SOLUCIONARIO TRIGONOMETRÍA 5

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1
SOLUCIONARIO 
CUADERNO DE TRABAJO
ACTIVIDADES CAP 01
ÁNGULO TRIGONOMÉTRICO
01 5p
8
 = 5
8
(180°) = 112,5° = 112°30'00''
Clave C
02 30x + 10(9°) = 270° – 90°
  x = 3°
Clave B
03 x2 – 3x – 10
9
 = x2 – 2x – 4
10
 10x2 – 30x – 100 = 9x2 – 18x – 36
 x2 – 12 – 64 = 0
 x – 16 = 0
 
x –16
x +4 ⇒ x = 4
Clave D
04 3C + 2S = 240 ; 10S = 9C
 15C + 10S = 240(5)
 15C + 9C = 240(5)
 ⇒ C = 50
Clave D
05 S + 9C = 330 ; 9C = 10S
 S + 10S = 330
 ⇒ S = 30
  R = 
p
6 Clave C
06 10

k3 – 1
19


 = 9

k3 + 1
19


 10k3 – 10
19
 = 9k3 + 9
19
 k3 = 19
19
 ⇒ k = 1
 S = 1 – 1
19
 = 18
19
 ⇒ 18
19
p = 180R  R = 
p
190
Clave C
07 1
2S
 – 1
3C
 = 1
20
 ; S
9
 = C
10
 ⇒ 5
9C
 – 3
9C
 = 1
20
 ⇒ C = 40
9
 Luego: 18
19
p = 200R
  R = 
p
45 Clave B
08 nC + nS = 3800

R
p


 n

200R
p


 + 

180R
p


n = 3800

R
p


 ⇒ 380n = 3800  n = 10
Clave C
09 (10k)2 + (9k)2 = 2(10k)3 – 5(9k)(10k)2 +
 4(9k)2(10k) – (9k)3 – 2(9k)(10k)
 ⇒ 181 = 11k – 180 ⇒ k = 361
11
  C = 10

361
11


 = 3610
11
Clave C
10 a° → (a – 3)
 180° → 120
 ⇒ 120a = 180(a – 3)
 2a = 3a – 9
 a = 9
  a – 3 = 6
Clave B
CUADERNO DE TRABAJO
01 b = 2°4'5''  b = 2(3600'') + 4(60'') + 5''
  b = 7200'' + 240'' + 5''  b = 7445''
Clave B
02 5x – 3 = 6x – 9
 x = 6
 
(5x – 3)°
(6x – 9)°
Clave C
03 Siendo S
C
 = 9
10
  7x + 2
8x
 = 9
10
  x = 10
 S = (7x + 2) = 72 × p
180
  El ángulo es: 2p rad
5 Clave B
04 Se cumple: 
 C + S + R = 2S + R + 4  C = S + 4
 10k = 9k + 4  k = 4
 ⇒ C = 40  R = 40p
200
 = p
5
 ∴ El ángulo es p
5
 rad
Clave E
05 10k(9k)2 + (9k)3 = (10k – 9k)2
  810k3 + 729k3 = k2  k = 1
1539
 R = 9k × p
180
 = 9 × 1
1539
 × 
p
180
 R = p
30780 Clave A
06 • 
n2
1
19
n2 –
 = 9
10
  10n2 – 10
19
 = 9n2
  n2 = 10
19
  C = 10
19
 • R = 10
19
 × p
200
 = p
380 Clave E
07 • xy° zw' = 50g50m
 • Pero 50g× 9°
10g = 45°
 50m = 


1
2
g
× 9°
10g×60'
1°
 = 27'  xy° zw' = 45°27'
  q = 4 + 5
2 + 7
 = 1
Clave A
08 180R
p
 + 200R
p
 + R = 95 + p
4
 380R
p
 + R = 380 + p
4
 
 R 


380 + p
p
 = 


380 + p
4
  R = p
4
Clave B
09 4R 9p pR +
Rp
 = 5  2R + 3p = 5 Rp
 2R – 5 Rp + 3p = 0  R = 9p
4
  R = p
 
2 R –3 p
 R –1 p  9p
4
 = 405°
Clave B
10 7N
5°
 = 63N
x
  x = 45°
  x = 45°×p rad
180°
 = p
4
 rad
Clave C
TAREA
01 230° + x = 40°
  x = – 190°
02 40ng + (24n)° = 90°
  40ng × 9°
10g + (24n)° = 90°
  36n + 24n = 90
  n = 3
2
03 prad
32
 × 180°
prad
 = 45°
8
 prad
32
 = 5° 5
8
 = 5° 300'
8
 prad
32
 = 5° 75'
2
 = 5° 37' 30''
 
  a + b – c = 12
↓ ↓ ↓
a b c
TRIGONOMETRÍA 5°
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
2
04 • 9k – 13
8
 = 10k – 2
12
 108k – 156 = 80k – 16
 28k = 140
 k = 5
 • S = 9(5) = 45
 • 45 – 13
8
 = x
2
4
  x = 4
REFORZANDO
01 –(2 – 2x)° + 5xg 

9°
10g


 = 180°
  – 2 + 2x + 9x
2
 = 180  x = 28
Clave D
02 Se tiene que a = 27k y b = 50k, entonces:
 H = 27

27k+100k
27k


 + 42 = 127+42 = 13
Clave D
03 Tenemos 10S = 9C  10 

1080
60


 =
 = 9(3x + 5)  20 = 3x + 5  x = 5
 Luego β = 20g + (11(5))g = 75g
 Entonces 75g = 75 

p
200


 rad = 3p
8
rad
Clave C
04 Tenemos S = 9k, C = 10k y R = pk
20
  162k2 – 10k = 100k2 + 18k
  62k = 28  k = 14
31
 Luego, C = 10 

14
31


 = 

140
31


Clave A
05 El ángulo C:
 180° – (6t)° > 0  180 > 6t  t < 30 
 Para que C tome su menor medida t = 29,
 Luego:
 C = 180° – 174° = 6° = 6p
180
rad = p
30
rad 
Clave E
06 ap
15
rad + 30(a + b)° + 100
3
 (a – b)g = 180°
  12a + 30(a + b) + 30(a – b) = 180
  72a = 180  a = 5
2
 Luego, A = 5
2
 

p
15


 = p
6
rad
Clave B
07 Tenemos 10S = 9C
  10 

18x – 18
p


 = 9 

10x + 30
p


  90 

2x – 2
p


 = 90 

x + 3
p


  x = 5
p
 Luego, S = 18 

x – 1
p


 = 18 

5
p
 – 1
p


 = 18 

4
p


 Por lo tanto, R = p
180
 (18) 

4
p


 = 4
10
 = 2
5
rad
Clave B
08 b – a = 2300  b = 2300 + a
 Luego:
 a
27
 = 2300 + a
50
  23a = 27 × 2300  a = 2700
 Por consiguiente, 

2700
60


°
 = 45° = p
4
rad 
Clave D
09 • a°a' + agam
 = 

61a
60


°
 + 9
10


101a
100


°
 = 

5777a
300


 
°
 = 11,554° 
  5777a = 3(11554)  a = 6
 Luego, agam = 6g6m = 

606
100


°
  R = p
200
 

606
100


 = p

303
10000


 = 0,0303p rad
Clave E
10 Como 5a + 4β = – 21° (1)
 4a – 5β = 180° (2)
 Resolviendo simultáneamente, las ecua-
ciones (1) y (2) se tiene que:
 a = 15° β = –24° 
 Luego:
 –a – β = –15° + 24° = 9° = 9 

p
180


 = p
20
rad
Clave D
11 Se tienen que:
 S3 + SC + S
C3 + SC + S
 = 9
10
  S
C
 

S2 + C + 1
C2 + S + 1


 = 9
10
  S2 + C + 1 = C2 + S + 1
  0 = 19k2 – k  k = 1
19
  Rrad = p
20
 

1
19


rad = p
380
rad
 Clave C
12 Tenemos S = 9k, C = 10k y R = pk
20
 Entonces:
 27
k3 + 27
k3 + 27
k3 = 3
64
  3(27)
k3 = 3
64
  k3 = 27(64)  k = 12
  R = p
20
 (12) = 3p
5
rad
 Clave D
13 a = (2a – 1)prad = (a – 1)°
  (2a – 1)p
p
 = a – 1
180
  2a – 1 = a – 1
180
  360a – 180 = a – 1  359a + 1 = 180
  Luego β = 180g = 9p
10
rad
 Clave C
14 (100,405)g = 100g + (0,4)g + (0,005)g 
  ((0,4)g = 40m, (0,005)g = 50s 
  (100,405)g = 100g40m50s
  a = 100, b = 40 y c = 50
 Por lo tanto, γ mide 60°45' ≅ 27p
80
rad
 Clave C
15 a = 162000'' = 45° = 50g y β = 3p
25
rad = 24g 
  a + β = 74g, luego Com(a + β) = 26g
 Clave D
ACTIVIDADES CAP 02
LONGITUD DE ARCO
01 L = 

40° × 
p
180°


18 ⇒ 4p m
Clave B
02 11 = q(14) ⇒ q = 11
14
 = 1
4
 

22
7


 = 1
4
p
Clave B
03 2
r
 = 4
r + 3
 ⇒ r + 3 = 2r ⇒ r = 3 
Clave C
04 x
a
 = 3x
b
 ⇒ a
b
 = 1
3 Clave E
05 2p
5
 = L
7
 ⇒ L = 14p
5 Clave A
06 x = 2(4) ⇒ x = 8
 y = 3(4) ⇒ y = 12
  y – x = 4 
Clave B
07 nv = 80p
2p (4)
 ⇒ nv = 10
Clave C
08 q = 2
r
 = 4
r + 4
 ⇒ r + 4 = 2r ⇒ r = 4 
Clave C
09 Propiedad: x = a(a) + b(b)
a + b
  x = a2 + b2
a + b
Clave A
10 L = 10(2pa) = 
p
3
(a2 + 62a – 3)
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
3
 60a = a2 + 62a – 3
  3 = a2 + 2a
Clave A
CUADERNO DE TRABAJO
01 L = qR  2p = 14×q  q = 2p
14
 = p
7
Clave C
02 20
x
 = 40
x + 30
 = 20
30
  x = 30
Clave A
03 L2
r
 = L
2r
 = L1
3r
  L1 + L2
4r
 = L
2r
 
  16p
4
 = L
2
  L = 8p
Clave B
04 OB = 2OD  OB = 2r y OD = r
 L2
2r
 = L1
r
  L2 – L1
r
 = L2
2r
 
A
B
C
D
r
r
L1 L2
  10
r
 = L2
2r
  L2 = 20
Clave B
05 a
180
 = b
200
 = q
p
 3bq = 720
p
 a 
3b
3b
a
 3



10
9
a q = 720
p
 a  q = 216
p
Clave B
06 • a + b = p
2
 (1)
 • L1 – L2 = 2  R(a – b) = 2 (2)
 • a
b
 = 2 (3)
 De (1) y (3): b = p
6
  a = p
3
 De (2): R p
6
 = 2  R = 12
p
Clave A
07 5L
r
 = 8L
a + r
 = 9L
18
  r = 10 y a = 6
 Pero a + b + r = 18 
  b = 2 
8L 9L5L
a
a
b
b
r
r
18
  M = a – b = 6 – 2 = 4
Clave B
08 OC = 132 – 52  OC = 12
 L1 = 12q  L2 = 13q  K = L1
L2
 = 12
13
Clave B
09 Long EC = long BE 
 = 4 p
6
 = 2p
3
  Perímetro: 
A
B
E
C
4
44
D
30°30°
60° 60°
 4p
3
 + 4 = 4



p
3
1 +
Clave A
10 5 = D1
2p(2)
 
  D1 = 20p D1 D2
2 2 · 6
 2 = D2
2p(6)
  D2 = 24p
 D = D1 + D2 + 4 3 
 D = 44p + 4 3 = 4(11p + 3)
Clave A
TAREA
01 • L = qR  56 = q4
  q = 14
02 q = 140°× p
180°
rad = 7p
9
 L = qR  L = 7p
9
⋅R
  L
R
 = 7p
9
03 
04 • 4L
x
 = 5L
x + 2 
 = 4L
x
 = L
2
  x = 8
REFORZANDO
01 L = qR  33 = 3R
  R = 11
Clave D
02 q = 30° = p
6
 R = 6
 L = qR = p
6
 ⋅ 6 = p m
Clave A03 4
R + 3
 = 2
R
 = 2
3
  R = 3
Clave C
04 L = 110 cm R = 70 cm
 L = qR  110 = q ⋅ 70
 3n + 1 = 25  n = 8
 q = n + 1
n
 = 9
8
n
n
n + 1q
 q = 11
7
 = 1
2
 

22
7


 q = p
2
rad
Clave D
05 L1
r
 = L2
2r
 = L3
3r
  L1
1
 = L2
2
 = L3
3
  L1 = q, L2 = 2q, L3 = 3q
  k = (1 + 4 + 9)q2
q(3q – q)
 = 7
Clave E
06 qg × 9°
10g = 9q°
10
 
 L1 = 9q°
10
 (3R)
 L2 = q°(R)
  k = L1
L2
 = 
qR
27qR
10
 = 10
27
Clave B
07 c = b
r
 = a
r + x
  c = a – b
x
  x = (a – b)c–1
Clave C
08 Perímetro:
 = 1
p + 3
 ⋅ p
6
 + 1
p + 3
 ⋅ p
6
 + 1
p + 3
 = 1
3
Clave D
09 
Clave B
10 
 Como los sectores circulares tienen el 
mismo radio y la región sombreada tiene 
 un arco de 7p
6
 
  Perímetro = 7p R
6
 
Clave E
11 nv = Lc
2p(1)
  20 = Lc
2p
  Lc = 40p y L = 40p
4
 + 2  L = 10p +2
Clave D
12 
 
  q = S
r
 = C 
2R + r
  q = C – S
2R
2 R
2 
R
2 R
120°12
0°
120°
1
3
 ⋅ V
1
3
 ⋅ V
L = 2pR + 6R
L = 2R(p + 3)
1
3
 ⋅ V
60° 60° 30°
60°
60°
60°
r
2R
2R
r
S Cq
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
4
 q = 
20R
p
2R
 = 10
p
Clave B
13 
 
 z
r
 = y 
a + r
 = x 
a + b + r
 y – z 
a
 = x – y 
b
  yb – zb = ax – ay
  ay + by = ax + bz
  M = ay + by 
ax + bz
 = 1
Clave B
14 
 
