16[Anal Add Math CD]

1Additional Mathematics SPM Chapter 16

Penerbitan Pelangi Sdn. Bhd.

1. (a)

370

y

x

Quadrant I(b)

520

y

x

Quadrant III (c)

420

y

x

73 =73

180 =420 Quadrant I(d)

420

y

x

73 =73

180 =420 Quadrant IV

2. (a)

30

y

x

36030=330

(b)

120

y

x

360120=240

(c) 43 =43 180

=240

43

y

x

pi

2 43 =23

CHAPTER

16 Trigonometric Functions

2 Additional Mathematics SPM Chapter 16


(d) 53 =53 180

=300

53

y

x

pi

2 53 =13

3.

3

4 5

0

yP(3, 4)

x

(a) sin q= 45(b) cos q= 35(c) tan q= 43(d) sec q= 53(e) cosec q= 5

4

(f) cot q= 34

4.

8

6

10

P(6, 8)

y

xA

(a) cot A= 68 = 3

4

(b) sin A= 810 = 45(c) sec A= 106 = 53

5.

30

6021

3

(a) sin120=sin60

=AB32

(b) cos210=cos30

=AB32

(c) cot(120)=cot240 =cot60

= 1AB3

(d) tan(30) =tan330 =tan30

= 1AB3

(e)

45

451

1

2

sin315=sin45

= 1AB2

6.

4

3 5

0

y

Ax

(a) cos A=cosa = 45(b) cot A=cota = 43(c) sec A=sec a = 5

4



7. 1

5

0

y

Bx

24

(a) tan B=tana =ABB24

(b) sin B =sina

=ABB245

(c) cosec B=coseca = 5

ABB24

8. (a) tan A= sin Acos A

=

35 45

=1 35 21 54 2

= 34

(b) sec A= 1cos A = 1

45 = 5

4

(c) cosec A= 1sin A = 135 = 53

9.

1 t 2

0

y

x

1

t

(a) cot q= ABBBB1 t2

t(b) sec(90q)=cosecq = 1t(c) sin (q)=sinq =t

10. (a) sin q= 12

q=30,150

(b) cos q= 12

q=120,240(c) 2tanq=5 tan q= 5

2 q=tan1 1 52 2 =6812,24812

(d) 4cotq=tanq

41 1tan q 2=tanq tan2 q=4 tan q=2 q=6326,11634,24326,29634

(e) 3 sin qcos(90q)=1 3 sin q sin q=1 2sinq=1 sin q= 1

2 q=210,330

(f) cot q+tan(90q)=4 cot q + cot q=4 2cotq=4 cot q=2 tan q= 1

2 q=2634,20634(g) 3sec(90q) cosec q=4 3 cosec q cosec q=4 2cosecq=4 cosec q=2 sin q= 1

2 q=210,330

(h) sin2 q + 3 sin q+2=0 (sin q + 1)(sin q+2)=0 sin q=1 , sinq =2 q=270 qundefined Therefore,q=270

(i) 2secq + cos q=1 2cos q + cos q=1 2+cos2 q=cosq cos2 q cos q2=0 (cos q2)(cosq+1)=0 cos q=2 , cosq=1 qundefined q=180 Therefore,q=180



(j) sin2q=1 2q=270,630 q=135,315

(k) 2cos2q+1=0 cos2q= 1

2 2q=120,240,480,600 q=60,120,240,300

(l) tan (q+45)=1 q+45=135,315 q=90,270(m) 2cot(2q30)=3 cot(2q30)= 3

2

tan(2q30)= 23 2q30=tan1 1 23 2 =3341, 21341, 39341,

57341 2q=6341, 24341, 42341,

60341 q=3151, 12151, 21151,

30151

(n) sec 1 q2 +102=2 cos 1 q2 +102=

12

q2+10=60

q2=50

q=100

(o) 2cosec1 q3 302=3 cosec 1 q3 302=

32

sin 1 q3 302=23

q3 30=4149

q3 =7149 q=21527

(p) cos (q) + cos2 q2=0 cos q + cos2 q2=0 cos2 q + cos q2=0 (cos q+2)(cosq1)=0 cos q =2 , cosq=1 qundefined q=0,360 Therefore,q=0,360

11. (a)

180 270 360

y

x

y = sin 2x

900

1

1

(b)

180 270 360

y

x

900

2

2

y = 2 sin x

(c)

180 270 360

y

x

900

3

1

1

y = 2 sin x + 1

(d) 180 270 360

y

x90

135 225 315450

1

2 y = sin 2x 1

(e)

pi 3pi 2pi

y

pi

220

1

1y = cos 2

(f)

pi 3pi 2pi

y

pi

220

2

2y = 2 cos



(g)

pi 2pi

y

0

3

1 y = 2 cos 1

(h)

pi 2pipi 3pi

2 2

y

0

2y = cos 2 + 1

(i)

90 180 270 360

13545 225 315

y y = tan 2x

x0

(j)

90 180 270 360

y y = 2 tan x

x0

(k)

90 180 270 360

y

y = |sin x |

x0

1

(l)

90 180 270 360

y

y = |cos 2x |

x0

1

(m)

90 180 270 360

y

y = |sin 2x| + 1

x0

2

1

(n)

90 180 270 360

y

y = |sin 2x|

x0

1

(o)

60 180120 240 360300

yy = sin x

x0

23

1

1

(p)

60 180120

240

360300

y

y = cos x

x0

23

1

1

(q)

120 240 360

y

y = 2 sin 3x

x0

2

2

(r)

360180

y

y = 3 sin x

x0

21

3



12. (a)

2pipi

pi

y

y =

y = sin x

x

x

0

1

1

(b) sinx=x sin x= x Thenumberofsolutionsis2.

