sol_4b13

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Chapter 13: Basic Trigonometry 147

Chapter 13 Basic Trigonometry

Skills Assessment (P.278)

1. (a) By Pythagoras theorem,

x2

+ 52

= 62

x = 6 52 2-

= 11

(b) sini=6

11

cosi= 65

tani=5

11

2. cos30ctan60c - sin2

45c

= ( )32

23

2

1-b bl l

=22

3 1-

= 1

3. (a) sin63c = .0 891, cor. to 3 sig. fig.

(b) cosx = 0.763

x = .40 3c, cor. to 3 sig. fig.

4. sin2

i+ cos2

i= 1

sin2

i+2

53a k = 1

sin2

i=2516

sini=45

tani=cos

sin

i

i

=

5354

=34

5. cos2

17c + cos2

73c

= cos2

17c + sin2(90c - 73c)

= cos2

17c + sin2

17c

= 1

6. sinitan(90c - i) =sin

tan i

i

=sin

cos

sin

i

i

i

= cosi

7. (a) Coordinates ofQ = ( ,- )3 4

(b) Coordinates ofT= ( , )2 5

8. (a) Coordinates ofB = ( , )2 3-

(b) Coordinates ofD = ( , )5 4-

(c) Coordinates ofN= ( , )1 2-

Exercise 13A (P.292)

1. (a) Since -234c is the corresponding negative

angle ofi,

-234c = i- 360c

i= 126c

(b) Since iis the corresponding negative angle

of 282c,

i= 282c - 360c

= 78c-

(c) Since -145c is the corresponding negative

angle ofi,

-145c = i- 360c

i= 215c

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148 Solutions

(d) When sini1 0, ilies in Quad. III or IV;

when cosi= 0.472 0, ilies in Quad. I or IV.

Hence, for sini1 0 and cosi= 0.47,

ilies in Quad. IV.

5. (a) sini=12

13-

=1213

-

cosi=135

tani=512-

=5

12-

(b) sini=3

2 2

cosi=31-

=31

-

tani=1

2 2

-= 2 2-

6. (a) sini=.1

0 6= .0 6

cosi=.1

0 8= .0 8

tani=..

0 60 8

= .0 75

(b) sini=1

54-

=5

4-

cosi=153

-=

53

-

tani=

5

54

3

-

-=

34

7. (a) OP= ( )8 152 2

- +

= 17

` sini= 1715

cosi=17

8-=

178

-

tani=8

15-

=815

-

2. (a) a 180c1 215c1 270c

` 215c lies in Quad. III.

(b) a 90c1 93c1 180c

` 93c lies in Quad. II.

(c) -128c = 232c - 360c

a 180c1 232c1 270c

` -128c lies in Quad. III.

(d) 380c = 20c + 360c

a 0c1 20c1 90c

` 380c lies in Quad. I.

(e) -250c = 110c - 360c

a 90c1 110c1 180c

` -250c lies in Quad. II.

(f) a 270c1 333c1 360c

` 333c lies in Quad. IV.

3. (a) a cosiis positive.

` ilies in Quad. I or IV.

(b) a taniis negative.

` ilies in Quad. II or IV.

(c) sini= 0.22 0

` ilies in Quad. I or II.

(d) cosi=14

- 1 0

` ilies in Quad. II or III.

4. (a) When sini1 0, ilies in Quad. III or IV;

when tani2 0, ilies in Quad. I or III.

Hence, for sini1 0 and tani2 0,

ilies in Quad. III.

(b) When cosi2 0, ilies in Quad. I or IV;

when sini= 0.52 0, ilies in Quad. I or II.

Hence, for cosi2 0 and sini= 0.5,

ilies in Quad. I.

(c) When tani1 0, ilies in Quad. II or IV;

when cosi1 0, ilies in Quad. II or III.

Hence, for tani1 0 and cosi1 0,

ilies in Quad. II.

