

4

3cosec

4

5sec

4

7cot

4

3tan

=

=

+

=

=

6216

1

3

242

3

1

2

1

3cosec4

4sec

3cot6sin

3

104cosec4

4sec

32cot

6sin

3

10cosec4

4

21sec

3

5cot

6

7sin

=

+

=

+

=

+

+

+

=

+

2cos

2sin

2cot

2sin

22tan

2sin

2tan

2sin

2tanL.H.S.

C

CC

CC

CC

CBA

=

=

=

=

+=

12. (a)

2

331

2

)3(

2

1

)30180(cosec

60tan45sin

)210360(cosec

)60360(tan)45180(sin

)210cosec(

420tan225sin

32

32

32

32

+=

+

=

+=

+++=

+

(b)

(c)

13. (a)

A

AAA

CBA

2

2

cosec

1cot1)(cotcot

1)(cotcotL.H.S.

=

==

+=

AA

A

A

A

A

AA

A

AA

A

CBA

2cosecsin

sin

cos

cos

1

sin

cotsec

sin

)(cotsec

sin

)(cotsecR.H.S.

=

=

=

=

+=

A

CBACBA

sin

)(cotsec1)(cotcot

R.H.S.L.H.S.

+=+

=

(b)

2cos

22sin

2sin

2sinR.H.S.

C

C

C

BA

=

=

=

+=

2sin

2sin

2tan

R.H.S.L.H.S.

BACBA +=

+

=

14.


.

+=

22cot

xy is a periodic function with period

2

.

15.

The graph repeats itself at intervals of .

y = 1 + cosec 2x is a periodic function with period .

8/2/2019 1254108407_M2V104_FS_01e

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70



185.2cot

2tancot

=

=

=

16. Draw the straight liney= 2 on the graph of2

secx

y = .

The two graphs intersect at3

2=x for 0 x 2.

The solution of 22

sec =x

for 0 x 2is3

2=x .

17.



975.0sin

7

4sinsin

=

=

=

18.



11cosec

=

=

19.

20.


(rejected)5

1or1cosec

0)cosec51)(cosec1(

0cosec5cosec41 2

=

=

=+

=

21.

0)2sin3)(1sin3(

02sin9sin9

011sin9cos9

2

2

=

=+

=+





3

2or

3

1sin

=

=

22.

0)2cosec3)(15cosec(

030cosec43cosec3

027cosec43cot3

2

2

=+

=

=



(rejected)3

2or15cosec

=

=

Level 2

23. Let OA=rcm and AOB=rad.

)2......(52

1

)1......(122

2 =

=+

r

rr

From (1), )3.....(221

=r

By substituting (3) into (2), we have

1or5

0)1)(5(

056

10212

5212

2

1

2

2

2

==

=+

=

=

r

rr

rr

rr

rr

When r= 5,5

22

5

12==

When r= 1, 1021

12== (rejected)

rad5

2=AOB

24. (a) (i) Area of 22 msin2

1xOCF=

(ii) Area ofABCDEF

222

222

msin2

1)(

2

120

msin2

1)(

2

1102

++=

++=

xyx

xyx

(b) x+y= 80In ODE,

=

=+

05.44

cos)80(2608080 2222

In OCF,

667.2

05.44cos22 2222

=

=+

x

xxx

Area ofABCDEF

fig.)sig.3to(cor.m2480

m

05.44sin)667.2(2

1

05.44180

)80(2

120

2

2

2

2

=

+

=

8/2/2019 1254108407_M2V104_FS_01e

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71

25. (a)


.

y= cot 2x+ 1 is a periodic function with period

2

.

(b) From the graph,y= cot 2x+ 1 does not have

maximum or minimum values. Thus, the range is

< cot 2x+ 1 < +.

26. (a)

The graph repeats itself at intervals of.

y= cosec 2x 1 is a periodic function with

period .

(b) From the graph,y= cosec 2x 1 does not have


any value in the interval (2, 0). Thus, the range is < cosec 2x 1 2 or 0 cosec 2x 1 < +.

27.

2sec

2

1cos

2

1cos1

2

1sin

cosecsin2

2

2

2

2

=

=

=

=

=

x

x

x

x

xx

Draw the straight liney= 2 on the graph ofy= sec2x.

The two graphs intersect at

4

5,

4

3,

4

=x and

4

7for 0 x 2.

The solutions of 2 sinx= cosecx for 0 x 2are

4

5,

4

3,

4

=x or

4

7.

28.

13

5or1cot

0)5cot13)(1(cot

05cot8cot13

0cosec13cot818

2

2

=

=+

=

=+




fig.),sig.3to(cor.rad0.785=

29.

22

22

sin4sin2894sin4545

sin4sin2894cos45

sin27cos45

7cos45sin2

sec745tan2

+=

+=

=

=+

=+

(rejected)2.85

orfig.)sig.3to(cor.290.0

7

2sin

0)2sin7(

04sin28sin49

2

2

=

=

=

=+

rad0.290=

30.

6

19or

6

17,

6

7,

6

52

2

32cos

3

22sec

3cosec2sec

=

=

=

=

12

19or

12

17,

12

7,

12

5 =

31. (a) 222 )2(44 yxyxyx +=++

(b)

L.H.S.

4cossin4sin3

3sincoscossin4sin3

3coscossin4sin4R.H.S.

2

222

22

=

++=

++++=

+++=

3coscossin4

sin44cossin4sin3

2

22

++

+=++

(c)

3)cossin(2

3coscossin4sin4

4cossin4sin3)(

2

22

2

++=

+++=

++=

f

Iff() attains its minimum value, (2 sin + cos )2

attain minimum value.i.e. 0)cossin(2 2 =+

8/2/2019 1254108407_M2V104_FS_01e

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cmtan4 r

OPOC

=

=


orfig.)sig.3to(cor.68.2

2

1tan

=

=

Minimum value of 3)( =f

32. (a)

2

1

2

2

1

tantan2

tan1

tancossin2

sincos

2

2

2

22

+=

=

=

=

t

tt

t

y

(b) (i)

3

41

22

1

2

2tancossin2

sincos

2

2

2

22

==

=+

=

t

t

t

The equation has2tancossin2

sincos22

=

no solutions.

(ii)

2or,0

0tan

0

0

2

1

2

1

2

2

1tan

cossin2

sincos

1tancossin

sincos

cossintan)sin(cos

2

2

22

22

22

=

=

=

=

=+

=

=

=

t

t

t

The equation

cossintan)sin(cos 22 =

has 3 solutions.

33. (a)

1

cosecsin

=

= n

(b)

4cosecsin

4)cosec(sin

cosecsin4)cosec(sin)cosec(sin

cosecsin

2

22

22

=

=

+=

=+

m

m

m

(c)

=

=+

)2......(4cosecsin

)1......(cosecsin

2m

m

(1) + (2) :2

4sin

2 +=

mm

(d) Ifm=3,

2

433sin

2 +=


orfig.)sig.3to(cor.rad53.3=

34. (a)

(b) Area of trapeziumABCD

)tan8

1(tan4

tan)tan4tan4(

2

1

)(2

1

22

+=

++=

+=

r

rrrr

OABCAD

(c) Area of trapeziumABCD= 64 cm2

)tan8

1)(tan4(464 22 +=

(rejected)16

2571or

16

2571tan

08tantan82

+=

=+

754.0= (cor. to 3 sig. fig.)

1254108407_m2v104_fs_01e

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