Matemática ciência e aplicações, Vol. 2, Cap 2

7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2

httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 115

Razotildees trigonomeacutetricas

na circunferecircncia 2

Exerciacutecios

1 y∙ 0983083 1 ndash (ndash1)

2 middot12

∙ 21

∙ 2

2 a) ndash1 c)3

2

b) 0 d) 12

e) ndash

22

g) 0

f ) ndash

32

h) ndash

12

3

3

5

3

4

3

2

3

sen π3

∙ sen 2π3

∙ 3

2

sen 4π3

∙ sen 5π3

∙ ndash

32

4 sen π3

∙ sen 2π3

∙ 32

sen

4π

3 ∙

sen

5π

3 ∙

ndash

3

2

sen 5π4

∙ sen 7π4

∙ ndash

22

5 a) sen 75deg lt sen 85deg

b) sen 100deg gt sen 170deg

170deg

sen 170deg

100deg

sen 100deg

c) sen 250deg gt sen 260deg

250deg260deg

sen 260deg

sen 250deg

sen 260deg

d) sen 300deg gt sen 290deg

6 a) sen 130deg∙ sen 50deg∙ 076604

b) sen 230deg∙ ndashsen 50deg∙ ndash 076604

c) sen 320deg∙ ndash sen 40deg ∙ ndash 064279

d) sen π5

∙ sen 36deg∙ 058779

e) sen 3π5

∙ sen 108deg∙ sen 72deg∙ 095106

7 a) 1o Qrarr seno eacute positivo

b) 3 lt 314 sen 3 eacute positivo pois 3 tem imagem no 2o Q

c) 5 gt 471 (ou 3π2

aproximadamente) Logo 5 tem

imagem no 4o Q e sen 5 eacute negativo

d) 2o Qrarr sen 100deg gt 0

e) 200ordm tem imagem no 3o Q sen 200deg lt 0

8 a) π7

tem imagem no 1o quadrante sen π7

983101 a gt 0

b) Observe que 8π

7

983101 π 983083 π

7

7

8

7

sen 8π7

983101 ndash sen π7

983101 ndasha

9 a)

=ndash

6

5

6

6

1

2

c)

3

2

1

S∙ π6

5π6

S∙ 3π2

b)

0O

d)

=ndash2

4

7

4=+

4

5

4

4

ndash 2

2

2

2

S∙ 0 π S∙ 5π

4 7π

4

10 a) y∙ 0 ndash (ndash1)12

middot 1983083 0 ∙

112

∙ 2

b) x∙ 2

2 middot 0983083 (ndash1) middot

3

2 ∙ ndash

3

2

MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2



11 5

6

7

6

11

6

6

ndash 3

2

3

2

cos π6

∙ cos 11π6

∙ 3

2

cos 5π6

∙ cos 7π6

∙ ndash

32

12 4

5

6

5

9

5

5

36deg

cos π5

∙ cos 9π5

gt 0

cos 4π5

∙ cos 6π5

lt 0

13 a)3

2 e) 0

b) 0 f ) ndash2

2

c) ndash

12

g) 12

d) ndash1 h) 1

14 a)65deg

85deg

cos 85deg

cos 65deg

cos 85ordm lt cos 65ordm

b) cos 89deg gt 0

cos 91deg lt 0 rArr cos 89deg gt cos 91deg

c)

50deg

20deg

cos 50deg

cos 340deg

cos 340ordm gt cos 50ordm

d)

10deg170deg

190deg cos 170ordm∙ cos 190ordm

15 k983101 0rarr cos 0983101 1

k983101 1rarr cos π

2

983101 0

k983101 2rarr cos π 983101 ndash1

k983101 3rarr cos 3π2

983101 0

16 a) 0 ndash 32

∙ ndash

32

860697 12

(F)

b)

sen π3

2

∙ 32

2

∙ 34

cos π3

2

∙ 12

2

∙ 14

34

983083 14

∙ 1 (V)

c) Como π2

cong 157 os nuacutemeros reais 1 e 2 tecircm ima-

gens respectivamente no 1o e 2o quadrantes

Daiacute cos 1 gt 0 e cos 2 lt 0 donde concluiacutemos que

cos 2 lt cos 1 (V )

d)100deg

sen 100deg

cos 100deg

Como |sen 100deg| gt |cos 100deg| segue que sen 100deg 983083

983083 cos 100deg gt 0 Observe que sen 100deg gt 0 e cos 100deg lt 0

(F)

e) 6 lt 2π (cong628) assim o nuacutemero real 6 tem imagem no

4o quadrante e cos 6 gt 0 (F)

f ) O raio da circunferecircncia trigonomeacutetrica eacute unitaacuterio (F)

17 OA983101 1

OB983101 cos π6

983101 32

AB983101 sen π6

983101 12

periacutemetro983101 1983083 32

983083 12

983101 3983083 32

UC

aacuterea983101 OB middot AB2

983101

32

middot 12

2 983101

38

UA

18 OB 983101 1 AB983101 12 983101 05BD983101 DE983101 sen α

OD983101 cos α

ABC 1057305 OBD rArr ABOB

983101 ACOD

983101 BCBD

051 983101 AC

cos α 983101 BCsen α rArr

AC∙ 12

cos α

BC∙ 12

sen α

A aacuterea do triacircngulo ABC eacute12

middot AC middot BC 983101 12

middot 12

cos α middot 12

sen α 983101 sen α middot cos α8

Ca piacutetulo 2 bull R az otildees trigonomeacutetricas





33

4

A

B

45deg

0

- AOB eacute isoacutesceles pois m(AOcircB)∙ m (ABO)∙ 45Logo OA∙ OB ∙ 1 (note que OA eacute o raio da circun-

ferecircncia trigonomeacutetrica)

-tg π

4 eacute a medida algeacutebrica de AB que eacute 1

Logo tg π4

∙ 1

b) F e) V

c) V

3

f ) F

tg 2π 983101 0

34 cos2 α ∙ 1 ndash sen2 α ∙ 1 ndash 13

2

∙ 1 ndash 19

∙ 89

α 2o Q

rArr cos α ∙ ndash 89 ∙ ndash 2 2

3

tg α ∙ sen αcos α ∙

1

3ndash 2 2

3

∙ ndash 12 2 middot 22 ∙ ndash 24

35 sen2 α ∙ 1 ndash 210

2

∙ 96100

α 4o Q

senα ∙ ndash 96100 ∙

∙ ndash

4 610

∙ ndash

2 65

tg α ∙ ndash

2 65

15

∙ ndash2 6

36 tg x∙ ndash3 rArr sen xcos x

∙ ndash3 rArr sen x∙ ndash3 cos x

sen2 x983083 cos2 x∙ 1 rArr (ndash3 cos x)2 983083 cos2 x∙ 1 rArr

cos2 x∙ 110

rArr cos x∙ 1010

e sen x∙ ndash 3 1010

OAOB

∙

1010

ndash

3 1010

∙ ndash

13

37 a) positivo e) negativo

b) negativo f ) positivo

c) negativo g) positivo

d) negativo h) positivo

38

11

6

2

53

4

3

011

10

-sec 2π

5 gt 0 cossec 2π

5 gt 0 e cotg 2π

5 gt 0

-sec 3π

4 lt 0 cossec 3π

4 gt 0 e cotg 3π

4 lt 0

-sec 3 lt 0 cossec 3 gt 0 e cotg 3 lt 0

-sec 11π

10 lt 0 cossec 11π

10 lt 0 e cotg 11π

10 gt 0

-sec

11π

6 gt 0 cossec

11π

6 lt 0 e cotg

11π

6 lt 0

39 a)1

cos π6

∙ 1

32

∙ 2 33

b) 1

tg 2π3

∙ 1

ndash tg π3

∙ 1ndash 3

∙ ndash

33

c)1

sen 5π6

∙ 1

sen π6

∙ 112

∙ 2

d) 1cos 210deg

∙ 1ndash cos 30deg

∙ 1

ndash 32

∙ ndash 2 33

e) 1sen 315deg

∙ 1ndash sen 45deg

∙ 1

ndash 22

∙ ndash 2

f ) 1tg 45deg

∙ 11

∙ 1

40 sec x983101 52

rArr cos x 983101 25

sen2 x 983101 1 ndash cos2 x rArrsen2 x∙

∙

1 ndash

4

25 rArr sen2

x983101

21

25 como xisin

4o

Q temos

sen x983101 ndash

215

tg x983101 sen xcos x

983101 ndash

215

25

983101 ndash

212

cotg x983101 ndash

2

21 ndash 2 21

21

cossec x983101 1sen x

983101 ndash 5

21 983101 ndash 5 21

21

sec x983101 1cos x

983101 52




41 sec2 x983101 1983083 tg2 xrArr 73

2

983101 1983083 tg2 xrArr tg2 x983101 409 rArr

x 4o Q tg x983101 ndash

2 103

cotg x983101 ndash 3

2 10 983101 15 middot ndash 1

10 logo m983101 ndash

1010

m

42 a) OQ983101 cossec α 983101 103

rArr sen α ∙ 310

cos2 α 983101 1 ndash sen2 α 983101 1 ndash 310

2

983101 1 ndash 9100

983101 91100

rArr

α 1o Q cos α 983101

9110

sen α 983083 cos α 983101 310 983083

9110

9831013983083 91

10

b) cotg2 α 983101 cos2 αsen2 α

983101

91100

9

100

983101 919

43 132 983101 52 983083 x2 rArr x983101 12 (o outro cateto mede 12)

tg α 983101 125 rArr cotg α 983101 5

12


sen β 983101 810

rArr cossec β 983101 108

983101 54

45 a) forall α isin [0 2π] temos que sec α ⩾ 1 ou sec α ⩽ ndash1 (F)

b) V

c) Como cossec2 α 983101 1983083 cotg2 α teriacuteamos 32 983101 1983083 32 rArr

rArr 9983101 10 o que eacute absurdo (F)

d) Como π2

lt 7π8 lt π temos que

cotg 7π8 lt 0

sec 7π8 lt 0

cotg 7π8 middot sec 7π

8 gt 0 (V )

51

B

H

h4 5

x 6 ndash x

6

A C

Temos

BH eacute a altura relativa do lado AC

42 983101 h2 983083 x2 (1)

52 983101 h2 983083 (6 ndash x)2 983101 h2 983083 36 ndash 12x 983083 x2 (2)

De (1) em (2) escrevemos

52 983101 42 983083 36 ndash 12x rArr 12x983101 27 rArr x983101 2712 983101 9

4

em (1) rArr 16983101 h2 983083 94

2

rArr h2 983101 16 ndash 8116 983101 175

16 rArr

rArr h983101 1754

cm

ABH cos α 983101 x

4 983101

94

4 983101 9

16 rArr sec α 983101 16

9

46 cos x983101 2

7 rArr sec x983101 7

2 983101 m

4 rArr 2 m983101 28rArr m983101 14

47 a) 2o quadrante c) 3o quadrante

b) 3o quadrante d) 4o quadrante

48 OB983101 sec π6

983101 1

cos π6

983101 1

32

983101 2 33

OA983101 cossec π6

983101 1

sen π6

983101 112

983101 2

AOB eacute retacircngulo em O sua aacuterea eacute

OA middot OB2

983101

2 33

middot 2

2 983101 2 33

(UA)

49 Q

O

P 30deg

P

a) Observe que P rsquo eacute a imagem do arco de 120deg

|PQ|983101 |cotg 120deg|983101 1tg 120deg

983101 minus 1tg 60deg

983101

983101 minus 1

3 983101 1

3 983101

33

A aacuterea do triacircngulo POQ eacute

12

middot OQ middot |PQ|983101 12

middot 1 middot 33

983101 36

(UA)

b) Aplicando Pitaacutegoras noPOQ vem

|OP|2 983101 |PQ|2 983083 |OQ|2 rArr |OP|2 983101 3

3

2

983083 12 983101

983101 13

983083 1983101 43

rArr OP983101 23

983101 2 33

(UC)

50 cossec x983101 3rArr sen x983101 13

cos2 x983101 1 ndash 13

2

983101 89

rArr

x 1o Q cos x983101 2 2

3 tg x983101

13

2 23

983101 1

2 2

983101 24




52 a2 983101 1cos2 x

ndash 1983101 1 ndash cos2 xcos2 x

983101 sen2 xcos2 x

983101 tg2 x

b2 983101 1sen2 x

ndash 1983101 1 ndash sen2 xsen2 x

983101 cos2 xsen2 x

983101 cotg2 x983101 1tg2 x

Daiacute a2 middot b2 983101 1 rArr (ab)2 983101 1 rArr a middot b983101 9832171

53 sec

2

x983101

4rArr

sec x983101

ndash2 (observe que xisin

3

o

Q e cos x lt 0)a) sec2 x 983101 1 983083 tg2 x rArr 4 983101 1 983083 tg2 x rArr tg2 x 983101 3 rArr

rArr tg x983101 3

b) Como sec x983101 ndash2 cos x 983101 ndash 12

e x983101 π 983083 π3

983101 4π3

54 Se tiveacutessemossen2 β ∙ 4

9

cos2 β ∙ 1625

rArr

rArr sen2 β 983083 cos2 β 983101 49

983083 1625

860697 1 rArr natildeo

55 a)

cotg2 x

1983083 cotg2 x 983101

cos2 xsen2 x

1983083 cos2 xsen2 x

983101

cos2 xsen2 x

sen2 x983083 cos2 xsen2 x

983101

∙

cos2 xsen2 x

1sen2 x

983101 cos2 x

b)

sen α middot tg α 983083 cos α 983101 sen2 αcos α

983083 cos α 983101

∙ sen2 α983083 cos2 α

cos α

983101 1

cos α

983101 sec α

c) tg x983083 cotg x983101 sen xcos x

983083 cos xsen x

983101

∙ sen2 x 983083 cos2 xsen x middot cos x

983101 1sen x middot cos x

983101 1sen x

middot 1cos x

983101

983101 cossec x middot sec x

d) tg x983083 cos x1983083 sen x

983101 sen xcos x

983083 cos x1983083 sen x

983101

sen x (1983083 sen x)983083 cos2 xcos x middot (1983083 sen x)

983101 sen x983083 sen2 x983083 cos2 x

cos x middot (1983083 sen x) 983101

983101 sen x983083 1cos x middot (1983083 sen x)

983101 1cos x

983101 sec x

56

57

58

59

x eacute um arco do 2ordm quadrante

tg x negativo

cotg x negativo

cotg x positivo2

π +

( )cotg x negativo+ π

sinal de y

( ) ( )

( ) ( ) ( )

minus sdot +

= = minusminus sdot minus


tg x positivo

cotg x positivo

cotg x negativo2

π +

( )cotg x positivo+ π

sinal de y ( ) ( )

( ) ( ) ( )

+ sdot minus= = minus

+ sdot +


sen x negativo

cos x negativo

sec x negativo

tg x positivo

( )sec x positivominus π

sinal de y ( ) ( ) ( )

( ) ( ) ( )

minus sdot minus sdot minus= = minus

+ sdot +

x 0 x2

πisin lt lt rArr x eacute do 1ordm quadrante

tg x 5=

( )sen x sen x+ π = minus

cos x sen x2

π minus =

( )tg x tg xπ + =

( )sen x sen xπ minus =

( )cos x cos xπ minus = minus

( )tg 2 x tg xπ minus = minus

senx senxy

minus sdot=

tg xsdot

senx sdot ( ) ( )cos x tg xminus sdot minus

senxtg x 5

cosx= minus = minus = minus

0 x2

πlt lt rArr x eacute do 1ordm quadrante

sen xcosx2

tg x2 sen x

cos x2

π minus π minus = = π minus




2 2sen x cos x 1+ =

221

cos x 110

+ =

2 1 99cos x 1

100 100= minus = rArr

60 a)

b)

61 sen

cotg1 cos

αα + =

+ α

cos sen

sen 1 cos

α α+ =

α + α

( )

( )

2cos 1 cos sen

sen 1 cos

α + α + α= =

α + α

2 2cos cos sen

sen sen cos

α + α + α= =

α + α sdot α

( )

1 cos 1cossec

sen 1 cos sen

+ α= = = α

α + α α

62 ( ) ( )cos x 2 tg x 2tg x 1sdot + sdot + =

( )( )2 cos x cos x tg x 2 tg x 1= + sdot + =

24 c os x tg x 2 cos x 2 cos x tg x cos x tg x= sdot + + sdot + sdot =

2

2

sen x sen x sen x4 cos x 2cos x 2 cos x cos x

cos x cos xcos x= sdot + + sdot + sdot =

63

64

65 2 2 2 2 2 2sen x sen y sen x sen y cos x cos y+ minus sdot + sdot =

( )2 2 2 2 2sen x sen y 1 sen x cos x cos y= + sdot minus + sdot =

2 2 2 2 2sen x sen y cos x cos x cos y= + sdot + sdot =

( )2 2 2 2sen x cos x sen y cos y= + sdot + =

2 2sen x cos x 1= + =

66

1cossec x ndash 1

+ 1cossec x+ 1

=67

3 11cosx

10rArr = rArr

3 11

10tg x 3 1112

10

π minus = =

sen tg cosαsdot α + α = sen sen

coscos

αsdot α+ α =

α

2 2sen cos

cos

α sdot α= =

α

1sec

cos= α

α

2tg

ecs1

1

α minus =α minus

2tg sec 1

sec 1

α minus α + =α minus

2

2

sen 11

coscos

11

cos

αminus +

αα= =

minus

α

2

2

sen cos cos

cos

1 cos

cos

α minus α + α

α=

minus α

α

2

1 cos

1cossec

1 cos cos

minus α

α= = = α

minus α α

2sen x4 sen x 2 cos x sen x

2cos x= + + + =

2 22 cos x 2 sen x

5senxcosx

+= + =

( )2 22 sen x cos x

5sen x cosx

+

= + = 5senx 2sec x+

sen x cos x

sen x cos x

+=

minus

senx cos x

senx

senx cos x

senx

+

=minus

1 cotg x

1 cotg x

+

minus

( ) ( )sen x tgx cosx cotgx+ sdot + =

senx cos x senx cotgx tgx cos x tgx cotgx= sdot + sdot + sdot + sdot

senx cos x senx= sdot + cosx

senxsdot

senx

cosx+ cosxsdot 1+ =

sen x cos x cos x sen x 1= sdot + + + =

( )sen x 1 c os x cos x 1= + + + =

( ) ( )1 cos x 1 sen x= + sdot +

=

sena senb cosa cosb

cosa cosb sena senb

+ ++ =

+ minus

( )( )