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11. (a) a 202=x

2+ 16

2

` x2

= 144

x = 12 orx = -12(rejected)

` sini=2016

=54

cosi=2012

=53

tani=1216

=34

(b) a 42

=x2

+ ( )72

-

` x2

= 9

x = -3 orx = 3(rejected)

` sini=4

7-=

47

-

cosi= 43- = 43-

tani=3

7

-

-=

3

7

12. (a) a 1.72

= (-0.8)2

+y2

` y2

= 2.25

y = 1.5 ory = -1.5(rejected)

` sini=..

1 51 7

=1715

cosi= ..

1 7

0 8-

= 178

-

tani=.

.0 8

1 5-

=815

-

(b) a 82

= 72

+y2

` y2

= 15

y = 15- ory = 15 (rejected)

` sini=8

15-=

8

15-

cosi=87

tani=715-

=715

-

(b) OQ = ( )4 332 2+ -

= 7

` sini=733-

=733

-

cosi= 74

tani=33

4-

=433

-

8. (a) cos123c = .0 545- , cor. to 3 d.p.

(b) sin340c = .0 342- , cor. to 3 d.p.

(c) tan(-121c) = .1 664, cor. to 3 d.p.

(d) cos(-65c) = .0 423, cor. to 3 d.p.

9. (a) tan276.8c = .8 386- , cor. to 3 d.p.

(b) sin(-188.5c) = .0 148, cor. to 3 d.p.

(c) cos450.5c = .0 009- , cor. to 3 d.p.

(d) tan(-192.8c) = .0 227- , cor. to 3 d.p.

10.

sini cosi tani

Quadrantin which

ilies

(a) - + - IV

(b) - - + III

(c) + - - II

(d) - + - IV

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150 Solutions

(c)

i

y

xO

270i

From the graph, (270c - i) lies in Quad. III.

(d)

i

y

xO

180+ i

From the graph, (180c + i) lies in Quad. III.

15. Let P(x ,y) be a point on the terminal side ofi

and OP= r.

a cosi10 and 0c1i1 180c,

` ilies in Quad. II.

We can letx = -3 and r= 5,

then 5 = ( ) y32 2

- +

y = 4

` sini=54

tani=3

4-

=34

-

13. (a)y

xO

P

50

i

From the graph, i= 180c - 50c = 130c

(b) y

xO

P

50

i

From the graph, i= -(180c + 50c)

= 230c-

14. a ilies in Quad. I.

` 0c1i1 90c

(a)

i

y

xO

90+ i

From the graph, (90c + i) lies in Quad. II.

(b)

i

y

xO

360 i

From the graph, (360c - i) lies in Quad. IV.

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4.sin cos

1tan

1

i i

+

+

i

=sin cos

1cos

sin

i i

+

+

i

i

=sin cos

sin

sin cos

i i+

i

i i+

= sini

5. sin(180c - i) : cos(90c + i)

= sin[180c + (-i)] : (-sini)

= sini: (-sini)

= sin2i-

6.cos

cos(360 ) tan (90 )$c c

i

i i- +=

cos

costan

1$

i

ii

-c m

=tan

1i

-

7. (a) sin210c = sin(180c + 30c)

= -sin30c

=21

-

(b) cos315c = cos(270c + 45c)

= sin 45c

=

2

1

(c) tan330c = tan(360c - 30c)

= -tan30c

=3

1-

(d) cos150c = cos(180c - 30c)

= -cos30c

=2

3-


and OP= r.

When ilies in Quad. I,

we can letx = 5 andy = 12,

then r= 5 122 2

+ = 13

` sini=1312

cosi=135

When ilies in Quad. III,

we can letx = -5 andy = -12,

then r= ( ) ( )5 122 2

- + -

= 13

` sini=1312

-

cosi=135

-

Exercise 13B (P.304)

1. 1 - sinicositani= 1 - sinicosi:cos

sin

i

i

= 1 - sin2

i

= cos2i

2. sini+ cos(270c - i)

= sini+ cos[270c + (-i)]

= sini+ sin(-i)

= sini- sini

= 0

3.2

( sin )cos

sin cos

1

1

i i

i i

+

+ -=

2

( sin )cos

sin ( sin )

1

1 1

i i

i i

+

+ --

=2

( sin )cos

sin sin

1 i i

i i

+

+

=(

(

sin ) cos

sin sin )

1

1

i i

i i

+

+

=cos

sin

i

i

= tani

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152 Solutions

12. (a)(

(

sin 1)(sin 1)

cos 1)(cos 1)

i i

i i

- +

- += 2

2

sin 1

cos 1

i

i

-

-

= 2

2

(

(

cos ) 1

sin ) 1

1

1

i

i

-

-

-

-

= 2

2

cossin

ii

= tan2i

tan2

i=2

512

-a k

=25

144

(b) Let P(x ,y) be a point on the terminal side

ofiand OP= r.