 
30° 60° 60°
L1L1
R R
R R
RRR
L2
L3
 
 L1 = p 
3
R L2 = p 
6
R L3 = R 3 – R
 Piden: 2L1 + L2 + L3 
  2 p 
3
R + p 
6
R + R 3 = R 

5p
6
 + 3 

Clave B
15 
 
 2(R1 + R2 + R3) = 100  R1 + R2 + R3 = 50
  ∑ Longitudes = p(R1 + R2 + R3) = 50p 
Clave E
CUADERNO DE TRABAJO CAP 03
ÁREA DEL SECTOR CIRCULAR
01 S = 1
2
 120g 
p
200g (8)2
  S = 19,2p m2
Clave A
02 Propiedad:
 S
b2 = 3S
n2 ⇒ n = 3b
 Longitud de arco:
 
A
O D
C
n
b
b
Sq
2S
a 3
n
B q( 3b) = a 3 
  q = a
b
 Clave B
r
r
a
a
b
b
yz x
R3 R2
R1
03 

L + 9
2


2 = 21
 ⇒ L = 12
 
A
O D
C L
r
r
2
2
9
q
21u2
B
 9
r
 = 12
r + 2
 ⇒ r = 6
 Perímetro: 2p
 2p= 6 + 9 + 6 = 21 u 
Clave E
04 S
x2 = 3S
y2 = 6S
42
 ⇒ x 3 = 2 2
 y = 2 2
 
A
O F
E
C
D
4
S
3S
2S y
x
B  y + x 3 = 4 2
Clave C
05 Propiedad:
 x = 3(2) + 18(1)
1 + 2
 ⇒ x = 8
 
E
O B 1
1
2
2
A
M
N
18
S3
r
r
x
F
a
 Área:
 S = 3 + 8
2
 (1)
 S = 11/2 cm2
Clave E
06 Propiedad:
 
S1
r2 = 
S1 + S2
(2r)2
 
C
O B
A
r
r
r
r
S1
S2
D  
S2
S1
 = 3
Clave C
07 1
2


2p
5


r2 = 45p ⇒ r = 15
 
2p
5
45p
r
r
 1
2


2p
5
 – a 

(30)2 = 30p
 
2p
5 – a
2r
2r 30p
  a = 
p
3
Clave B
08 S = (2R)2
2
 
p
3
 – sen60°
 S = 2R2 
p
3
 – 3
2
  S = R2(2p – 3 3)/3
 
30°
60°
2R S
R R BQ
P
O
A
30°
Clave C
09 
 
 
P
O
q
Q
10q10
10
10 108
6
6
⇒
2p(6)
 ⇒ 10q = 2p(6)  q = 6p
5
Clave C
10 16(w + x + z) = 5(9x)
 w + x + z = 45
16
 x
 
5na
2
3nb
2
 = 25x
9 45
16x
 
D
C
A
Q
P
2n
2n
3n
3n R
16x ab
16w
9w
9x
9z
16z
S
O
  a
b = 16
27
Clave B
CUADERNO DE TRABAJO
01 Área = 11×14
2
 
 
14
14
11 Área = 77
Clave A
02 Área = 


2 + 4
2
 3 = 9
Clave E
03 3x + 1 = (2x – 1)(x – 1)  x = 3
 Luego: S = 1
2
 (2)(5)2 = 25 cm2
Clave B
04 Sabemos: 2r + L = 20 + p y L = p
10
 r
  2r + p
10
 r = 20 + p 
  r = 10 20g L
r
  S = 1
2
× p
10
×(10)2 = 5p cm2
Clave C
05 1) qa = 1; 2) q(a + b) = 2a; 3) q(3a + b) = b
 De 1) y 2): a + b
a
 = 2a  a + b = 2a2 
  b = 2a2 – a
 De 1) y 3): 3a + b
a
 = b  2a + 2a2
a
 = 2a2 – a
  2a2 – 3a – 2 = 0  (a – 2)(2a + 1) = 0
  a = 2, b = 6, q = 1
2
 rad
 Área del sector circular COD = 1
2
×1
2
×82 = 16 m2
Clave E
06 
9
Long AB
 = 
16 + 9
45  Long AB = 27
Clave C
07 6
9
 = 
9 + SX
10 
  6
3
 = 
9 + SX
10 
9 cm2 6
A
BD
C
SXO 10
  9 + SX = 25  SX = 16
Clave B
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
5
08 S1 = p
2
R2 – S
 S2 = q
2
(4R)2 – S RR 2R
S2S1
S
 
 S1 = S2  q
2
(4R)2 = p
2
R2  q = p
16
Clave C
09 LAB = p
3
 R
 # vueltas = 
p
3
×R
2p(0,2)
 = 45
  R = 54 
  S = 1
2
×p
3
×(54)2 = 486p
B
O
A
R
R
Clave D
10 r2 + b2 = (a – b)2 ⇒ r2 = a(a – 2b)
 S = 1
2
qa(a – 2b)
 S = qa
2
(a – 2b) 
a –b
a
b b
r
 
Clave A
TAREA
01 q = 2rad  L = 4
  A = 42
2(2)
 = 4
02 A = 16 ⋅ 18
2
 ⇒ A = 144 m2
03 
 
• Perímetro:
 10 + 5q = 100 = q = 2
• Área:
 S = (2)52
2
 = 25
5
5
5qq S
04 
 
•A = 

2 + 3
2


 ⋅ 4 = 10 m2
3 m2 m A
4 m
4 m
REFORZANDO
01 
 • 1
2
 

p
12


 (4R2) = 21p
5
  R2 = 126
5
 • Área = 1
2
 

5p
12


 ⋅ R2
 Área = 1
2
 

5p
12


 ⋅ 126
5
 = 21p
4
 
Clave D
2R
p/12
R
A
O
D
B
CR R
5p
12
02 • 4l = pr ⇒ r = 
4l
p
 
 S = 1
2
 (l) 

4l
p


 = 8
p
  l = 2
Clave D
03 4
O
A
C
B4
a q
 • 4q = 29p
30
  q = 
29p
120
 • 4
2(q – a)
2
 = p
3
  8 

29p
120
 – a 

 = p
3
 
  29p
120
 – a = p
24
  a = p
5
 rad
Clave D
04 
 • S1
(2L)2 = S1 + S2
(3L)2  S1
4
 = 
S2
5
  S1 = 4
5
 (S2) = 4
5
 

p
2


 = 2p
5
  S1
r2 = S1 + S2
(r + 4)2  
2p
5
r2 = 
2p
5
 + p
2
 
(r + 4)2
  r = 8 cm
Clave C
05 
Clave E
06 • S = 25 L = (3x + 4) q = (4 – x) r = ?
  (3x + 4)2
2(4 – x)
 = 25  x = 2
  L = qr  r = L
q
 = 
3x + 4
4 – x
 
  r = 5 m
Clave E
07 
 
q = 
p
12
 r = 12
 S = 122
2
 

p
12


 = 6p
O B
A
12
12
12 C
p
4
p
3
p
3 – p4 = p
12
Clave D
08 q = 50g prad
200g = p
4
 S = 2p r = ?
 S = qr2
2
  2p = p
8
 ⋅ r2  r = 4
O
A
r
r
r
S
M
N
l
B
4l
p
Si L1 = 2L
 L2 = 3L
4
r
S1 2L 3LS2
• 
Long AB
S
 = 4p
S + 3S
 Long AB = 2p
S 3SO
A
C
D
B
4p
 L = 2p(4)  L = 8p  L
p
 = 8p
p
 = 8
Clave C
09 
 • a
2
S
 = (a + x)2
2S
 = (a + x + y)2
3S
  a
1
 = a + x
2
 = a + x + y
3
  a + x + y = 3a
  3S = (a + x + y)
2
 ⋅ a = a
2 3
2
Clave D
10 
 • S1
r2 = S1 + S2
(2r)2  S1
1
 = 
S1 + S2
4
 
  S1
1
 = S2
3
  S2
S1
 = 3
Clave A
11 
 
A
F
E
D
r
r120°
60°
B
C
O
 • El ángulo del área sombreada es: 
 360° – 120° = 240° = 4p
3
  1
2
 

4p
3


 r2 = 8p
3
  r = 2 
 • Por triángulo notable:
 OC = 2r ∧ OF = 3r 
  Área sector AOB = 1
2
 

p
3


 ⋅ (3r)2 = 6p 
Clave A
12 
 • La longitud del arco AB es: 
 r 

p
2


 = 9p  r = 18
 • La longitud del arco OC mas la 
 longitud del arco AC es igual a la longi- 
 tud de AB: 9p
  Perímetro es: 9p + 18 = 9(2 + p) 
Clave E
S
E
F
C
D
A
B
O S S a
a
x
y
r
r
r
r
S1 S2
A
B
C
O
r
r
r
r
p
3
p
3 p
6
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
6
13 
 Asomb = 4p – A 
2
 – A 
4
 – A 
 A = 3(22) 3
4
 = 3 3
 A 
2
 = 
22
2
 

p
3


 – 2
2 3
4
 = 2p
3
 – 3
 A 
4
 = 
42
2
 

p
3


 – 4
2 3
4
 = 8p
3
 – 4 3
  A sombreada: 
 4p – 

10p
3
 – 2 3 

 = 2p
3
 + 2 3
Clave C
14 
 • S1
(2r)2 = S1 + S2
(5r)2 = 
S1 + S2 + S3
(6r)2
  S1
4
 = S1 + S2
25
 = 
S1 + S2 + S3
36
  S1
4
 = S2
21
 = S3
11
 = k
  S1 = 4k; S2 = 21k; S3 = 11k;
  S2
S1 + S3
 = 
21k
4k + 11k 
 = 7
5
Clave D
15 
 
A
C
P
Q
O1
O2
BO
r
r
r
r r r
r
r
p
3
2p
3
2p
3
 • Siendo O1A = O2B = r = 3 µ 
 • L AB = 3r 

p
3


 = 3p
 • Long AC + Long BC = r 

2p
3


 + r 

2p
3


 = 4p 
  Perímetro = 4p + 3p = 7p
Clave A
ACTIVIDADES CAP 04
R.T. DE ÁNGULOS AGUDOS
A
B
O
Q
O1
P
2
2
2
2
2
2
2
p
3
3r
r
2r
S1 S2 S3
01 
3
3
5
q
4
a
 
 ∴ tanq = 3
4
 = 0,75
Clave E
02 
A P
a q β
Ta a a B
N
n
n
C
 ∴ P = 3a
n
 + a
2n
 2n
2a
 = 3 + 1
2
 = 3,5
Clave B
03 
 senq = 9
2b
 = b
7
 ⇒ b = 3 
7
2
 Reemplazando:
 P = 
3 
7
2
 
2
7
= 9
14
 
Clave A
04 
2nn
q
n 5 
 
 secq = n 5
n
 ⇒ secq = 5 
Clave E
05 
q
85k 84k
13k
65
A
C B
 Perímetro:
 2p = 182k .......(1)
 13k = 65k ⇒ k = 5
 En (1): 2p =910
Clave C
06 
a
13 12
5
 
 ⇒ 
3 12
13
 – 4 5
13
5 12
13
 + 4 5
13
 = 1
5
Clave E
07 
β
2 1
3
a
2
3
7 secβ = 2
3
 E = 
12 3
2
 
2
+ 9 1
3
2
3 7
3
2
 – 2
1
2
 ⇒ E = 4
Clave B
b bH
N
A C
B
q
q 7
2
08 
20 3 30 3 
20
40
x° y°
60120°
60° 60°
30
 coty – cotx = 8
2 3
 – 7
3 3
 = 5 3
9
Clave C
09 B P C
4a
2a
2a
4a
a 3a
N
A D
φ
a
2a 5 
 E = 
1
4
 + 4
2
 
2
5
 + 1
5
 ⇒ E = 53 3
40
Clave C
10 
n
an + 1
BA
C
 En dato:
 a
n
 + n + 1
n
 = 3
2
 ⇒ a = n – 2
2
 Teorema de Pitágoras:
 n2 + n – 2
2
 
2
 = (n + 1)2 ⇒ n = 12
 ⇒ 2 12
5
 + 13
5
 = 10
Clave D
CUADERNO DE TRABAJO
01 • senB = b
c
 • cotA = b
a
 • cscB = c
b
 
A
B
Cb
ac
  V = c



b
c
 – a



b
a
 + b



c
b
 = c
Clave E
02 3tanA = 2cscC  3



a
c
 = 2



b
c
  3a = 2b
2k 3k
  
A B
C
b a
c
 
 Como a2 + c2 = b2  4k2 + c2 = 9k2 
  c = 5k
 M = 5 



a
c
 + 6



b
a
 = 5
2k
5 k




 + 6



3k
2k
  M = 2 + 9 = 11
Clave D
03 • tanq = 6 • senq = 6
7
 P = ( 6)2 + 42 



6
7
 
1
6
7
  P = 12
Clave B
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
7
04 • tanq = a
b
 • tana = b
2a
 
b
a
a
  tana · tanq = b
2a
 · a
b
 = 1
2
Clave B
05 2,4 = 24
10
 = 12
5
 Perímetro: 
 30k = 180  k = 6 
5k
12k
13k
  13k = 78
Clave D
06 • secA = c
b
 • cotB = a
b
 
 P = c
2
b2 – a
2
b2 + 1 = c
2 – a2
b2 + 1 
  P = b
2
b2 + 1 = 2 
A
B
Cb
ac
Clave A
07 • senA = a
c
 • senB = b
c
 • tanA = a
b
 • tanB = b
a
 
 
A
BC
b
a
c
 
  E = 3



a
c




c
a
 – 



a
b




b
a
 = 2
Clave D
08 Se ve: s ≅s
 q + q
2
 = 90°  q = 60°
 
2 2
2
a a
  tanq = 
3
1
 = 3
Clave E
09 2x + x + 2 = 8  x = 2
 
x + 22x
5  tanq = 5
4
Clave A
10 En dato: c
b
 = 1
2
 · b
a
  2ac = b2 
 
 N = 
c2
a2 – 2 c
a
 + 1
2 + a
b
 · b
c
 · c
a
 = c
2 – 2ac + a2
 3a2
 
B
A Cb
ac
 Reemplazando: N = a
2 + b2 – b2 + a2
 3a2 = 2a2
3a2
  N = 2
3 Clave B
TAREA
01 
 E = 2 