13. (a)

2pipi 3pipi

piy y = 2

y = 2 |sin 2x |

x

3x

0

2 2

21

2

(b) sin2x= 3x2

1

2sin2x= 3x 2 Thereare3solutions.

14. y=3sinx

x 0 63

2

2356

y 0 1.5 2.6 3 2.6 1.5 0

y=2 x

x 0 2

y 2 1.5 1

0

2

3

1

x

y = 3 sin x

y

pi pi pi 2pi 5 pi pi6 3 2 3 6

xy = 2 pi

sin x + x3

= 23 3 sin x + x =2

3 sin x=2 xThesolutionsare0.7and2.8.

15. y=1+cosxx 0 0.5 1 1.5 2 2.5 3.0y 2 1.88 1.54 1.07 0.58 0.20 0.01

y= 34

x

x 0 1 2

y 0 3432

0

2

1

x

y = 1 + cos x

y

1 32

3y = x 4

4cosx+4=3x cos x+1= 3

4x

Thesolutionis1.5.

16. (a) a2 + c2=b2

a2

b2 + c2b2 =

b2b2

1 ab 22

+ 1 cb 22

=1

sin2 q + cos2 q=1(b) sin2 q + cos2 q=1 sin

2 qcos2 q + cos2 qcos2 q =

1cos2 q tan2 q+1=sec2 q(c) sin2 q + cos2 q=1 sin

2 qsin2 q + cos2 qsin2 q =

1sin2 q 1 + cot2 q=cosec2 q

17. (a) 1 + sin2 A =1+(1cos2 A) =2cos2 A(b) tan A(tan A cot A) =tan2 A 1 =(sec2 A 1) 1 =sec2 A2



(c) cot2 A sin2 A = cos2 Asin2 A sin2 A

=cos2 A =1sin2 A(d) cosec2 A + cot2 A=cosec2 A + (cosec2 A 1) =2cosec2 A 1

(e) 1sin2 A + 1cos2 A =

cos2 A + sin2 Asin2 A cos2 A =

1sin2 A cos2 A =1 1sin2 A 21

1cos2 A 2 =cosec2 A sec2 A

(f) 3 sin22Acos22A + sec22A=3tan22A + sec22A

=3(sec22A 1) + sec22A =3sec22A 3 + sec22A =4sec22A 3

18. (a) sin2 q + cos q=1 1 cos2 q + cos q=1 cos2 q cos q=0 cos q (cos q1)=0 cos q=0 ,cosq=1 q=90,270 q=0,360 Therefore,q=0,90,270,360

(b) tan q + cot q=2 tan q + 1tan q =2 tan2 q+1=2tanq tan2 q2tanq+1=0 (tan q 1)(tan q1)=0 tan q=1 q=45,225(c) sec2 q + tan2 q=4 1 + tan2 q + tan2 q=4 2tan2 q=3 tan2 q= 3

2

tan q=ABB32 q=5046, 12914, 23046,

30914

(d) 3 cosec q4= 1sin2 q 3sin q 4=

1sin2 q 3 sin q4sin2 q=1 4sin2 q 3 sin q1=0 (4sinq + 1)(sin q1)=0

sin q= 14 , sinq=1

q=19429,34531 q=90 Therefore,q=90,19429,34531

(e) cos22q+3sin2q=3 1 sin22q+3sin2q=3 sin22q3sin2q+2=0 (sin2q2)(sin2q1)=0 sin2q2=0 , sin2q1=0 sin2q=2 sin2q=1 qundefined 2q=90,450 q=45,225 Therefore,q=45,225

(f) tan(2q30)= 2cos(2q30)

sin(2q30)cos(2q30) =

2cos(2q30)

sin(2q30)cos(2q30)=2cos(2q30) sin(2q30)cos(2q30)2cos(2q30)=0 cos(2q30)[sin(2q30)2]=0 cos(2q30)=0 2q30=90,270,450,630 2q=120,300,480,660 q=60,150,240,330 sin(2q30)2=0 sin(2q30)=2 qundefined Therefore,q=60,150,240,330

19. (a) cos B(sin A cos A) + sin B(cos A + sin A) =cos B sin A cos B cos A + sin B cos A

+ sin B sin A = sin A cos B + sin B cos A cos B cos A

+ sin B sin A =sin(A + B) (cos A cos B sin A sin B) =sin(A + B) cos (A + B) (b) cos (A + B)sin B cos A =

cos A cos B sin A sin Bsin B cos A = cos A cos Bsin B cos A

sin A sin Bsin B cos A =

cos Bsin B sin Acos A

=cotB tan A(c) sec A sin (A + B) sec B =

1cos A (sin A cos B + cos A sin B)1cos B

= 1cos A cos B (sin A cos B + cos A sin B)