When ilies in Quad. II,

we can letx = -5 andy = 12,

then r= ( )5 122 2

- + = 13

` sini=1213

cosi=135

-

(

(

sin 1) (sin 1)

cos 1) (cos 1)

i i

i i

- +

- +

=1 1

1 1

1312

1312

13

5

13

5

- +

- +- -

a aa ak kk k

=25144

When ilies in Quad. IV,

we can letx = 5 andy = -12,

then r= ( )5 122 2+ - = 13

` sini=1312

-

cosi= 135

(

(

sin 1) (sin 1)

cos 1) (cos 1)

i i

i i

- +

- +

=1 1

1 1

1312

1312

135

135

- +

- +

- -a aa a

k kk k

=25144

(c) The method in (a) is more convenient and

quicker.

8. 2

2

tan

tan

1i

i

+=

2

2

2

2

cos

sin

cos

sin

1i

i

i

i

+

=

2

2 2

2

2

cos

sin cos

cos

sin

i

i i

i

i

+

= sin2i

= (-0.6)2

= .0 36

9.2

cos sin

cos sin sin2

i i

i i i-=

2

2 2

2

cos

cos sin

cos

cos sin

cos

sin2-

i

i i

i

i i

i

i

=2

2

cos

sin

cos

sin

cos

sin2-

i

i

i

i

i

i

=2

tan

tan tan2

i

i i-

=

2

2

43

43

43

- b l

= 21

-

10. 1 - sin(270c - i)cos(-i) = 1 + cos2

i

= 1 +2

54

-a k

=4125

11.sin( )

cos( ) tan ( )

360

90 180$

c

c c

i

i i

-

- +=

sin

sin tan$

i

i i

-

= tani-

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15. (a)(

cos

)sin 180c

i

i+# tan(90c - i)

=cos

sin

i

i-#

tan

1

i

= cos

sin

i

i-

#

sin

cos

i

i

= 1-

(b) cos(180c + i) : cos(360c - i) + 1

= -cos i: cos i+ 1

= 1 - cos2i

= sin2i

16.cos( )

tan sin( )

1 180c i

i i

- -

- -=

cos

tan sin

1 i

i i++

=cos

sin

1

cossin

i

i

+

+ii

=cos1

cos

sin sin cos

i+

i

i i i+

=cos1

)(

cos

sin cos1

i+

i

i i+

=cos

sin

i

i

= tani

17.tan ( )

sin ( ) cos( )

180

270 90$

c

c c

i

i i- +

+

=(

tan

cos ) ( sin )$

i

i i- -

=cos sin

cos

sin

$i i

i

i

= cos2i

18. cos(90c - i) : sin(360c - i) +

cos(180c - i) : sin(90c - i)

= sini: (-sini) + (-cosi) : cosi

= -sin2

i- cos2

i

= 1-

13. (a) sin240c = sin(180c + 60c)

= -sin 60c

=23

-

sin240c

= sin(270c

- 30c)

= -cos30c

=23

-

(b) tan120c = tan(90c + 30c)

=tan30

1-

c

= 3-

tan120c = tan(180c - 60c)

= -tan 60c

= 3-

14. (a)cos

cos

1 i

i

+-

cos

cos

1 i

i

-

= 2cos

cos (1 cos ) cos (1 cos )