3
2 2
 – 1
2 2


 = 2 

2
2 2


 = 1
cscq = 3
2 2
 
cotq = 1
2 2
 
3
1
2 2
q
02 
tanB = b
a
 cotB = a
b
cotA = b
a
 tanA = a
b
A
C Ba
b
c
 H = 

b
a
 + a
b
 

2
 – 

b
a
 – a
b


2
 = 2 + 2 = 4
03 Se cumple:
 • tanq + cotq = a
b
 + b
a
 • (a + b)2 – (a – b)2 = a2 + b2
 4ab = a2 + b2
 4 = a
2 + b2
ab
 = a
b
 + b
a
  E = a
b
 + b
a
 = = 4
04 • m2 + n2 = (2 mn)2
 m2 + n2 = 4mn
 m
n
 + n
m
 = 4
 • tana + 1
tana
 = 4
  tan2a – 4tana + 1 = 0
  tan2a – 4tana + 4 = 3
  (tana – 2)2 = 3
REFORZANDO
01 
Clave C
02
 
M = a
2
b2 + c
2
b2 + b
2
c2 – a
2
c2
M = a
2 + c2
b2 + b
2 – a2
c2
M = b
2
b2 + c
2
c2 = 2
C
B Ac
a
b
Clave B
03 R = 30°  cotR = 3
 K = 32 + 2 = 5
Clave B
04 
Clave A
W = 17 

3
34


2
 – 1 
W = 7
2
 = 3,5
3
5
34
q
K = 

a
c
 + c
a


 a
b
 ⋅ c
b
K = 

a2 + c2
ac


 · ac
b2
K = b2
ac
 ⋅ ac
b2 = 1
A
B Ca
c
b
05 
 Perímetro: 910
Clave C
06 
 Si CD = AB
  (a + 2)2 + b2 = (b + 2)2 + a2
  a = b  q = β
  cosq = cosβ
 senβ = senq
  M = 1
Clave A
07 
 
A
B Ca
c
b
• b2 = 5
2
 ac
 a2 + c2 = 5
2
 ac
2a2 – 5ac + 2c2 = 0
2a – c ⇒ c = 2a
 a – 2c
 • Ĉ > Â  c > a
  cotC = a
c
 = 1
2
Clave B
08 
 
A
N
q
q
r
r
2r 2rO
tanq = 2r
r
tanq = 2 
Clave E
09 cotq = 3r – r
r
 
 
3
 r – r
r
q
 cotq = 3 – 1
Clave D
10 tana = 3
1
  tana = 1
 
A
B
45°
D
CML
3
3
2
1
2L
L
a
q
Clave E
11  ACE
 a = 53°
2
 + 53°
2
 a = 53°
 A
B C
DO
E
r r
rr
53°/2 a
53°
2
Clave E
13K = 65
 K = 5
 AB = 420
 AC = 425
 BC = 65
A
B C65 = 13K
85K84K
q
β
A D E
C
B
2
2 b
a
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
8
12 
Clave D
13 • 78k + 77 = 89k  k = 7
 • Área = 78k ⋅ 80k
2
 B C
A
H 39k
80 k
89k
39k
78k + 77
 Área = 152 880 cm2
Clave A
14
 •12k = 48  k = 4
 • Perímetro:
 18k + 72 = 144 cm
 
48 12k
5k
13k
12 cm
12 cm
q
Clave B
15 
 ABC: (2b)2 + (2c)2 = (2a)2
 b2 + c2 = a2
	 CBM: b2 + (2c)2 = CM2
	 ABN: (2b)2 + c2 = AN2
5(b2 + c2) = CM2 + AN2
  CM2 + AN2 = 5a2 
Clave B
ACTIVIDADES CAP 05
R.T. DE ÁNGULOS NOTABLES
01 I – V; II – F; III – F; IV – V
Clave A
02 E = 32 (2)(1)
2 1
3
2
 + ( 2)2
 ⇒ E = 9
4
 = 2,25
Clave B
03 x 
1
2
 – 2 = x 3 
 ∴ x = –4(1 + 2 3)
11
Clave A
04 8 3
4
 – x(1) = 3csc32°sen32°
 ∴ x = 3
Clave E
F
4 4K
11K
11K
4k
7k
7
A D
CB
E
q
q
tanq = 7k
11k
tanq = 7
11
A
B
C
NM
b c
cb
2a
05 
 
30°A
B
H
C
45°
n
n
b
2n = 3 
n 2 
n 3 
 2n = 3 ⇒ n = 3
2
 ⇒ b = 3
2
 3 + 3
2
 ⇒ b = 3 + 3
2
Clave B
06 x – 30° = 70° – x
 2x = 100°
 ∴ x = 50°
Clave C
07 3x + 30° + 5x – 4° = 90°
 8x = 64°
 ∴ x = 8°
Clave D
08 sec(3x + 43°) = csc(8x – 30°)
 ⇒ 3x + 43° + 8x – 30° = 90°
 11x = 77°
 ∴ x = 7°
Clave B
09 
53° 3 H
2
45
O B
E
A
q
37°
 cotq = 4
2
 
 ∴ cotq = 2
Clave B
10 
2
1 H
1
O
C
B
D
E
A
1
1
1 1
q q 3 – 1
 
 
 OED: ED = 3 
 CHD: cotq = 3 – 1
1 Clave E
CUADERNO DE TRABAJO
01 
 
8



3
4
 – x(1)
1
2
 = 3



2
1
  6 – x = 3  x = 3
Clave C
02 
  tanq = 9
31
 
B
15
31 12
9
A C
37°
H 
Clave B
03 cota = 5
4
 E = 41 · 4
41
 + 8

5
4


 = 14 
4
5
41
Clave D
04 60° – x + 70° – 3x = 90°  x = 10°
Clave D
05 90° – 2x = 3x  x = 18°
  csc30° · tan60° = 2 · 
3
1
 = 2 3
Clave B
06 (a + b + c) + (a – b + c) = 90°  a + c = 45°
 M = tan60° · sen45°  M = 
3
1
×
2
2
 = 
6
2
Clave E
07 sen3x · sec6x = 1  sen3x = cos6x
 3x + 6x = 90°  x = 10°  x = p
18
Clave C
08 E = 
1
32 + 
1
32
 + 
2
32 
 E = 
6
319 
4
1
1
A C
B
2 3
60°
Clave A
09 tan(30° – q) = cot(30° + 3q)
 30° – q + 30° + 3q = 90° 
  q = 15° 
 30°
15°
15°
20
20x
  x = 10
Clave B
10 csc(90° – a) = 5
4
  sen(90° – a) = 4
5
  90° – a = 53°  a = 37° 
 k = sen37° · tan53° + cos37° = 3
5
 · 4
3
 + 4
5
 = 8
5
Clave D
TAREA
01 E: 1
2
 ⋅1 ⋅ 4
5
 = 4
10
 = 2
5
02 • sen(2x – 10°) = cos(4x + 20°)
 (2x – 10°) + (4x + 20°) = 90°
 3x = 40°
 • sen(3x + 5°) = sen45° = 
2
2
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
9
03 
 
6 2
12 = a
6 36
45° 30°
6
b
B
A H C
  a + b = 18 + 6 3
04 • sec(2x + 15°) = csc(3x + 20°)
  2x + 15° + 3x + 20° = 90°
 x = 11°
 • tan(4x + 1°) = tan45° = 1
REFORZANDO
01 
Clave B
02 2x + 3x = 90°  x = 18° 
 sen(36° – 6°) + 2ctg(54° – 9°) – cos (72° – 12°)
 sen30°+ 2ctg45° – cos60° = 2
Clave B
03 secx = cscy  x + y = 90° 
 2° + 2β + 4a – 2° = 90°
 2β + 4a = 90°
 β + 2a = 45°
 R = sec2β – tg45°
csc4a – ctg45°
 = csc4a – 1
csc4a – 1
 = 1
Clave A
04 sena = cosβ  2x2 + 5x – 1 = 2 – 3x – x2 
  3x2 + 8x – 3 = 0  x = 1
3
 
 sena = 2x2 + 5x – 1  sena = cosβ = 8
9
  tana = 8
17
, senβ = 17
9
 
  M = 2 17 

8
17
 + 17
9


 + 2
9
  M = 20
Clave A
05 • cosx = 2
3
 
  tanx = 5
2
 cscx = 3
5
 • M = 5
2
 ⋅ 3
5
 + 
2 ⋅ 1
2
2
 = 2
Clave D
ctga + 2tgβ = 6
ctga + 2ctga = 6
ctga = 2
5 

5 + 2
5


 = 7
1
2
5
a
β
17
8
9
a
β
5
2
3
x
a
06 
 
β
a
1
2
5
 tgβ = 2ctgβ ⋅ ctgβ
3ctg2β + ctg2β
  tgβ = 1
2
 Luego:
 5 (secβ – sena) = 5

5
2
 – 2
5


 = 5
2
 – 2 = 1
2
 
Clave D
07 cos6x = sen(8y + 10°)
 6x + 8y = 80°
 sen(40° – x – y) csc(40° – x – y) + 2tg(30° + 
x+ 3y)ctg(30° + x + 3y) + 3sen46°sec(80° – 
36°)
 ⇒ 1 + 2(1) + 3sen46°csc46° = 6 
Clave E
08 
 sec33°tga – sec33° = 2sec33°sen(β – 8°)
csc(β – 8°) – sec33°tga
 tga – 1 = 2 – tga
 2tga = 3  tga = 3
2
 ( 13 – 2) 

13 + 2
3


 + 

13
2


2
 
3 + 13
4
 = 25
4 Clave D
09 
 4x – 30° = 40° – 3x
 x = 10°
 sen20° – cos70°
 cos70° – cos70° = 0
Clave E
10 
 1
2
( 17a)(5a)sena = 1
2
 ⋅ 4a ⋅ 4a
 sena = 16
5 17
, csca = 5 17
16
 256
5
 

25(17)
256


 – 4 = 81
Clave B
aa/2
3
2
13
13
3tg(4x – 30°)sen32°
sen32° ⋅ sen60° ⋅ tg(40° – 3x)
 = 2
3tg(4x – 30°) = 2 3
2
tg(40° – 3x)
a
A D
CB
4a 5a 4a
3aa
17
a
E
11 tanx + 2tanx ⋅ cotx – 6 ⋅ 1
2
 ⋅ cotx = 0
 1
  cotx = 1 ⇒ x = 45°
 Piden: 2 2 (cos45° + sec45°)
 2 2 

1
2
 + 2 

 = 6
Clave C
12 sec(a + 2b)cos(25° – c) = csc(b – 2a)cos(25° – c)
  sec(a + 2b) = csc(b – 2a)
  a + 2b + b – 2a = 90°  3b – a = 90° 
 M = sec(90°–30°) + tan

90°+60°
2




90°
3
 – 15° 

 
 M = sec60° + tan75° ⋅ tan15°
 M = 2 + (2 + 3 )(2 – 3 ) = 3
Clave D
13 
 cot C
2
 = a + b
c
 = a
c
 + cscC
 cot C
2
 – cscC = a
c
 cot C
2
 = cot (2C – 10°)
 C
2
 = 2C – 10°  C = 20°
3
 
 sen9C + cos(3C + 10°) 
  sen60° + cos30° = 3
Clave C
14 sena ⋅ x2 + 2x ⋅ sena + cosβ = 0 
 sena ⋅ x2 + 2sena ⋅ x + cosβ = 0 
 x = –2sena ± 4sen2a – 4senacosβ
2sena
 
  4sen2a = 4senacosβ  sena = cosβ 
  a + β = 90°
Clave A
15 
 (sec2A)° – (csc2B)g ⋅ 9°
10g = p
18
rad ⋅ 180°
prad
 sec2A – 9
10
 csc2B = 10
 10sec2A – 9csc2B = 100
 10csc2B – 9csc2B = 100
 csc2B = 100  cscB = 10
  tanA = 3 11
Clave C
A
B EC
c b
a b
C
2
C
2
A
BC
1
10
3 11
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
10
ACTIVIDADES CAP 06
RESOLUCIÓN DE TRIÁNGULOS RECTÁNGULOS
01 B
H
20
A hcotq
h
htanq
q
q
C
 htanq + hcotq = 20
 h 
25
12
 = 20 ⇒ h = 9,6 
Clave E
02 ABC: EC = m
2
 secq
 ECD: DC = m
2
 secqcota
Clave C
03 
n b
n
B
q
q
A
N
C tanq = b
n
 = n
n + b
 b2 + bn = n2
 ⇒ b2 – n2 = –bn
 P = b
n
 – n
b
 = b
2 – n2
bn
 = –bn
bn
 ⇒ P = –1
Clave E
04 
CA
B
MS
q
a S
n
a
b
 
 S = an
2
 senq = nb
2
 sena
 ⇒ senq
sena
 = b
a
Clave B
05 
A C
H
B
L
2
 secq L
2
 secq
L/2 L/2
q q
 
 
 Perímetro de ABC:
 2p = L + Lsecq
 2p = L(1 + secq)
Clave B
06 Sea: x la longitud del lado del cuadrado 
PQRS
 APQ: AP = xcotq
 RSC: SC = xcotq
 ⇒ 2xcotq + x = L ⇒ x = L
2cotq + 1
Clave C
07 CB = 2cotq = 7tana
 ⇒ 2
7
 1
tana
 = 1
cotq
 ⇒ tanq = 2
7 cota
Clave B
08 
B
q
A
m
mtanqmsecq
C 
 Perímetro:
 2p = m + mtanq + msecq
 ∴ 2p = m(1 + tanq + secq)
Clave B
09 B
E
DA acota acota
a
x
a β
C
 ABC: x = 2acotasenβ
Clave B
10 
k
x
xcos6q
xc
os
q
xc
os
3 q
xc
os
2 q
xc
os
4 q
xc
os
5 q
q
q
q
qq
q
q
 xcos6q = ktanq
 ∴ x = ksec6qtanq
Clave C
CUADERNO DE TRABAJO
01 ED = m(cosq – senq)
 
 
A E
B C
D
m
 Clave C
02 • BC = mtanq 
 • CD = mtanq · cosa
Clave D
03 tana = 
5
5
2 = 1
2
 tanβ = 0,5
1
 = 1
2
 
 tana + tanb = 1 2
1
1
0,51,5
5
2
5
β
β
Clave D
04
 Rcosq + x = R
 x = R(1 – cosq)
 
R
xO E
A
B
Clave A
05 AB = ncota
 Área = AB · BC
2
 
A
B Cn
 Área = n
2cota
2 Clave C
06 
 CD = L tan a
2
 
 Área = L2 tan a
2
 A
M
B C
D
L
L 2
Clave B
07 2hcotq + hcota = d
  h = d
2cotq + cota 
M
h
h
Clave A
08 2
3
13
x
3
 x
3
 = 
13
3  x = 
13
9
 