( )( )

( )( )

( )( )

sena senb sena senb

cos a cos b sena senb

cos a cos b cos a cosb


+ minus +=

+ minus

+ minus +=

+ minus

( )( )

2 2 2 2sen a sen b cos a cos b1 1 0

cos a cosb sena senb

minus + minus= = minus =

+ minus




=cossec x cossec x(cossec x ndash 1) middot (cossec x+ 1)

= 2 cossec xcossec2 x ndash 1

=

=2 cossec x

cotg2 x =

2sen x

middot sen2 xcos2 x

=

= 2 middot 1cos x

middot sen xcos x

= 2 middot sec x middot tg x

tg x+ 1tg x

middot 1cos x

ndash cos x middot 1sen x

ndash sen x =

= tg2 x+ 1

tg x middot 1 ndash cos2 x

cos x middot 1 ndash sen2 x

sen x =

= sec2 xtg x

middot sen 2 xcos x

middot cos 2 xsen x

= sec2 xtg x

middot sen x middot cos x =

= 1cos2 x

middot cos xsen x

middot sen x middot cos x = 1

68

+ 1 + ndash 1

69

70 Como tg x middot cotg x = 1 escrevemos

(m ndash 2) middot3

= 1 rArr m2 ndash 2m ndash 3 = 0 rArr m = 3 ou

m = ndash1

m

Eacute preciso resolver o sistema

3 cos x + sen x = ndash1 1

cos2 x + sen2 x = 1 2

De 1 vem sen x = ndash1 ndash 3 cos x Em 2 cos2 x + (ndash1 ndash

ndash 3 cos x)2 = 1rArr

rArr 10 cos2 x + 6 cos x = 0rArr cos x = 0 ou cos x = ndash35

bull Se cos x= 0 em 1 obtemos sen x= ndash1 02 + (ndash1)2 = 1

71

Daiacute os resultados procurados satildeo

y = 0 ndash (ndash1) = 1 ou y = ndash 35

ndash 45

= ndash 75

72 a)

b) 12 5 23

y cotg sen cos7 11 12

π π π = sdot +

12 14 2 22

7 7 7 7

π π π π= minus = π minus rArr 3ordm quadrante

12cotg

7

πrArr negativo

5 5 180ordm82ordm

11 11

π sdot= cong rArr 1ordm quadrante

5sen

11

πrArr

positivo

23 242

12 12 12 12


23cos

12

πrArr positivo

sinal de y

( ) ( ) ( )( ) ( )times + + + == minus minus

Exerciacutecios complementares

1 Se sen x= 1 x= π

2 cos x= cos π

2 = 0 assim

sen x+ cos x= 1+ 0= 1

Se cos x = 1 x = 0 sen x = sen 0 = 0 e deste modo

sen x+ cos x= 0+ 1= 1

2 a) Pela relaccedilatildeo fundamental cos2 x= 1 ndash sen2 x daiacute

2(1 ndash sen2 x)= 3 ndash 3 sen x

2 sen2 x ndash 3 sen x+ 1= 0

sen x= ndash (ndash3)plusmn 9 ndash 4 middot 2 middot 1

2 middot 2 = 3plusmn 1

4

sen x= 1 ou sen x= 1

2

2 2tgx cosxy1 cos x

sdot= rArr

minus 2tgx cos xy

sen x

sdot= rArr

senx

cosxyrArr =

cosxsdot

2sen xrArr

1y cossec x

senx= =

Como2 2cossec x 1 cotg x= + vem

ox 3 Qisin2

2 24 576 625cossec x 1 1

7 49 49

= + = + = rArr

25cossec x7

rArr = minus e desse modo

25

y cossec x 7

= = minus

9 7x sec tg cotg

8 6 7

π π π = sdot +

9 8

8 8 8 8

π π π π= + = π + rArr 3ordm quadrante

9sec

8

πrArr

negativo

7 6

6 6 6 6


7tg

6

πrArr

positivo

7

π eacute do 1ordm quadrante cotg positivo

6

πrArr

sinal de x ( ) ( ) ( )( ) ( )+ + + == minus sdot minus

bull Se cos x= ndash5

em 1 obtemos sen x= 5

ndash5

+

+ 4

5

2

= 1

3 4 3 2




b) Se sen x= 1 entatildeo x= π2

Se sen x= 12

podemos ter x= π6

ou

x= π ndash π6

= 5π6

3 Da relaccedilatildeo fundamental sen2 x+ cos2 x= 1 vem

m+ 13

2 + m ndash 1

32 = 1 rArr 2m2 + 2

9 = 1 rArr

rArr 2m2 = 79

rArr m2 = 718

rArr m= plusmn 7

3 2 middot

2

2 =

= plusmn 146

(cong 0623)

Observe que esses valores de m garantem que

ndash 1 sen x 1 e ndash1 cos x 1

4 a) Sabemos que

OS= sec α e OC= cossec α

Usando Pitaacutegoras no

COS vemCS2 = OS2 + OC2 rArr CS2 = sec2 α + cossec2 α =

= 1cos2 α

+ 1sen2 α

=

= sen2 α + cos2 αcos2 x middot sen2 x

= 1cos2 x middot sen2 x

= 1cos2 α

middot 1sen2 α

Daiacute

CS= 1

cos2 α middot 1

sen2 α como sen α gt 0 e

cos α gt 0 vem

CS= 1cos α

middot 1sen α

= sec α middot cossec α

b) α =

π

6 OS=

sec

π

6 =

sec 30deg=

1

cos 30deg =

= 23

middot 33

= 2 33

POS eacute retacircngulo em P

OS2 = OP2 + PS2 rArr 2 33

2

= 12 + PS2 rArr 43

ndash 1=

= PS2 rArr PS2 = 13

rArr PS= 33

O periacutemetro doPOS eacute2 3

3 + 1+ 3

3 = ( 3 + 1) UC

5 a) sec x= cos x rArr 1cos x

= cos x rArr cos2 x= 1 rArr

rArr cos x= 1(x= 0) ou cos x= ndash1(x= π)

b) sec x= tg x rArr 1cos x

= sen xcos x

cos x∙ 0 rArr

rArr sen x= 1 rArr x= π2

Mas observe que se x= π2

cos x= cos π2

= 0 Logo

natildeo existe x que satisfaccedila a equaccedilatildeo

c) tg x= cotg x rArr tg x= 1

tg x

rArr tg2 x= 1 rArr

rArr tg x= 1 x= π4

ou x= 5π4

ou

tg x= ndash1 x= 3π4

ou x= 7π4

6 sec x middot cossec x ndash sec2 x

cotg x ndash 1 =

1cos x

middot 1sen x

ndash 1cos2 x

cos xsen x

ndash 1 =

=

cos x ndash sen xcos2 x middot sen2 x

cos x ndash sen xsen x

= cos x ndash sen xcos2 x middot sen x

middotsen x

cos x ndash sen x =

= 1cos2 x

= 1

14

2 = 16

7 a)

α

0 A

T

OA= 1 (raio do ciclo)

AT= tg α

OT 2 = OA2 + AT 2 rArr OT 2 = 12 + (tg α)2 = sec2 α rArr

rArr OT= sec α

b) anaacuteloga

8 a)

α β

x 800 ndash x

800

h

cotg α = 5 rArr tg α = 5

rArr x

= 5

rArr x= 5 h (1)

cotg β = 15 rArr tg β = 1

15 rArr h

800 ndash x = 1

15 rArr

rArr 15 h= 800 ndash x (2)

Substituindo (1) em (2) vem (1)

15 h = 800 ndash 5 h rArr 20 h = 800 rArr h = 40 m rArr

rArr(1)

x= 200 m

b) Os observadores estatildeo a 200 m e a 600 m do edifiacutecio

1 h 1

9 Note inicialmente que se a equaccedilatildeo eacute de 2o grau entatildeo

sen α ∙ 0

a= sen α b= ndash2 cos α e c= ndashsen α




∆ = (ndash2 cos α)2 ndash 4 middot sen α middot (ndashsen α)=

= 4 cos2 α + 4 sen2 α = 4

Daiacute x= ndash (ndash2 cos α) plusmn 2

2 middot sen α =

cos α plusmn 1sen α

10 a) sen2 3π20

+ cos2 3π20

= 1 rArr a2 + cos2 3π20

= 1 rArr

rArr cos2 3π20 = 1 ndash a2 rArr cos 3π

20 = 1 ndash a2 note que

cos 3π20

= cos 27deggt 0

b) Notando que 3π20

+ 17π20

= π temos que as imagens

de 3π20

e 17π20

satildeo simeacutetricas em relaccedilatildeo ao eixo dos

senos

3π

20

17π

20

Logo sen 17π20

= sen 3π20

= a

c) Como 3π20

+ 7π20

= 10π20

= π2

concluiacutemos que 3π20

e 7π20

satildeo complementares e portanto

cos 7π20

= sen 3π20

= a

d) Como 2π ndash 3π

20

= 37π

20

temos que as imagens de 3π

20e 37π20

satildeo pontos simeacutetricos em relaccedilatildeo ao eixo dos

cossenos

3π

20

37π

20

Daiacute sen 37π20

= ndash sen 3π20

= ndasha

11 (a b c) eacute PA rArr b= 2

cos x= a+ c2 rArr cos x= 2b

5 (1)

rArr a =(3)

2b

sec x= 5a

rArr 1cos x

= 5a

rArr cos x= a5

(2)

sen2 x+ cos2 x= 1 rArr b3

2

+ 2b5

2

= 1 rArr

rArr 61 b2 = 25 middot 9 rArr b2 = 25 middot 961

rArrbgt 0

rArr b=

5 middot 3

61 =

15 61

61

a+ c

Em (3) a= 30 61

61

A razatildeo da PA eacute

b ndash a= 15 61

61 ndash 30 61

61 = ndash 15 61

61Assim

c ndash b= ndash 15 61

61

rArr c ndash 15 61

61

= ndash 15 61

61

rArr c= 0

12 a) sen α = 15

ndash cos α sen2 α + cos2 α = 1 rArr

rArr 15

ndash cos α2

+ cos2 α = 1 rArr

rArr 2 cos2 α ndash 25

cos α ndash 2425

= 0 rArr

rArr 50 cos2 α ndash 10 cos α ndash 24= 0 rArr

rArr cos α =10 plusmn 70

100

45

ndash35

Se cos α = 45

entatildeo sen α = 15

ndash 45

= ndash 35

e α tem

imagem no 4o quadrante

Se cos α = ndash 35


+ 35

= 45

e α tem


13 a) F sen 300deg= ndash sen 60deg= ndash 32

lt 0

b) V sen 160deg= sen(180deg ndash 160deg)= sen 20deg= cos 70deg

complementares

portanto sen2 70deg+ sen2 160deg= sen2 70deg+ cos2 70= 1c) V devemos ter

ndash1 cos x 1 rArr ndash1 5 ndash m3

1 rArr

rArr ndash3 5ndash m 3 rArr ndash8 ndashm ndash2 rArr 8 m 2

14 1o membro=

= (cossec x ndash cotg x)2 = 1sen x

ndash cos xsen x

2

= (1 ndash cos x)2

sen2 x =

= (1 ndash cos x)2

1 ndash cos2 x =

(1 ndash cos x)2

(1 ndash cos x) middot (1+ cos x) =

1 ndash cos x

1+ cos x =

= 2o membro

15 tg x ndash sen x

sen3 x =

sen x 1cos x ndash 1

sen3 x =

1cos x ndash 1

sen2 x =

=

1cos x ndash 1

1 ndash cos2 x =

1 ndash cos x

cos x =

1(1 ndash cos x) middot (1+ cos x)

=

=1

cos x middot

1

1+ cos x =

sec x

1+ cos x

| Ca piacutetulo 2 bull R az otildees trigonomeacutetricas





Comoπ5

+ 3π10

= 2π + 3π10

= π2

temos

tgπ5

= 1

tg 3π10

Daiacute em () temos log 1 = 0

26 AB= AC e BC = 10 cm a) Acirc= 2(B + C )= 2 middot 2B = 4B

Acirc+ B + C = 180deg rArr A+ 2B = 180deg

rArr

rArr B = C = 30deg e Acirc = 120deg com Acirc2

= 60deg

Daiacute cos B =

102

AB rArr

32

= 5AB

rArr

rArr AB= AC= 10

3 =

10 33

Assim

2p= 10+ 2 middot10 3

3 = 10+

20 33

cm

b) sen x+ cos x= k

sen2 x+ cos2 x+ 2 sen x cos x= k 2

1+ 2 sen x cos x= k 2

sen x cos x= k 2 ndash 1

2

sen3 x+ cos3 x=

= (sen x+ cos x) middot (sen2 x ndash sen x cos x+ cos2 x)=

= k middot 1 ndash k 2 ndash 1

2 = k middot

2 ndash k 2 + 12

=

= k

2

(3 ndash k 2)

27

1

y

O D

α

1 x

BAE

C

a) AE= cotg α= 1

tg α = 1tg π3

= 1

3 = 33

AB= EB ndash AE= 1 ndash 33

DC= tg α = tgπ3

= 3

BD= 1rArr BC= DC ndash BD= 3 ndash 1

aacuterea= AB middot BC2

= 12

middot 1 ndash 33

middot ( 3 ndash 1)=

= 12

middot 3 ndash 1 ndash 1+ 33

aacuterea=

1

2 middot

4 3

3 ndash

2=

2 3

3 ndash

1

b) bull AB= EB ndash AE= 1 ndash cotg α

bull BC= DC ndash BD= tg α ndash 1

bull Aacuterea= (1 ndash cotg α) middot (tg α ndash 1)2

=

= 1

= tg α ndash 1 ndash cotg α middot tg α + cotg α2

=

= tg α + cotg α ndash 22

()= 12 middot sen αcos α + cos αsen α ndash 2 =

= 12 middot

sen2 α+ cos2 α ndash 2 sen α cos αsen α middot cos α

=

= 12 middot

1 ndash 2 sen α cos αsen α cos α

= 12 middot

1sen α cos α ndash 2 =

= 1

2 sen α cos α ndash 1 que eacute uma expressatildeo equiva-

lente a ()

Desafio

O maior valor possiacutevel para a soma eacute 36 quando todos os

algarismos satildeo iguais a 9 e o nuacutemero formado eacute 9999

A soma 35 eacute obtida quando trecircs algarismos satildeo iguais a 9 e o outro

eacute igual a 8 Com essa configuraccedilatildeo haacute quatro possibilidades

9998

9989

9899

8999

Assim a resposta eacute 5

Testes

1 Os acircngulos satildeo 45ordm e 135ordm

45ordm 135ordm 180ordm+ =

Resposta c

3

2 Resposta d

2sen x com x

5 2

π= lt lt π rArr 2o quadrante

2 2sen x cos x 1+ =

24cos x 1

25+ = rArr

2 4 21cos x 1

25 25= minus = rArr

21cos x

5rArr = minus

2sen x 2 215tg xcos x 2121

5

= = = minus

minus

Resposta d




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus

II Verdadeira pois1 1

sec 60ordm 21cos60ordm

2

= = = e

( )sen 90ordm cos 180ordm 1 1 2minus = minus minus =

III Verdadeira pois 316ordm e 314ordm satildeo acircngulos

complementares do 4o quadrante

( )316ordm 314ordm 630ordm 7 90ordm + = = sdot

IV Falsa pois1 1

cossec 30ordm 21sen30ordm2

= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c

6 Como cossec2 x= 1+ cotg2 x podemos escrever

3 middot (1+ cotg2 x) ndash 4 cotg x= 3

3 cotg2 x ndash 4 cotg x= 0rArr cotg x middot (3 cotg x ndash 4)= 0

Daiacute

cotg x= 0rArr xnotin 0π2

cotg x= 43

rArr cos xsen x

= 43

rArr sen x= 3 cos x4

9 cos2 x16

+ cos2 x= 1rArr cos x= 45

rArr sen x= 5

Resposta a

3

7 sen xcos x

= a rArr sen x= a middot cos x

(a cos x)2 + cos2 x= 1rArr cos2 x middot (1 + a2)= 1rArr

rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2

sen x+ cos x= ndash a

1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8

Acircngulo do 2o quadrante sen 0 e cos 0rArr gt gt

2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13

= minus = minus = rArr = minus

Resposta c

9 ( )22 2A cos x sen y sen x cos y= + + minus

x y sen x cos y2π

= minus rArr = e seny cos x=

2 2 2A cos x cos x 2 cos x= + =

10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor

valor assumido pela expressatildeo

Resposta a

11 2 sen θ = 3 tg2 θ rArr

rArr 2 sen θ = 3 sen2 θ

cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ

1 ndash sen2 θ rArr 3 sen θ = 2 ndash 2 sen2 θ rArr

rArr 2 sen2 θ + 3 sen θ ndash 2= 0rArr

rArr sen θ = ndash3plusmn 9+ 164

rArr

rArr sen θ = ndash 3plusmn 54

1o Q

sen θ = 12

cos2 θ = 1 ndash sen2 θ = 1 ndash 14 = 34 rArr cos θ = 32

Resposta b

12 2senx

y sen x tg x sen x sen x sec xcosx

= sdot = sdot = sdot =

( )21 cos x sec x= minus sdot = 2sec x cos x sec xminus sdot =

sec x cos x= minus

Resposta b

| MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2



13 200ordm eacute do 3 o quadrante sen 200ordm 0rArr lt cos 200ordm 0lt

tg200ordm 0gt e cos200ordm sen200ordmlt

Assim cos 200ordm sen200ordm tg200ordmlt lt

Resposta b

14 Da figura sen θ = br

e cos θ = ar

Como sen (π ndash θ)= sen θ =br

concluiacutemos

b= r middot sen (π ndash θ)

cos (π + θ)= ndash cos θ = ndash ar

rArr r cos (π + θ)= ndasha

Resposta d

15 a cos x= 1+ sen x

b cos x= 1 ndash sen xrArr (a cos x) middot (b cos x)=

= (1+ sen x) middot (1 ndash sen x)rArr ab cos2 x= 1 ndash sen2 x rArr

rArr ab middot cos2 x= cos2 xrArr ab= 1

Resposta d

16 Resposta b

17 ∆ = 0rArr 22 ndash 4 middot 1 middot sen α = 0rArr sen α = 1rArr α = π2

Resposta d

19 sec2 x= 1+ tg2 x

2 tg2 x ndash (1+ tg2 x)= 3 rArr tg2 x= 1+ 3 rArr

rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x

= 1+ 3 rArr (1+ 3 ) middot cos2 x= 1 ndash cos2 xrArr

rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x

cossec x tg x sen x cos x sen x cos xcosx sen x cos xsen x cotg x senxsenx senx

minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b

21 C(x)= 2 ndash cosxπ6 V(x)= 3 2 middot sen

xπ12 0⩽ x⩽ 6

Lucro L(x)= V(x) ndash C(x)