1 i

i i i i

-

- - +

= 2

2

sin

cos2

i

i-

= 2tan

2-

i

(b) 2

2

tan

tan

1

1

i

i

+

-+ sin

2i

=

2

2

2

2

1

1

cos

sin

cos

sin

+

-

i

i

i

i

+ sin2

i

=

2

2 2

2

2 2

cos

cos sin

cos

cos sin

i

i i

i

i i-

+

+ sin2i

= cos2

i- sin2

i+ sin2

i

= cos2i

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154 Solutions

21. tan100ctan190c + tan110ctan200c +

tan120c tan210c + g + tan170ctan260c

= tan(90c + 10c)tan(180c + 10c) +

tan(90c + 20c)tan(180c + 20c) +

tan(90c + 30c) tan(180c + 30c) +

tan(90c + 40c) tan(180c + 40c) +

tan(90c + 50c)tan(180c + 50c) +

tan(90c + 60c)tan(180c + 60c) +

tan(90c + 70c)tan(180c + 70c) +

tan(90c + 80c)tan(180c + 80c)

=tan10

1-

cb l tan10c +

tan20

1-

cb l tan 20c +

tan30

1-

cb l tan30c +

tan40

1-

cb l tan40c +

tan501

- cb l tan50c

+ tan601

- cb l tan60c

+

tan70

1-

cb l tan70c +

tan 80

1-

cb l tan80c

= 8-

Exercise 13C (P.317)

1. Reference angle a = 45c

a cosiis positive, ` ilies in Quad. I or Quad. IV.

` i= 45cori= 360c - 45c

i.e. i= 45cor315c

2. y

x

0

1

1

90 180 270 360

y = sin x

From the graph of the sine function,

when sini= -1, we see that i= 270c.

19. (a) tan300c + sin120c

= tan(360c - 60c) + sin(180c - 60c)

= -tan60c + sin60c

= - 3 +23

=23

-

(b) sin225c - cos135c

= sin(180c + 45c) - cos(180c - 45c)

= -sin 45c + cos 45c

= -2

1

2

1+

= 0

(c) tan150c#

sin300c

= tan(180c - 30c)# sin(360c - 60c)

= (-tan30c)# (-sin60c)

=3

1-b l #

2

3-b l

=21

(d) cos240c' tan225c

= cos(180c + 60c)' tan(180c + 45c)

= (-cos60c)' (tan45c)

=21

- ' 1

=21

-

20. sin2

10c + sin2

20c + sin2

30c + g + sin2

80c

= sin2

10c + sin2

20c + sin2

30c + sin240c +

sin2(90c - 40c) + sin

2(90c - 30c) +

sin2(90c - 20c) + sin

2(90c - 10c)

= sin2

10c + sin2

20c + sin2

30c + sin240c +

cos2

40c + cos2

30c + cos2

20c + cos2

10c

= 4

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9. 2cosi= 1

cosi=21

[Reference angle a = 60c

a cosiis positive,

` ilies in Quad. I or Quad. IV.]

` i= 60cori= 360c - 60c

i.e. i= 60cor300c

10. 3tani= -4

tani=34

-

[Reference angle a = 53.130c, cor. to the nearest

0.001c

a taniis negative,

` ilies in Quad. II or Quad. IV.]

` i= 180c - 53.130cori= 360c - 53.130c

i.e. i= .126 9cor .306 9c, cor. to the nearest 0.1c

11. -2tani= 3

tani=32

-

` i= 180c - 56.310c ori= 360c - 56.310c

= 123.690c(rejected) or303.690c

` i= .303 7c, cor. to the nearest 0.1c

12. 4cosi= -3

cosi=43

-

` i= 180c - 41.410cori= 180c + 41.410c

= 138.590cor221.410c(rejected)

` i= .138 6c, cor. to the nearest 0.1c

13. sini= sin34c

a sin34c = sin(180c - 34c)

i.e. sin34c = sin146c

` When 0cGiG 360c,

i= 34cor146c

3. Reference angle a = 21.801c, cor. to the nearest

0.001c

a taniis positive,

` ilies in Quad. I or Quad. III.

` i= 21.801cori= 180c + 21.801c


4. For any angle i,

-1G cos iG 1,

so cosicannot be equal to 1.2.

` ihas no solutions.