3
3
3
1
11
2
x
  x–1 = 13
9
Clave B
09 Área ABC = (6)(8)senq
2
 = 24senq
 Área BDE = (2)(7)senq
2
 = 7senq
  Asombreada = 24senq – 7senq = 17senq
Clave A
10 Área = 
px
2
 = 
pqcosa
2
 + 
xqsena
2
 
  px = pqcosa + xqsena
  x(p – qsena) = pqcosa
  x = 
pqcosa
p – qsena
 D AC
B
xp q
Clave C
TAREA
01 BC: mtana  CD: mtana ⋅ cosβ
02 
 x = mtana ⋅ secβ
 
mtana
m
x
a
β
03 B
A
4
C
2
3
8
3
60°
30
°
• Área:
 (8 3)(2 3)
2
 = 24
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
11
04 
 
 Perímetro = n(1 + cscq + cotq)
REFORZANDO
01 
 y = 4secq
 x = ytanq
 
  x = 4secq

senq
cosq


  x = 4sec2q ⋅ senq
Clave D
02 
 • ED = 6sena 
 • AD = 3csca
  AE = AD – ED = 3(csca – 2sena)
Clave B
03 BD = n⋅tan53° = 4n
3
 CD
BD
 = 3
2
  CD = 4n 3
3⋅2
  CD = 2n 3
3
Clave C
04 x = 16cot37°
 x = 16 ⋅ 4
3
 x = 64
3
Clave C
05 • BC = 4tana
 • CD
BC
 = cotq
  CD = 4tana ⋅ cotq
Clave B
06 
 
A DQ
O 3
3
aa
a
B CN
T
M
P
6
 NT = TP = 3sena
 ND = 6csca
  PD = 6csca – 6sena 
ncotq
ncscqA C
B
n
q
A
B y
x
E
C
q
q
4
B
C
3
3
EA D
a
a
CB
A
D
12
20 16
53°
37
°
37°
x
 = 6 

1
sena
 – sena 

 = 6(1 – sen2a)csca
Clave B
07 
 • Tenemos DC = 2 ⋅ AC
 Si AC = x  DC = 2x
 • T.P. (2x + 10)2 + x2 = 625
  4x2 + 40x + 100 + x2 – 625 = 0
  x = 7
  ctga + 5 csca + ctgβ + 4
7
 = 2 + 5 ( 5) + 24
7
 + 4
7
 = 7 + 4 = 11 
Clave D
08 
 Área = 1
 ⋅ k
2
 = 1
 k = 2
 tga = 1
2
 = BC
AC
 = 4(2)
AC
  AC = 16
 EC = 14
  ED = 2
  FD = 5 
 Además FB = 7 5  tgq = 7
Clave A
09 
 tga = 4
3
 = 4t
3t
 ; 3k = 5t  k = 5t
3
 
 3ctgβ = 3 

8t
3
t
 = 8 
Clave C
10 
7 5
A
B CD
25
10 2x = 14 
x = 7 
β a
B
A CDEk
1
F
6p
7p
4k
qa
8 5
C
A D
a a
β
E
B
t
4t
3t
2k
5t
k
4t
3
8t
3
4sena
2sena
3cosa
B
3
1
2
M P
Q
H
N
A
C
D
a
a
a
 En ADN: 4sena = CD = PQ
 4sena = 2sena + 3cosa
 2sena = 3cosa
 tga = 3
2
  tga + ctga = 3
2
 + 2
3
 = 13
6
Clave A
11 
 seca = 5
3
; t = 3k
5
 cotq = 
5k + 12
5
 k
9
5
 k
 = 37
9
 
  9ctgq – 5 = 37 – 5 = 32
Clave A
12
 
 secq + tgq = 5
3
 + 4
3
 = 3
 
17k
A 4k
4k
5k3k
k
D
T
B
Ca
q
a
Clave C
13 
 T.P.: DE = 2 2; AD = 2 6
  M = 
3 + 2 

4
2 2


 
2 

2 6 + 4
2 2


 
 = 5
6 + 2
 
  M = 5
2
 6 – 1
Clave A
14 
 
25k25
k
25k
7k 7k
k
q18k
24k
16°
B C
A D
⇒
q 18k
k
 2cot2q = ?
 Entonces:
 tanq = k
18k
  tanq = 1
18
 
 Por identidad del ángulo doble
 tan2q = 2tanq
1 – tan2q
 
 Reemplazando: 
  2cot2q = 323
18
 
Clave C
3k
5t
4k
3k
a
a
q
5k
A
B C
E
D
9
5
k = 3t 4t = 12
5
C
AM
D
3
q
q
aa/2
a
3
21O BE
2
2 6
2 2
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
12
15 
 
3 3 53°
a
a
a
q
q
A
B
T
C
PE8k
3k
8
3k
6–8k
D
 • 9ABC (notable 37 y 53)
 Sea AB = 8  BC = 6
 • 9ABD ∼ 9TBE: BT = 3k ∧ TE = 8k
 • tana = 3
8
 = 8k
8 + 3k
  k = 24
55
 
 • ECP: tanq = 3k
6 – 8k
 Reemplazando el valor de k:
 
  tanq = 
3 ⋅ 24
25
6 – 8 ⋅ 24
55
 
  tanq = 12
23
Clave A
ACTIVIDADES CAP 07
ÁNGULOS VERTICALES Y HORIZONTALES
01 
37°
45°
x 9/4
3
3 3
4
4 3
4
 x + 9
4
 = 12
4
 
 ⇒ x = 3
4
 = 0,75 m
Clave B
02 
3
3
60°
1,73
3,46
a
2 3
3 3
33
 ⇒ cota = 2 3 + 3
3 3
 ∴ cota = 2 + 3
3 Clave D
03 
2n 10 km2n – d
(2n – d) 3
d30°
30°
a
a
2d
n 3 d 3A
B
	 De	la	figura:	(2n – d) 3 = n 3 + d 3
 2n – d = n + d 
 ⇒ n = 2d = 5 km
 Se pide: AB = 2d = n = 5 km
Clave A
04 
6 3
T
H
B
A
4n
3n
30° 53°
 ⇒ 6 3 + 3n = 4n 3
 n = 
6(4 + 3)
13
 ∴ TH = 24(4 + 3)
13 Clave B
05 
x 3
9 3
33
x/ 3
x x
30°
30°
60°
 x 3 + 3 + x
3
 = 9 3
 ⇒ 4 3x = 24 3
 ∴ x = 6 m
Clave D
06 
h
hcotφ H
A C
B
hcota
d
φ a
 Dato: cotφ + cota = 1
2
 
 hcotφ + hcota = d
 
 ⇒ h 1
2
 = d ∴ h = 2d
Clave A
07 
H
A B
P
M
d
a β
h
2
cota h
2
cotβ
h
2
h
2
 ⇒ h
2
 cota + h
2
 cotβ = d
 ∴ h = 2d
 cota + cotβ 
Clave D
08 
S2 S1
h
β
β
hcota
hcotβ
a
a
 S1 + S2 = 4(cota + cotβ)
 ⇒ hcota + hcotβ = 4(cota + cotβ)
 ∴ h = 4
Clave B
09 
12 16
37°
37°
45°
45°
P
H
A B
d
12
 ⇒ d = 12 + 16
 ∴ h = 28 m
Clave C
10 
d1 d2
a
a
β
β
γ
γ
h
a a 2a
P
H A B C
 ⇒ 
d1
d2
 = a
2a
 
 ∴ 
d1
d2
 = 1
2
Clave ACUADERNO DE TRABAJO
01 	 Del	gráfico	se	observa:
 hcota – hcotb = 20
 h(cota – cotb) = 20 
  h(0,25) = 20 
  h = 80 
h
20
Clave B
02 	 Del	gráfico:
 x + 36 = 48
  x = 12
53°45°
x 3(12) = 36
48 Clave C
03  D = h(cota – cotb)
 
h
a b
hcotbD hcota Clave D
04 tanb – tana = 3
10
 h
20
 – h
50
 = 9
10
  h = 30 
h
30 20AB
Clave B
05 
30°
2040  H = 60
Clave E
06 24a = 300 + 14a 
  a = 30
 h = 7a = 7(30) 
7a
10a
300 100
14a
h
53°/216°
 
  h = 210
Clave C
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
13
07 cot15° = 3d
1,5
 30° 60°
15°
2d
d
2d
1,5
  2d = 2 + 3
Clave B
08 Htan2q = H – h
 
(H – h)
H
h
90°– q
  h
H
 = 1 – tan2q
Clave C
09 
  3h = 3(6) = 24
 
h = 6
3h
3h
45°
53°
Clave C
10 mBAOB = 90° – (22°30' + 11°15')
 mBAOB = 56°15'
 a = 180° – mBAOB
2
 
  a = 61°52'30''
 q = a – 11°15' 
  q = 50°37'30'' 
A
B
O
N
N
11
°15
'
11
°15
'
56°15'
22°30'
E
E
d
d
  B se encuentra en dirección S50°37'30''E 
respecto de A.
Clave C
TAREA
01
  D = h(cota – cotb)
 
a
D
b
h
02 
 • 3
4
 = 24 – h
24
 
  h = 6
 
 
53°
(24 – h)
(24 – h)
h
24
(24 – h)
45
°
45°
53°
03 
 4k = 16
 k = 4
  x = 4
 
 
45°
4k
x k
3k
16
37°
37°45°
04
  x = h(cotb – cota)
 
 
a
h
b
x hcota
hcotb
REFORZANDO
01 
  x = 225 3
 
Clave D
02 
  x = 80
Clave A
03 
  H = 18
Clave A
04 
 H = 25k 
  H = 25 

3,6
9


 
  H = 10
 
Clave A
05 
  x = 8
 
Clave D
06 
 
 
Clave B
07 
 • cot2q = x
25
 
 x
25
 = 12
5
 
 
13
B
12
12
A
C
x
5
2q
q
q  x = 60
 
Clave E
08 
 
  cotq = 4 5
3
 
37°
8d
4d
5d
3d
S
E
E
q
4 5d
N
 
Clave B
x
225
30°
30°
x
60
37°
37°
15°
15° 30°
36 H
36
53° 12k
37°
3,6 9k
16k
H
x
24
24
32
37° 45°
150 m
200 m
53°
53°
09 
 
 H = 6(1,7) + 1,5  H = 11,9 
Clave C
10 
 3h = 3(6) = 18
 
 
Clave A
11 
 
 
 
a
a90 – a
1 m 1 m1 mx x
2 m
5 m
 tana = x
5
 = 2
x
  x = 10  cota = 10
2
 Clave E
12 
  x = dtanq
 d
d
x
q q
q
 
Clave A
13 
 
  cota – tana = 2  tana + cota = 2 2
 
Clave B
14 
 
  cotβ = 4cota + 3tana
1
  cotβ = 4 3 
 
Clave D
15 
 17
6cota 7tanaa
a
q
q
6 7
  7tana + 6
tana
 = 17
  7tan2a – 17tana + 6 = 0
30° 60°
12
12
1,5 1,5
H
6
6 3
3h
45° 53°
6 = h
3h
37
°
a
a
h
2(H–h) (H–h)tana
(H–h)cota
h
H
(H – h)
β
a
a
4cota 3tana
1
11
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
14
 
  tana = 3
7
 ∨ tana = 2
  tanq = 7
3
 ∨ tanq = 1
2
 
Clave A
ACTIVIDADES CAP 08
PLANO CARTESIANO
01 PQ = [a – 2 – (a + 3)]2 + [b + 1 – (b – 1)]2
 PQ = (–5)2 + 22 ⇒ PQ = 29
Clave D
02 G –4 + 2 + 8
3
; 1 + 5 + 3
3
 ⇒ G(2;3)
Clave A
03 S = 1
2
 
1 1
–3 5
5 0
1 1
 = 1
2
|10 + 22| ⇒ S = 16
Clave B
04 
h S
H
A(–1; 1)
C(5; –3)B(3; 7)
 BC = (3 – 5)2 + (7 – (–3)2
 ⇒ BC = 2 26
 S = 1
2
 
–1 1
5 –3
3 7
–1 1
 ⇒ S = 26
 (2 26)h
2
 = 26 ⇒ h = 26
Clave B
05 L1: y + 2 = 4 – (–2)
7 – (–1)
 (x + 1) ⇒ y = 3
4
x – 5
4
 L2: y – 3 = –1 – 3
8 – (–2)
 (x + 2) ⇒ y = – 2x
5
 + 11
5
 Igualando: 3
4
x – 5
4
 = – 2x
5
 + 11
5
 ⇒ x = 3
 y = 1
 ∴ P = (3; 1) 
Clave C
06 Q(4; –2)
P(–2; 10)
2n
n
A(x; y)
 
 x = –2(2) + (4)1
2 + 1
 = 0
 y = 10(2) + (–2)1
2 + 1
 = 6
 ∴ A(0; 6) 
Clave D
07 x + (–2) = 1 + 0 ⇒ x = 3
 y + 6 = 4 + 8 ⇒ y = 6
 ∴ x + y = 9
Clave B
7tana – 3
 tana – 2
08 C(x; 8)
M(a; b)
B(–1; 5)
A(1; –7)
n
n
 Pendiente:
 ⇒ 5 – (–7)
–1 – 1
 = 8 – 5
x – (–1)
 
 ⇒ x = – 3
2
 M: punto medio de BC
 2a = –1 – 3
2
 ⇒ a = – 5
4
 2b = 5 + 8 ⇒ b = 13
2
 ∴ M – 5
4
; 13
2
 
Clave C
09 P(1; 0); Q = 5
2
; 0 ⇒ PQ = 3
2
 Área: SPAQ = 1
2 
3
2
 (1) ⇒ SPAQ = 3
4
Clave A
10 
 
 Menor distancia:
 ⇒ 1 + d = 32 + 42
 
1
d
1
0
Y
X
P(3; 4)
∴ d = 4
Clave B
CUADERNO DE TRABAJO
01 x = –8 + 2
2
 = –6
2
 = –3 ; y = 6 – 4
2
 = 2
2
 = 1
  M(–3; 1)
Clave E
02 	 Graficando:
 
2
3
A(2; –3)
B(–4; 1)
P(x0; y0)
 x0 = 2(–4) + 3(2)
3
 = –8 + 6
5
 = –2
5
 y0 = 2(1) + 3(–3)
2 + 3
 = 2 – 9
5
 = –7
5
  P(x0; y0) = P



2
5
7
5
– ; –
Clave B
03 x0 = 2(6) + 5(2)
2 + 5
 = 12 + 10
7
 = 22
7
 y0 = 2(–2) + 5(7)
2 + 5
 = –4 + 35
7
 = 31
7
  P(x0; y0) = P



22
7
31
7
;
Clave D
04 G es baricentro.
 
