Matemática ciência e aplicações, Vol. 2, Cap 2
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Transcript of 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
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 215
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
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 315
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 415
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 515
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
44 102 983101 62 983083 x2 rArr x983101 8 (o outro cateto mede 8)
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
MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 615
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
x eacute um arco do 1ordm quadrante
tg x positivo
cotg x positivo
cotg x negativo2
π +
( )cotg x positivo+ π
sinal de y ( ) ( )
( ) ( ) ( )
+ sdot minus= = minus
+ sdot +
x eacute um arco do 3ordm quadrante
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 715
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
cos a cos b sena senb
+ 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
MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 815
=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
π π π π= minus = π minus rArr 3ordm quadrante
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
π π π π= + = π + rArr 3ordm quadrante
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
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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
MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
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∆ = (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
entatildeo sen α = 15
+ 35
= 45
e α tem
imagem no 2o quadrante
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
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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
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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
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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
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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
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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
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
44 102 983101 62 983083 x2 rArr x983101 8 (o outro cateto mede 8)
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
MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
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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
x eacute um arco do 1ordm quadrante
tg x positivo
cotg x positivo
cotg x negativo2
π +
( )cotg x positivo+ π
sinal de y ( ) ( )
( ) ( ) ( )
+ sdot minus= = minus
+ sdot +
x eacute um arco do 3ordm quadrante
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
cos a cos b sena senb
+ 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
MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
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=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
π π π π= minus = π minus rArr 3ordm quadrante
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
π π π π= + = π + rArr 3ordm quadrante
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
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∆ = (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
entatildeo sen α = 15
+ 35
= 45
e α tem
imagem no 2o quadrante
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
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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
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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
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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
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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
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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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
44 102 983101 62 983083 x2 rArr x983101 8 (o outro cateto mede 8)
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
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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
x eacute um arco do 1ordm quadrante
tg x positivo
cotg x positivo
cotg x negativo2
π +
( )cotg x positivo+ π
sinal de y ( ) ( )
( ) ( ) ( )
+ sdot minus= = minus
+ sdot +
x eacute um arco do 3ordm quadrante
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
cos a cos b sena senb
+ 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
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=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
π π π π= minus = π minus rArr 3ordm quadrante
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
π π π π= + = π + rArr 3ordm quadrante
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
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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
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∆ = (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
entatildeo sen α = 15
+ 35
= 45
e α tem
imagem no 2o quadrante
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
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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
| Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
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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
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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
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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
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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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
44 102 983101 62 983083 x2 rArr x983101 8 (o outro cateto mede 8)
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
MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
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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
x eacute um arco do 1ordm quadrante
tg x positivo
cotg x positivo
cotg x negativo2
π +
( )cotg x positivo+ π
sinal de y ( ) ( )
( ) ( ) ( )
+ sdot minus= = minus
+ sdot +
x eacute um arco do 3ordm quadrante
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
cos a cos b sena senb
+ 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
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=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
π π π π= minus = π minus rArr 3ordm quadrante
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
π π π π= + = π + rArr 3ordm quadrante
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
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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
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∆ = (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
entatildeo sen α = 15
+ 35
= 45
e α tem
imagem no 2o quadrante
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
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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
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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
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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
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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
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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
44 102 983101 62 983083 x2 rArr x983101 8 (o outro cateto mede 8)
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
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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
x eacute um arco do 1ordm quadrante
tg x positivo
cotg x positivo
cotg x negativo2
π +
( )cotg x positivo+ π
sinal de y ( ) ( )
( ) ( ) ( )
+ sdot minus= = minus
+ sdot +
x eacute um arco do 3ordm quadrante
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
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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
cos a cos b sena senb
+ 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
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=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
π π π π= minus = π minus rArr 3ordm quadrante
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
π π π π= + = π + rArr 3ordm quadrante
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
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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
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∆ = (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
entatildeo sen α = 15
+ 35
= 45
e α tem
imagem no 2o quadrante
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
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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
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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
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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
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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
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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
x eacute um arco do 1ordm quadrante
tg x positivo
cotg x positivo
cotg x negativo2
π +
( )cotg x positivo+ π
sinal de y ( ) ( )
( ) ( ) ( )
+ sdot minus= = minus
+ sdot +
x eacute um arco do 3ordm quadrante
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
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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
cos a cos b sena senb
+ 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
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=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
π π π π= minus = π minus rArr 3ordm quadrante
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
π π π π= + = π + rArr 3ordm quadrante
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
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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
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∆ = (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
entatildeo sen α = 15
+ 35
= 45
e α tem
imagem no 2o quadrante
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
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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
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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
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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
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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
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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
cos a cos b sena senb
+ 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
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=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
π π π π= minus = π minus rArr 3ordm quadrante
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
π π π π= + = π + rArr 3ordm quadrante
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
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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
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∆ = (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
entatildeo sen α = 15
+ 35
= 45
e α tem
imagem no 2o quadrante
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
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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
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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
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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
| Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
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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
| MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
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=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
π π π π= minus = π minus rArr 3ordm quadrante
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
π π π π= + = π + rArr 3ordm quadrante
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
Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
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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
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∆ = (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
entatildeo sen α = 15
+ 35
= 45
e α tem
imagem no 2o quadrante
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
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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
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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
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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
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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
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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
MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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∆ = (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
entatildeo sen α = 15
+ 35
= 45
e α tem
imagem no 2o quadrante
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
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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
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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
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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
| Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
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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
| MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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∆ = (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
entatildeo sen α = 15
+ 35
= 45
e α tem
imagem no 2o quadrante
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
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
| Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
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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
| Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
| MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
| Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 1315
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
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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
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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
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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
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7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
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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
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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
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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
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7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
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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
| Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 1515
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
| MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 1415
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
| Ca piacutetulo 2 bull R az otildees trigonomeacutetricas
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 1515
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
| MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2
7232019 Matemaacutetica ciecircncia e aplicaccedilotildees Vol 2 Cap 2
httpslidepdfcomreaderfullmatematica-ciencia-e-aplicacoes-vol-2-cap-2 1515
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
| MATEMAacuteTICA CIEcirc NCIA E AP LICACcedilOtildeE S 2