0.001c

a cosiis negative, ` ilies in Quad. II.

` i= 180c - 83.108c

i.e. i= .96 9c, cor. to the nearest 0.1c


0.001c

a taniis positive,

` ilies in Quad. III.

` i= 180c + 78.690c

i.e. i= .258 7c, cor. to the nearest 0.1c


0.001c

a siniis negative,

` ilies in Quad. III or Quad. IV.

` i= 180c + 17.458cori= 360c - 17.458c


8. 5sini= 2

sini=25

[Reference angle a = 23.578c, cor. to the nearest

0.001c

a siniis positive,

` ilies in Quad. I or Quad. II.]

` i= 23.578cori= 180c - 23.578c

i.e. i= .23 6cor .15 46 c, cor. to the nearest 0.1c

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156 Solutions

20. 3cosi+ 4sini= 0

3 +cos

sin4

i

i= 0

` 3 + 4tani= 0

tani= 43

-

` i= 180c - 36.870cori= 360c - 36.870c

i= .143 1cor .323 1c, cor. to the nearest 0.1c

21. tani(3sini+ 2) = 0

` tani= 0 or3sini+ 2 = 0

If tani= 0,

i= 0cor180c

If 3sini+ 2 = 0,

sini

= 3

2

-

` i= 180c + 41.810cor360c - 41.810c

= 221.8cor318.2c, cor. to the nearest 0.1c

` i= 0cor180cor .221 8cor .318 2c

22. 4cosisini+ cosi= 0

cosi(4sini+ 1) = 0

` cosi= 0 or4sini+ 1 = 0

If cosi= 0,

i= 90cor270c

If 4sini+ 1 = 0,

sini=41

-

` i= 180c + 14.478cor360c - 14.478c

= 194.5cor345.5c, cor. to the nearest 0.1c

` i= 90cor .194 5cor270cor .345 5c

23. sini- 3 tani= 0

sini- 3 :cos

sin

i

i= 0

sinicosi- 3 sini= 0

sin i(cosi- 3 ) = 0

` sini= 0 orcosi- 3 = 0

If sini= 0,

i= 0cor180cor360c(rejected)

If cosi- 3 = 0,

cosi= 3 (rejected)

` i= 0cor180c

14. tani= -tan28c

a -tan28c = tan(180c - 28c)

= tan(360c - 28c)

i.e. -tan 28c = tan 152c = tan 332c

` When 0cGiG 360c,

i= 152cor332c

15. cosi= sin67c

a sin67c = cos(90c - 67c)

= cos(270c + 67c)

i.e. sin67c = cos23c = cos337c

` When 0cGiG 360c,

i= 23cor337c

16. 5 cosi= -2

cosi=5

2-

i= 180c - 26.565cori= 180c + 26.565c

` i= .153 4cor .206 6c, cor. to the nearest 0.1c

17. 2 sini= 1

sini=2

1

i= 45cori= 180c - 45c

` i= 54 cor 513 c

18. 3 tani= -4

tani=3

4-

i= 180c - 66.587cori= 360c - 66.587c

` i= .113 4cor .293 4c, cor. to the nearest 0.1c

19. sinicosi= 0

cos

sin

i

i- 1 = 0

` tani= 1

` i= 45cori= 180c + 45c

i= 45cor225c

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(f) Since cosisini1 0,

2cos

cos sin

i

i i1 0

cos

sin

i

i1 0

tani1 0

` The terminal side ofilies in Quad. II

or IV.

2. (a) OP= ( )4 32 2

- +

= 5

` sini=53

cosi= 54- = 54-

tani=4

3-

=43

-

(b) OP= ( ) ( )3 72 2

- + -

= 4

` sini=4

7-=

47

-

cosi=43-

=43

-

tani=3

7

-

-=

3

7

(c) a 502

=x2

+ 482

` x2

= 196

x = 14 orx = -14(rejected)

` sini=5048

=2524

cosi=5014

=257

tani=1448

=724

24. 2 sini= tani

2 sini=cos

sin

i

i

2 sinicosi= sini

2

sini

cosi

- sini

= 0 sini( 2 cosi- 1) = 0

` sini= 0 or 2 cosi- 1 = 0

If sini= 0,

i= 0cor180cor360c(rejected)

If 2 cosi- 1 = 0,

cosi=2

1

i= 45cor315c

` i= 0cor45cor180cor315c

Supp. Exercise 13 (P.321)

1. (a) a 270c1 272c1 360c

` The terminal side ofilies in Quad. IV.