L(3)= 3 2 middot sen3π12

ndash 2 ndash cos3π6

= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c

22 ∆TABbase TB= cotgα

altura (relativa a TB)1 ndash sen α

aacuterea TAB

S= 12

middot cotg α middot (1 ndash sen α)

Resposta d

23 x+ y= 90deg rArr cos x= sen y e cos2 x= 3 cos2 y

Assim temos

sen2 y= 3 cos2 yrArr 3 cos2 y+ cos2 y= 1rArr cos2 y= 14

rArr

rArr cos y= + 12

y= 60deg e x = 30deg rArr y ndash x= 30deg

Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +

2 2 2x x 2x 1 x 2x 1rArr + minus + = + + rArr

2x 4x 0minus = rArr

( )x x 4 0minus = rArr x 0= ou x 4=

Resposta d




26 BC eacute tangenterArr BCperp OPrArr ∆OPB eacute retacircngulorArr

rArr tg α = PBOP

= PB

∆OPC tambeacutem eacute retacircngulorArr OC2 = 1+ PC2 rArr

rArr cossec2 α = 1+ PC2 rArr PC2 = cotg2 α

PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )

sen x 1 sen x sen x 1 sen x

1 sen x 1 sen x

minus + += =

+ minus

2 2

2

sen x sen x sen x sen x

1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c

29 x= 1 verifica a equaccedilatildeo

(cos2 α) middot 12 ndash (4 cos α sen β) middot 1+ 32

sen β = 0

cos2 α ndash 4 cos α sen β + 32

sen β = 0

Como α e β satildeo complementares sen β = cos α

Temos

cos2 α ndash 4 cos α middot cos α + 32

cos α = 0rArr

rArr ndash3 cos2 α + 32

cos α = 0rArr

rArr cos α middot ndash3 cos α + 32

= 0 rArr cos α = 0 natildeo serve

pois α ne 90deg ou ndash3 cos α + 32

= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O

bull O ponto P ao percorrer a distacircncia d no sentido anti-

-horaacuterio ldquoatingerdquo o ponto R

bull O comprimento do arco PR eacute d e a medida em radia-

nos do acircngulo POR eacute α = dr

bull Temos OPeixo x RSeixo y Desse modo a distacircncia

percorrida no eixo x pelo pontoQ pode ser represen-

tada pela medida de PS

bull No ∆OSR temos cosdr

= OSOR

rArr OS= r middot cosdr

bull Daiacute PS= OP ndash OS= r ndash r middot cosdr

= r 1 ndash cosdr

Resposta b




11 5

6

7

6

11

6

6

ndash 3

2

3

2

cos π6

∙ cos 11π6

∙ 3

2

cos 5π6

∙ cos 7π6

∙ ndash

32

12 4

5

6

5

9

5

5

36deg

cos π5

∙ cos 9π5

gt 0

cos 4π5

∙ cos 6π5

lt 0

13 a)3

2 e) 0

b) 0 f ) ndash2

2

c) ndash

12

g) 12

d) ndash1 h) 1

14 a)65deg

85deg

cos 85deg

cos 65deg

cos 85ordm lt cos 65ordm

b) cos 89deg gt 0

cos 91deg lt 0 rArr cos 89deg gt cos 91deg

c)

50deg

20deg

cos 50deg

cos 340deg

cos 340ordm gt cos 50ordm

d)

10deg170deg

190deg cos 170ordm∙ cos 190ordm

15 k983101 0rarr cos 0983101 1

k983101 1rarr cos π

2

983101 0

k983101 2rarr cos π 983101 ndash1

k983101 3rarr cos 3π2

983101 0

16 a) 0 ndash 32

∙ ndash

32

860697 12

(F)

b)

sen π3

2

∙ 32

2

∙ 34

cos π3

2

∙ 12

2

∙ 14

34

983083 14

∙ 1 (V)

c) Como π2

cong 157 os nuacutemeros reais 1 e 2 tecircm ima-

gens respectivamente no 1o e 2o quadrantes

Daiacute cos 1 gt 0 e cos 2 lt 0 donde concluiacutemos que

cos 2 lt cos 1 (V )

d)100deg

sen 100deg

cos 100deg

Como |sen 100deg| gt |cos 100deg| segue que sen 100deg 983083

983083 cos 100deg gt 0 Observe que sen 100deg gt 0 e cos 100deg lt 0

(F)

e) 6 lt 2π (cong628) assim o nuacutemero real 6 tem imagem no

4o quadrante e cos 6 gt 0 (F)

f ) O raio da circunferecircncia trigonomeacutetrica eacute unitaacuterio (F)

17 OA983101 1

OB983101 cos π6

983101 32

AB983101 sen π6

983101 12

periacutemetro983101 1983083 32

983083 12

983101 3983083 32

UC

aacuterea983101 OB middot AB2

983101

32

middot 12

2 983101

38

UA

18 OB 983101 1 AB983101 12 983101 05BD983101 DE983101 sen α

OD983101 cos α

ABC 1057305 OBD rArr ABOB

983101 ACOD

983101 BCBD

051 983101 AC

cos α 983101 BCsen α rArr

AC∙ 12

cos α

BC∙ 12

sen α

A aacuterea do triacircngulo ABC eacute12

middot AC middot BC 983101 12

middot 12

cos α middot 12

sen α 983101 sen α middot cos α8






33

4

A

B

45deg

0



-tg π


Logo tg π4

∙ 1

b) F e) V

c) V

3

f ) F

tg 2π 983101 0


2

∙ 1 ndash 19

∙ 89

α 2o Q


3


1

3ndash 2 2

3


35 sen2 α ∙ 1 ndash 210

2

∙ 96100

α 4o Q


∙ ndash

4 610

∙ ndash

2 65

tg α ∙ ndash

2 65

15

∙ ndash2 6




cos2 x∙ 110

rArr cos x∙ 1010


OAOB

∙

1010

ndash

3 1010

∙ ndash

13





38

11

6

2

53

4

3

011

10

-sec 2π

5 gt 0 cossec 2π

5 gt 0 e cotg 2π

5 gt 0

-sec 3π

4 lt 0 cossec 3π

4 gt 0 e cotg 3π

4 lt 0


-sec 11π

10 lt 0 cossec 11π

10 lt 0 e cotg 11π

10 gt 0

-sec

11π

6 gt 0 cossec

11π

6 lt 0 e cotg

11π

6 lt 0

39 a)1

cos π6

∙ 1

32

∙ 2 33

b) 1

tg 2π3

∙ 1

ndash tg π3

∙ 1ndash 3

∙ ndash

33

c)1

sen 5π6

∙ 1

sen π6

∙ 112

∙ 2

d) 1cos 210deg


∙ 1

ndash 32

∙ ndash 2 33

e) 1sen 315deg


∙ 1

ndash 22

∙ ndash 2

f ) 1tg 45deg

∙ 11

∙ 1

40 sec x983101 52



∙

1 ndash

4

25 rArr sen2

x983101

21

25 como xisin

4o

Q temos

sen x983101 ndash

215


983101 ndash

215

25

983101 ndash

212

cotg x983101 ndash

2

21 ndash 2 21

21


983101 ndash 5

21 983101 ndash 5 21

21

sec x983101 1cos x

983101 52




41 sec2 x983101 1983083 tg2 xrArr 73

2

983101 1983083 tg2 xrArr tg2 x983101 409 rArr


2 103


2 10 983101 15 middot ndash 1


1010

m

42 a) OQ983101 cossec α 983101 103

rArr sen α ∙ 310


2

983101 1 ndash 9100

983101 91100

rArr

α 1o Q cos α 983101

9110

sen α 983083 cos α 983101 310 983083

9110

9831013983083 91

10


983101

91100

9

100

983101 919


tg α 983101 125 rArr cotg α 983101 5

12


sen β 983101 810


983101 54


b) V



d) Como π2


cotg 7π8 lt 0

sec 7π8 lt 0


8 gt 0 (V )

51

B

H

h4 5

x 6 ndash x

6

A C

Temos


42 983101 h2 983083 x2 (1)




4

em (1) rArr 16983101 h2 983083 94

2

rArr h2 983101 16 ndash 8116 983101 175

16 rArr

rArr h983101 1754

cm

ABH cos α 983101 x

4 983101

94

4 983101 9

16 rArr sec α 983101 16

9

46 cos x983101 2


2 983101 m

4 rArr 2 m983101 28rArr m983101 14



48 OB983101 sec π6

983101 1

cos π6

983101 1

32

983101 2 33

OA983101 cossec π6

983101 1

sen π6

983101 112

983101 2


OA middot OB2

983101

2 33

middot 2

2 983101 2 33

(UA)

49 Q

O

P 30deg

P


|PQ|983101 |cotg 120deg|983101 1tg 120deg


983101

983101 minus 1

3 983101 1

3 983101

33


12


middot 1 middot 33

983101 36

(UA)


|OP|2 983101 |PQ|2 983083 |OQ|2 rArr |OP|2 983101 3

3

2

983083 12 983101

983101 13

983083 1983101 43

rArr OP983101 23

983101 2 33

(UC)


cos2 x983101 1 ndash 13

2

983101 89

rArr

x 1o Q cos x983101 2 2

3 tg x983101

13

2 23

983101 1

2 2

983101 24




52 a2 983101 1cos2 x


983101 sen2 xcos2 x

983101 tg2 x

b2 983101 1sen2 x


983101 cos2 xsen2 x

983101 cotg2 x983101 1tg2 x


53 sec

2

x983101

4rArr

sec x983101


3

o


rArr tg x983101 3


e x983101 π 983083 π3

983101 4π3


9

cos2 β ∙ 1625

rArr

rArr sen2 β 983083 cos2 β 983101 49

983083 1625


55 a)

cotg2 x

1983083 cotg2 x 983101

cos2 xsen2 x


983101

cos2 xsen2 x


983101

∙

cos2 xsen2 x

1sen2 x

983101 cos2 x

b)


983083 cos α 983101

∙ sen2 α983083 cos2 α

cos α

983101 1

cos α

983101 sec α


983083 cos xsen x

983101



983101 1sen x

middot 1cos x

983101


d) tg x983083 cos x1983083 sen x

983101 sen xcos x

983083 cos x1983083 sen x

983101


983101 sen x983083 sen2 x983083 cos2 x



983101 1cos x

983101 sec x

56

57

58

59


tg x negativo

cotg x negativo

cotg x positivo2

π +


sinal de y

( ) ( )

( ) ( ) ( )

minus sdot +



tg x positivo

cotg x positivo

cotg x negativo2

π +


sinal de y ( ) ( )

( ) ( ) ( )


+ sdot +


sen x negativo

cos x negativo

sec x negativo

tg x positivo


sinal de y ( ) ( ) ( )

( ) ( ) ( )


+ sdot +

x 0 x2


tg x 5=


cos x sen x2

π minus =

( )tg x tg xπ + =




senx senxy

minus sdot=

tg xsdot


senxtg x 5


0 x2


sen xcosx2

tg x2 sen x

cos x2





2 2sen x cos x 1+ =

221

cos x 110

+ =

2 1 99cos x 1


60 a)

b)

61 sen

cotg1 cos

αα + =

+ α

cos sen

sen 1 cos

α α+ =

α + α

( )

( )

2cos 1 cos sen

sen 1 cos

α + α + α= =

α + α

2 2cos cos sen

sen sen cos

α + α + α= =

α + α sdot α

( )

1 cos 1cossec

sen 1 cos sen

+ α= = = α

α + α α




2

2



63

64





2 2sen x cos x 1= + =

66

1cossec x ndash 1

+ 1cossec x+ 1

=67

3 11cosx

10rArr = rArr

3 11

10tg x 3 1112

10

π minus = =


coscos

αsdot α+ α =

α

2 2sen cos

cos

α sdot α= =

α

1sec

cos= α

α

2tg

ecs1

1

α minus =α minus

2tg sec 1

sec 1


2

2

sen 11

coscos

11

cos

αminus +

αα= =

minus

α

2

2

sen cos cos

cos

1 cos

cos

α minus α + α

α=

minus α

α

2

1 cos

1cossec

1 cos cos

minus α

α= = = α

minus α α


2cos x= + + + =

2 22 cos x 2 sen x

5senxcosx

+= + =

( )2 22 sen x cos x

5sen x cosx

+

= + = 5senx 2sec x+

sen x cos x

sen x cos x

+=

minus

senx cos x

senx

senx cos x

senx

+

=minus

1 cotg x

1 cotg x

+

minus




senxsdot

senx

cosx+ cosxsdot 1+ =


( )sen x 1 c os x cos x 1= + + + =

( ) ( )1 cos x 1 sen x= + sdot +

=

sena senb cosa cosb

cosa cosb sena senb

+ ++ =

+ minus

( )( )

( )( )

( )( )

( )( )

sena senb sena senb




+ minus +=

+ minus

+ minus +=

+ minus

( )( )




+ minus






=

=2 cossec x

cotg2 x =

2sen x

middot sen2 xcos2 x

=

= 2 middot 1cos x

middot sen xcos x


tg x+ 1tg x

middot 1cos x


ndash sen x =

= tg2 x+ 1



sen x =

= sec2 xtg x

middot sen 2 xcos x

middot cos 2 xsen x

= sec2 xtg x


= 1cos2 x

middot cos xsen x


68

+ 1 + ndash 1

69


(m ndash 2) middot3


m = ndash1

m



cos2 x + sen2 x = 1 2





71



ndash 45

= ndash 75

72 a)

b) 12 5 23


π π π = sdot +

12 14 2 22

7 7 7 7


12cotg

7

πrArr negativo

5 5 180ordm82ordm

11 11


5sen

11

πrArr

positivo

23 242

12 12 12 12


23cos

12

πrArr positivo

sinal de y



1 Se sen x= 1 x= π

2 cos x= cos π

2 = 0 assim

sen x+ cos x= 1+ 0= 1


sen x+ cos x= 0+ 1= 1






4


2

2 2tgx cosxy1 cos x

sdot= rArr

minus 2tgx cos xy

sen x

sdot= rArr

senx

cosxyrArr =

cosxsdot

2sen xrArr

1y cossec x

senx= =


ox 3 Qisin2

2 24 576 625cossec x 1 1

7 49 49

= + = + = rArr

25cossec x7


25

y cossec x 7

= = minus

9 7x sec tg cotg

8 6 7

π π π = sdot +

9 8

8 8 8 8


9sec

8

πrArr

negativo

7 6

6 6 6 6


7tg

6

πrArr

positivo

7


6

πrArr




ndash5

+

+ 4

5

2

= 1

3 4 3 2





Se sen x= 12

podemos ter x= π6

ou

x= π ndash π6

= 5π6


m+ 13

2 + m ndash 1

32 = 1 rArr 2m2 + 2

9 = 1 rArr

rArr 2m2 = 79

rArr m2 = 718

rArr m= plusmn 7

3 2 middot

2

2 =

= plusmn 146

(cong 0623)



4 a) Sabemos que




= 1cos2 α

+ 1sen2 α

=



= 1cos2 α

middot 1sen2 α

Daiacute

CS= 1

cos2 α middot 1


cos α gt 0 vem

CS= 1cos α

middot 1sen α


b) α =

π

6 OS=

sec

π

6 =

sec 30deg=

1

cos 30deg =

= 23

middot 33

= 2 33



2

= 12 + PS2 rArr 43

ndash 1=

= PS2 rArr PS2 = 13

rArr PS= 33


3 + 1+ 3

3 = ( 3 + 1) UC





= sen xcos x

cos x∙ 0 rArr



cos x= cos π2

= 0 Logo



tg x

rArr tg2 x= 1 rArr

rArr tg x= 1 x= π4

ou x= 5π4

ou


ou x= 7π4


cotg x ndash 1 =

1cos x

middot 1sen x

ndash 1cos2 x

cos xsen x

ndash 1 =

=




middotsen x

cos x ndash sen x =

= 1cos2 x

= 1

14

2 = 16

7 a)