= sin A cos Bcos A cos B + cos A sin Bcos A cos B

=sin Acos A +

sin Bcos B =tanA + tan B



(d) 31 sin A sin Bcos A cos B 43cos A cos Bsin (A + B) 4

=3 cos A cos B sin A sin Bcos A cos B 43cos A cos Bsin (A + B) 4

= cos (A + B)sin (A + B) =cot(A + B)

20. sin (A + B)=sinA cos B + cos A sin BLet B=Asin2A=sinA cos A + cos A sin A =2sinA cos A

21. cos (A + B)=cosA cos B sin A sin BLet B=Acos2A=cosA cos A sin A sin A =cos2A sin2 A =cos2A (1 cos2 A) =cos2A 1 + cos2 A =2cos2 A 1

22. tan (A + B)= tan A + tan B1 tan A tan BLet B=Atan2A= tan A + tan A1 tan A tan A = 2tanA1 tan2 A

23. (a) cos2 A sin2 Asin A cos A =

2cos2A2sinA cos A

= 2cos2Asin2A

=2cot2A(b) cos2 A1=cos2 A (sin2 A + cos2 A) =cos2 A sin2 A cos2 A =sin2 A =cos2A cos2 A(c) (tan A tan B)(sin A sin B) =1 sin Acos A

sin Bcos B 2(sin A sin B) = sin A cos B sin B cos Acos A cos B (sin A sin B)

=sin(A B)1 sin A sin Bcos A cos B 2 =sin(A B) tan A tan B

(d) cosec2A tan A =

1sin2A

sin Acos A = 1

2sinA cos A sin Acos A

= 12sin2 A

2sinA cos A =

cos2Asin2A

=cot2A

24. (a) 6 sin x cos x=1 2sinx cos x= 13 sin2x= 13 2x=1928,16032,37928,52032 x=944,8016,18944,26016(b) 2sin2x + sin2 x=cos2 x 2sin2x=cos2 x sin2 x 2sin2x=cos2x sin2x

cos2x =12

tan2x= 12

2x=2634, 20634, 38634, 56634

x=1317, 10317, 19317, 28317

(c) sin (x+30)=sin(x30) sin xcos30+cosxsin30 =sinxcos30cosxsin30 cos xsin30+cosxsin30 =sinxcos30sinxcos30 2cosxsin30=0 cos x=0 x=90,270

25. (a) cos xcos60sinxsin60=sin30 cos (x+60)=sin30 = 1

2 x+60=60,300,420 x=0,240,360(b) 2sin2 x+3sin2x=1 3sin2x=12sin2 x =cos2x sin2x

cos2x =13

tan2x= 13 2x=1826, 19826, 37826,

55826 x=913, 9913, 18913,

27913



(c) 2sinxcos45=2cosxsin451 2sinxcos452cosxsin45=1 sin xcos45cosxsin45= 1

2

sin (x45)= 12

x45=210 x=255

26. (a) sin (x + y)=sinx cos y + cos x sin y = 25 +

15 = 35(b) sin (x y)=sinx cos y cos x sin y = 25

15 = 15

27. (a) cos (x + y)=cosx cos y sin x sin y = 1

4 1

2

= 14

(b) cos (x y)=cosx cos y + sin x sin y = 1

4 + 1

2

= 34

28. (a)

1

2

y

xA

B

40

5 3

3

tan (A B) = tan A tan B1 + tan A tan B

=

34

(AB3 )1 + 1 34 2(AB3 )

=

34

+ AB3 1 3AB3

4

=

3+4AB34

43AB34

= 3+4AB3

43AB3

(b) sin (A + B)=sinA cos B + cos A sin B =1 35 21

12 2 + 1

45 21AB32 2

= 310 + 4AB310

= 4AB3 310

(c) cos2A=12sin2 A =121 35 2

2

=121 925 2 =1 18

25

= 725

(d) sin2B=2sinB cos B =21 AB32 21

12 2

=AB32

29. Given tan A= 12

and sin B= 13

1

2

y

x

A

0

51

3

y

x

B

08

\ A and Barein2ndquadrant,where90, A ,180and90, B ,180.

(a) (i) tan2A= 2tanA1 tan2 A

=21 12 2

1 1 12 22

= 134

= 43

10

Additional Mathematics SPM Chapter 16


(ii) tan4A=tan2(2A) = 2tan2A1 tan22A

=21 43 2

1 1 43 22

= 83

1 169

=1 83 21 97 2

= 247

(iii) tan 3A=tan(A+2A) = tan A+tan2A1 tan Atan2A

=1 12 2 + 1

43 21 1 12 21

43 2

= 1

2 43

1 23 = 116

31 = 11

2

(iv) tan A =tan21 A2 2

12 =

2tanA2

1 tan2 A2

4tanA2 =tan2 A

2 1

tan2 A24tan

A21=0

tan A2 =

(4)ABBBBBBBBBBB(4)24(1)(1)2(1)