(b) a 90c1 175c1 180c

` The terminal side ofilies in Quad. II.

(c) -146c = 214c - 360c

a 180c1 214c1 270c

` The terminal side ofilies in Quad. III.

(d) Sincesin

tan i

i1 0,

sin

cos

sin

i

i

i1 0

cosi1 0

` The terminal side ofilies in Quad. II

or III.

(e) Sincetan

cosi

i2 0,

cos

sincos

i

i

i

2 0

2cos

sin

i

i2 0

a cos2

iH 0

i.e. sini2 0

` The terminal side ofilies in Quad. I

or II.

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158 Solutions

(c) tan210c = tan(180c + 30c)

= tan 30c

=3

1

(d) sin135c

= sin(180c

- 45c)

= sin45c

=1

2

6.sin

sin cos

2

3 4

i

i i+=

cos

sin

cos

sin

cos

cos

2

3 4+

i

i

i

i

i

i

=tan2

3 4tan

i

i+

=2

3 4

512

5

12+

-

-

bb ll

=32

7. 2sin ( )

cos( )

1 2 180

270

c

c

i

i

- -

-= 2

sin

sin

1 2 i

i

-

-

= 21 2

54

54

-

-

b

b

l

l

=720

8. (a) Reference angle a = 66.50c, cor. to the

nearest 0.01c

a taniis positive,

` ilies in Quad. I or Quad. III.

` i= 66.50cori= 180c + 66.50c

i.e. i= 67cor247c, cor. to the nearest

degree

(d) a 172

= 152

+y2

` y2

= 64

y = -8 ory = 8(rejected)

` sini=17

8-=

178

-

cosi=1715

tani=15

8-=

158

-

3. (a) 2

2

cos

cos

2

1

i

i-= 2

2

cos

sin

2 i

i

=2

tan2

i

(b) cositani sinsin

1i-

ib l

= cosi: sincos

sin

sin

1i-

i

i

ib l

= sini sinsin

1i-

ib l

= 1 - sin2

i

= cos2i

4. (a) sin(90c + i) - cos(180c - i)

= cosi+ cosi

= cos2 i

(b)tan ( )

tan( )180

270c

c

i

i

-

-=

tan

tan

1

i-

i

= tan2i-

5. (a) cos300c = cos(360c - 60c)

= cos60c

= 21

(b) tan(-225c) = -tan225c

= -tan(180c + 45c)

= -tan45c

= 1-

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(c) 4sini=37

sini=127

` i= 35.69cori= 180c - 35.69c

` i= 36cor144c, cor. to the nearest

degree


and OP= r.

We can lety = -12, r= 13,

then 13 = ( )x 122 2+ -

x = -5 or5(rejected)

` tani=5

12--

=125

cosi=13

5-=

135

-


and OP= r.

We can letx = 21,y = -20,

then r= ( )21 202 2+ -

= 29 or-29(rejected)

` sini=

29

20-=

29

20-

cosi=2921


and OP= r.

When ilies in Quad. I,

we can letx = 5 and r= 7,

then 7 = y52 2+

y = 24 or 24- (rejected)

` sini=247

, tani=24

5

When ilies in Quad. IV,

we can letx = 5 and r= 7,

then 7 = y52 2+

y = 24 (rejected) or 24-

` sini=724

- , tani=5

24-

(b) y

x0

90

180 270

360

1

1

y = cos x

From the graph of the cosine function,

when cosi= 0, we see that i= 90cor

270c.

(c) sini= -sin54c

a -sin54c = sin(180c + 54c)

= sin(360c - 54c)

i.e. -sin54c = sin234c = sin306c

` When 0c1i1 360c,

i= 234cor306c

(d) Reference angle a = 74.93c, cor. to the

nearest 0.01c

a cosiis negative,

` ilies in Quad. III.

