Exercicios de concurso publico (1641)

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212
Conexões com a Matemática
	 b)	
π
π
cossec cos
sec cos
cos cos
x
x x
x
x x
x
x x
x x
2
2
1
1
2
2 8
5 8
5
8
5 5
cotg
tg
tg tg cotg
=
c
c
m
m
14. a) (1 1 cotg2 x) 8 (1 2 cos2  x) 5 cossec2 x 8 sen2 x 5
sen
1sen5 5
x
x1 8 2
2
    b) 
1 cossec2
5
cossec x x
1
1
11
1
(cossec ) (cossec )
cossec cossec5
2 1
1 1 2 5
x x
x x
1 1
1 1
8
cossec 1
cossec
2
sen
2
2
5
cotg
sen
x
x
x
x
x x2 55 8 8 5tg
2
2
2
2 sen8
sen cos
sen
cosx x
x
x
x2 8 55 5
2
22
5 2 8 tg x 8 sec x
15. a)  cos cos
cotg
A x x
x
11 85 2 2
2
 V
V A 5 cos2 x 1 cos2 x 8 tg2 x V
V A 5 cos2 x 8 (1 1 tg2 x) V
V A 5 cos2 x 8 sec2 x V A 5 1
    b) 
tg
x x
x
B 11 85 sensen2 2
2
V B 5 sen2 x 1 sen2 x 8 cotg2 x V
V B 5 sen2 x 8 (1 1 cotg2 x) V
V B 5 sen2 x 8 cossec2 x	V B 5 1
    c)  
cotg1
cossec secx x
x x
C 5 8
tg
 V
V 
1
cos
cos
cos
x x
x
x
x
x
C
1 1
5
8sen
sen
sen V
V  cos
cos
cos
cos
cosx x
x x
x x
x x
x xC
1 1
1
5
8
8
1
5
8
8
sen sen
sensen
sen2 2
 V
V C 5 1
16. 1 1 cotg2 x 5 cossec2 x V 1 x
x
151
sen
cotg2
2
 V
V  1
2
2x
5
3
15cotg
2
2
e o
 V  12x
9
255cotg2  V
V  x
9
165cotg2  V  6x
3
45cotg
Como 
3
2 ,
π
, , πx
2
 então:  x
3
425cotg
17. Como sec2 x 5 1 1 tg2 x e  ,sencossec x x
15 temos:
s 1 tg 1 tgec
cotg8
cos
cossec cosx x
x
x x
x
x
x
1
5
8
1
5
1
5
sen
2 2 2
tg
tg
(1 tg ) tg5 5 8 5
x
x
x x
1
11
1
2
2
1 15 1 5 1 8 5
2
3
2
3
4
9
2
38
2
e o> =H G
5 8 513
2
3
8
39
4
18. 
1 cotg
cotg
1
sen
cos
co
1
5 5sen
s
x
x
x
x
x
x
1
2
2
2
2
2
2
 
sen
sen cos
sen
cos
1
5
x
x x
x
x
2
2 2
2
2
sen
cos sen5 8 cos
x
x x x
1 3
2
9
25 5 5
2
2
2 2
2 f p
19. 
2 cos c
sen tg
cotg
cotg
5
8 8
8 5 8
1cossec
otg
y
x x
x x x
x
x
2
1 2
 V
V 
2
tg
1
tg
5
1
tg
y
x
x
x
12
1
8  V
V 
22
tg
1 9 1
5
1
5 5
tg
y
x
x
x
2
9
3 2
3 10
2
8 8
1tg
20.  (2 s )2 5sec en
cotg
x x
x
18 22
2
2
= (tg2 x 1 1) (2 2 sen2 x) 2 tg2 x 5
5 2tg2 x 2 tg2 x sen2 x 1 2 2 sen2 x 2 tg2 x 5
5 tg2 x (1 2 sen2 x) 1 2 2 sen2 x 5
5 
sen cos 2 sen 28 1
cos x
x x x2 5
2
2
2 2
21. Temos:
x = 7tg t V tg t x
7
5  V cotg t x
75
y = 2cossec t V cossec t
y
2
5
Como cossec2 t 5 1 1 cotg2 t, resulta:
11
y
x2
75
2 2
e co m  V  11
y
x4
495
2
2
 V 
V	 5 1 V
y
x
x
4
49
2
2
2
	x2y2 5 4x2 1 196 V	
V x2y2 2 4x2 5 196
Portanto, uma relação entre x e y, independente de t, é: 
x2y2 2 4x2 5 196