=4ABB20

2

=42AB5

2

=2AB5 =2AB5,2+AB5

(b) (i) sec2A= 1cos2A

=1

2cos2 A 1 =

121 2AB5 2

2

1

= 135 = 53

(ii) cos2A= 35 cos4A=cos2(2A) =2cos22A 1 =21 35 2

2

1

= 1825

1

= 725

(iii) cos 3A=cos(A+2A) =cosAcos2A sin Asin2A =cosAcos2A sin A(2sinA cos A) =1 2AB5 21

35 2 1AB5

2 1AB5

1 2AB5 2 = 6

5AB5 + 4

5AB5 = 2

5AB5

(iv) cos A=2cos2 A2 1

2AB5

=2cos2 A2

1

2cos2 A2=0.1056

cos2 A2=0.0528

Since45, A2

,90,

then cos A2=0.2298

(c) (i) sin2B=2sinB cos B =21 13 21

AB83 2 =21 13 21

2AB23 2 = 4

AB29

11



(ii) sin4B=2sin2Bcos2B =2sin2B(12sin2 B) =21 4AB29 23121

13 22

4 = 8AB29 1

79 2 = 56AB281 (iii) sin 3B=sin(B+2B) =sinBcos2B + cos Bsin2B =1 13 21

79 2 + 1 AB83 21

4AB29 2 = 7

27 + 16

27

= 2327

(iv) cos B=cos21 B2 2 AB83 =12sin

2 B2

2sin2 B2=1.9428

sin2 B2=0.9714

Since45, B2

,90,

then sin B2=0.9856

1.

1

0

y

x

k

1 k 2

(a) tan q= kABBBB1 k2

(b) sec q = 1cos q

= 1ABBBB1 k2

(c) cos(90q)=sinq =k

2. sin2 q +2sinq=3 sin2 q+2sinq3=0 (sin q + 3)(sin q1)=0 sin q =3 , sinq=1 q undefined q=90

3. 6 cosec2 x=13cotx 6(1 + cot2 x)=13cotx 6 + 6 cot2 x=13cotx 6 cot2 x 13 cot x+6=0(2cotx 3)(3 cot x2)=0cot x= 3

2 , cotx= 23

tan x= 23 tan x=32

x=3341,21341 x=5619,23619Therefore,x=3341,5619,21341,23619

4. (a) cot xsin2x = cos xsin x (2sinx cos x) = 2cos2 x = 1+cos2x

(b) cot xsin2x = 12

1+cos2x = 12

cos2x = 12

2x = 120,240,480,600 x = 60,120,240,300

5. 2+2cos2q=3cosq 2+2(2cos2 q1)=3cosq 2+4cos2q2=3cosq 4cos2 q 3 cos q=0 cos q(4cosq 3)=0 cos q=0 , 4cosq3=0 q=90,270 cosq= 3

4 q=4125,31835Therefore,q=4125,90,270,31835

6. (a)

2pipi

pi

y

y =

y = 2 sin x

x

2x

0

2

2

3pipi

22

(b) 2sinx + 2x =0

2sinx= 2x

A suitable graph to be added is y= 2x

Thereare2solutions.

12



7. tan x 2sinx=0 sin xcos x 2sinx=0 sin x2sinx cos x=0 sin x(12cosx)=0sin x=0 , 12cosx=0 x=0,180,360 cos x= 1

2 x=60,300Therefore,x=0,60,180,300,360.

8. (a)

18090 x

y

y = 1

y = sin 2x

x

1800

1

1

(b) sin x cos x= 12

x360

2sinx cos x=1 x180

sin2x=1 x180

The line to be drawn is y=1 x180

.

Thereare3solutions.

9. (a) (sin x + cos x)2=sin2 x+2sinx cos x + cos2 x =sin2 x + cos2 x+sin2x =1+sin2x(b)

3pipi

22

2pipi

pix

y

y = 2

y = cos 2x

x0

1

2

1

2sin2 x3= x 2sin2 x1= x +2 12sin2 x= x 2 cos2x= x 2 The straight line to be drawn is y= x 2. Thereare2solutions.

10. (a) cot q tan q= cos qsin q sin qcos q

= cos2 q sin2 qsin q cos q

=cos2qsin q cos q

= 2cos2q2sinq cos q

=2cos2qsin2q

=2cot2q(b) (i)

2pipi

pi3y y = x 1

y = 2 sin x

x0

23

2

21

(ii) sin 32

x= 3x2

12

2sin 32

x= 3x 1 The straight line is y= 3 x1. Thereisonlyonesolution.

1. 2cos(2q15)=1 cos(2q15)= 1

2 2q15=60,300,420,660 q=3730,15730,21730,33730

2.