 
A(2; 5)
C(8; 3)
B(–4; 1)
M
N
G(x0; y0)
 x0 = –3 + 6 + 9
3
 = 12
3
 = 4
 y0 = 4 – 5 – 11
3
 = – 12
3
 = –4
  G(x0; y0) = G(4; –4)
Clave B
05 
 
+
=
+–8
4
–15
–19
3
20
8
31
3
–4
4
3
2
1
–5
2
A
B
C
A
 Finalmente:
 S = 31 – (–19)
3
 
 S = 31 + 19
2
 = 50
2
 
Y
X
C(4; –5)
A(3; 2)
B(–4; 1)
S
  S = 25
Clave C
06 l = (–1 – 0)2 + (2 – 3)2
 l = 1 + 1  l = 2
 Hallando el perímetro:
 2p = 4l 
  2p = 4 2 B(0; 3)A(–1; 2)
Clave C
07 P(–12 + 5; –1 + 2) = P(–7; 1)
 Q(–7 + 5; 1 + 2) = Q(–2; 3)
 
 B(–12; –1)
A(3; 5)
5
P
Q
5
5 2
2
2
15
Clave D
08 • c – b
2
 = b
3
 = c
5
  b = 3
5
 c
 • b + c = 32
3
  3
5
 c + c = 32
3
 
Y
X
(–3; 0)
2
3
(2; c)
(0; b)
c – b
b
  c = 20
3
  b = 4
Clave D
09 • AD = DB 
  (a + 8)2 + (0 – 0)2 = (a – 10)2 + 122
 (a + 8)2 = (a – 10)2 + 122
 (a + 8)2 – (a – 10)2 = 144
 18(2a – 2) = 144  a = 5
Clave B
10 
 
S/4
S/4
S/4
S/4
C(4; –6)
B(6; 8) B
C
A
Triángulo
mediano
A(–2; 4)
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
15
 S = 1
2
 
A
C
B
A
 = 1
2
 
+ +16
–36
–16
–36
12
32
24
68
–2
4
6
–2
4
–6
8
4
  S = 52  S
4
 = 52
4
 = 13
Clave C
TAREA
01 El punto medio es:
 M = 

7 + 5
2
; 3 + 1(–1)
2


 M = (6; 1)
02 Si el baricentro es (x; y)
 x = –2 + 1 + 4
3
 = 1
 y = 3 – 3 + 2
3
 = 2
3
 
  Las coordenadas 

1; 2
3


 
03 AB = (2 + 4)2 + (5 + 1)2 = 6 2
 BC = (2 – 5)2 + (5 + 1)2 = 3 5
 AC = (5 + 4)2 + (–1 + 1)2 = 9
  Perímetro = 6 2 + 3 5 + 9
04 
 
 
1 3
5 0
7 1
1 3
A
B
C
A
S = 1
2
 = 1
2
⇒ S = 1
2
 |26 – 16| 
  S = 5
REFORZANDO
01 I. (V) II. (F) III. (V) IV. (V)
  VFVV
Clave B
02 OA = (–5)2 + (3)2 = 34
 OB = 32 + 42 = 25
 OC = (–3)2 + (–2)2 = 13
 OD = (–4)2 + 12 = 17
 OE = 52 + 22 = 29
  El punto C esta mas cerca.
Clave C
03 • a + 3
2
 = 8  a = 13
 • b + 5
2
 = 9  b = 13
  M = a – b = 0
Clave A
A(1; 3)
C(7; 1)
B(5; 0)
04 
Clave E
05 d(AB) = (7 – 1)2 + (5 – 2)2
 d(AB) = 362 + 9 = 3 5
  Área = (3 5)2 = 45
Clave B
06 • S1 = 3 ⋅ 6
2
 = 9 
 • S2 = 4 ⋅ 2
2
 = 4
 • S3 = 4 ⋅ 2
2
 = 4
 
 
 
B
D
A
6
6
S
S2
S3
S4
S1
EO
Y
X
 • S4 = 2 ⋅ 3
2
 = 3
 S + S1 + S2 + S3 + S4 = 6 ⋅ 6
 S + 9 + 4 + 4 + 3 = 36
  S = 16
Clave A
07 
 
 
 
B(3; 7)
C(5; 5)
A(1; 2)
D(x; y)
Y
X
 1 + 5 = 3 + x  x = 3
 7 + y = 2 + 5  y = 0
  xy = 3 · 0 = 0
Clave E
08 1 + (–1) + x
3
 = 0  x = 0
 1 + 3 + y
3
 = 0  y = –4
  C(0; –4)
Clave E
09 
M 

– 1 + 1
2
; 5 + 1
2
 

 = M(0; 3)
MC = (4 – 0)2 + (5 – 3)2 = 2 5
Clave E
(7; 7)
6
2
22
222
2
2
2
2
2
2
6
(5; 5)
(3; 3)
(1; 1)
C(4; 5)
M
A(–1; 5)
B(1; 1)
10 AB = 92 + 122 = 15 
 AC = 252 + 02 = 25
 BC = 162 + 122 = 20
  El 9 es rectángulo
Clave D
11 
 
C(7; 9)
P(x; y)
3k
4k
A(0; 0)
 • x = 0(3) + 7(4)
4 + 3
 = 4
 • y = 0(3) + 9(4)
4 + 3
 = 36
7
  x + y = 36
7
 + 4 = 64
7 Clave C
12 El baricentro del 9ABC es el mismo que 
el baricentro del 9MNP
 x = 1 + 2 + (–6)
3
 = –1
 y = 3 + (–1) + (–5)
3
 = –1
 El baricentro es: (–1; –1)
  (–1) + (–1) = –2
Clave E
13 Siendo:
 G 

–5 + 3 – 1
3
; 3 + 2 – 5
3


 = G 

–1; 1
3


 
 
 
 2 13C(–1; –4) B(3; 2)
h
q
G 

–1; 1
3


 
 
–1 1
3
–1 –4
3 2
–1 1
3
S = 1
2
 = 1
2
 |3 + 43
3
|= 52
6
 = 26
3
  2 13 ⋅ h
2
 = 26
3
 h = 2 13
3
Clave B
14 
 • –a
–b +r
 = x
r
  x = ar
r – b
 • S = 1
2
(xr)  S = ar2
2(r – b)
Clave B
20
15
25
(–b; –a)
(a; b)
r
x–a
–b
r
r
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
16
15 Siendo C(x; y)
 • –5 + x = –3 + 2  x = 4 
 • –1 + 2 = –5 + y  y = 6 
 
 S = 1
2
|–42 – 40|= 41
  S1 + S2 = 41
2 Clave B
ACTIVIDADES CAP 09
R.T. DE ÁNGULOS DE CUALQUIER MEDIDA
01 I. (+)
(–)(–)
 = (+) II. (–)(–)
(–)
 = (–)
 III. (+)(+)
– (–)
 = (+)
Clave A
02 270° < a < 360° ⇒ 135° < a
2
 < 180°
 90° < a
3
 < 120°
 I. T = (+)(–) = (–)
 II. M = [–(–)][–(+)] = (–)
 III. M + T = (–) + (–) = (–)
Clave E
03 x = –3; y = –6; r = 3 5
 G = 5 –6
3 5
 – –3
3 5
 = –1
Clave B
04 x = –3; y = 4 r = 5
 seca + tana = 5
–3
 + 4
–3
 = –3
Clave B
05 
β
φ
r
r
0
Y
X
(–b; –a)
(a; b)
 ⇒ tanφtanβ = –a
–b
 b
a
 
 ∴ tanφtanβ = 1
Clave E
06 a = 5k; β = 2k
 5k – 2k = 360°n
 k = 120°n
 a = 600°n; β = 240°n
 1441 < 600n – 240n < 2159
 4,01 ... < n < 5,99 ...
 ⇒ n = 5
 ∴ a = 600°(5) = 3000°
Clave C
A
D
C
B
A
S = 1
2
 = 1
2
–5 –5
–3 2
4 6
2 –1
–5 –5
07 
r
r
0 q
Y
P(–3; 5)
(0; 0)
H(3; –5)
X
 x = 3; y = –5
 r2 = 32 + (–5)2 
 ⇒ r = 34
 ⇒ F = 3 34
3
2
 – –5
3
 ∴ F = 13
Clave A
08 
0
a
–β
Y
(–2; 4)
X
2 5
 a – β = 270°
 –β = 270° – a
 sec(–β) = sec(270° – a)
 secβ = – csca
 tanβ = cota
 ∴ sec2β – tanβ = – 2 5
4
2
 – –2
4
 = 7
4
Clave C
09 Y
(–4; –3)
(–3; –4)
X
–β
ω
–q
a a – β = 270°
 –β = 270° – a
 cot(–β) = cot(270° – a)
 cotβ = –tana
 ω – q = –270°
 q = 270° + ω
 tanq = –cotω
 ∴ tanq·cotβ = – 4
3
 – 4
3
 = 16
9
Clave C
10 
ω
–a
–β
φ
Y
(–1; 2)
X
(–1; – 3)
 tanβ = –cotφ
 secβ = cscφ
 sena = cosω
 ⇒ 5 –1
5
 – 3 2
– 3
 + 3 –1
3
 
 ∴ 1+ 2 – 1 = 2
Clave C
CUADERNO DE TRABAJO
01 • csca < 0  a  IIC o IIIC
 • tana > 0  a  IC o IIIC
  a  IIIC
Clave C
02 • csca = r
y
  csca = 5
–4
 ; r2 = 32 + 42
 • cota = x
y
  cota = –3
–4
 = 3
4
  E = –5
4
 + 3
4
 = –2
4
 = –1
2
 
Clave E
03 El ángulo coterminal es: 360°k + 20°
 750° < 360°k + 20° < 8000°
 37,5° < 18°k + 1 < 400°
 21; 22
 • 360°(21) + 20 = 7580°
 • 360°(22) + 20 = 7940°
 15520° Clave E
04 a  IIIQ b  IVQ
 P



a
2
a
2
– –3; Q(a 3; –a)
 tana = 
3
1
 tanb = 
3
–1
  E = tana + tanb = 0 = 0
Clave A
05 Como a y b son coterminales se cumple
  a – b = 360°k
 E = 
tan(3(a – b) + 45°) + csca · sena
sen(2(a – b) + 30°)
  E = tan45° + 1
sen30°
 = 
2
1
2
 = 4
Clave C
06 I(F) II(V) III(F) IV(F)
  Solo II. Clave C
07 tanq = 
y
x
 = 4
–7
 
 tanq = –4
7
 
Y
X
(–7; 4)
4
7
–7
(4; 7)
Clave B
08 • tana = –1
3
  a  IIC
 
 K = 
3
10
–3 + 5 
10
1 



2(–3)
 
Y
X–3
1
10
 
  K = 1
–2 10
2 



 = – 10
10 Clave B
09 tanq = –4
3
; (q  IVQ)
 r = 5
 • cscq = –5
4
 • secq = 3
5
 
Y
X
(3; –4)
3
5
–4
  E = 12



5
3
+–5
4
 = 5
Clave A
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
17
10 (4k + 1) p
2
; (4k – 1) p
2
; (2k – 1) p
2
 son ángulos 
 cuadrantales y lo reemplazamos por sus 
coterminales:
 E = a
3cot90° + sen270°
atan180° + cos180°
 = a
3(0) + (–1)
a(0) + (–1)
 = 1
Clave C
TAREA
01 x = – 12 y = 5 r = 13
 E = 13
–12
 + 5
–12
 = –18
12
 = – 3
2
02 senq = –1
3
 = y
r
  y = – 1; r = 3: x = –2 2 
 secq = 3
–2 2
 tanq = –1
–2 2
 E = 2 

–3
2 2
 – 1
2 2


 
 = 2 

–4
2 2


 = –2
03 Q = (+) ⋅ (+)
(–) ⋅ (+)
 = (–)
04 • a = 5k ∧ q = 3k
  5k – 3k = 360n
 k = 180n ∧ a = 900n; q = 540n
 • 4032 < 5k + 3k < 4608
  4032 < 1440n < 4608
 n = 3
  a = 2700°; q = 1620°
REFORZANDO
01 E = –sen270° + cos90°
–csc90° – tanp
 E = –(–1) + 0
–1 – 0
 = 1
–1
 = –1
Clave A
02 sen 3p
2
 = –1; cos2p = 1 
 sen p
2
 = 1; csc p
2
 = 1
 sec p = –1
 E = –(x – y)2 – 4xy(1)
y2(1) + x2(1) – 2xy(–1)
 = –(x + y)2
(x + y)2 = –1
Clave A
03 • a2 + b2 = c2
 • a 

b
c


 + b 

a
c


 = c
  2ab = c2  2ab = a2 + b2 
  (a – b)2 = 0  a = b
  M = b
a
 + a
b
 = 1 + 1 = 2
Clave C
04 • T = (+) + (+) = (+)
 • R = (–) + (–) = (–)
 • C = [(+) + (+)][(–) + (–)] = (–)
  (+)(–)(–)
Clave C
05 
  tanβ = –2
–3
 = 2
3
 
 Clave B
06 
 r = (–3)2 + 42 = 5
 E = senq · cosq
  E = 

4
5


 

–3
5
 

 = – 12
25
 
Clave A
07 E = sen(–a) · cos(–a)
 E = –sena ⋅ cosa
X
Y
(–1; –2)
a
  E = – 

–2
5


 

–1
5


 = – 2
5
 
Clave B
08 5(1 – 2sen2x) + 3senx = 4
 10sen2x – 3senx – 1= 0
 
 
 senx = 1
2
 ∨ senx = – 1
5
 Como x ∈ III C  senx = – 1
5
  y = – 1; r = 5
  cosx = – 2 6
5
Clave D
09 senφ = – 3
5
  y = –3; r = 5 ∧ x = 4
 k = r
x
 + y
x
  k = 5
4
 – 3
4
 = 2
4
 = 1
2
Clave C
10 E = sen90° + cos180°
sen270°
  E = 1 – 1
 –1
 = 0
Clave C
11 a < β y cosa = cscβ
X
Y
(–3; –2) (2; –3)
β
X
Y(–3; 4)
r = 5 q
2senx – 1
5senx 1
 Piden:
  E = 2seca
senβ
 = 
2
cosa
1
cscβ
 = 2
Clave B
12 Siendo a el menor: 304° < a < 430°
 El mayor ángulo es: 360°k + a
 La suma es: 360°k + 2a = 2480° ......(3)
  a = 1240° – 180°k
 
