1Additional Mathematics SPM Chapter 16
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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
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(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
3Additional Mathematics SPM Chapter 16
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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
4 Additional Mathematics SPM Chapter 16
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(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
5Additional Mathematics SPM Chapter 16
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(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
6 Additional Mathematics SPM Chapter 16
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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
7Additional Mathematics SPM Chapter 16
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(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
8 Additional Mathematics SPM Chapter 16
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(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
9Additional Mathematics SPM Chapter 16
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(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
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(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
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(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
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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
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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
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=
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
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Additional Mathematics SPM Chapter 16
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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
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Additional Mathematics SPM Chapter 16
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(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
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Additional Mathematics SPM Chapter 16
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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
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Additional Mathematics SPM Chapter 16
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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
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Additional Mathematics SPM Chapter 16
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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
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Additional Mathematics SPM Chapter 16
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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.
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Additional Mathematics SPM Chapter 16
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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
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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