10

y

x

k

1 + k 2

(a) sin q= kABBBB1 + k2

(b) sec(90q)= 1cos(90q)

= 1sin q =

ABBBB1 + k2k 3. 2sinq=tanq

= sin qcos q2sinq cos q sin q=0 sin q(2cosq1)=0 sin q=0 , 2cosq1 =0 q=0,180,360 cosq = 1

2 q =60,300Therefore,q=0,60,180,300,360

13



4. cot q=4tanq 1tan q =4tanq

tan2 q= 14

tan q= 12

q=2634,15326,20634,33326

5. 3 + sec2 x=tanx 3 + 1 + tan2 x=tanx tan2 x tan x2=0(tan x2)(tanx+1)=0tan x=2 , tanx=1 x=6326,24326 x=135,315Therefore,x=6326,135,24326,315

6. 2sin2 x + cos x=1 2(1cos2 x) + cos x=122cos2 x + cos x1=0 2cos2 x cos x1=0(2cosx + 1)(cos x1)=02cosx+1=0 , cosx=1 cos x= 1

2 x=0,360

x=120,240Therefore,x=0,120,240,360

7. 3 cot2 x + cosec x1=03(cosec2 x 1) + cosec x1=0 3 cosec2 x 3 + cosec x1=0 3 cosec2 x + cosec x4=0 (3 cosec x+4)(cosecx1)=0cosec x= 43 , cosec x=1 sin x=1

x=90

sin x= 34

x=22835,31125Therefore,x=90,22835,31125

8. 4sinq= 12cosq

2sinq cos q= 14

sin2q= 14

2q=1429,16531,37429,52531 q=715,8246,18715,26246

9. cos2q=sinq 12sin2 q=sinq 2sin2 q + sin q1=0(2sinq 1)(sin q+1)=0

sin q= 12

, sin q=1 q=270

q=30,150Therefore,q=30,150,270

10. sin4qcos2q

=1

sin4q=cos2q2sin2qcos2qcos2q=0 cos2q(2sin2q1)=0cos2q=0 2q=90,270,450,630 q=45,135,225,3152sin2q1=0 sin2q= 1

2 2q=30,150,390,510 q=15,75,195,255Therefore,q=15,45,75,135,195,225,255,315

11. sin2(2x+10)=2cos(2x+10)21 cos2(2x+10)=2cos(2x+10)2 cos2(2x+10)+2cos(2x+10)3=0[cos(2x+10)+3][cos(2x+10)1]=0cos(2x+10)+3=0 cos(2x+10) =3 x undefinedcos(2x+10)=1 2x+10=0,360,720 2x=10,350,710 x=5,175,355Therefore,x=175,355

12.

1

0

1

y

x

k

tA

B

1 k2

1 t 2

(a) tan (A B) = tan A tan B1 + tan A tan B

=

kABBBB1 k2

+ ABBBB1 t2

t1 + 1 kABBBB1 k2 21

ABBBB1 t2t 2

14



=

kt + ABBBBBBBBBB(1 k2)(1 t2)(ABBBB1 k2)(t)

tABBBB1 k2 kABBBB1 t2(ABBBB1 k2)(t)

=kt + ABBBBBBBBBB(1 k2)(1 t2)tABBBB1 k2 kABBBB1 t2

(b) cosec2B= 1sin2B

=1

2sinB cos B =

12(ABBBB1 t2 )(t)

= 12tABBBB1 t2

13. (sin q + cos q)(sin q cos q)=sin2 q cos2 q=(1cos2 q) cos2 q=12cos2 q

14. cos2q+2sin2 q=(12sin2 q)+2sin2 q =1

15.

2pi

2pi

yy = |tan x |

x0

16.

180 360

y

y = |cos x | + 2

x0

1

2

3

17.

180 360

y

y = sin 2x

x0

1

1

18. (a) sec qcos q + cosec qsin q

= 1cos2 q + 1sin2 q

= sin2 q + cos2 qcos2 q sin2 q

= 1(cos q sin q)(cos q sin q) =

4(2cosq sin q)(2cosq sin q)

=4sin22q

=4cosec22q

(b)

pi

pi

pi 2 2

yy = + 1

y = tan x

x

x

0

2

1

x cos x + sin x=cosx

x + sin xcos x =1

x + tan x=1 x +1=tanx The suitable straight line is y= x + 1 Thereisonlyonesolution.

19. (a) cot x + tan x = cos xsin x + sin xcos x

=cos2 x + sin2 xsin x cos x

=1sin x cos x

=2

2sinx cos x =

2sin2x

=2cosec2x(b)

pi 2pi

y

y = (x pi)2

y = |sin 2x |x

0

1

15



x2+2=sin2x+2x sin2x=x22x+2 =(x)2 The suitable graph is y=(x)2 There are 3 solutions

20. (a) 2cos2 x sin2 x=cos2 x + cos2 x sin2 x =cos2 x+cos2x =(1sin2 x)+cos2x =1+cos2x sin2 x(b) y=cos2x

x 0 45 90 135 180y 1 0 1 0 1

0

0.5

0.5

1.0

x60 120 180

y

y = |cos 2x|

x120

12y =

2cos2x= x60

1

2cos2x= x60

1

cos2x= x120

12

The suitable line is y= x120

12

The solution is x=120

21. (a) sin 3A =sin(A+2A)

=sinAcos2A + cos Asin2A =sinA(12sin2 A) + cos A(2sinA cos A) =sinA2sin3 A+2sinA cos2 A =sinA(12sin2 A+2cos2 A) =sinA[12(sin2 A cos2 A)] =sinA(1+2cos2A)(b)