Reemplazo: 304° < 1240° – 180°k < 430°
(2)
(1) 
 De (1) y (2): k = 5 De(3): a = 340°
  El mayor ángulo es: 2140°
Clave D
13 E = cotβ – cscβ
 E = x
y
 – r
y
 E = 15
–20
 – 

25
–20
 

 = –10
–20
 = 1
2
Clave A
14 361° ∈ IC  sen361° = (+)
 455° ∈ IIC  cos455° = (–)
 I. (+) ⋅ (–) = (–)
 II. (–) – (–) = (–)
 III. (+) ⋅ (+) = (+)
  (–) (–)(+)
Clave C
15 (3sena – 4)(5sena + 2) = 0  sena = –2
5
 Como a ∈ IIIC  y = –2; r = 5; x = – 21
  cota ⋅ cosa = 

– 21
–2




– 21
5


 = –2,1
 
Clave D
ACTIVIDADES CAP 10
CIRCUNFERENCIA TRIGONOMÉTRICA
01 
 
Y
C(senq;cosq)
B(5;0)A(–4;0)
q X
senq
 AB = 5 – (–4) = 9 ; senq = cosq ⇒ q = 
p
4
 Área: 1
2
(9)senq = 9
2
cosq
 Pero: 9
2
cosq > 9
4
 ⇒ q ∈ 0 ; 
p
4
 
Clave B
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
18
02 Área: 1
2
(2)senq = 2
3
 ⇒ senq = 2
3
 Pitágoras:
 

2
3


2
 + cos2q = 12
 
Y
X
B
A C
CT
1
11
senq
q
  cosq = 5
2
Clave D
03 Área S = (2cos)(1 – senq)
  S = cos(1 – senq)
 
Y
X
S
CT
1
senqse
nq
cosqcosq
q
Clave B
04
 Área:
 S = 1
2
(p–φ)12– 1
2
(–cosφ)senφ
 
 
 
Y
X
CT
1
1
S
senq
cosq
φ
S = 2p – 2φ + sen2φ
4
Clave D
05 AC = 2sena
 CB = 1
 Área de la región ABCD:
 S = 1
2
(1)(2sena)
 
Y
X
D
B
A
S
C
1
1O
1
a
  = sena
Clave B
06 cosa – n
n
 = 1 + cosa
1
 n = cosa
2 + cosa
 Área de la región sombreada
 S = 1
2


1 + cosa
2 + cosa


(1)
CT
Y
X
cosa – n
1 +cosa
S
1
O45°
1
n
a
 S = 1 +cosa
2 + cosa
Clave B
07 1
–2x
 = –sena
–x + cosa
 –x + cosa = 2xsena
 cosa = x(1 + 2sena)
 
CT
Y
X
–sena
cosaQ(x;0)
1
2
1
2
1
a
  x = cosa
1 + 2sena
Clave A
08 Área de la región
 triángularAMB:
 S = 1
2
( 2)

1 + 2
2


 S = 1 + 2
2
 
Y
XB
A
M
1
2/2
2/2
2/2
ω
ω
 S = 1
2
tan

3p
8


Clave B
09 MT = –secq – 1
 Área de región AMT
 S = 1
2
(–secq – 1)(–senq)
 
–sen
q
Y
XA
T
M
O1 1
1
A'
P
q
  S = 1
2
[tanq + senq]
Clave D
10 n
1 – n
 = 1
–cosq
 
 n = 1
1 – cosq
 Área de la región APN
 S = 1
2
(1)(1) + 1
2
(1) 1
1 – cosq
Y
XA
P
N
A
H O
1
qsenq
cosq
45°
1–n n
n
  S = 1
2
2 – cosq
1 – cosq
Clave C
CUADERNO DE TRABAJO
01 sena < 0
  AB = –sena
 
Y
XA O
B
Clave B
02 cosq < 0
  PQ = –cosq
 
 
Y
X
P Q
Clave D
03 Tenemos OA = sena, OB =|senb|
 Luego, AB = OA + OB = sena +|senb|= 
sena – senb, pues b es un ángulo del ter-
cer cuadrante.
Clave C
04 Los BDO y CAO son congruentes, lue-
go OC= OB.
  OC = –seca.
Clave A
05 	 De	 la	figura,	 los	 triángulo	AOM	y	AHP	
son semejantes, luego:
 OM
1
 = 
|senq|
1 +|cosq|
 
 OM = 
senq
1 – cosq
 
Y
X
P
O A1H
M
Clave A
06 OB =|cosa|, OD = sena
 OB = PD 
 AD = AO + OD 
Y
X
P
O
A
1
B
D
 AD = 1 + sena
 PA = cos2a + (1 + sena)2
 PA = 2 + 2sena
Clave B
07 		 Del	gráfico,	PQ	=	–sena
 Luego, área sombreada:
  1
2
 (PQ)(1) 
P
B
A' A
B'
Q O 1
  – 1
2
 sena
Clave C
08 	 En	 la	 figura,	 OA	 =	 secq, BC = senq, 
OC = cosq.
 Área OAB = 1
2
 secq · senq = 1
2
 tanq
 Área OCB = 1
2
 cosq · senq
  Área sombreada = 1
2
 (tanq – cosq · senq)
 = 1
2
 tanq(1 – cos2q) = 1
2
 sen3q · secq m2
Clave A
09 	 De	la	figura:	
 S



2p
3
sec , 0 = S(–2; 0)  base = 1 m
 T



cos , senp
3
p
3
 = T




3
2
– 1
2
 , 
  altura = 
3
2
 m
  Área sombreada = 1
2
 (base)(altura)
  1
2
 (1)




3
2
 = 
3
4
 m2
Clave E
10 P(cosq, senq) 
 Área de la región
 sombreada: 
 S = 
–(cosq + senq)
2
 m2
 
Y
X
O
P
1
1
1
Clave B
TAREA
01 AB = –cosa
02 EF = –sena
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
19
03 
 Área = 1
2
 (2)(–senq) = –senq
04 
 Área = 1
2
 (–cosq) = – cosq
2
 
REFORZANDO
01 AB = –senq
Clave E
02 PQ = –cosa
Clave D
03 OB = cosa
 OA = –cosβ
  AB = cosa – cosβ
Clave B
04 
 5p
4
 = p + p
4
 
X
Y
O
(1; 0)
cos 5p
4
sen 5p
4
P(a)
a
 cos 5p
4
 = cos

p + p
4


 = –cosp
4
 = – 2
2
 sen 5p
4
 = sen

p + p
4


 = –senp
4
 = – 2
2
 P(a) = P 

– 2
2
 ; – 2
2


 Luego, la suma de coordenadas es:
 – 2
2
 + – 2
2
 = – 2
Clave B
05 	 Del	gráfico	se	tiene	que	el	área:	
 = sena
2
 = 2
4
  a = 3p
4
 
 Luego arco = 3p
4
 
Clave C
06 
 Área:
 1
2
 (2)(senq) = senq
 
Clave A
senq
2
–cosq
1
A
B
0
X
A' 1 1 A
Y
O
senq
q
07 Nótese que:
 OM = tan q
2
 
XA 1 A
Y
C
qq/2
O
M B
 Clave D
 
08 x2 + y2 =1
  1
4
 + b2 = 1 
  b = ± 3
2
 
 ÁreaABO
 = 3
2
 seca
2
 
 = 3
4
 seca µ2
 Clave A
09 	 Del	gráfico:
 a = 360° – (φ – β), OB = OA = 1µ
 Área del 9AOB: 
 = 1
2
 OA ⋅ OB ⋅ sena 
 = 1
2
 (1)(1) ⋅ sen(–(φ – β))
 = – 1
2
 sen(φ – β)
 Clave A
10 
 
 = 1
2
 1 – 

– 3
2
 + 1
2


 = 1
4
 + 3
4
 
 Clave A
11 OC = OB = secq
 A = ABOC – AAOD
 = 1
2
 qsec2q – 1
2
 q(1)2 
 = q
2
 (sec2q – 1) = q
2
 tg2q
 Clave A
12 Coordenadas del punto P:
 

– 4
5
; 3
5


, C es una circunferencia unitaria
 AB = |tga|
 OA = CB = 1
  Perímetro buscado:
 = 2 + 2 |tga|= 2 + 2 – 3
4
 
Aseca
a
O
B 

– 1
2
; b 

X
A
B
Y
O
φa
β
X
(0; 1)
1
(1; 0)
Y
30°
 

– 3
2
; 1
2


3
2
1
2
S = 1
2
 = 
0 1
– 3
2
1
2
1 0
0 1
 = 2 + 3
2
 = 7
2
 = 3,5 µ
 Clave A
13 
 
 
X
Y
O
A
B
C
cosβ
cosa
a β
 Área del triángulo AOC = 1
2
(1)(–cosa)
 Área del triángulo OCB = 1
2
(1)(cosβ)
 Área buscada = 1
2
 (cosβ – cosa) µ2
 Clave C
14 	 De	la	figura:
 S 

sec 2p
3
 ; 0 

 = S(–2; 0)  base = 1µ
 T 

cos p
3
 ; sen p
3
 

 = T 

– 1
2
 ; 3
2
 

  altura = 3
2
 µ
  Area sombreada = 1
2
 (base)(altura)
 = 1
2
(1) 

3
2
 

 = 3
4
 µ2
 Clave E
15 
 AS = (–cosq) ⋅ 1
2
 + senq ⋅ 1
2
 
 AS = senq – cosq
2
 AS = 2 ⋅ sen(q – 45°)
2
 AS = Máximo  [sen(q – 45°)]Max = 1
 Por lo tanto el área máxima es = 2
2
 µ2
 Clave E
ACTIVIDADES CAP 11
C.T. REPRESENTACIÓN DE SENO Y COSENO
01
 
 sen70° > sen140° > 
 sen50° > sen210° > sen300°
Y
X
se
n3
00
°sen210°
se
n7
0°
se
n5
0°
sen140°
Clave B
X
A'
B
B'
A
1
–1
Y
Y
O
senq
cosq
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
20
02 cot70° > cos310° > cos130°> cos220° > 
cos160°
Clave C
03 FVV
Clave E
04 VVF
Clave E
05 FVV
Clave E
06 VFF
Clave C
07 FFV
Clave B
08 FVV
Clave B
09
 cosx < 3
2
 cosx < 0,86
 
Y
XO
IVCIIIC
ICIIC
x
0,86
 30° < x < 330°
 Por lo tanto la desigualdad presenta solu-
ciones en los cuatro cuadantes.
Clave E
10 x = 
p
6
 = 30°
 y = 5p
6
 = 150°
 
Y
X
5p
6
10p
6
p
6
y =
x =
z =
B
C
FE
A
D
 z = 10p
6
 = 300°
 senx = 1
2
 ; cosx = 3
2
 seny = 1
2
 ; cosy = – 3
2
 senz = – 3
2
 ; cosz = 1
2
 
 senx + seny + senz + cosx + cosy + cosz = 3
2
 – 3
2
Clave A
CUADERNO DE TRABAJO
01 
  El mayor es 
 sen190°
 
260°
330°
190°
210°
230°
Y
X
Clave A
02 
  El mayor es 
 sen255°.
 
255°
240°
200°
300°
340°
Y
X
Clave C
03 
  VFV. 
 
 
0,5
5,10
1,52
4
1,57
4,71
3,14 6,28
Y
X
Clave B
04 
  El mayor es 
 sen1.
 
0,4
0
0,8
11,57
4,71
3,14
4,8
5,5
Y
X
Clave A
05 I. (F)
 II. (F)
 III. (V)
 IV. (F)
 
cosx
cosx
senx
senx
senx
Y
X
Clave B
06  FFV Clave B
07 I. cos80° > cos100° II. cos200° < cos300°
 III. cos50° > cos70°
  >; <; > Clave C
08 –1  senq  1  –4  4senq  4
  –5  4senq – 1  3
  [–5; 3] Clave D
09 –1 < cosa < 0  –2 < 2cosa < 0
  –1 < 2cosa + 1 < 1
Clave D
10 p
8
 < x < 5p
24
  p
4
 < 2x < 5p
12
  0 < 2x – p
4
 < p
6
 
 0 < sen



p
4
2x – < 1
2
  0 < 2sen



p
4
2x – < 1
 1 < 2sen



p
4
2x – + 1 < 2 
  1
2
 < 
2sen



p
4
2x – + 1
1
 < 1
  I  2; 4
Clave C
TAREA
01 
  cos70° < cos310° < cos40°
 
Y
X
70°
40°
310°
02 
  sen100°
03 –1  senq  1
 –10  10senq  10
 –4  10senq + q  16 
 [–4; 16]
04 p < a < 3p
2
 
 
 –1 < cosa < 0
 3 > –3cosa > 0
 3 > M > 0
REFORZANDO
01 OP = senq
 PB = 1 – senq
Clave B
02 OP = –cosq
 A'P = 1 – (–cosq) = 1 + cosq
Clave B
03 I. (V) II.(V)
Clave A
04 
 I. (F)
 II.(F)
 
Clave D
05
 
 |cos160°|> sen160°
 –cos160° > sen160°
 0 > sen160° + cos160° 
 I. (F) sen160° < cos160°
 II.(F)
Clave B
06 
 
Y
X
75°
100°
60°
Y
X
X
Y
340°
100°
190°
70°
X
Y
160°
X
Y
100°
200°
300°
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
21
 • |cos200°|> |sen200°|
 • |sen100°|> |cos100°|
 • |cos300°|< |sen300°|
 I. (V) II.(V) III.(F)
Clave C
07 –1 < cosβ < 0
 –2 < 2cosβ < 0
 –1 < 2cosβ + 1 < 1
 –1 < L < 1
 L ∈ 〈–1; 1〉
Clave E
08 
 
 I. senx1 > senx2 (V) 
 II. cosx1 > cosx2 (F) 
 III. 2p < 2x1 < 2x2 < 3p (F)
 sen2x1? sen2x2 
Clave E
09 p
2
 < x
2
 < p  x
2
 ∈ IIC
 –1 < cos x
2
 < 0
 –3 < 3cos x
2
 < 0
 –4 < 3cos x
2
 –1 < –1
 L ∈ 〈–4; –1〉
Clave D
10 • 17p
24
  x  21p
24
 17p
12
 – p
12
  2x – p
12
  21p
12
 – p
12
 4p
3
  2x – p
12
  5p
3
 