α

0 A

T


AT= tg α


rArr OT= sec α

b) anaacuteloga

8 a)

α β

x 800 ndash x

800

h


rArr x

= 5

rArr x= 5 h (1)


15 rArr h

800 ndash x = 1

15 rArr




rArr(1)

x= 200 m


1 h 1


sen α ∙ 0






= 4 cos2 α + 4 sen2 α = 4


2 middot sen α =


10 a) sen2 3π20

+ cos2 3π20

= 1 rArr a2 + cos2 3π20

= 1 rArr



cos 3π20

= cos 27deggt 0


+ 17π20


de 3π20

e 17π20


senos

3π

20

17π

20

Logo sen 17π20

= sen 3π20

= a

c) Como 3π20

+ 7π20

= 10π20

= π2


e 7π20


cos 7π20

= sen 3π20

= a


20

= 37π

20


20e 37π20


cossenos

3π

20

37π

20

Daiacute sen 37π20

= ndash sen 3π20

= ndasha



5 (1)

rArr a =(3)

2b

sec x= 5a

rArr 1cos x

= 5a

rArr cos x= a5

(2)


2

+ 2b5

2

= 1 rArr


rArrbgt 0

rArr b=

5 middot 3

61 =

15 61

61

a+ c

Em (3) a= 30 61

61


b ndash a= 15 61

61 ndash 30 61

61 = ndash 15 61

61Assim


61

rArr c ndash 15 61

61

= ndash 15 61

61

rArr c= 0

12 a) sen α = 15


rArr 15

ndash cos α2

+ cos2 α = 1 rArr


cos α ndash 2425

= 0 rArr



100

45

ndash35

Se cos α = 45


ndash 45

= ndash 35

e α tem




+ 35

= 45

e α tem



lt 0


complementares



1 rArr


14 1o membro=


ndash cos xsen x

2

= (1 ndash cos x)2

sen2 x =

= (1 ndash cos x)2

1 ndash cos2 x =

(1 ndash cos x)2


1 ndash cos x

1+ cos x =

= 2o membro

15 tg x ndash sen x

sen3 x =


sen3 x =

1cos x ndash 1

sen2 x =

=

1cos x ndash 1

1 ndash cos2 x =

1 ndash cos x

cos x =


=

=1

cos x middot

1

1+ cos x =

sec x

1+ cos x






Comoπ5

+ 3π10

= 2π + 3π10

= π2

temos

tgπ5

= 1

tg 3π10




rArr


= 60deg

Daiacute cos B =

102

AB rArr

32

= 5AB

rArr

rArr AB= AC= 10

3 =

10 33

Assim

2p= 10+ 2 middot10 3

3 = 10+

20 33

cm

b) sen x+ cos x= k




2

sen3 x+ cos3 x=



2 = k middot

2 ndash k 2 + 12

=

= k

2

(3 ndash k 2)

27

1

y

O D

α

1 x

BAE

C

a) AE= cotg α= 1

tg α = 1tg π3

= 1

3 = 33


DC= tg α = tgπ3

= 3



= 12

middot 1 ndash 33


= 12


aacuterea=

1

2 middot

4 3

3 ndash

2=

2 3

3 ndash

1




=

= 1


=



= 12 middot


=

= 12 middot


= 12 middot


= 1


lente a ()

Desafio





9998

9989

9899

8999


Testes



Resposta c

3

2 Resposta d

2sen x com x

5 2


2 2sen x cos x 1+ =

24cos x 1

25+ = rArr

2 4 21cos x 1

25 25= minus = rArr

21cos x

5rArr = minus


5

= = = minus

minus

Resposta d




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus



2

= = = e





IV Falsa pois1 1


= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c




Daiacute


cotg x= 43

rArr cos xsen x

= 43


9 cos2 x16


rArr sen x= 5

Resposta a

3

7 sen xcos x



rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2


1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8


2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13


Resposta c


x y sen x cos y2π



10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor


Resposta a



cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ




rArr


1o Q

sen θ = 12


Resposta b

12 2senx




sec x cos x= minus

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b






33

4

A

B

45deg

0



-tg π


Logo tg π4

∙ 1

b) F e) V

c) V

3

f ) F

tg 2π 983101 0


2

∙ 1 ndash 19

∙ 89

α 2o Q


3


1

3ndash 2 2

3


35 sen2 α ∙ 1 ndash 210

2

∙ 96100

α 4o Q


∙ ndash

4 610

∙ ndash

2 65

tg α ∙ ndash

2 65

15

∙ ndash2 6




cos2 x∙ 110

rArr cos x∙ 1010


OAOB

∙

1010

ndash

3 1010

∙ ndash

13





38

11

6

2

53

4

3

011

10

-sec 2π

5 gt 0 cossec 2π

5 gt 0 e cotg 2π

5 gt 0

-sec 3π

4 lt 0 cossec 3π

4 gt 0 e cotg 3π

4 lt 0


-sec 11π

10 lt 0 cossec 11π

10 lt 0 e cotg 11π

10 gt 0

-sec

11π

6 gt 0 cossec

11π

6 lt 0 e cotg

11π

6 lt 0

39 a)1

cos π6

∙ 1

32

∙ 2 33

b) 1

tg 2π3

∙ 1

ndash tg π3

∙ 1ndash 3

∙ ndash

33

c)1

sen 5π6

∙ 1

sen π6

∙ 112

∙ 2

d) 1cos 210deg


∙ 1

ndash 32

∙ ndash 2 33

e) 1sen 315deg


∙ 1

ndash 22

∙ ndash 2

f ) 1tg 45deg

∙ 11

∙ 1

40 sec x983101 52



∙

1 ndash

4

25 rArr sen2

x983101

21

25 como xisin

4o

Q temos

sen x983101 ndash

215


983101 ndash

215

25

983101 ndash

212

cotg x983101 ndash

2

21 ndash 2 21

21


983101 ndash 5

21 983101 ndash 5 21

21

sec x983101 1cos x

983101 52




41 sec2 x983101 1983083 tg2 xrArr 73

2

983101 1983083 tg2 xrArr tg2 x983101 409 rArr


2 103


2 10 983101 15 middot ndash 1


1010

m

42 a) OQ983101 cossec α 983101 103

rArr sen α ∙ 310


2

983101 1 ndash 9100

983101 91100

rArr

α 1o Q cos α 983101

9110

sen α 983083 cos α 983101 310 983083

9110

9831013983083 91

10


983101

91100

9

100

983101 919


tg α 983101 125 rArr cotg α 983101 5

12


sen β 983101 810


983101 54


b) V



d) Como π2


cotg 7π8 lt 0

sec 7π8 lt 0


8 gt 0 (V )

51

B

H

h4 5

x 6 ndash x

6

A C

Temos


42 983101 h2 983083 x2 (1)




4

em (1) rArr 16983101 h2 983083 94

2

rArr h2 983101 16 ndash 8116 983101 175

16 rArr

rArr h983101 1754

cm

ABH cos α 983101 x

4 983101

94

4 983101 9

16 rArr sec α 983101 16

9

46 cos x983101 2


2 983101 m

4 rArr 2 m983101 28rArr m983101 14



48 OB983101 sec π6

983101 1

cos π6

983101 1

32

983101 2 33

OA983101 cossec π6

983101 1

sen π6

983101 112

983101 2


OA middot OB2

983101

2 33

middot 2

2 983101 2 33

(UA)

49 Q

O

P 30deg

P


|PQ|983101 |cotg 120deg|983101 1tg 120deg


983101

983101 minus 1

3 983101 1

3 983101

33


12


middot 1 middot 33

983101 36

(UA)


|OP|2 983101 |PQ|2 983083 |OQ|2 rArr |OP|2 983101 3

3

2

983083 12 983101

983101 13

983083 1983101 43

rArr OP983101 23

983101 2 33

(UC)


cos2 x983101 1 ndash 13

2

983101 89

rArr

x 1o Q cos x983101 2 2

3 tg x983101

13

2 23

983101 1

2 2

983101 24




52 a2 983101 1cos2 x


983101 sen2 xcos2 x

983101 tg2 x

b2 983101 1sen2 x


983101 cos2 xsen2 x

983101 cotg2 x983101 1tg2 x


53 sec

2

x983101

4rArr

sec x983101


3

o


rArr tg x983101 3


e x983101 π 983083 π3

983101 4π3


9

cos2 β ∙ 1625

rArr

rArr sen2 β 983083 cos2 β 983101 49

983083 1625


55 a)

cotg2 x

1983083 cotg2 x 983101

cos2 xsen2 x


983101

cos2 xsen2 x


983101

∙

cos2 xsen2 x

1sen2 x

983101 cos2 x

b)


983083 cos α 983101

∙ sen2 α983083 cos2 α

cos α

983101 1

cos α

983101 sec α


983083 cos xsen x

983101



983101 1sen x

middot 1cos x

983101


d) tg x983083 cos x1983083 sen x

983101 sen xcos x

983083 cos x1983083 sen x

983101


983101 sen x983083 sen2 x983083 cos2 x



983101 1cos x

983101 sec x

56

57

58

59


tg x negativo

cotg x negativo

cotg x positivo2

π +


sinal de y

( ) ( )

( ) ( ) ( )

minus sdot +



tg x positivo

cotg x positivo

cotg x negativo2

π +


sinal de y ( ) ( )

( ) ( ) ( )


+ sdot +


sen x negativo

cos x negativo

sec x negativo

tg x positivo


sinal de y ( ) ( ) ( )

( ) ( ) ( )


+ sdot +

x 0 x2


tg x 5=


cos x sen x2

π minus =

( )tg x tg xπ + =




senx senxy

minus sdot=

tg xsdot


senxtg x 5


0 x2


sen xcosx2

tg x2 sen x

cos x2





2 2sen x cos x 1+ =

221

cos x 110

+ =

2 1 99cos x 1


60 a)

b)

61 sen

cotg1 cos

αα + =

+ α

cos sen

sen 1 cos

α α+ =

α + α

( )

( )

2cos 1 cos sen

sen 1 cos

α + α + α= =

α + α

2 2cos cos sen

sen sen cos

α + α + α= =

α + α sdot α

( )

1 cos 1cossec

sen 1 cos sen

+ α= = = α

α + α α




2

2



63

64





2 2sen x cos x 1= + =

66

1cossec x ndash 1

+ 1cossec x+ 1

=67

3 11cosx

10rArr = rArr

3 11

10tg x 3 1112

10

π minus = =


coscos

αsdot α+ α =

α

2 2sen cos

cos

α sdot α= =

α

1sec

cos= α

α

2tg

ecs1

1

α minus =α minus

2tg sec 1

sec 1


2

2

sen 11

coscos

11

cos

αminus +

αα= =

minus

α

2

2

sen cos cos

cos

1 cos

cos

α minus α + α

α=

minus α

α

2

1 cos

1cossec

1 cos cos

minus α

α= = = α

minus α α


2cos x= + + + =

2 22 cos x 2 sen x

5senxcosx

+= + =

( )2 22 sen x cos x

5sen x cosx

+

= + = 5senx 2sec x+

sen x cos x

sen x cos x

+=

minus

senx cos x

senx

senx cos x

senx

+

=minus

1 cotg x

1 cotg x

+

minus




senxsdot

senx

cosx+ cosxsdot 1+ =


( )sen x 1 c os x cos x 1= + + + =

( ) ( )1 cos x 1 sen x= + sdot +

=

sena senb cosa cosb

cosa cosb sena senb

+ ++ =

+ minus

( )( )

( )( )

( )( )

( )( )

sena senb sena senb




+ minus +=

+ minus

+ minus +=

+ minus

( )( )




+ minus






=

=2 cossec x

cotg2 x =

2sen x

middot sen2 xcos2 x

=

= 2 middot 1cos x

middot sen xcos x


tg x+ 1tg x

middot 1cos x


ndash sen x =

= tg2 x+ 1



sen x =

= sec2 xtg x

middot sen 2 xcos x

middot cos 2 xsen x

= sec2 xtg x


= 1cos2 x

middot cos xsen x


68

+ 1 + ndash 1

69


(m ndash 2) middot3


m = ndash1

m



cos2 x + sen2 x = 1 2





71



ndash 45

= ndash 75

72 a)

b) 12 5 23


π π π = sdot +

12 14 2 22

7 7 7 7


12cotg

7

πrArr negativo

5 5 180ordm82ordm

11 11


5sen

11

πrArr

positivo

23 242

12 12 12 12


23cos

12

πrArr positivo

sinal de y



1 Se sen x= 1 x= π

2 cos x= cos π

2 = 0 assim

sen x+ cos x= 1+ 0= 1


sen x+ cos x= 0+ 1= 1






4


2

2 2tgx cosxy1 cos x

sdot= rArr

minus 2tgx cos xy

sen x

sdot= rArr

senx

cosxyrArr =

cosxsdot

2sen xrArr

1y cossec x

senx= =


ox 3 Qisin2

2 24 576 625cossec x 1 1

7 49 49

= + = + = rArr

25cossec x7


25

y cossec x 7

= = minus

9 7x sec tg cotg

8 6 7

π π π = sdot +

9 8

8 8 8 8


9sec

8

πrArr

negativo

7 6

6 6 6 6


7tg

6

πrArr

positivo

7


6

πrArr




ndash5

+

+ 4

5

2

= 1

3 4 3 2





Se sen x= 12

podemos ter x= π6

ou

x= π ndash π6

= 5π6


m+ 13

2 + m ndash 1

32 = 1 rArr 2m2 + 2

9 = 1 rArr

rArr 2m2 = 79

rArr m2 = 718

rArr m= plusmn 7

3 2 middot

2

2 =

= plusmn 146

(cong 0623)



4 a) Sabemos que




= 1cos2 α

+ 1sen2 α

=



= 1cos2 α

middot 1sen2 α

Daiacute

CS= 1

cos2 α middot 1


cos α gt 0 vem

CS= 1cos α

middot 1sen α


b) α =

π

6 OS=

sec

π

6 =

sec 30deg=

1

cos 30deg =

= 23

middot 33

= 2 33



2

= 12 + PS2 rArr 43

ndash 1=

= PS2 rArr PS2 = 13

rArr PS= 33


3 + 1+ 3

3 = ( 3 + 1) UC





= sen xcos x

cos x∙ 0 rArr



cos x= cos π2

= 0 Logo



tg x

rArr tg2 x= 1 rArr

rArr tg x= 1 x= π4

ou x= 5π4

ou


ou x= 7π4


cotg x ndash 1 =

1cos x

middot 1sen x

ndash 1cos2 x

cos xsen x

ndash 1 =

=




middotsen x

cos x ndash sen x =

= 1cos2 x

= 1

14

2 = 16

7 a)

α

0 A

T


AT= tg α


rArr OT= sec α

b) anaacuteloga

8 a)

α β

x 800 ndash x

800

h


rArr x

= 5

rArr x= 5 h (1)


15 rArr h

800 ndash x = 1

15 rArr




rArr(1)

x= 200 m


1 h 1


sen α ∙ 0






= 4 cos2 α + 4 sen2 α = 4


2 middot sen α =


10 a) sen2 3π20

+ cos2 3π20

= 1 rArr a2 + cos2 3π20

= 1 rArr



cos 3π20

= cos 27deggt 0


+ 17π20


de 3π20

e 17π20


senos

3π

20

17π

20

Logo sen 17π20

= sen 3π20

= a

c) Como 3π20

+ 7π20

= 10π20

= π2


e 7π20


cos 7π20

= sen 3π20

= a


20

= 37π

20


20e 37π20


cossenos

3π

20

37π

20

Daiacute sen 37π20

= ndash sen 3π20

= ndasha



5 (1)

rArr a =(3)

2b

sec x= 5a

rArr 1cos x

= 5a

rArr cos x= a5

(2)


2

+ 2b5

2

= 1 rArr


rArrbgt 0

rArr b=

5 middot 3

61 =

15 61

61

a+ c

Em (3) a= 30 61

61


b ndash a= 15 61

61 ndash 30 61

61 = ndash 15 61

61Assim


61

rArr c ndash 15 61

61

= ndash 15 61

61

rArr c= 0

12 a) sen α = 15


rArr 15

ndash cos α2

+ cos2 α = 1 rArr


cos α ndash 2425

= 0 rArr



100

45

ndash35

Se cos α = 45


ndash 45

= ndash 35

e α tem




+ 35

= 45

e α tem



lt 0


complementares



1 rArr


14 1o membro=


ndash cos xsen x

2

= (1 ndash cos x)2

sen2 x =

= (1 ndash cos x)2

1 ndash cos2 x =

(1 ndash cos x)2


1 ndash cos x

1+ cos x =

= 2o membro

15 tg x ndash sen x

sen3 x =


sen3 x =

1cos x ndash 1

sen2 x =

=

1cos x ndash 1

1 ndash cos2 x =

1 ndash cos x

cos x =


=

=1

cos x middot

1

1+ cos x =

sec x

1+ cos x






Comoπ5

+ 3π10

= 2π + 3π10

= π2

temos

tgπ5

= 1

tg 3π10




rArr


= 60deg

Daiacute cos B =

102

AB rArr

32

= 5AB

rArr

rArr AB= AC= 10

3 =

10 33

Assim

2p= 10+ 2 middot10 3

3 = 10+

20 33

cm

b) sen x+ cos x= k




2

sen3 x+ cos3 x=



2 = k middot

2 ndash k 2 + 12

=

= k

2

(3 ndash k 2)

27

1

y

O D

α

1 x

BAE

C

a) AE= cotg α= 1

tg α = 1tg π3

= 1

3 = 33


DC= tg α = tgπ3

= 3



= 12

middot 1 ndash 33


= 12


aacuterea=

1

2 middot

4 3

3 ndash

2=

2 3

3 ndash

1




=

= 1


=



= 12 middot


=

= 12 middot


= 12 middot


= 1


lente a ()