1

0

y

xk

A1 k 2

(i) cos A=k 2cos2 A

21=k

cos2 A2=

1 + k2

Since0, A2

,45,

then cos A2=ABBBB1 + k2

(ii) cos2A=2cos2 A 1 =2k2 1 (iii) cos 3A =cos(A+2A) =cosAcos2A sin Asin2A =k(2k2 1) ABBBB1 k2 (2sinA cos A) =k(2k2 1) ABBBB1 k2 (2 ABBBB1 k2 k) =k(2k21)2k(1 k2) =2k3 k2k+2k3 =4k3 3k =k(4k2 3)

22. (a) 1cos2 A + 2sinAcos A

= 1cos2 A + 2sinA cos Acos2 A

=sec2 A + 2sinA cos Acos A cos A =sec2 A+2tanA =1+tan2 A+2tanA =tan2 A+2tanA + 1 =(1+tanA)2

(b)

1

y

xt2B

1 t 2

(i) cos2B =t 2cos2 B1=t cos2 B = 1 + t

2

Since270, 2B ,360 135, B ,180, then cos B =ABBBBB1 + t2

16



(ii) cos B=12sin2 B2

ABBBBB1 + t2 =12sin2 B2 2sin2 B

2=1+ABBBBB1 + t2

sin2 B2=

1 + ABBBBB1 + t22

Since 67 12, 1

2B ,90,

then sin B2=ABBBBB1 + ABBBBB1 + t2

2

23. (a) sin (x y)=sinx cos y cos x sin y =k p(b) sin2y=2siny cos y =

2siny cos y2sinx cos x 2sinx cos x

=4(sinx cos y)(cos x sin y)

sin2x =4kpcosec2x(c) sin x sin y =sinx sin y + cos x cos y cos x cos y =cosx cos y + sin x sin y cos x cos y =t (cos x cos y sin x sin y) =t cos (x + y)

24. (a) 12sinx cos x

sin xcos x =12sin2 x2sinx cos x

= cos2xsin2x

=cot2x(b) tan (x + y)=sec2 x tan2 x =1+tan2 x tan2 x =1 x + y=225........................1 tan (x y)=1 x y=135........................2 1 + 2,2x=360 x=180 Substitute x=180into1, y=225180 =45

25. (a) sin (x+30)+cos(x60)=0 sin xcos30+cosxsin30+

(cos xcos60+sinxsin60)=0 AB3

2 sin x + 1

2 cos x + 1 12 cos x +

AB32

sin x2=0 cos x + AB3 sin x=0 AB3 sin x=cosx sin xcos x =

1AB3

tan x= 1AB3

x=150,330(b)

1

1

y

x

k

tA

B1 + k 2

1 t 2

sin (A B) =sinA cos B cos A sin B =1 kABBBB1 + k2 2(t) 1

1ABBBB1 + k2 2(

ABBBB1 t2 )

=kt

ABBBB1 + k2 ABBBB1 t

2

ABBBB1 + k2

26. (a) 1 + 1sin2 x =3tan x

sin2 x+1=sin2 x 1 3 cos xsin x 2 sin2 x+1=3sinx cos x sin2 x + sin2 x + cos2 x=3sinx cos x 2sin2 x 3 sin x cos x + cos2 x=0 (2sinx cos x)(sin x cos x)=0 2sinx cos x=0 tan x= 1

2

x=2634,20634 sin x cos x=0 tan x1=0 tan x=1 x=45,225 Therefore,x=2634,45,20634,225(b) cos (A B)=sin(A C) cos A cos B + sin A sin B=sinA cos C cos A sin C sin A sin B sin A cos C= cosA sin C cos A cos B

17



sin A(sin B cos C)=cosA(sin C cos B) sin Acos A =

sin C cos Bsin B cos C tan A= (cos B + sin C) (cos C sin B) = cos B + sin C cos C sin B

27. (a) cos q=sin1q 4 2 cos q=sinq cos

4 cos q sin

4

cos q + cos q sin 4=sinq cos

4

sin qcos q =1 + sin

4cos

4

tan q=1 + sin

4cos

4 q=6730,24730

(b) 1+cos2B1cos2B =

1+(2cos2 B 1)1(12sin2 B)

= 2cos2 B

2sin2 B =1cos Bsin B 2

2

=cot2 B =cosec2 B 1

28. (a) (i) cos70=cos2(35) =2cos2351 =2g2 1 (ii) sin50=sin2(25) =2sin25cos25

=2fABBBBBBBB1 sin225 =2fABBBBB1 f 2

(iii) cos25=ABBBBBBBB1 sin225 =ABBBBB1 f 2 cos25=2cos212 1

21

ABBBBB1 f 2 =2cos212 121

cos212 12=

1 + ABBBBB1 f 22

cos12 12=ABBBBBBB1 + ABBBBB1 f 2

2

(b) cos45=12sin2(22.5)

1AB2

=12sin2(22.5)

sin2(22.5)= 12 11

1AB2

2

sin22.5=ABBBBBBBB12 11 1AB2 2 29. (a) 1+sec2A=1+ 1

cos2A = cos2A + 1

cos2A

=

cos2A + 1sin2Acos2Asin2A

= cos2A + 1sin2A

1cot2A

= 2cos2 A 1 + 1

2sinA cos A (tan2A)