X
Y
60° 60°
 – 1
2
  cos 

2x – p
12


  1
2
  1  L  5
 
Clave C
11 0  cos2  1
 0  2cos2q  2
 3  2cos2q + 3  5
 A ∈ [3; 5]
Clave A
12 
 
X
Y
x1
p
2x1
2x1
sen2x1
senx2
senx1
sen2x2
sen2x2 sen2x2
cosx1 cosx2
x2
2x2
2x2
3p
2
X
Y
100°
cos100°
sen100°
cos250°
cos340°sen200°
200°
290°250°
 I. (F) II.(F) III.(F)
Clave E
13 –1  cosq < 0 ∨ 0 < cosq  1
 secq  –1 ∨ secq ≥ 1
 2k – 3
5
  –1 ∨ 2k – 3
5
 ≥ 1
 k  –1 ∨ k ≥ 4
 
–1
  –1 < k < 4
Clave C
14 11
12
  x  35
12
  11p
24
  px
2
  35p
24
  11p
24
 – p
8
  px
2
 – p
8
  35p
24
 – p
8
 p
3
  px
2
 – p
8
  4p
3
 –1  cos

px
2
 – p
8


  1
2
 –4  4cos

px
2
 – p
8


  2
 –3  4cos

px
2
 – p
8


 + 1 3
  C ∈ [–3; 3]
Clave B
15 I. (V) 
 II. (V) 
 III. (V) cosx2 < cosx1 
Clave E
ACTIVIDADES CAP 12
C.T. REPRESENTACIÓN DE TANGENTE Y COTANGENTE
01 tan5 <tan2 < tan3 < tan1 < tan4
Clave D
02 FFF
Clave D
03 I-V ; II-F ; III-V
Clave A
X
Y
60°
240°
1
2
– 1
2
X
x2
x1
senx1 > senx2
Y
p
2
X
x2
x1 |cosx2| > |cosx1|
Y
04 VFV
Clave D
05 Área de la
 región TPA
 S = 1
2
(–tanq)(1)
 
Y
XtanqS
1M
O
A
P
T
q
 S = – 1
2
tanq
Clave C
06 I-V ; II-F ; III-V ; IV-V
Clave D
07 I-F ; II-V ; III-F
Clave B
08 Área de la 
 región THO:
 S = 1
2
(–cosq)(–tanq)
 
Y
Xcosq
tanq
S
1
M
O A
H
T
q
  S = 1
2
senq 
Clave E
09 I-F ; II-F ; III-V
Clave D
10 AD = 1 – cosq
 Área de la región ABCD
 S = 1
2
(senq + tanq)(1 – cosq)
Y
X
senq
cosq
tanqM
O D
S
A
B
C
q
  S = tanqsen2q
2
Clave E
CUADERNO DE TRABAJO
01 Se tiene: q > 0, a < 0
 OA = 1, 
 AT = tanq
  T(1; tanq) 
Y
XA(1; 0)
B(0; 1)
D
P
O
Q
 OB = 1, BC = cota
  C(cota; 1)
   = 1 + tanq + cota + 1 = 2 + tanq + cota 
Clave C
02 I. (V)
 II. (F)
 III. (V)
 
tan135° = tan315°
tan300°
tan100°
tan200°
tan50°Y
X
Clave B
03 Expresando en función de la tangente:
 I. cot20° > cot50° (V)
 II. cot120° < cot160° (F)
 III. cot220° > cot260° (V)
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
22
 
 
 
co
t2
60
°
co
t1
60
°
co
t1
20
°
co
t5
0°
co
t2
20
°
co
t2
0°Y
X
Clave C
04 tan1 > tan0,5 
 > tan3,5 > tan3 
 > tan2,5
 
tan3
tan2,5
2,5 0,51,57
4,71
3,5
3,14
1
3
6,28
0
tan3,5
tan0,5
tan1
Y
X  El mayor 
 es tan1.
Clave D
05 PQ
2
 = sen(180° – q)  PQ = 2senq
 PO = cosq
  Área = 
cosq · 2senq
2
 = senq · cosq
Clave D
06 A = 1
2
 (–tanq) + 1
2
 (–cotq) – 
p
4
  4A = –2tanq – 2 cotq – p
  4A + 2cotq = –2tanq – p 
Clave C
07 0  tan2x < 1 
 0  3tan2x < 3
 1  3tan2x + 1 < 4 
1
–1
Y
X
 1  L < 4
  [1; 4
Clave D
08 
 
Y
X
tanx1
tanx2
 I. tanx1 < tanx2 (V)
 
 II. tanx1 < tanx2 (V) 
Y
X
tanx1
tanx2
x2
x1
 
 
Y
X
tanx1
tanx2
 III.|tanx1|<|tanx2| (V)
 
Clave A
09 Área de la región sombreada:
 = – 1(–tana)
2
 – 
(–a)
2
 + 




p
2
+ a
2
 = 
tana
2
 + 
a
2
 + 
a
2
 + 
p
4
 
Y
XO
1
 = 
– tana
2
 + a + 
p
4
 m2
Clave B
10 
AD = –cota
AC = 1 + cosb 


 
  ÁreaABC = – cota
2
 (1 + cosb)
 = –cota · cos2



b
2
Clave C
TAREA
01 
 
X
Y
tan80° = c
tan65° = b
tan135° = a
  a < b < c
02 
X
Y
tan160°
tan140°
tan110°
tan105° = tan285°
tan100°
 
 
 
03 
 Área = 1 · 1
2
 + (1) 

– tanq
2


 = 

1 – tanq
2


04 Hallando el orden de las tangentes:
 
 tan200° < tan230° < tan260°
 1
tan200°
 > 1
tan230°
 > 1
tan260°
 cot200° > cot230° > cot260°
  El mayor es cot200°
REFORZANDO
01 
 El mayor es tan2p
5 Clave E
02 I. (F) II. (F) III. (V)
Clave D
I. (F)
II. (F)
III.(V)
X1 1
Y
–tanq
tanqq
X
Y
tan230°
230°
tan260°
260°
tan200°
200°
X
Y
tan 2p/5
tan p/4
tan 2p/9
tan p/5
tan 2p/5
03 
 
  El menor es cot100°
Clave C
04 
 I. (V) II. (F) III. (V)
Clave C
05 
  Área = 1
2
 (tanq – tana) 
Clave D
06 
 
X1O
N
AA'
T
M
1
Y
q –tanq
 –tanq
2
 = ON
1
  ON = –tanq
2
  Área = 1
 ⋅ ON
2
 = – tanq
4
Clave C
07 a = –cosq b = 1 + cosq
 
 
 
X
a b
Y
(0; 1)
(–cosq; – senq)
cosq
senq
1; tanq
S2
S1
–tanq
q
 S1 = 1(–cosq)
2
 S2 = –tanq(1 + cosq)
2
  S1 + S2 = 1
2
 (–senq – cosq – tanq)
 = – 1
2
 (senq + cosq + tanq)
Clave A
08 	 Analizando	la	gráfica	se	tiene:
 a = cosa, b = sena, c = 0, d = csca, e = 1,
 f = tga
X
Y
cot100°
cot80°
cot70°
cot205°cot245°
X
Y
cot290°
cot310°
cot50°
cot40°
cot190°cot210°
X1
Y
tanq
a
q
–tana
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
23
 
 Luego: bf + dc
|ae|
 = sena ⋅ tga + 0
|cosa|
 = sen2a
–cos2a
 = –tan2a
 
Clave C
09 
 x = 1
2
 tanq
 
O
x
1 1
tanq
q
A
P
M
A'
 A = 1
 ⋅ x
2
 
  A = tanq
4
 
Clave C
10 
 
O q – p
2
q
A
M
N
Q
Y
P
 AN = tan 

q – p
2


 = cotq
 MQ = 1 – cos 

q – p
2


 = 1 – senq
  S = 0,5 cotq(1 – senq) 
Clave C
11 I. (V) II. (F) III. (F) 
Clave D
12 F = tan2q – 1 + 2tanq = (tanq + 1)2 – 2
 – 1
2
  senq  2
2
  – sen 
p
6
  senq  sen 
p
4
  – p
6
  q  p
4
  tan

– p
6

 
 tanq  tan 
p
4
  – 3
3
  tanq  1 1 – 3
3
  tanq + 1  2
  (tanq + 1)2  4  (tanq + 1)2 – 2  2
  F  2  FMAX = 2 
Clave C
13 • CD = senq OD = cosq 
 AB = tanq
 A = 

senq + tanq
2

 
(1 – cosq)
 A = senq
2 ⋅ 
sen2q
cosq
 = tanqsen2q
2 
 
Clave E
O
A
B'
(c; d)
(e; f)
(a; b)
tga
sena
cosa
AA'
tanqsenq
1 – cosq
14 
 
O 1
C
AA'
Y
M
T
senq
q
P
 S = PM ⋅ PA
2
 S = senq(1 – cosq) 
2
 
Clave E
15 	 –1	≤	cosa	≤	1
 – p
3
	≤	p
3
 cosa	≤	p
3
 cot 

– p
3

 
≤	cot 

p
3
 cosa 
 
≤	cot 

p
3

 
 – 1
3
	≤	cot 

p
3
 cosa 
 
≤	 1
3
 	0	≤	cot2

p
3
 cosa 
 
		≤1
3
 
 
Clave C
ACTIVIDADES CAP 13
REDUCCIÓN AL PRIMER CUADRANTE I
01 F = –sena
sena
 + –cosa
cosa
 + tana
tana
 F = –1 – 1 + 1 = –1
Clave A
02 K = tan17° + cot17°
tan17° + cot17°
 (–tan17°)
 K = –tan17°
Clave A
03 C = (–sen60°)tan45°
(–cos60°)tan30°
 = 


3
2


(1)


1
2


 

1
2


 C = 2 3
Clave B
04 C = –sena
sena
 + cosa
cosa
 
 C = –1 + 1 = 0
Clave E
05 senq = –2cosq ⇒ tanq = –2
 R = (–cotq)(–secq)(tanq)
(–senq)(–cscq)(–cosq)
 = – secq
cosq
 R = –sec2q = –[1 + tan2q] = –[1 + (–2)2]
 R = –5
Clave B
06 K = sen60°(–cos60°)(–tan60°)
(–sec45°)
 K = – 
3
2
 

1
2


 3
2
 = – 3 2
8
Clave D
07 J = (–cotx)(tanx)(–cscx)
 J = cscx
Clave B
08 K = (–senx)(–cscx) + (–tanx)cotx
 K = 1 – 1 = 0
Clave E
09
 
ω
q
ω
1212
12
12 5
A'(–12;5) A(12;5)
 tanq = –5/12
Clave C
10 a = q + 180°
 tana = tan(180° + q)
 tana = –tanq
q
a B
3
A 2 2
C
37° tana = – 3/2
Clave D
CUADERNO DE TRABAJO
01 
 E = 
cos

p + 
p
9


 –sen

p + 2p
3


 –cot

2p + 2p
9


 
cos
p
9
 –sen

p
6


 –cot

2p
9


 
 E = 
–cos

p
9


cos
p
9
 · 
–sen

p + 2p
3


sen
p
6
 · 
cot

2p
9


cot

2p
9


 E = (–1) · 
1
2
3
2
–
 · 1 = 3
Clave E
02 cos170° = –cos10°  cos150° = –cos30°
  cos130° = –cos50°  cos110° = –cos70°
  cos90° = 0
  cos10° + cos30° + ... + cos170° = 0
Clave E
03 L = (–cos45°)(–cot60°)(sec60°)
(cos30°)(tan37°)(–tan60°)
 L = –
2
2
3
3
2
3
2
3
4
3
⋅ ⋅
⋅ ⋅
= – 
9 3 3
8 2 3 = – 
27
8 6
 
Clave C
04 
 E =
4 2 1
2
2
2
3
4
3
2
1
2
( ) 
( )
−
= 
( 3 – 1)( 3 + 1)
3( 3 + 1)
 
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
24
 E = 
2
3( 3 + 1)
Clave B
05 E = –senx
–senx
 – cotx
–cotx
 + secx
secx
 E = 1 + 1 + 1 = 3
Clave E
06 
 E = 
2x2(cos60°) + 3
2
 x(tan53°) + 2(sen30°)
x(+tan45°) – (–1)
 E = 
2x2 · 1
2
 + 3
2
 x · 4
3
 + 2 · 1
2
x(1) – (–1)
 = (x + 1)2
x + 1
 = x + 1
Clave D
07 
sen(–q) – cosq
–senq – sen(q)
 = 
–senq – cosq
–2senq
 = 1
2
 (1 + cotq)
Clave B
08 tana = – a
b
 
 
(b; a)
(a; b) Clave C
09 E = –2(–cotx)
cotx(–secx)(–cosx)
 = 2
Clave E
10 E = +senx
–senx
 · –senx
–senx
 = –1
Clave E
TAREA
01 L = (cos30°)(sen60°)
cos225°
 L = (cos30°)(sen60°)
–cos45°
 = 
3
2
 ⋅ 3
2
– 2
2
 = –6
4 2
 = – 3 2
4
02 Procedemos, por partes:
 • sen240° = –sen60° = – 3
2
 • cos120° = –cos60°  cos120° = – 1
2
 Reemplazando: C = 
– 3
2
– 1
2
  C = 3 
03 Por partes:
 • tan225° = +tan45°  tan225° = 1
 • sen330° = –sen30°  sen330° = – 1
2
 Luego: C = 1
– 1
2
  C = –2 
 
04 
 tanq = – 5
–2
 = 5
2
 
 
REFORZANDO
01 E = cos2x
(–cscx)2 – 1
 + (–senx)2
sec2x – 1
 E = cosx2
cot2x
 + senx2
tan2x
 E = sen2x + cos2x
 E = 1
Clave C
02 E = –sen(180° + 20°) – cos(270° – 20°)
sen(180° – 20°)
 E = – (–sen20°) – (–sen20°)
sen20°
 E = 2sen20°
sen20°
 = 2
Clave B
03 sena = sen(180°– 42°+q) ⋅ tan(270°+28°–q)
2sen(42° – q) ⋅ cot(q – 28°)
 sena = –sen(q – 42°)(–cot (28° – q))
2sen(42° – q) ⋅ cot(q – 28°)
 sena = 1
2
  y = 1 r = 2 ∧ x = – 3 
  E = 2
– 3
 + 1
– 3
 = 3
– 3
 = – 3 
Clave B
04 C = (–cos45°)⋅ (–sen60°) ⋅ (–cot60°) 
(–cos30°) ⋅ (sen30°)
 C = cos45° ⋅ cos30° ⋅ tan30° 
cos30° ⋅ sen30°
 C = cos45° 
cos30°
 = 
2
2
3
2
 = 6
3
Clave A
05 M = –sen60° ⋅ (–sen37°) ⋅ (sen45°)2
 M = – 3
2
 ⋅ –3
5
 ⋅ 