` i= 180c + 74.93c

i.e. i= 255c, cor. to the nearest degree

(e) Reference angle a = 23.58c, cor. to the

nearest 0.01c

a siniis negative,

` ilies in Quad. III or Quad. IV.

` When 0c1i1 180c, ihas no

solutions.

9. (a) 2cosi= 2-

cosi= 22

-

` i= 180c - 45cori= 180c + 45c

` i= 135cor225c

(b) 2tani= 5

tani=25

` i= 68.20cori= 180c + 68.20c

` i= 68cor248c, cor. to the nearest

degree

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160 Solutions

(b) sin(-120c)' cos240c

= -sin120c' cos240c

=( )

( )

cos

sin

180 60

180 60-

c c

c c

+

-

= cossin

60

60- c

c

-

= tan60c

= 3

(c) cos210c# tan330c

= cos(180c + 30c)# tan(360c - 30c)

= -cos30c#cos30

sin30-

c

cb l = sin30c

=2

1

(d) tan240c + tan(-480c) - tan120c

= tan240c - tan480c - tan120c

= tan240c - tan(360c + 120c) - tan120c

= tan240c - tan120c - tan120c

= tan(180c + 60c) - 2tan(180c - 60c)

= tan60c + 2tan60c

= 3tan60c

= 3 3

16. (a) cos2

10c + cos2

20c + cos2

30c + g +

cos2

170c

= cos2

10c + cos2

20c + g + cos2

80c +

cos2

90c + cos2(90c + 10c) + g +

cos2(90c + 70c) + cos

2(90c + 80c)

= cos2

10c + cos2

20c + g + cos2

80c +

cos2

90c + sin2

10c + g + sin2

70c +

sin2

80c

= 8

(b) tan95ctan105ctan115cg tan175c

= tan95cg tan125ctan135ctan(270c -

125c) g tan(270c - 95c)

=( )

tan tan tan

tan tan tan

tan

tan

115 105 95

105 115 125

125

95 1

c c c c

c c c c -

= 1-

13. (a) 2sin

1

i- 2

tan

1

i

= 2sin

1

i- 2

2

sin

cos

i

i

= 2

2

sincos1

i

i-

= 2

2

sin

sin

i

i

= 1

(b)sin1

1

i-+

sin1

1

i+

= 2( (

sin

sin ) sin )

1

1 1

i

i i

-

+ + -

= 2cos

2

i

14. (a) 2

2

cos ( ) tan( )

tan( ) cos ( )

180 90

270 90$

$c c

c c

i i

i i

+ -

+ +

= 2

2

(

[ ]

cos ) tan ( )

tan ( ) sin

90

360 90 $

$ c

c c

i i

i i

-

-

-

-

= 2

2

cos tan( )

tan ( ) sin

90

90 $

$ c

c

i i

i i-

-

-

= 2

2

cos

sin

i

i-

= tan2i-

(b) cos(90c - i)sin(-i) -

cos(360c + i)sin(90c - i)

= sini(-sin i) - cosicosi

= -sin2

i- cos2

i

= 1-

(c) sin(180c + i)cos2(360c - i) +

sin3(360c - i)

= -sinicos2

i+ (-sini)3

= -sinicos2

i- sin3

i

= -sini(cos2

i+ sin2

i)

= sini-

15. (a) cos150c - sin300c

= cos(180c - 30c) - sin(270c + 30c)

= -cos30c + cos30c

= 0

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If 3cosi+ 1 = 0,

cosi=31

-

i= 180c - 70.529cor180c + 70.529c

= 109.5cor250.5c, cor. to 1 d.p.

` i= 0cor .109 5cor180cor .250 5c

23. 4sini- 3tani= 0

4sini-cos

sin3

i

i= 0

4sinicosi- 3sini= 0

sini(4cosi- 3) = 0

` sini= 0 or4cosi- 3 = 0

If sini= 0, i= 0cor180c

If 4cosi- 3 = 0,

cosi=43

i= 41.410cor360c - 41.410c

= 41.4cor318.6c, cor. to 1 d.p.