=1 cos Asin A 2(tan2A) =cotAtan2A = tan2Atan A(b) tan (x + y)=3 tan x + tan y1 tan x tan y =3 tan x + tan y=33tanx tan y.........1 2tany=3tanx2 tan y= 3

2 tan x 1..............................2

Substitute 2 into 1, tan x + 3

2 tan x1=33tanx1 32 tan x 12

=3 92

tan2 x + 3 tan x

92

tan2 x 12

tan x4=0 9 tan2 x tan x8=0 (9 tan x + 8)(tan x1)=0 tan x= 89 , tanx=1 x=13822,31822 x=45,225

Therefore,x=45,13822,225,31822

18



30. (a) sin (x y)sin (x + y) =35

5 sin x cos y 5 cos x sin y =3sinx cos y + 3 cos x sin y 2sinx cos y =8cosx sin y sin xcos x =

8 sin y2cosy

tan x =4tany.............................1 tan y tan (x + y)=1 tan y1 tan x + tan y1 tan x tan y 2=1 tan y tan x + tan2 y=1tanx tan y 2tany tan x + tan2 y1=0...................2 Substitute 1 into 2, 2tany(4tany) + tan2 y1=0 9 tan2 y1=0 (3 tan y 1)(3 tan y+1)=0 tan y= 13 ,tany=

13 Since90, y ,180,thentany= 13 .

(b) sin2x= 12tanx

2sinx cos x= cos x2sinx

4sin2 x cos x cos x=0 cos x(4sin2 x1)=0 cos x=0 x=90,270 4sin2 x1=0 sin2 x= 1

4

sin x= 12

x=30,150,210,330 Therefore,x=30,90,150,210,270,330

31. (a) 32

cos x= cos2 xsin x

32

cos x sin x=cos2 x 3 cos x sin x=2cos2 x 2cos2 x 3 cos x sin x=0 cos x(2cosx 3 sin x)=0 cos x=0 x=90,270 2cosx 3 sin x=0 2cosx =3sinx sin xcos x =

23 tan x= 23 x=3341,21341 Therefore,x=3341,90,21341,270

(b) tan A 11 cot A =tan A 11 1tan A

= tan A 1tan A 1tan A =(tanA 1)1 tan Atan A 1 2 =tanA

32. (a) 4cosxcos2x=sinxsin2x 4cosxcos2x=sinx(2sinx cos x) 4cosx(12sin2 x)=2sin2 x cos x 4cosx 8 cos x sin2 x2sin2 x cos x=0 4cosx 10 cos x sin2 x=0 cos x(410sin2 x)=0 cos x=0 x=90,270 410sin2 x=0 sin2 x= 410 sin x=ABBB410 x=3914,14046,21914,32046 Therefore,x=3914, 90, 14046, 21914,

270,32046

(b) Given sin A cos B= 12

and cos A sin B= 13 , 0, A ,90,0, B ,90 sin (A + B)=sinA cos B + cos A sin B sin (A + B)= 1

2 + 13

sin (A + B)= 56 , 0, A + B ,180 A + B=5627,12333 sin (A B)=sinA cos B cos A sin B sin (A B)= 1

2 13

sin (A B)= 16 ,0, A B ,90 A B=936 A + B=5627......................1 A B =936........................2 1 + 2,2A=663 A=332 Substitute A=332 into 1, B=5627332 =2325

OR A + B=12333....................3 A B=936........................4

19



3 + 4,2A=1339 A=6635 Substitute A=6635 into 3, B=123336635 =5658

Therefore,A=332,B=2325 or A=6635,B=5658

33. (a) sin x cos x= 12

(sin x cos x)2= 14

sin2 x + cos2 x2sinx cos x= 14

1sin2x= 14

sin2x= 34

2x=4835,13125 x=2418,6543(b) tan A+cot2A =

sin Acos A + cos2Asin2A

= sin Asin2A + cos Acos2Acos Asin2A =

2sin2 A cos A + cos A(12sin2 A)cos Asin2A = 2sin

2 A cos A + cos A2sin2 A cos Acos Asin2A =

cos Acos Asin2A =

1sin2A

=cosec2A tan A+cot2A=cosec2A Let 2A=90 A=45 tan45+cot90=cosec90 1+cot90=1 cot90=0

34. (a) 4cos2 A1=sin2A 4cos2 Asin2A1=0 4cos2 A2sinA cos A (sin2 A + cos2 A)=0 3 cos2 A2sinA cos A sin2 A=0 (3 cos A + sin A)(cos A sin A)=0 3 cos A + sin A=0 tan A=3 A=10826,28826 cos A sin A=0 cos A=sinA tan A=1 A=45,225 Therefore,A=45,10826,225,28826

(b) sin (x+45)cos(x+45) = (sinx cos45+cosx sin45)(cosx cos45

sin xsin45) =1 1AB2 sin x +

1AB2

cos x21 1AB2 cos x 1AB2

sin x2 = 1

2sin x cos x 1

2sin2 x + 1

2cos2 x 1

2cos x sin x

= 12

cos2 x 12

sin2 x

= 12

(cos2 x sin2 x)

= 12

(1 sin2 x sin2 x)

= 12(12sin2 x)