Desafio





9998

9989

9899

8999


Testes



Resposta c

3

2 Resposta d

2sen x com x

5 2


2 2sen x cos x 1+ =

24cos x 1

25+ = rArr

2 4 21cos x 1

25 25= minus = rArr

21cos x

5rArr = minus


5

= = = minus

minus

Resposta d




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus



2

= = = e





IV Falsa pois1 1


= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c




Daiacute


cotg x= 43

rArr cos xsen x

= 43


9 cos2 x16


rArr sen x= 5

Resposta a

3

7 sen xcos x



rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2


1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8


2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13


Resposta c


x y sen x cos y2π



10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor


Resposta a



cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ




rArr


1o Q

sen θ = 12


Resposta b

12 2senx




sec x cos x= minus

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b




33

4

A

B

45deg

0



-tg π


Logo tg π4

∙ 1

b) F e) V

c) V

3

f ) F

tg 2π 983101 0


2

∙ 1 ndash 19

∙ 89

α 2o Q


3


1

3ndash 2 2

3


35 sen2 α ∙ 1 ndash 210

2

∙ 96100

α 4o Q


∙ ndash

4 610

∙ ndash

2 65

tg α ∙ ndash

2 65

15

∙ ndash2 6




cos2 x∙ 110

rArr cos x∙ 1010


OAOB

∙

1010

ndash

3 1010

∙ ndash

13





38

11

6

2

53

4

3

011

10

-sec 2π

5 gt 0 cossec 2π

5 gt 0 e cotg 2π

5 gt 0

-sec 3π

4 lt 0 cossec 3π

4 gt 0 e cotg 3π

4 lt 0


-sec 11π

10 lt 0 cossec 11π

10 lt 0 e cotg 11π

10 gt 0

-sec

11π

6 gt 0 cossec

11π

6 lt 0 e cotg

11π

6 lt 0

39 a)1

cos π6

∙ 1

32

∙ 2 33

b) 1

tg 2π3

∙ 1

ndash tg π3

∙ 1ndash 3

∙ ndash

33

c)1

sen 5π6

∙ 1

sen π6

∙ 112

∙ 2

d) 1cos 210deg


∙ 1

ndash 32

∙ ndash 2 33

e) 1sen 315deg


∙ 1

ndash 22

∙ ndash 2

f ) 1tg 45deg

∙ 11

∙ 1

40 sec x983101 52



∙

1 ndash

4

25 rArr sen2

x983101

21

25 como xisin

4o

Q temos

sen x983101 ndash

215


983101 ndash

215

25

983101 ndash

212

cotg x983101 ndash

2

21 ndash 2 21

21


983101 ndash 5

21 983101 ndash 5 21

21

sec x983101 1cos x

983101 52




41 sec2 x983101 1983083 tg2 xrArr 73

2

983101 1983083 tg2 xrArr tg2 x983101 409 rArr


2 103


2 10 983101 15 middot ndash 1


1010

m

42 a) OQ983101 cossec α 983101 103

rArr sen α ∙ 310


2

983101 1 ndash 9100

983101 91100

rArr

α 1o Q cos α 983101

9110

sen α 983083 cos α 983101 310 983083

9110

9831013983083 91

10


983101

91100

9

100

983101 919


tg α 983101 125 rArr cotg α 983101 5

12


sen β 983101 810


983101 54


b) V



d) Como π2


cotg 7π8 lt 0

sec 7π8 lt 0


8 gt 0 (V )

51

B

H

h4 5

x 6 ndash x

6

A C

Temos


42 983101 h2 983083 x2 (1)




4

em (1) rArr 16983101 h2 983083 94

2

rArr h2 983101 16 ndash 8116 983101 175

16 rArr

rArr h983101 1754

cm

ABH cos α 983101 x

4 983101

94

4 983101 9

16 rArr sec α 983101 16

9

46 cos x983101 2


2 983101 m

4 rArr 2 m983101 28rArr m983101 14



48 OB983101 sec π6

983101 1

cos π6

983101 1

32

983101 2 33

OA983101 cossec π6

983101 1

sen π6

983101 112

983101 2


OA middot OB2

983101

2 33

middot 2

2 983101 2 33

(UA)

49 Q

O

P 30deg

P


|PQ|983101 |cotg 120deg|983101 1tg 120deg


983101

983101 minus 1

3 983101 1

3 983101

33


12


middot 1 middot 33

983101 36

(UA)


|OP|2 983101 |PQ|2 983083 |OQ|2 rArr |OP|2 983101 3

3

2

983083 12 983101

983101 13

983083 1983101 43

rArr OP983101 23

983101 2 33

(UC)


cos2 x983101 1 ndash 13

2

983101 89

rArr

x 1o Q cos x983101 2 2

3 tg x983101

13

2 23

983101 1

2 2

983101 24




52 a2 983101 1cos2 x


983101 sen2 xcos2 x

983101 tg2 x

b2 983101 1sen2 x


983101 cos2 xsen2 x

983101 cotg2 x983101 1tg2 x


53 sec

2

x983101

4rArr

sec x983101


3

o


rArr tg x983101 3


e x983101 π 983083 π3

983101 4π3


9

cos2 β ∙ 1625

rArr

rArr sen2 β 983083 cos2 β 983101 49

983083 1625


55 a)

cotg2 x

1983083 cotg2 x 983101

cos2 xsen2 x


983101

cos2 xsen2 x


983101

∙

cos2 xsen2 x

1sen2 x

983101 cos2 x

b)


983083 cos α 983101

∙ sen2 α983083 cos2 α

cos α

983101 1

cos α

983101 sec α


983083 cos xsen x

983101



983101 1sen x

middot 1cos x

983101


d) tg x983083 cos x1983083 sen x

983101 sen xcos x

983083 cos x1983083 sen x

983101


983101 sen x983083 sen2 x983083 cos2 x



983101 1cos x

983101 sec x

56

57

58

59


tg x negativo

cotg x negativo

cotg x positivo2

π +


sinal de y

( ) ( )

( ) ( ) ( )

minus sdot +



tg x positivo

cotg x positivo

cotg x negativo2

π +


sinal de y ( ) ( )

( ) ( ) ( )


+ sdot +


sen x negativo

cos x negativo

sec x negativo

tg x positivo


sinal de y ( ) ( ) ( )

( ) ( ) ( )


+ sdot +

x 0 x2


tg x 5=


cos x sen x2

π minus =

( )tg x tg xπ + =




senx senxy

minus sdot=

tg xsdot


senxtg x 5


0 x2


sen xcosx2

tg x2 sen x

cos x2





2 2sen x cos x 1+ =

221

cos x 110

+ =

2 1 99cos x 1


60 a)

b)

61 sen

cotg1 cos

αα + =

+ α

cos sen

sen 1 cos

α α+ =

α + α

( )

( )

2cos 1 cos sen

sen 1 cos

α + α + α= =

α + α

2 2cos cos sen

sen sen cos

α + α + α= =

α + α sdot α

( )

1 cos 1cossec

sen 1 cos sen

+ α= = = α

α + α α




2

2



63

64





2 2sen x cos x 1= + =

66

1cossec x ndash 1

+ 1cossec x+ 1

=67

3 11cosx

10rArr = rArr

3 11

10tg x 3 1112

10

π minus = =


coscos

αsdot α+ α =

α

2 2sen cos

cos

α sdot α= =

α

1sec

cos= α

α

2tg

ecs1

1

α minus =α minus

2tg sec 1

sec 1


2

2

sen 11

coscos

11

cos

αminus +

αα= =

minus

α

2

2

sen cos cos

cos

1 cos

cos

α minus α + α

α=

minus α

α

2

1 cos

1cossec

1 cos cos

minus α

α= = = α

minus α α


2cos x= + + + =

2 22 cos x 2 sen x

5senxcosx

+= + =

( )2 22 sen x cos x

5sen x cosx

+

= + = 5senx 2sec x+

sen x cos x

sen x cos x

+=

minus

senx cos x

senx

senx cos x

senx

+

=minus

1 cotg x

1 cotg x

+

minus




senxsdot

senx

cosx+ cosxsdot 1+ =


( )sen x 1 c os x cos x 1= + + + =

( ) ( )1 cos x 1 sen x= + sdot +

=

sena senb cosa cosb

cosa cosb sena senb

+ ++ =

+ minus

( )( )

( )( )

( )( )

( )( )

sena senb sena senb




+ minus +=

+ minus

+ minus +=

+ minus

( )( )




+ minus






=

=2 cossec x

cotg2 x =

2sen x

middot sen2 xcos2 x

=

= 2 middot 1cos x

middot sen xcos x


tg x+ 1tg x

middot 1cos x


ndash sen x =

= tg2 x+ 1



sen x =

= sec2 xtg x

middot sen 2 xcos x

middot cos 2 xsen x

= sec2 xtg x


= 1cos2 x

middot cos xsen x


68

+ 1 + ndash 1

69


(m ndash 2) middot3


m = ndash1

m



cos2 x + sen2 x = 1 2





71



ndash 45

= ndash 75

72 a)

b) 12 5 23


π π π = sdot +

12 14 2 22

7 7 7 7


12cotg

7

πrArr negativo

5 5 180ordm82ordm

11 11


5sen

11

πrArr

positivo

23 242

12 12 12 12


23cos

12

πrArr positivo

sinal de y



1 Se sen x= 1 x= π

2 cos x= cos π

2 = 0 assim

sen x+ cos x= 1+ 0= 1


sen x+ cos x= 0+ 1= 1






4


2

2 2tgx cosxy1 cos x

sdot= rArr

minus 2tgx cos xy

sen x

sdot= rArr

senx

cosxyrArr =

cosxsdot

2sen xrArr

1y cossec x

senx= =


ox 3 Qisin2

2 24 576 625cossec x 1 1

7 49 49

= + = + = rArr

25cossec x7


25

y cossec x 7

= = minus

9 7x sec tg cotg

8 6 7

π π π = sdot +

9 8

8 8 8 8


9sec

8

πrArr

negativo

7 6

6 6 6 6


7tg

6

πrArr

positivo

7


6

πrArr




ndash5

+

+ 4

5

2

= 1

3 4 3 2





Se sen x= 12

podemos ter x= π6

ou

x= π ndash π6

= 5π6


m+ 13

2 + m ndash 1

32 = 1 rArr 2m2 + 2

9 = 1 rArr

rArr 2m2 = 79

rArr m2 = 718

rArr m= plusmn 7

3 2 middot

2

2 =

= plusmn 146

(cong 0623)



4 a) Sabemos que




= 1cos2 α

+ 1sen2 α

=



= 1cos2 α

middot 1sen2 α

Daiacute

CS= 1

cos2 α middot 1


cos α gt 0 vem

CS= 1cos α

middot 1sen α


b) α =

π

6 OS=

sec

π

6 =

sec 30deg=

1

cos 30deg =

= 23

middot 33

= 2 33



2

= 12 + PS2 rArr 43

ndash 1=

= PS2 rArr PS2 = 13

rArr PS= 33


3 + 1+ 3

3 = ( 3 + 1) UC





= sen xcos x

cos x∙ 0 rArr



cos x= cos π2

= 0 Logo



tg x

rArr tg2 x= 1 rArr

rArr tg x= 1 x= π4

ou x= 5π4

ou


ou x= 7π4


cotg x ndash 1 =

1cos x

middot 1sen x

ndash 1cos2 x

cos xsen x

ndash 1 =

=




middotsen x

cos x ndash sen x =

= 1cos2 x

= 1

14

2 = 16

7 a)

α

0 A

T


AT= tg α


rArr OT= sec α

b) anaacuteloga

8 a)

α β

x 800 ndash x

800

h


rArr x

= 5

rArr x= 5 h (1)


15 rArr h

800 ndash x = 1

15 rArr




rArr(1)

x= 200 m


1 h 1


sen α ∙ 0






= 4 cos2 α + 4 sen2 α = 4


2 middot sen α =


10 a) sen2 3π20

+ cos2 3π20

= 1 rArr a2 + cos2 3π20

= 1 rArr



cos 3π20

= cos 27deggt 0


+ 17π20


de 3π20

e 17π20


senos

3π

20

17π

20

Logo sen 17π20

= sen 3π20

= a

c) Como 3π20

+ 7π20

= 10π20

= π2


e 7π20


cos 7π20

= sen 3π20

= a


20

= 37π

20


20e 37π20


cossenos

3π

20

37π

20

Daiacute sen 37π20

= ndash sen 3π20

= ndasha



5 (1)

rArr a =(3)

2b

sec x= 5a

rArr 1cos x

= 5a

rArr cos x= a5

(2)


2

+ 2b5

2

= 1 rArr


rArrbgt 0

rArr b=

5 middot 3

61 =

15 61

61

a+ c

Em (3) a= 30 61

61


b ndash a= 15 61

61 ndash 30 61

61 = ndash 15 61

61Assim


61

rArr c ndash 15 61

61

= ndash 15 61

61

rArr c= 0

12 a) sen α = 15


rArr 15

ndash cos α2

+ cos2 α = 1 rArr


cos α ndash 2425

= 0 rArr



100

45

ndash35

Se cos α = 45


ndash 45

= ndash 35

e α tem




+ 35

= 45

e α tem



lt 0


complementares



1 rArr


14 1o membro=


ndash cos xsen x

2

= (1 ndash cos x)2

sen2 x =

= (1 ndash cos x)2

1 ndash cos2 x =

(1 ndash cos x)2


1 ndash cos x

1+ cos x =

= 2o membro

15 tg x ndash sen x

sen3 x =


sen3 x =

1cos x ndash 1

sen2 x =

=

1cos x ndash 1

1 ndash cos2 x =

1 ndash cos x

cos x =


=

=1

cos x middot

1

1+ cos x =

sec x

1+ cos x






Comoπ5

+ 3π10

= 2π + 3π10

= π2

temos

tgπ5

= 1

tg 3π10




rArr


= 60deg

Daiacute cos B =

102

AB rArr

32

= 5AB

rArr

rArr AB= AC= 10

3 =

10 33

Assim

2p= 10+ 2 middot10 3

3 = 10+

20 33

cm

b) sen x+ cos x= k




2

sen3 x+ cos3 x=



2 = k middot

2 ndash k 2 + 12

=

= k

2

(3 ndash k 2)

27

1

y

O D

α

1 x

BAE

C

a) AE= cotg α= 1

tg α = 1tg π3

= 1

3 = 33


DC= tg α = tgπ3

= 3



= 12

middot 1 ndash 33


= 12


aacuterea=

1

2 middot

4 3

3 ndash

2=

2 3

3 ndash

1




=

= 1


=



= 12 middot


=

= 12 middot


= 12 middot


= 1


lente a ()

Desafio





9998

9989

9899

8999


Testes



Resposta c

3

2 Resposta d

2sen x com x

5 2


2 2sen x cos x 1+ =

24cos x 1

25+ = rArr

2 4 21cos x 1

25 25= minus = rArr

21cos x

5rArr = minus


5

= = = minus

minus

Resposta d




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus



2

= = = e





IV Falsa pois1 1


= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c




Daiacute


cotg x= 43

rArr cos xsen x

= 43


9 cos2 x16


rArr sen x= 5

Resposta a

3

7 sen xcos x



rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2


1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8


2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13


Resposta c


x y sen x cos y2π



10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor


Resposta a



cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ




rArr


1o Q

sen θ = 12


Resposta b

12 2senx




sec x cos x= minus

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b




41 sec2 x983101 1983083 tg2 xrArr 73

2

983101 1983083 tg2 xrArr tg2 x983101 409 rArr


2 103


2 10 983101 15 middot ndash 1


1010

m

42 a) OQ983101 cossec α 983101 103

rArr sen α ∙ 310


2

983101 1 ndash 9100

983101 91100

rArr

α 1o Q cos α 983101

9110

sen α 983083 cos α 983101 310 983083

9110

9831013983083 91

10


983101

91100

9

100

983101 919


tg α 983101 125 rArr cotg α 983101 5

12


sen β 983101 810


983101 54


b) V



d) Como π2


cotg 7π8 lt 0

sec 7π8 lt 0


8 gt 0 (V )

51

B

H

h4 5

x 6 ndash x

6

A C

Temos


42 983101 h2 983083 x2 (1)




4

em (1) rArr 16983101 h2 983083 94

2

rArr h2 983101 16 ndash 8116 983101 175

16 rArr

rArr h983101 1754

cm

ABH cos α 983101 x

4 983101

94

4 983101 9

16 rArr sec α 983101 16

9

46 cos x983101 2


2 983101 m

4 rArr 2 m983101 28rArr m983101 14



48 OB983101 sec π6

983101 1

cos π6

983101 1

32

983101 2 33

OA983101 cossec π6

983101 1

sen π6

983101 112

983101 2


OA middot OB2

983101

2 33

middot 2

2 983101 2 33

(UA)

49 Q

O

P 30deg

P


|PQ|983101 |cotg 120deg|983101 1tg 120deg


983101

983101 minus 1

3 983101 1

3 983101

33


12


middot 1 middot 33

983101 36

(UA)


|OP|2 983101 |PQ|2 983083 |OQ|2 rArr |OP|2 983101 3

3

2

983083 12 983101

983101 13

983083 1983101 43

rArr OP983101 23

983101 2 33

(UC)


cos2 x983101 1 ndash 13

2

983101 89

rArr

x 1o Q cos x983101 2 2

3 tg x983101

13

2 23

983101 1

2 2

983101 24




52 a2 983101 1cos2 x


983101 sen2 xcos2 x

983101 tg2 x

b2 983101 1sen2 x


983101 cos2 xsen2 x

983101 cotg2 x983101 1tg2 x


53 sec

2

x983101

4rArr

sec x983101


3

o


rArr tg x983101 3


e x983101 π 983083 π3

983101 4π3


9

cos2 β ∙ 1625

rArr

rArr sen2 β 983083 cos2 β 983101 49

983083 1625


55 a)

cotg2 x

1983083 cotg2 x 983101

cos2 xsen2 x


983101

cos2 xsen2 x


983101

∙

cos2 xsen2 x

1sen2 x

983101 cos2 x

b)