` i= 0cor .41 4cor180cor .318 6c

Lively Maths Problem

24. (a) Let the rise in height be y m when the

passenger capsule has moved from point P

to point Q.

OQ =2

135m = 67.5 m

sin 150c =.

y

67 5

y = 33.75

` The rise in height is 33.75 m when the

passenger capsule has moved from

point Pto point Q.

(b) (i) The vertical distance between the

passenger capsule and the lowest

point Sof the wheel

= (67.5 - 56) m

= 11.5 m

(ii) sini=.67 5

56-

` i= 180c + 56.06cor

i= 360

c- 56.06

c

` i= .236 06cor .303 94c, cor. to

2 d.p.

17. 5tani= 3-

tani=35

-

` i= 180c - 19.107cor360c - 19.107c


` i= .160 9c, cor. to 1 d.p.

18. 6 sini= -2

sini=6

2-

` i= 180c + 54.736cor360c - 54.736c


` i= .234 7c, cor. to 1 d.p.

19. 3 sinicosi= 0

cos

sin3

i

i- 1 = 0

3 tani- 1 = 0

tani=3

1

` i= 30cori= 180c + 30c

i= 30cor210c

20. 2 sini+ 9sin(90c - i) = 0

2 sini+ 9cosi= 0

cos

sin2

i

i+ 9 = 0

2 tani+ 9 = 0

tani=2

9-

` i= 180c - 81.070c

i= .98 9c, cor. to 1 d.p.

21. tanisini- sini= 0

sini(tani- 1) = 0

` sini= 0 ortani- 1 = 0

If sini= 0, i= 0cor180c

If tani- 1 = 0, tani= 1

i= 45cor180c + 45c

` i= 0cor45cor180cor225c

22. 3tanicosi= -tani

3tanicosi+ tani= 0

tani(3cosi+ 1) = 0

`

tani= 0 or3cosi+ 1 = 0 If tani= 0, i= 0cor180c

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162 Solutions

III. When 90c1i1 135c,

tani1 0

sini2 0

` tani1 sini

` III is correct.

` Only I and III are correct.

4. B 2sin(270c - i)cos 240c -

2sin120ccos(180c - i)

= 2sin[270c + (-i)]cos(180c + 60c) -

2sin(90c + 30c) cos[180c + (-i)]

= 2(-cosi)(-cos60c) -

2(cos30c)(-cosi)

= -2

cosi

2

1-

a k + 23

2b lcosi

= cosi+ 3 cosi

= 3(1 )cosi+

5. C 2cosi+ 3 = 0

cosi=23

-

` i= 180c - 30cori= 180c + 30c

` i= 150cor210c

6. C sinx cosx21

-a k = 0

` sinx = 0 orcosx -21

= 0

If sinx = 0,

x = 0cor180cor360c(rejected)

If cosx -21

= 0,

cosx =2

1

x = 60cor300c

` x = 0cor60cor180c or300c

` The equation has 4 distinct roots.

HKCEE Questions

7. A 8. B 9. B

10. A 11. D 12. A

Revision Test (P.327)

1. A Let P(x ,y) be a point on the terminal side

ofiand OP= r.

When ilies in Quad. II,

we can letx = -3,y = 4,

then r= ( ) 432 2+-

= 5

` sini=54

cos(90c + i) + sin(-i) = -sini- sini

= -2sini

= 254

- a k

= 58

-

2. D 1 + tan2(180c -x) = 1 + 2

2

cos (180 )

sin (180 )

x

x

c

c

-

-

= 1 + 2

2

( )

( )

cos

sin

x

x

-

= 1 + 2

2

cos

sin

x

x

= 2

2 2

cos

cos sin

x

x x+

= 2cos x

1

3. C I.y

i0 45 90 135 180

1

1

y = tan i

y = cos i

From the graph of the tangent and

cosine functions, when

90c1i1 135c,

tani1 cosi

` I is correct.

II. When 90c1i1 135c,

sini2 0

cosi1 0

` sini2 cosi

` II is incorrect.

sol_4b13

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