= 12

sin2 x

1. 3sin2x=2tanx 3(2sinx cos x)=2 sin xcos x 3 sin x cos2 x=sinx 3 sin x cos2 x sin x=0 sin x(3 cos2 x1)=0sin x=0 x=0,180,3603 cos2 x1 =0 cos2 x = 13 cos x = 1

AB3 x =5444,12516,23444,30516Therefore,x=0, 5444, 12516, 180, 23444, 30516,360

2. 2cos x =tan2 x2 2secx=(sec2 x1)2 =sec2 x 3 sec2 x2secx3=0 (sec x + 1)(sec x3)=0sec x+1=0 , sec x3=0

sec x=3 cos x= 13

x=7032,28928

sec x=1 cos x=1 x=180

Therefore,x =7032,180,28928

3. q=sin2q =2sinq cos q =2 ABBBBBBB1 cos2 q cos q =2pABBBBB1 p2

20



4. Givensin2q= 35\ cos2q= 45

x

y

04

3 5 2

Alternative cos22q=1sin22q =11 35 2

2

= 1625

cos2q= 45 Since2qisanobtuseangle,therefore, 45 isignored.

(a) cos2q = 45 12sin2 q = 45 2sin2 q =1+ 45 = 95 sin2 q = 910 sin q = 3

ABB10

(b) sin 3q=sin(q+2q) =sinqcos2q + cos qsin2q =

3ABB10

1 45 2 + ABBBBBBB1 sin2 q 135 2

=12

5ABB10 + ABBBBBB1 910 1 35 2

=12

5ABB10 + ABBB110 1 35 2

=12

5ABB10 + 3

5ABB10 =

95ABB10

5. (a) Since cos A is positive and sin B is negative,therefore A and Bisin4thquadrant.

x

y

A0

2

3 5

x

y

B0

23

5

tan (A B)= tan A tan B1 + tan A tan B

=

AB52

1 2AB5 21 + 1 AB52 21

2AB5 2

=

AB52

+ 2AB51 + 1

=

5+42AB52

= 14AB5

(b) cos (A B) =cosA cos B + sin A sin B =1 23 21

AB53 2 + 1 AB53 21

23 2

=2AB59 +

2AB59

=4AB59

Since q is an acute angle,therefore, 3

ABB10

isignored.

21



6. tan2A=1 2tanA1 tan2 A =1 2tanA=1tan2 A tan2 A +2tanA1=0

tan A= 2ABBBBBBBBB44(1)(1)2(1)

=2AB8

2

=22AB2

2 =1AB2Since Aisanacuteangle,thereforetanA=1+AB2.

7. x 0 p

2p

3p2

2p

y 2 3 2 3 2

y

y = |sin x | + 2

0

1

2

3

pi pi 3pi 2pi 2 2

x

|sin x|=t2 |sin x|+2=tFor y=|sin x|+2tohavenosolutions,y=t cannot intersect with y=|sin x|+2.Therefore, t , 2,t . 3

8. 2cosec2A cot A=

2sin2A

cos Asin A=

22sinA cos A

cos Asin A=

1sin A cos A cos Asin A

=1 cos2 Asin A cos A

=sin2 Asin A cos A

=sin Acos A

=tanA

9. cos (A+30)+sin(A30)=cosAcos30sinAsin30+sinAcos30 cos Asin30=AB32

cos A 12

sin A + AB32

sin A 12

cos A

=AB32

cos A 12

cos A + AB32

sin A 12

sin A

=1 AB32 12 2 cos A + 1

AB32

12 2 sin A

=AB3 1

2(cos A + sin A)

10. cos2 y+2siny cos y =3sin2 y cos2 y+2siny cos y 3 sin2 y =0 (cos y sin y)(cos y + 3 sin y) =0 cos y sin y=0cos ycos y

sin ycos y =0 1 tan y=0 tan y=1 y=45,225 cos y + 3 sin y=0 cos ycos y +

3 sin ycos y =0 1 + 3 tan y=0 3 tan y=1 tan y= 13 y=1801826,3601826 =16134,34134

Therefore,y=45,16134,225,34134

11. tan (P + Q + R)=tan180 tan (P + Q) + tan R1 tan (P + Q) tan R =0

tan (P + Q) + tan R=0 tan P + tan Q1 tan P tan Q + tan R=0

1+21(1)(2)

+ tan R=0 3 + tan R=0 tan R=3

22



12. 2cosx sin x=R cos (x + a) =R cos x cos a R sin x sin a R cos a =2............................1 R sin a =1............................221

, R sin aR cos a = 1

2

tan a = 12

a =263412 + 22, (R cos a)2 + (R sin a)2=22 + 12 R2 cos2 a + R2 sin2 a=5 R2(cos2 a + sin2 a)=5 R2=5 R=AB5

Therefore,2cosx sin x =AB5 cos (x+2634) 2cosx sin x =1 AB5 cos (x+2634) =1 cos (x+2634) = 1

AB5 x+2634 =6326 x =63262634 =3652

Given R > 0

13.

2

15

f (x)=2sinx cos x =AB51 2AB5 sin x

1AB5

cos x2 =AB5(cos q sin x sin q cos x) =AB5[sin(x q)] =AB5 sin (x q)Since maximum value of sin (x q)is1,therefore the maximum value of f(x)=AB5

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