983083 cos α 983101

∙ sen2 α983083 cos2 α

cos α

983101 1

cos α

983101 sec α


983083 cos xsen x

983101



983101 1sen x

middot 1cos x

983101


d) tg x983083 cos x1983083 sen x

983101 sen xcos x

983083 cos x1983083 sen x

983101


983101 sen x983083 sen2 x983083 cos2 x



983101 1cos x

983101 sec x

56

57

58

59


tg x negativo

cotg x negativo

cotg x positivo2

π +


sinal de y

( ) ( )

( ) ( ) ( )

minus sdot +



tg x positivo

cotg x positivo

cotg x negativo2

π +


sinal de y ( ) ( )

( ) ( ) ( )


+ sdot +


sen x negativo

cos x negativo

sec x negativo

tg x positivo


sinal de y ( ) ( ) ( )

( ) ( ) ( )


+ sdot +

x 0 x2


tg x 5=


cos x sen x2

π minus =

( )tg x tg xπ + =




senx senxy

minus sdot=

tg xsdot


senxtg x 5


0 x2


sen xcosx2

tg x2 sen x

cos x2





2 2sen x cos x 1+ =

221

cos x 110

+ =

2 1 99cos x 1


60 a)

b)

61 sen

cotg1 cos

αα + =

+ α

cos sen

sen 1 cos

α α+ =

α + α

( )

( )

2cos 1 cos sen

sen 1 cos

α + α + α= =

α + α

2 2cos cos sen

sen sen cos

α + α + α= =

α + α sdot α

( )

1 cos 1cossec

sen 1 cos sen

+ α= = = α

α + α α




2

2



63

64





2 2sen x cos x 1= + =

66

1cossec x ndash 1

+ 1cossec x+ 1

=67

3 11cosx

10rArr = rArr

3 11

10tg x 3 1112

10

π minus = =


coscos

αsdot α+ α =

α

2 2sen cos

cos

α sdot α= =

α

1sec

cos= α

α

2tg

ecs1

1

α minus =α minus

2tg sec 1

sec 1


2

2

sen 11

coscos

11

cos

αminus +

αα= =

minus

α

2

2

sen cos cos

cos

1 cos

cos

α minus α + α

α=

minus α

α

2

1 cos

1cossec

1 cos cos

minus α

α= = = α

minus α α


2cos x= + + + =

2 22 cos x 2 sen x

5senxcosx

+= + =

( )2 22 sen x cos x

5sen x cosx

+

= + = 5senx 2sec x+

sen x cos x

sen x cos x

+=

minus

senx cos x

senx

senx cos x

senx

+

=minus

1 cotg x

1 cotg x

+

minus




senxsdot

senx

cosx+ cosxsdot 1+ =


( )sen x 1 c os x cos x 1= + + + =

( ) ( )1 cos x 1 sen x= + sdot +

=

sena senb cosa cosb

cosa cosb sena senb

+ ++ =

+ minus

( )( )

( )( )

( )( )

( )( )

sena senb sena senb




+ minus +=

+ minus

+ minus +=

+ minus

( )( )




+ minus






=

=2 cossec x

cotg2 x =

2sen x

middot sen2 xcos2 x

=

= 2 middot 1cos x

middot sen xcos x


tg x+ 1tg x

middot 1cos x


ndash sen x =

= tg2 x+ 1



sen x =

= sec2 xtg x

middot sen 2 xcos x

middot cos 2 xsen x

= sec2 xtg x


= 1cos2 x

middot cos xsen x


68

+ 1 + ndash 1

69


(m ndash 2) middot3


m = ndash1

m



cos2 x + sen2 x = 1 2





71



ndash 45

= ndash 75

72 a)

b) 12 5 23


π π π = sdot +

12 14 2 22

7 7 7 7


12cotg

7

πrArr negativo

5 5 180ordm82ordm

11 11


5sen

11

πrArr

positivo

23 242

12 12 12 12


23cos

12

πrArr positivo

sinal de y



1 Se sen x= 1 x= π

2 cos x= cos π

2 = 0 assim

sen x+ cos x= 1+ 0= 1


sen x+ cos x= 0+ 1= 1






4


2

2 2tgx cosxy1 cos x

sdot= rArr

minus 2tgx cos xy

sen x

sdot= rArr

senx

cosxyrArr =

cosxsdot

2sen xrArr

1y cossec x

senx= =


ox 3 Qisin2

2 24 576 625cossec x 1 1

7 49 49

= + = + = rArr

25cossec x7


25

y cossec x 7

= = minus

9 7x sec tg cotg

8 6 7

π π π = sdot +

9 8

8 8 8 8


9sec

8

πrArr

negativo

7 6

6 6 6 6


7tg

6

πrArr

positivo

7


6

πrArr




ndash5

+

+ 4

5

2

= 1

3 4 3 2





Se sen x= 12

podemos ter x= π6

ou

x= π ndash π6

= 5π6


m+ 13

2 + m ndash 1

32 = 1 rArr 2m2 + 2

9 = 1 rArr

rArr 2m2 = 79

rArr m2 = 718

rArr m= plusmn 7

3 2 middot

2

2 =

= plusmn 146

(cong 0623)



4 a) Sabemos que




= 1cos2 α

+ 1sen2 α

=



= 1cos2 α

middot 1sen2 α

Daiacute

CS= 1

cos2 α middot 1


cos α gt 0 vem

CS= 1cos α

middot 1sen α


b) α =

π

6 OS=

sec

π

6 =

sec 30deg=

1

cos 30deg =

= 23

middot 33

= 2 33



2

= 12 + PS2 rArr 43

ndash 1=

= PS2 rArr PS2 = 13

rArr PS= 33


3 + 1+ 3

3 = ( 3 + 1) UC





= sen xcos x

cos x∙ 0 rArr



cos x= cos π2

= 0 Logo



tg x

rArr tg2 x= 1 rArr

rArr tg x= 1 x= π4

ou x= 5π4

ou


ou x= 7π4


cotg x ndash 1 =

1cos x

middot 1sen x

ndash 1cos2 x

cos xsen x

ndash 1 =

=




middotsen x

cos x ndash sen x =

= 1cos2 x

= 1

14

2 = 16

7 a)

α

0 A

T


AT= tg α


rArr OT= sec α

b) anaacuteloga

8 a)

α β

x 800 ndash x

800

h


rArr x

= 5

rArr x= 5 h (1)


15 rArr h

800 ndash x = 1

15 rArr




rArr(1)

x= 200 m


1 h 1


sen α ∙ 0






= 4 cos2 α + 4 sen2 α = 4


2 middot sen α =


10 a) sen2 3π20

+ cos2 3π20

= 1 rArr a2 + cos2 3π20

= 1 rArr



cos 3π20

= cos 27deggt 0


+ 17π20


de 3π20

e 17π20


senos

3π

20

17π

20

Logo sen 17π20

= sen 3π20

= a

c) Como 3π20

+ 7π20

= 10π20

= π2


e 7π20


cos 7π20

= sen 3π20

= a


20

= 37π

20


20e 37π20


cossenos

3π

20

37π

20

Daiacute sen 37π20

= ndash sen 3π20

= ndasha



5 (1)

rArr a =(3)

2b

sec x= 5a

rArr 1cos x

= 5a

rArr cos x= a5

(2)


2

+ 2b5

2

= 1 rArr


rArrbgt 0

rArr b=

5 middot 3

61 =

15 61

61

a+ c

Em (3) a= 30 61

61


b ndash a= 15 61

61 ndash 30 61

61 = ndash 15 61

61Assim


61

rArr c ndash 15 61

61

= ndash 15 61

61

rArr c= 0

12 a) sen α = 15


rArr 15

ndash cos α2

+ cos2 α = 1 rArr


cos α ndash 2425

= 0 rArr



100

45

ndash35

Se cos α = 45


ndash 45

= ndash 35

e α tem




+ 35

= 45

e α tem



lt 0


complementares



1 rArr


14 1o membro=


ndash cos xsen x

2

= (1 ndash cos x)2

sen2 x =

= (1 ndash cos x)2

1 ndash cos2 x =

(1 ndash cos x)2


1 ndash cos x

1+ cos x =

= 2o membro

15 tg x ndash sen x

sen3 x =


sen3 x =

1cos x ndash 1

sen2 x =

=

1cos x ndash 1

1 ndash cos2 x =

1 ndash cos x

cos x =


=

=1

cos x middot

1

1+ cos x =

sec x

1+ cos x






Comoπ5

+ 3π10

= 2π + 3π10

= π2

temos

tgπ5

= 1

tg 3π10




rArr


= 60deg

Daiacute cos B =

102

AB rArr

32

= 5AB

rArr

rArr AB= AC= 10

3 =

10 33

Assim

2p= 10+ 2 middot10 3

3 = 10+

20 33

cm

b) sen x+ cos x= k




2

sen3 x+ cos3 x=



2 = k middot

2 ndash k 2 + 12

=

= k

2

(3 ndash k 2)

27

1

y

O D

α

1 x

BAE

C

a) AE= cotg α= 1

tg α = 1tg π3

= 1

3 = 33


DC= tg α = tgπ3

= 3



= 12

middot 1 ndash 33


= 12


aacuterea=

1

2 middot

4 3

3 ndash

2=

2 3

3 ndash

1




=

= 1


=



= 12 middot


=

= 12 middot


= 12 middot


= 1


lente a ()

Desafio





9998

9989

9899

8999


Testes



Resposta c

3

2 Resposta d

2sen x com x

5 2


2 2sen x cos x 1+ =

24cos x 1

25+ = rArr

2 4 21cos x 1

25 25= minus = rArr

21cos x

5rArr = minus


5

= = = minus

minus

Resposta d




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus



2

= = = e





IV Falsa pois1 1


= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c




Daiacute


cotg x= 43

rArr cos xsen x

= 43


9 cos2 x16


rArr sen x= 5

Resposta a

3

7 sen xcos x



rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2


1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8


2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13


Resposta c


x y sen x cos y2π



10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor


Resposta a



cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ




rArr


1o Q

sen θ = 12


Resposta b

12 2senx




sec x cos x= minus

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b




52 a2 983101 1cos2 x


983101 sen2 xcos2 x

983101 tg2 x

b2 983101 1sen2 x


983101 cos2 xsen2 x

983101 cotg2 x983101 1tg2 x


53 sec

2

x983101

4rArr

sec x983101


3

o


rArr tg x983101 3


e x983101 π 983083 π3

983101 4π3


9

cos2 β ∙ 1625

rArr

rArr sen2 β 983083 cos2 β 983101 49

983083 1625


55 a)

cotg2 x

1983083 cotg2 x 983101

cos2 xsen2 x


983101

cos2 xsen2 x


983101

∙

cos2 xsen2 x

1sen2 x

983101 cos2 x

b)


983083 cos α 983101

∙ sen2 α983083 cos2 α

cos α

983101 1

cos α

983101 sec α


983083 cos xsen x

983101



983101 1sen x

middot 1cos x

983101


d) tg x983083 cos x1983083 sen x

983101 sen xcos x

983083 cos x1983083 sen x

983101


983101 sen x983083 sen2 x983083 cos2 x



983101 1cos x

983101 sec x

56

57

58

59


tg x negativo

cotg x negativo

cotg x positivo2

π +


sinal de y

( ) ( )

( ) ( ) ( )

minus sdot +



tg x positivo

cotg x positivo

cotg x negativo2

π +


sinal de y ( ) ( )

( ) ( ) ( )


+ sdot +


sen x negativo

cos x negativo

sec x negativo

tg x positivo


sinal de y ( ) ( ) ( )

( ) ( ) ( )


+ sdot +

x 0 x2


tg x 5=


cos x sen x2

π minus =

( )tg x tg xπ + =




senx senxy

minus sdot=

tg xsdot


senxtg x 5


0 x2


sen xcosx2

tg x2 sen x

cos x2





2 2sen x cos x 1+ =

221

cos x 110

+ =

2 1 99cos x 1


60 a)

b)

61 sen

cotg1 cos

αα + =

+ α

cos sen

sen 1 cos

α α+ =

α + α

( )

( )

2cos 1 cos sen

sen 1 cos

α + α + α= =

α + α

2 2cos cos sen

sen sen cos

α + α + α= =

α + α sdot α

( )

1 cos 1cossec

sen 1 cos sen

+ α= = = α

α + α α




2

2



63

64





2 2sen x cos x 1= + =

66

1cossec x ndash 1

+ 1cossec x+ 1

=67

3 11cosx

10rArr = rArr

3 11

10tg x 3 1112

10

π minus = =


coscos

αsdot α+ α =

α

2 2sen cos

cos

α sdot α= =

α

1sec

cos= α

α

2tg

ecs1

1

α minus =α minus

2tg sec 1

sec 1


2

2

sen 11

coscos

11

cos

αminus +

αα= =

minus

α

2

2

sen cos cos

cos

1 cos

cos

α minus α + α

α=

minus α

α

2

1 cos

1cossec

1 cos cos

minus α

α= = = α

minus α α


2cos x= + + + =

2 22 cos x 2 sen x

5senxcosx

+= + =

( )2 22 sen x cos x

5sen x cosx

+

= + = 5senx 2sec x+

sen x cos x

sen x cos x

+=

minus

senx cos x

senx

senx cos x

senx

+

=minus

1 cotg x

1 cotg x

+

minus




senxsdot

senx

cosx+ cosxsdot 1+ =


( )sen x 1 c os x cos x 1= + + + =

( ) ( )1 cos x 1 sen x= + sdot +

=

sena senb cosa cosb

cosa cosb sena senb

+ ++ =

+ minus

( )( )

( )( )

( )( )

( )( )

sena senb sena senb




+ minus +=

+ minus

+ minus +=

+ minus

( )( )




+ minus






=

=2 cossec x

cotg2 x =

2sen x

middot sen2 xcos2 x

=

= 2 middot 1cos x

middot sen xcos x


tg x+ 1tg x

middot 1cos x


ndash sen x =

= tg2 x+ 1



sen x =

= sec2 xtg x

middot sen 2 xcos x

middot cos 2 xsen x

= sec2 xtg x


= 1cos2 x

middot cos xsen x


68

+ 1 + ndash 1

69


(m ndash 2) middot3


m = ndash1

m



cos2 x + sen2 x = 1 2





71



ndash 45

= ndash 75

72 a)

b) 12 5 23


π π π = sdot +

12 14 2 22

7 7 7 7


12cotg

7

πrArr negativo

5 5 180ordm82ordm

11 11


5sen

11

πrArr

positivo

23 242

12 12 12 12


23cos

12

πrArr positivo

sinal de y



1 Se sen x= 1 x= π

2 cos x= cos π

2 = 0 assim

sen x+ cos x= 1+ 0= 1


sen x+ cos x= 0+ 1= 1






4


2

2 2tgx cosxy1 cos x

sdot= rArr

minus 2tgx cos xy

sen x

sdot= rArr

senx

cosxyrArr =

cosxsdot

2sen xrArr

1y cossec x

senx= =


ox 3 Qisin2

2 24 576 625cossec x 1 1

7 49 49

= + = + = rArr

25cossec x7


25

y cossec x 7

= = minus

9 7x sec tg cotg

8 6 7

π π π = sdot +

9 8

8 8 8 8


9sec

8

πrArr

negativo

7 6

6 6 6 6


7tg

6

πrArr

positivo

7


6

πrArr




ndash5

+

+ 4

5

2

= 1

3 4 3 2





Se sen x= 12

podemos ter x= π6

ou

x= π ndash π6

= 5π6


m+ 13

2 + m ndash 1

32 = 1 rArr 2m2 + 2

9 = 1 rArr

rArr 2m2 = 79

rArr m2 = 718

rArr m= plusmn 7

3 2 middot

2

2 =

= plusmn 146

(cong 0623)



4 a) Sabemos que




= 1cos2 α

+ 1sen2 α

=



= 1cos2 α

middot 1sen2 α

Daiacute

CS= 1

cos2 α middot 1


cos α gt 0 vem

CS= 1cos α

middot 1sen α


b) α =

π

6 OS=

sec

π

6 =

sec 30deg=

1

cos 30deg =

= 23

middot 33

= 2 33



2

= 12 + PS2 rArr 43

ndash 1=

= PS2 rArr PS2 = 13

rArr PS= 33


3 + 1+ 3

3 = ( 3 + 1) UC





= sen xcos x

cos x∙ 0 rArr



cos x= cos π2

= 0 Logo



tg x

rArr tg2 x= 1 rArr

rArr tg x= 1 x= π4

ou x= 5π4

ou


ou x= 7π4


cotg x ndash 1 =

1cos x

middot 1sen x

ndash 1cos2 x

cos xsen x

ndash 1 =

=




middotsen x

cos x ndash sen x =

= 1cos2 x

= 1

14

2 = 16

7 a)

α

0 A

T


AT= tg α


rArr OT= sec α

b) anaacuteloga

8 a)

α β

x 800 ndash x

800

h


rArr x

= 5

rArr x= 5 h (1)


15 rArr h

800 ndash x = 1

15 rArr




rArr(1)

x= 200 m


1 h 1


sen α ∙ 0






= 4 cos2 α + 4 sen2 α = 4


2 middot sen α =


10 a) sen2 3π20

+ cos2 3π20

= 1 rArr a2 + cos2 3π20

= 1 rArr



cos 3π20

= cos 27deggt 0


+ 17π20


de 3π20

e 17π20


senos

3π

20

17π

20

Logo sen 17π20

= sen 3π20

= a

c) Como 3π20

+ 7π20

= 10π20

= π2


e 7π20


cos 7π20

= sen 3π20

= a


20

= 37π

20


20e 37π20


cossenos

3π

20

37π

20

Daiacute sen 37π20

= ndash sen 3π20

= ndasha



5 (1)

rArr a =(3)

2b

sec x= 5a

rArr 1cos x

= 5a

rArr cos x= a5

(2)


2

+ 2b5

2

= 1 rArr


rArrbgt 0

rArr b=

5 middot 3

61 =

15 61

61

a+ c

Em (3) a= 30 61

61


b ndash a= 15 61

61 ndash 30 61

61 = ndash 15 61

61Assim


61

rArr c ndash 15 61

61

= ndash 15 61

61

rArr c= 0

12 a) sen α = 15


rArr 15

ndash cos α2

+ cos2 α = 1 rArr


cos α ndash 2425

= 0 rArr



100

45

ndash35

Se cos α = 45


ndash 45

= ndash 35

e α tem




+ 35

= 45

e α tem



lt 0


complementares



1 rArr


14 1o membro=


ndash cos xsen x

2

= (1 ndash cos x)2

sen2 x =

= (1 ndash cos x)2

1 ndash cos2 x =

(1 ndash cos x)2


1 ndash cos x

1+ cos x =

= 2o membro

15 tg x ndash sen x

sen3 x =


sen3 x =

1cos x ndash 1

sen2 x =

=

1cos x ndash 1

1 ndash cos2 x =

1 ndash cos x

cos x =


=

=1

cos x middot

1

1+ cos x =

sec x

1+ cos x






Comoπ5

+ 3π10

= 2π + 3π10

= π2

temos

tgπ5

= 1

tg 3π10




rArr


= 60deg

Daiacute cos B =

102

AB rArr

32

= 5AB

rArr

rArr AB= AC= 10

3 =

10 33

Assim

2p= 10+ 2 middot10 3

3 = 10+

20 33

cm

b) sen x+ cos x= k




2

sen3 x+ cos3 x=



2 = k middot

2 ndash k 2 + 12

=

= k

2

(3 ndash k 2)

27

1

y

O D

α

1 x

BAE

C

a) AE= cotg α= 1

tg α = 1tg π3

= 1

3 = 33


DC= tg α = tgπ3

= 3



= 12

middot 1 ndash 33


= 12


aacuterea=

1

2 middot

4 3

3 ndash

2=

2 3

3 ndash

1




=

= 1


=



= 12 middot


=

= 12 middot


= 12 middot


= 1


lente a ()

Desafio





9998

9989

9899

8999


Testes



Resposta c

3

2 Resposta d

2sen x com x

5 2


2 2sen x cos x 1+ =

24cos x 1

25+ = rArr

2 4 21cos x 1

25 25= minus = rArr

21cos x

5rArr = minus


5

= = = minus

minus

Resposta d




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus



2

= = = e





IV Falsa pois1 1


= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c




Daiacute


cotg x= 43

rArr cos xsen x

= 43


9 cos2 x16


rArr sen x= 5

Resposta a

3

7 sen xcos x



rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2


1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8


2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13


Resposta c


x y sen x cos y2π



10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor


Resposta a



cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ




rArr


1o Q

sen θ = 12


Resposta b

12 2senx




sec x cos x= minus

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b




2 2sen x cos x 1+ =

221

cos x 110

+ =

2 1 99cos x 1


60 a)

b)

61 sen

cotg1 cos

αα + =

+ α

cos sen

sen 1 cos

α α+ =

α + α

( )

( )

2cos 1 cos sen

sen 1 cos

α + α + α= =

α + α

2 2cos cos sen

sen sen cos

α + α + α= =

α + α sdot α

( )

1 cos 1cossec

sen 1 cos sen

+ α= = = α

α + α α




2

2



63

64





2 2sen x cos x 1= + =

66

1cossec x ndash 1

+ 1cossec x+ 1

=67

3 11cosx

10rArr = rArr

3 11

10tg x 3 1112

10

π minus = =


coscos

αsdot α+ α =

α

2 2sen cos

cos

α sdot α= =

α

1sec

cos= α

α

2tg

ecs1

1

α minus =α minus

2tg sec 1

sec 1


2

2

sen 11

coscos

11

cos

αminus +

αα= =

minus

α

2

2

sen cos cos

cos

1 cos

cos

α minus α + α

α=

minus α

α

2

1 cos

1cossec

1 cos cos

minus α

α= = = α

minus α α


2cos x= + + + =

2 22 cos x 2 sen x

5senxcosx

+= + =

( )2 22 sen x cos x

5sen x cosx

+

= + = 5senx 2sec x+

sen x cos x

sen x cos x

+=

minus

senx cos x

senx

senx cos x

senx

+

=minus

1 cotg x

1 cotg x

+

minus




senxsdot

senx

cosx+ cosxsdot 1+ =


( )sen x 1 c os x cos x 1= + + + =

( ) ( )1 cos x 1 sen x= + sdot +

=

sena senb cosa cosb

cosa cosb sena senb

+ ++ =

+ minus

( )( )

( )( )

( )( )

( )( )

sena senb sena senb




+ minus +=

+ minus

+ minus +=

+ minus

( )( )




+ minus






=

=2 cossec x

cotg2 x =

2sen x

middot sen2 xcos2 x

=

= 2 middot 1cos x

middot sen xcos x


tg x+ 1tg x

middot 1cos x


ndash sen x =

= tg2 x+ 1



sen x =

= sec2 xtg x

middot sen 2 xcos x

middot cos 2 xsen x

= sec2 xtg x


= 1cos2 x

middot cos xsen x


68

+ 1 + ndash 1

69


(m ndash 2) middot3


m = ndash1

m



cos2 x + sen2 x = 1 2





71



ndash 45

= ndash 75

72 a)

b) 12 5 23


π π π = sdot +

12 14 2 22

7 7 7 7


12cotg

7

πrArr negativo

5 5 180ordm82ordm

11 11


5sen

11

πrArr

positivo

23 242

12 12 12 12


23cos

12

πrArr positivo

sinal de y



1 Se sen x= 1 x= π

2 cos x= cos π

2 = 0 assim

sen x+ cos x= 1+ 0= 1


sen x+ cos x= 0+ 1= 1






4


2

2 2tgx cosxy1 cos x

sdot= rArr

minus 2tgx cos xy

sen x

sdot= rArr

senx

cosxyrArr =

cosxsdot

2sen xrArr

1y cossec x

senx= =


ox 3 Qisin2

2 24 576 625cossec x 1 1

7 49 49

= + = + = rArr

25cossec x7


25

y cossec x 7

= = minus

9 7x sec tg cotg

8 6 7

π π π = sdot +

9 8

8 8 8 8


9sec

8

πrArr

negativo

7 6

6 6 6 6


7tg

6

πrArr

positivo

7


6

πrArr




ndash5

+

+ 4

5

2

= 1

3 4 3 2





Se sen x= 12

podemos ter x= π6

ou

x= π ndash π6

= 5π6


m+ 13

2 + m ndash 1

32 = 1 rArr 2m2 + 2

9 = 1 rArr

rArr 2m2 = 79

rArr m2 = 718

rArr m= plusmn 7

3 2 middot

2

2 =

= plusmn 146

(cong 0623)



4 a) Sabemos que




= 1cos2 α

+ 1sen2 α

=



= 1cos2 α

middot 1sen2 α

Daiacute

CS= 1

cos2 α middot 1


cos α gt 0 vem

CS= 1cos α

middot 1sen α


b) α =

π

6 OS=

sec

π

6 =

sec 30deg=

1

cos 30deg =

= 23

middot 33

= 2 33



2

= 12 + PS2 rArr 43

ndash 1=

= PS2 rArr PS2 = 13

rArr PS= 33


3 + 1+ 3

3 = ( 3 + 1) UC





= sen xcos x

cos x∙ 0 rArr



cos x= cos π2

= 0 Logo



tg x

rArr tg2 x= 1 rArr

rArr tg x= 1 x= π4

ou x= 5π4

ou


ou x= 7π4


cotg x ndash 1 =

1cos x

middot 1sen x

ndash 1cos2 x

cos xsen x

ndash 1 =

=




middotsen x

cos x ndash sen x =

= 1cos2 x

= 1

14

2 = 16

7 a)

α

0 A

T


AT= tg α


rArr OT= sec α

b) anaacuteloga

8 a)

α β

x 800 ndash x

800

h


rArr x

= 5

rArr x= 5 h (1)


15 rArr h

800 ndash x = 1

15 rArr




rArr(1)

x= 200 m


1 h 1


sen α ∙ 0






= 4 cos2 α + 4 sen2 α = 4


2 middot sen α =


10 a) sen2 3π20

+ cos2 3π20

= 1 rArr a2 + cos2 3π20

= 1 rArr



cos 3π20

= cos 27deggt 0


+ 17π20


de 3π20

e 17π20


senos

3π

20

17π

20

Logo sen 17π20

= sen 3π20

= a

c) Como 3π20

+ 7π20

= 10π20

= π2


e 7π20


cos 7π20

= sen 3π20

= a


20

= 37π

20


20e 37π20


cossenos

3π

20

37π

20

Daiacute sen 37π20

= ndash sen 3π20

= ndasha



5 (1)

rArr a =(3)

2b

sec x= 5a

rArr 1cos x

= 5a

rArr cos x= a5

(2)


2

+ 2b5

2

= 1 rArr


rArrbgt 0

rArr b=

5 middot 3

61 =

15 61

61

a+ c

Em (3) a= 30 61

61


b ndash a= 15 61

61 ndash 30 61

61 = ndash 15 61

61Assim


61

rArr c ndash 15 61

61

= ndash 15 61

61

rArr c= 0

12 a) sen α = 15


rArr 15

ndash cos α2

+ cos2 α = 1 rArr


cos α ndash 2425

= 0 rArr



100

45

ndash35

Se cos α = 45


ndash 45

= ndash 35

e α tem




+ 35

= 45

e α tem



lt 0


complementares



1 rArr


14 1o membro=


ndash cos xsen x

2

= (1 ndash cos x)2

sen2 x =

= (1 ndash cos x)2

1 ndash cos2 x =

(1 ndash cos x)2


1 ndash cos x

1+ cos x =

= 2o membro

15 tg x ndash sen x

sen3 x =


sen3 x =

1cos x ndash 1

sen2 x =

=

1cos x ndash 1

1 ndash cos2 x =

1 ndash cos x

cos x =


=

=1

cos x middot

1

1+ cos x =

sec x

1+ cos x






Comoπ5

+ 3π10

= 2π + 3π10

= π2

temos

tgπ5

= 1

tg 3π10




rArr


= 60deg

Daiacute cos B =

102

AB rArr

32

= 5AB

rArr

rArr AB= AC= 10

3 =

10 33

Assim

2p= 10+ 2 middot10 3

3 = 10+

20 33

cm

b) sen x+ cos x= k




2

sen3 x+ cos3 x=



2 = k middot

2 ndash k 2 + 12

=

= k

2

(3 ndash k 2)

27

1

y

O D

α

1 x

BAE

C

a) AE= cotg α= 1

tg α = 1tg π3

= 1

3 = 33


DC= tg α = tgπ3

= 3



= 12

middot 1 ndash 33


= 12


aacuterea=

1

2 middot

4 3

3 ndash

2=

2 3

3 ndash

1




=

= 1


=



= 12 middot


=

= 12 middot


= 12 middot


= 1


lente a ()

Desafio





9998

9989

9899

8999


Testes



Resposta c

3

2 Resposta d

2sen x com x

5 2


2 2sen x cos x 1+ =

24cos x 1

25+ = rArr

2 4 21cos x 1

25 25= minus = rArr

21cos x

5rArr = minus


5

= = = minus

minus

Resposta d




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus



2

= = = e





IV Falsa pois1 1


= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c




Daiacute


cotg x= 43

rArr cos xsen x

= 43


9 cos2 x16


rArr sen x= 5

Resposta a

3

7 sen xcos x



rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2


1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8


2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13


Resposta c


x y sen x cos y2π



10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor


Resposta a



cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ




rArr


1o Q

sen θ = 12


Resposta b

12 2senx




sec x cos x= minus

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b






=

=2 cossec x

cotg2 x =

2sen x

middot sen2 xcos2 x

=

= 2 middot 1cos x

middot sen xcos x


tg x+ 1tg x

middot 1cos x


ndash sen x =

= tg2 x+ 1



sen x =

= sec2 xtg x

middot sen 2 xcos x

middot cos 2 xsen x

= sec2 xtg x


= 1cos2 x

middot cos xsen x


68

+ 1 + ndash 1

69


(m ndash 2) middot3


m = ndash1

m



cos2 x + sen2 x = 1 2





71



ndash 45

= ndash 75

72 a)

b) 12 5 23


π π π = sdot +

12 14 2 22

7 7 7 7


12cotg

7

πrArr negativo

5 5 180ordm82ordm

11 11


5sen

11

πrArr

positivo

23 242

12 12 12 12


23cos

12

πrArr positivo

sinal de y



1 Se sen x= 1 x= π

2 cos x= cos π

2 = 0 assim

sen x+ cos x= 1+ 0= 1


sen x+ cos x= 0+ 1= 1






4


2

2 2tgx cosxy1 cos x

sdot= rArr

minus 2tgx cos xy

sen x

sdot= rArr

senx

cosxyrArr =

cosxsdot

2sen xrArr

1y cossec x

senx= =


ox 3 Qisin2

2 24 576 625cossec x 1 1

7 49 49

= + = + = rArr

25cossec x7


25

y cossec x 7

= = minus

9 7x sec tg cotg

8 6 7

π π π = sdot +

9 8

8 8 8 8


9sec

8

πrArr

negativo

7 6

6 6 6 6


7tg

6

πrArr

positivo

7


6

πrArr




ndash5

+

+ 4

5

2

= 1

3 4 3 2





Se sen x= 12

podemos ter x= π6

ou

x= π ndash π6

= 5π6


m+ 13

2 + m ndash 1

32 = 1 rArr 2m2 + 2

9 = 1 rArr

rArr 2m2 = 79

rArr m2 = 718

rArr m= plusmn 7

3 2 middot

2

2 =

= plusmn 146

(cong 0623)



4 a) Sabemos que




= 1cos2 α

+ 1sen2 α

=



= 1cos2 α

middot 1sen2 α

Daiacute

CS= 1

cos2 α middot 1


cos α gt 0 vem

CS= 1cos α

middot 1sen α


b) α =

π

6 OS=

sec

π

6 =

sec 30deg=

1

cos 30deg =

= 23

middot 33

= 2 33



2

= 12 + PS2 rArr 43

ndash 1=

= PS2 rArr PS2 = 13

rArr PS= 33


3 + 1+ 3

3 = ( 3 + 1) UC





= sen xcos x

cos x∙ 0 rArr



cos x= cos π2

= 0 Logo



tg x

rArr tg2 x= 1 rArr

rArr tg x= 1 x= π4

ou x= 5π4

ou


ou x= 7π4


cotg x ndash 1 =

1cos x

middot 1sen x

ndash 1cos2 x

cos xsen x

ndash 1 =

=




middotsen x

cos x ndash sen x =

= 1cos2 x

= 1

14

2 = 16

7 a)

α

0 A

T


AT= tg α


rArr OT= sec α

b) anaacuteloga

8 a)

α β

x 800 ndash x

800

h


rArr x

= 5

rArr x= 5 h (1)


15 rArr h

800 ndash x = 1

15 rArr




rArr(1)

x= 200 m


1 h 1


sen α ∙ 0






= 4 cos2 α + 4 sen2 α = 4


2 middot sen α =


10 a) sen2 3π20

+ cos2 3π20

= 1 rArr a2 + cos2 3π20

= 1 rArr



cos 3π20

= cos 27deggt 0


+ 17π20


de 3π20

e 17π20


senos

3π

20

17π

20

Logo sen 17π20

= sen 3π20

= a

c) Como 3π20

+ 7π20

= 10π20

= π2


e 7π20


cos 7π20

= sen 3π20

= a


20

= 37π

20


20e 37π20


cossenos

3π

20

37π

20

Daiacute sen 37π20

= ndash sen 3π20

= ndasha



5 (1)

rArr a =(3)

2b

sec x= 5a

rArr 1cos x

= 5a

rArr cos x= a5

(2)


2

+ 2b5

2

= 1 rArr


rArrbgt 0

rArr b=

5 middot 3

61 =

15 61

61

a+ c

Em (3) a= 30 61

61


b ndash a= 15 61

61 ndash 30 61

61 = ndash 15 61

61Assim


61

rArr c ndash 15 61

61

= ndash 15 61

61

rArr c= 0

12 a) sen α = 15


rArr 15

ndash cos α2

+ cos2 α = 1 rArr


cos α ndash 2425

= 0 rArr



100

45

ndash35

Se cos α = 45


ndash 45

= ndash 35

e α tem




+ 35

= 45

e α tem



lt 0


complementares



1 rArr


14 1o membro=


ndash cos xsen x

2

= (1 ndash cos x)2

sen2 x =

= (1 ndash cos x)2

1 ndash cos2 x =

(1 ndash cos x)2


1 ndash cos x

1+ cos x =

= 2o membro

15 tg x ndash sen x

sen3 x =


sen3 x =

1cos x ndash 1

sen2 x =

=

1cos x ndash 1

1 ndash cos2 x =

1 ndash cos x

cos x =


=

=1

cos x middot

1

1+ cos x =

sec x

1+ cos x






Comoπ5

+ 3π10

= 2π + 3π10

= π2

temos

tgπ5

= 1

tg 3π10




rArr


= 60deg

Daiacute cos B =

102

AB rArr

32

= 5AB

rArr

rArr AB= AC= 10

3 =

10 33

Assim

2p= 10+ 2 middot10 3

3 = 10+

20 33

cm

b) sen x+ cos x= k




2

sen3 x+ cos3 x=



2 = k middot

2 ndash k 2 + 12

=

= k

2

(3 ndash k 2)

27

1

y

O D

α

1 x

BAE

C

a) AE= cotg α= 1

tg α = 1tg π3

= 1

3 = 33


DC= tg α = tgπ3

= 3



= 12

middot 1 ndash 33


= 12


aacuterea=

1

2 middot

4 3

3 ndash

2=

2 3

3 ndash

1




=

= 1


=



= 12 middot


=

= 12 middot


= 12 middot


= 1


lente a ()

Desafio





9998

9989

9899

8999


Testes



Resposta c

3

2 Resposta d

2sen x com x

5 2


2 2sen x cos x 1+ =

24cos x 1

25+ = rArr

2 4 21cos x 1

25 25= minus = rArr

21cos x

5rArr = minus


5

= = = minus

minus

Resposta d




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus



2

= = = e





IV Falsa pois1 1


= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c




Daiacute


cotg x= 43

rArr cos xsen x

= 43


9 cos2 x16


rArr sen x= 5

Resposta a

3

7 sen xcos x



rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2


1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8


2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13


Resposta c


x y sen x cos y2π



10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor


Resposta a



cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ




rArr


1o Q

sen θ = 12


Resposta b

12 2senx




sec x cos x= minus

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b





Se sen x= 12

podemos ter x= π6

ou

x= π ndash π6

= 5π6


m+ 13

2 + m ndash 1

32 = 1 rArr 2m2 + 2

9 = 1 rArr

rArr 2m2 = 79

rArr m2 = 718

rArr m= plusmn 7

3 2 middot

2

2 =

= plusmn 146

(cong 0623)



4 a) Sabemos que




= 1cos2 α

+ 1sen2 α

=



= 1cos2 α

middot 1sen2 α

Daiacute

CS= 1

cos2 α middot 1


cos α gt 0 vem

CS= 1cos α

middot 1sen α


b) α =

π

6 OS=

sec

π

6 =

sec 30deg=

1

cos 30deg =

= 23

middot 33

= 2 33



2

= 12 + PS2 rArr 43

ndash 1=

= PS2 rArr PS2 = 13

rArr PS= 33


3 + 1+ 3

3 = ( 3 + 1) UC





= sen xcos x

cos x∙ 0 rArr



cos x= cos π2

= 0 Logo



tg x

rArr tg2 x= 1 rArr

rArr tg x= 1 x= π4

ou x= 5π4

ou


ou x= 7π4


cotg x ndash 1 =

1cos x

middot 1sen x

ndash 1cos2 x

cos xsen x

ndash 1 =

=




middotsen x

cos x ndash sen x =

= 1cos2 x

= 1

14

2 = 16

7 a)

α

0 A

T


AT= tg α


rArr OT= sec α

b) anaacuteloga

8 a)

α β

x 800 ndash x

800

h


rArr x

= 5

rArr x= 5 h (1)


15 rArr h

800 ndash x = 1

15 rArr




rArr(1)

x= 200 m


1 h 1


sen α ∙ 0






= 4 cos2 α + 4 sen2 α = 4


2 middot sen α =


10 a) sen2 3π20

+ cos2 3π20

= 1 rArr a2 + cos2 3π20

= 1 rArr



cos 3π20

= cos 27deggt 0


+ 17π20


de 3π20

e 17π20


senos

3π

20

17π

20

Logo sen 17π20

= sen 3π20

= a

c) Como 3π20

+ 7π20

= 10π20

= π2


e 7π20


cos 7π20

= sen 3π20

= a


20

= 37π

20


20e 37π20


cossenos

3π

20

37π

20

Daiacute sen 37π20

= ndash sen 3π20

= ndasha



5 (1)

rArr a =(3)

2b

sec x= 5a

rArr 1cos x

= 5a

rArr cos x= a5

(2)


2

+ 2b5

2

= 1 rArr


rArrbgt 0

rArr b=

5 middot 3

61 =

15 61

61

a+ c

Em (3) a= 30 61

61


b ndash a= 15 61

61 ndash 30 61

61 = ndash 15 61

61Assim


61

rArr c ndash 15 61

61

= ndash 15 61

61

rArr c= 0

12 a) sen α = 15


rArr 15

ndash cos α2

+ cos2 α = 1 rArr


cos α ndash 2425

= 0 rArr



100

45

ndash35

Se cos α = 45


ndash 45

= ndash 35

e α tem




+ 35

= 45

e α tem



lt 0


complementares



1 rArr


14 1o membro=


ndash cos xsen x

2

= (1 ndash cos x)2

sen2 x =

= (1 ndash cos x)2

1 ndash cos2 x =

(1 ndash cos x)2


1 ndash cos x

1+ cos x =

= 2o membro

15 tg x ndash sen x

sen3 x =


sen3 x =

1cos x ndash 1

sen2 x =

=

1cos x ndash 1

1 ndash cos2 x =

1 ndash cos x

cos x =


=

=1

cos x middot

1

1+ cos x =

sec x

1+ cos x






Comoπ5

+ 3π10

= 2π + 3π10

= π2

temos

tgπ5

= 1

tg 3π10




rArr


= 60deg

Daiacute cos B =

102

AB rArr

32

= 5AB

rArr

rArr AB= AC= 10

3 =

10 33

Assim

2p= 10+ 2 middot10 3

3 = 10+

20 33

cm

b) sen x+ cos x= k




2

sen3 x+ cos3 x=



2 = k middot

2 ndash k 2 + 12

=

= k

2

(3 ndash k 2)

27

1

y

O D

α

1 x

BAE

C

a) AE= cotg α= 1

tg α = 1tg π3

= 1

3 = 33


DC= tg α = tgπ3

= 3



= 12

middot 1 ndash 33


= 12


aacuterea=

1

2 middot

4 3

3 ndash

2=

2 3

3 ndash

1




=

= 1


=



= 12 middot


=

= 12 middot


= 12 middot


= 1


lente a ()

Desafio





9998

9989

9899

8999


Testes



Resposta c

3

2 Resposta d

2sen x com x

5 2


2 2sen x cos x 1+ =

24cos x 1

25+ = rArr

2 4 21cos x 1

25 25= minus = rArr

21cos x

5rArr = minus


5

= = = minus

minus

Resposta d




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus



2

= = = e





IV Falsa pois1 1


= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c




Daiacute


cotg x= 43

rArr cos xsen x

= 43


9 cos2 x16


rArr sen x= 5

Resposta a

3

7 sen xcos x



rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2


1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8


2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13


Resposta c


x y sen x cos y2π



10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor


Resposta a



cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ




rArr


1o Q

sen θ = 12


Resposta b

12 2senx




sec x cos x= minus

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b





= 4 cos2 α + 4 sen2 α = 4


2 middot sen α =


10 a) sen2 3π20

+ cos2 3π20

= 1 rArr a2 + cos2 3π20

= 1 rArr



cos 3π20

= cos 27deggt 0


+ 17π20


de 3π20

e 17π20


senos

3π

20

17π

20

Logo sen 17π20

= sen 3π20

= a

c) Como 3π20

+ 7π20

= 10π20

= π2


e 7π20


cos 7π20

= sen 3π20

= a


20

= 37π

20


20e 37π20


cossenos

3π

20

37π

20

Daiacute sen 37π20

= ndash sen 3π20

= ndasha



5 (1)

rArr a =(3)

2b

sec x= 5a

rArr 1cos x

= 5a

rArr cos x= a5

(2)


2

+ 2b5

2

= 1 rArr


rArrbgt 0

rArr b=

5 middot 3

61 =

15 61

61

a+ c

Em (3) a= 30 61

61


b ndash a= 15 61

61 ndash 30 61

61 = ndash 15 61

61Assim


61

rArr c ndash 15 61

61

= ndash 15 61

61

rArr c= 0

12 a) sen α = 15


rArr 15

ndash cos α2

+ cos2 α = 1 rArr


cos α ndash 2425

= 0 rArr



100

45

ndash35

Se cos α = 45


ndash 45

= ndash 35

e α tem




+ 35

= 45

e α tem



lt 0


complementares



1 rArr


14 1o membro=


ndash cos xsen x

2

= (1 ndash cos x)2

sen2 x =

= (1 ndash cos x)2

1 ndash cos2 x =

(1 ndash cos x)2


1 ndash cos x

1+ cos x =

= 2o membro

15 tg x ndash sen x

sen3 x =


sen3 x =

1cos x ndash 1

sen2 x =

=

1cos x ndash 1

1 ndash cos2 x =

1 ndash cos x

cos x =


=

=1

cos x middot

1

1+ cos x =

sec x

1+ cos x






Comoπ5

+ 3π10

= 2π + 3π10

= π2

temos

tgπ5

= 1

tg 3π10




rArr


= 60deg

Daiacute cos B =

102

AB rArr

32

= 5AB

rArr

rArr AB= AC= 10

3 =

10 33

Assim

2p= 10+ 2 middot10 3

3 = 10+

20 33

cm

b) sen x+ cos x= k




2

sen3 x+ cos3 x=



2 = k middot

2 ndash k 2 + 12

=

= k

2

(3 ndash k 2)

27

1

y

O D

α

1 x

BAE

C

a) AE= cotg α= 1

tg α = 1tg π3

= 1

3 = 33


DC= tg α = tgπ3

= 3



= 12

middot 1 ndash 33


= 12


aacuterea=

1

2 middot

4 3

3 ndash

2=

2 3

3 ndash

1




=

= 1


=



= 12 middot


=

= 12 middot


= 12 middot


= 1


lente a ()

Desafio





9998

9989

9899

8999


Testes



Resposta c

3

2 Resposta d

2sen x com x

5 2


2 2sen x cos x 1+ =

24cos x 1

25+ = rArr

2 4 21cos x 1

25 25= minus = rArr

21cos x

5rArr = minus


5

= = = minus

minus

Resposta d




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus



2

= = = e





IV Falsa pois1 1


= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c




Daiacute


cotg x= 43

rArr cos xsen x

= 43


9 cos2 x16


rArr sen x= 5

Resposta a

3

7 sen xcos x



rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2


1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8


2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13


Resposta c


x y sen x cos y2π



10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor


Resposta a



cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ




rArr


1o Q

sen θ = 12


Resposta b

12 2senx




sec x cos x= minus

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b






Comoπ5

+ 3π10

= 2π + 3π10

= π2

temos

tgπ5

= 1

tg 3π10




rArr


= 60deg

Daiacute cos B =

102

AB rArr

32

= 5AB

rArr

rArr AB= AC= 10

3 =

10 33

Assim

2p= 10+ 2 middot10 3

3 = 10+

20 33

cm

b) sen x+ cos x= k




2

sen3 x+ cos3 x=



2 = k middot

2 ndash k 2 + 12

=

= k

2

(3 ndash k 2)

27

1

y

O D

α

1 x

BAE

C

a) AE= cotg α= 1

tg α = 1tg π3

= 1

3 = 33


DC= tg α = tgπ3

= 3



= 12

middot 1 ndash 33


= 12


aacuterea=

1

2 middot

4 3

3 ndash

2=

2 3

3 ndash

1




=

= 1


=



= 12 middot


=

= 12 middot


= 12 middot


= 1


lente a ()

Desafio





9998

9989

9899

8999


Testes



Resposta c

3

2 Resposta d

2sen x com x

5 2


2 2sen x cos x 1+ =

24cos x 1

25+ = rArr

2 4 21cos x 1

25 25= minus = rArr

21cos x

5rArr = minus


5

= = = minus

minus

Resposta d




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus



2

= = = e





IV Falsa pois1 1


= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c




Daiacute


cotg x= 43

rArr cos xsen x

= 43


9 cos2 x16


rArr sen x= 5

Resposta a

3

7 sen xcos x



rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2


1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8


2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13


Resposta c


x y sen x cos y2π



10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor


Resposta a



cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ




rArr


1o Q

sen θ = 12


Resposta b

12 2senx




sec x cos x= minus

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b




Comoπ5

+ 3π10

= 2π + 3π10

= π2

temos

tgπ5

= 1

tg 3π10




rArr


= 60deg

Daiacute cos B =

102

AB rArr

32

= 5AB

rArr

rArr AB= AC= 10

3 =

10 33

Assim

2p= 10+ 2 middot10 3

3 = 10+

20 33

cm

b) sen x+ cos x= k




2

sen3 x+ cos3 x=



2 = k middot

2 ndash k 2 + 12

=

= k

2

(3 ndash k 2)

27

1

y

O D

α

1 x

BAE

C

a) AE= cotg α= 1

tg α = 1tg π3

= 1

3 = 33


DC= tg α = tgπ3

= 3



= 12

middot 1 ndash 33


= 12


aacuterea=

1

2 middot

4 3

3 ndash

2=

2 3

3 ndash

1




=

= 1


=



= 12 middot


=

= 12 middot


= 12 middot


= 1


lente a ()

Desafio





9998

9989

9899

8999


Testes



Resposta c

3

2 Resposta d

2sen x com x

5 2


2 2sen x cos x 1+ =

24cos x 1

25+ = rArr

2 4 21cos x 1

25 25= minus = rArr

21cos x

5rArr = minus


5

= = = minus

minus

Resposta d




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus



2

= = = e





IV Falsa pois1 1


= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c




Daiacute


cotg x= 43

rArr cos xsen x

= 43


9 cos2 x16


rArr sen x= 5

Resposta a

3

7 sen xcos x



rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2


1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8


2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13


Resposta c


x y sen x cos y2π



10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor


Resposta a



cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ




rArr


1o Q

sen θ = 12


Resposta b

12 2senx




sec x cos x= minus

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b




4 I Verdadeira pois

3cos

6 2cotg 316

sen6 2

π

π= = =

πe

4 3sen

4 3 2tg 34 1

3 cos 3 2

π minus

π= = =

π

minus



2

= = = e





IV Falsa pois1 1


= = =

1 1sec120ordm 2

1cos120ordm2

= = = minusminus

Resposta c

5

1cos x

ndash cos x

1

sen x

ndash sen x

=

1 ndash cos2 xcos x

1 ndash sen2 x

sen x

= sen2 xcos x

middotsen xcos2 x

=

= sen3 xcos3 x

= tg3 x

Resposta c




Daiacute


cotg x= 43

rArr cos xsen x

= 43


9 cos2 x16


rArr sen x= 5

Resposta a

3

7 sen xcos x



rArr cos x= ndash 1

1+

a

2

sen x= ndash a

1 + a2


1 + a2 ndash

1

1 + a2 =

ndash1 ndash a

1 + a2

Resposta a

8


2 2sen x cos x 1+ =

2

2 12 144 25 5cos x 1 1 cos x

13 169 169 13


Resposta c


x y sen x cos y2π



10 Quando sen x= 13

5+ sen x =

35+ 1

= 05 eacute o menor


Resposta a



cos2 θ

senone 0 2=

3 sen θ

cos2 θ

2= 3sen θ




rArr


1o Q

sen θ = 12


Resposta b

12 2senx




sec x cos x= minus

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b







Resposta b


e cos θ = ar


concluiacutemos




Resposta d





Resposta d

16 Resposta b


Resposta d

19 sec2 x= 1+ tg2 x


rArr sen2 xcos2 x

= 1+ 3

1 ndash cos2 x

cos2

x


rArr cos2 x (1+ 3 + 1)= 1

cos2 x= 1

2+ 3 rArr cos x=

1

2 + 3Resposta c

20

2

2

1 sen x cos x sen x


minusminus

minus sdot= = =

minusminus minus

( )

2

2

cos x sen x

cosx

cos x sen x

minus

= =minus minus

1secx

cosxminus = minus

Resposta b


xπ12 0⩽ x⩽ 6




= 3 2 middot senπ4

ndash

ndash 2+ cosπ2

= 3 2 middot2

2 ndash 2+ 0= 3 ndash 2= 1

Resposta c



aacuterea TAB

S= 12


Resposta d


Assim temos


rArr

rArr cos y= + 12


Respostab

24

45deg

h

h 8

45deg θ

cotg θ = 76

rArr tg θ = 67

Assimh

h+ 8 =

67

rArr h= 48 m

Resposta b

18 π2 + π6 + π18 + hellip= a

11 ndash q =

π

21 ndash

13

=

π

223

= 3π4

cos3π4

= ndash cosπ4

= ndash2

2

Resposta b

25 2 2

sen cos 1α + α =

2 2x x 1

1x 1 x 1

minus + = rArr

+ +

2 2

2 2

x x 2x 11

x 2x 1 x 2x 1

minus ++ = rArr

+ + + +


2x 4x 0minus = rArr


Resposta d





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b





rArr tg α = PBOP

= PB



PC= cotg α

Assim BC=

PB+

PC=

tg α +

cotg αResposta a

27 sen x sen x

y1 sen x 1 sen x

= + =+ minus

( ) ( )

( )( )


1 sen x 1 sen x

minus + += =

+ minus

2 2

2


1 sen x

minus + += =

minus

2

2 sen x2 sec x tgx

cos x= = sdot sdot

Resposta a

28 0 x2

πlt lt e tg x 4=

cos x sec x

sen x cossec x

minus=

minus

2

2

cos x 1

cosx

sen x 1

senx

minus

=

minus

2

2

sen x senx

cos x cos xsdot =

3tg x 64= =

Resposta c



sen β = 0


sen β = 0


Temos


cos α = 0rArr


cos α = 0rArr




= 0rArr cos α = 12

rArr

rArr α = 60degπ3

e β = 30degπ6

Resposta d

30 y

xQ

P S

R

O









= OSOR



= r 1 ndash cosdr

Resposta b


Documents

Matemática ciência e aplicações, Vol. 2, Cap 2