2
2


2
 M = 3 3
20
 = 0,15 3 
Clave C
X
Y
3
3
2
2
q
5 
(–2; – 5) 
5 
06 L = 
(–senx)(–cotx) ⋅ tan(2p – x) 
sen 

3p
2
 – x 

 sen 

3p
2
 + x 

 L = 
senx ⋅(cotx)(–tanx)
–cosx ⋅ –cosx 
 L = –secxtanx
Clave D
07 A = 
(+cosx)(–tanx) tan(180° + x) 
(–cosx) 

1
cos(360° – x)


 sen(270° + x) 
 A = 
–cosx ⋅ tanx ⋅ tanx 
(–cosx) 

1
cosx


 (–cosx) 
 = –tan2x
Clave E
08 P = 
cos(2p – x) ⋅ cos(p – x) ⋅ tan(2p + x) 
sen

p
2
 – x 

 ⋅ 1
tan(p + x)
 ⋅ sen

3p
2
 + x 

 P = 
cosx ⋅ (–cosx) ⋅ tanx 
(cosx) ⋅ 

1
tanx


(–cosx)
 = tan2x
Clave E
09 tan(–β) = – 3
4
 –tanβ = – 3
4
  tanβ = 3
4
Clave D
10 
  cotq = x
y
 = – 4
1
 = –4
Clave D
11 –sen(180° – x)
cos(90° – x)
 = –senx
senx
 = –1
Clave B
12 x = 180° – y ⇒ tanx = – tany
 y = 270° – z ⇒ tany = cotz
 M = (–tany)(coty) + (cotz)(tanz)
 M = –1 + 1 = 0
Clave A
13 • A + B + C = 180°  B + C = 180° – A
  sen(B + C) = sen(180° – A) = senA
 E = senA + senA + sen(180°)
 E = 2senA
Clave B
37°
3
1
4
4
(–4 ; 1)
1
4
4
X
Y
q
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
25
14 
  tanq = y 
x
 = – 3
7
 
Clave D
15 
 E = sec 

 3p
2
 – a 

 = –csca
 E = –1 
sena
 = –1 
– 4
5
 = 5
4
 
 (–3; –4)
Y
X
5
a
Clave E
ACTIVIDADES CAP 14
REDUCCIÓN AL PRIMER CUADRANTE II
01 tan2933° = tan(8(360°) + 53°) = tan53°
 tan2933° = 4
3
Clave C
02 E = sen180° + cos2180°
cos4270°
 = 0 + (–1)2
(–1)4 
 E = 1
1
 = 1
Clave A
03 C = senx
cos(–x)
 + tan(–x)
 C = tanx – tanx = 0
Clave C
04 M = sen140°
sen40°
 + cos20°
cos160°
 M = sen40°
sen40°
 + cos20°
–cos20°
 = 1 – 1 = 0
Clave E
05 E = sen70°cos160°tan(–110°)
cot20°sen(–70°)sec160°
 E = –cos20°tan70°
–70°(–sec20°)
 = –cos220°
Clave D
06 E = |sen(p + q)|+|tan(4p/3 + q)|+|tan(p/2 + q)|
 E = |sen150°|+|tan210°|+|sec60°|
 E = |sen30°|+|tan30°|+|csc30°|
 E = 1
2
 + 
3
2
 + 2 = 
6
15 + 2 3
Clave C
07 cos(90°+a)cos(270°–a)–sen(180°–a)sen(–a)
 –sena(–sena)–sena(–sena)
 sen2a+sen2a
 2sen2a
Clave A
3
3
(7; 3)
(7; –3)
4
3
3
4
Y
q
08 sen[1 + (–senq)] = – 1
 senq – sen2q = –1
 Calculamos la segunda expresión:
 1 – (–senq) + cos2q
 1 + senq + 1 – sen2q
 2 + senq+ – sen2q = 2 – 1 = 1
Clave D
09 cos

3p
2
 – x 

(–sen(10p – x)) +
 sen(3p – x)cos

p
2
 – x 

 –senxsenx + senxsenx = 0
Clave E
10 sec cot

13p
2
 + y 

 – sec(tany) + sen(x – y)
 sec cot

p
2
 + y 

 – sec(tany) + sen13p
2
 sec(–tany) – sec(tany) + sen
p
2
 sec(tany) – sen(tany) + 1 = 1
Clave B
CUADERNO DE TRABAJO
01 sen(1800° + 60°) = sen60°
 sen(1860°) = 
3
2 Clave D
02 tan(14p – x) = tan(7(2p) – x) 
 tan(14p – x) = tan(–x) 
 tan(14p – x) = –tanx
Clave B
03 E = 
(–sen300° – 2tan60°)
sec30°
 E = (sen60° – 2tan60°)cos30°
 E = 

3
2
 – 2 3 



3
2


 ⇒ E = – 9
4
Clave B
04 
 E = 
a2sen
p
2
 + 2abcos2p – b2sen3p
2
b2tan2p – 4absen3p
2
 + (a – b)2cos2p
 E = 
a2 + 2ab – b2(–1)
b2(0) – 4ab(–1) + (a – b)2 = 
a2 + 2ab + b2
4ab + (a – b)2 
 E = 
a2 + 2ab + b2
a2 + 2ab + b2 = 1
Clave E
05 E = 2sen30° + tan45° + 2 3cos30°
 E = 2



1
2
 + 1 + 2 3 · 
3
2
 = 1 + 1 + 3 = 5
Clave D
06 
 E = 
x2(sen30°) + xcos300°
x – (–tan45°)
 = 
x2
2
 + x
2
x + 1
 = x
2
Clave C
07 
 E = 
cos 

– 
p
4


 tan 

– 
p
4


 sec 

– 2p
3


cot 

– 7p
6


 sen 

– 
p
3


 csc 

– 
p
6


 E =
2
2
1 2
3
3
2
2




−( ) −




( )( )
= 
2
3
Clave D
08 
 N = 




p
2
q –sec(q) · sen
cot(p – q) · cos 



3p
2
+ q
 = 
sec(q) (–cosq)
(–cotq) (senq)
 
 N = secq · 1  Ncosq = 1
Clave B
09 
 E = 
senq – tanq
–cscq + cotq
 = 
 sen 

p
3


 – tan 

p
3


–csc 

p
3


 + cot 

p
3


 
 E =
3
2
3
2 3
3
3
3
−
− +
= 3
2
Clave D
10 
 H = 
tan(p + x) sen 

p
2
 – x 

 secx
cot 

3p
2
 + x 

 senx tan 

p
4


 H = 
tanx · senx · secx
–tanx · senx · (1)
 = –secx
Clave B
TAREA
01 • tan(2580°) = tan(360° ⋅ 7 + 60°) 
 = tan60° = 3
02 • sen(36p + x) = sen(18(2p) + x) = senx 
 
03 • cos(14p – x)
 cos(7(2p) – x) = cos(–x) = cosx
04 • cot2580° 
 cot(360° ⋅ 7 + 60°) = cot60° = 3
3
 REFORZANDO
01 • cos(4005°) 
 cos(360° ⋅ 11 + 45°) = cos45° = 2
2
Clave E
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
26
02 L = sen(24p + x)
sen(14p – x)
 L = sen(x)
sen(– x)
 = senx
–senx
 = –1
Clave B
03 U = (2sec120° – 1) ⋅ (2sen143° – 1)
2cos240° – 1
 U = 
(2(–2) – 1) ⋅ (2

3
5


 – 1) 
2

–1
2


 – 1
 = 
(–5) ⋅ 

1
5


–2 = 1
2
 
Clave A
04 cos 157p
11
 = cos 

14p + 3p
11


 = cos 3p
11
 Como: cos 

157p
11


 ∈ IIC
 = cos 

p – 8p
11


 = – cos 8p
11
Clave C
05 • sen87p + 4 ⋅ cosp + 5 ⋅ sec 2p
 0 –1 1
  – 4 + 5 = 1
Clave C
06 tan 941p
4
 = tan 

234p + 5p
4


 = tan 5p
4
 = 1
 cot 199p
4
 = cot 

50p – p
4


 = –cot p
4
 = –1
  E = 1 – 1 = 0
Clave C
07 csc 

9p
2
 + q 

 + cot 

11p
2
 + q 

 = 1
2
 IIC IVC
 secq + (–tanq) = 1
2
  secq – tanq = 1
2 Clave C
08 cos 

2p
7


 + cos 

4p
7


 + cos 

6p
7


 = 
sen

3
2
 × 2p
7


sen

1
2
 × 2p
7


 = cos

2p
7
 + 6p
7


1
2
 = 
sen 3p
7
sen p
7
 ⋅ cos 4p
7
 = 
–2sen 3p
7
 ⋅ cos 3p
7
2sen p
7
 = 
–sen 6p
7
2sen 6p
7
 = – 1
2
Clave A
09 a2⋅csc(90°)+4ab sen30°⋅cos(180°)+b2 sen90°
a sec2180°⋅tan260°+b tan(–60°)⋅cot30°
  
a2 + 4ab 

1
2


 (– 1) + b2 
a(–1)2( 3)2 + b( 3)( 3)
  (a – b)2
3(a – b)
 = a – b
3
Clave D
10 P = sen p
6
 + cos

2p
3


 + tan p
4
 
 P = 1
2
 – cos

p
3
 

 + 1  P = 3
2
 – 1
2
 = 1
Clave B
11 
sen 

7p
12


cos 

p
12


 + 
sen 

p
12


cos 

7p
12


 
 
cos 

p
12


cos 

p
12


 + 
sen 

p
12


–sen 

p
12


 = 1 – 1 = 0
Clave A
12 G = 
2tan(p + q) + 3cot

p
2
 – q 

 
4sen(–q) + 5cos

3p
2
 + q 

 G = 2(tanq) + 3tanq
–4senq + 5(senq)
 = 5tanq
senq
 
  G = 5secq
Clave C
13 • a ∈ IV cuadrante
 tana = 
8sen(p – a) cos(– a) 
csc

p
2
 + a 

 
 tana = 8sena ⋅ cosa
seca 
 sena
cosa = 8sena ⋅ cosa
seca 
  sec3a = 8  seca = 2
Clave E
14 G

p
8


 = cot 9p
8
 – tan 46
3
 p
8
 + 2sen 16
3
 p
8
 + sec10 p
8
 = cot p
8
 + tan p
12
 + 2sen 2p
3
 – sec p
4
 = 2 + 1 + 2 – 3 + 2 

3
2


 – 2
  G

p
8


 = 3
Clave E
15 E = sen 

–p
3


 ⋅ cot 

p
4


 = – 3
2
 ⋅ (1) 
  E = – 3
2 Clave B
ACTIVIDADES CAP 15
IDENTIDADES TRIGONOMÉTRICAS FUNDAMENTALES
01 M = 

cosx
senx
 senx 

2
 + 

senx
cosx
 cosx 

2
 M = cos2x + sen2x = 1
Clave A
02 W = 
(1 – cos2x)senx
(1 – sen2x)cosx
3 = 
sen3x
cos3x
3
 W = senx
cosx
 = tanx
Clave D
03 senx + 1 – cos2x = 1 ⇒ senx = cos2x
 E = 1 + tan2x – 1 = tan2x = sen2x
cos2x
 E = sen2x
senx
 = senx
Clave E
04 E = 2sen2x – sen2x – cos2x
 E = (sen2x – cos2x)(sen2x + cos2x)
1
 E = sen4x – cos4x
Clave B
05 secx = tany = 7
 P = 1 + tan2y – sec2x + 1
 P = 1 + 72 – 72 + 1 = 2
Clave A
06 senx
cosx
 + cosx
senx
 = sen2x + cos2x
senx · cosx
 = 4
 1 = 4senxcos ⇒ senxcosx = 1
4
 C = 1
4
Clave B
07 L = sec2x – tan2x – 1 + 2tanx
 L = 1 – 1 + 2tanx = 2tanx
Clave D
08 (senx + cosx)(sen2x + cos2x – senxcosx)
(senx + cosx)
 = 7
8
 1 – senxcosx = 7
8
 ⇒ senxcosx = 1
8
 M = sen2x + cos2x
senxcosx
 = 1
1
8
 = 8
Clave E
09 9sen2x + cos2x + 6senxcosx = m2 ... (I)
 sen2x + 9cos2x – 6senxcosx = n2 ... (II)
 (I) + (II): 10(sen2x + cos2x) = m2 + n2
 m2 + n2 = 10
Clave B
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
27
10 R = 3(2(sec2x + tan2x))
sec2x – tan2x= 6(1 + tan2x + tan2x)
1
 R = 6 + 12tan2x = 6 + 12(cotx)–2
 a + 3b(cotx) – c = 6 + 3(4)(cotx)–2
 a = 6 ; b = 4 y c = 2
 a + b + c = 12
Clave B
CUADERNO DE TRABAJO
01 E = 
(sen2a + cos2a)2
cos2a
 = 
1
cos2a
 = sec2a
Clave A
02 
sena(1 – sen2a)
cosa
 = 
sena cos2a
cosa
 = sena cosa
Clave A
03 L = 
(senx – cosx)(senx + cosx)(sen2x + cos2x)
(senx – cosx)
 – cosx
 L = (senx + cosx)(sen2x + cos2x)
1
 – cosx 
 L = senx + cosx – cosx = senx
Clave C
04 senx
cosx
 + senx = 2(1 + cosx)
 
senx(1 + cosx)
cosx
 = 2(1 + cosx)
 tanx = 2 (x  III C)
  secx = – 5, cscx = 
2
5–
  C = (– 5)
2
5– 


 = 5
2
Clave B
05 E = 
(secq – 1)(1 + cosq)cosq
sen2q
 E = 
sen2q
1
cosq




(1 + cosq)cosq– 1
 E = 
(1 – cos2q)cosq
sen2q cosq
 = 
sen2q cosq
sen2q cosq
 = 1
Clave A
06 E = 
cos2x(4cos2x – 1)
sen2x(3 – 4sen2x)
 = 
cos2x(4cos2x – 1)
sen2x(3 – 4(1 – cos2x))
 E = 
cos2x(4cos2x – 1)
sen2x(4cos2x – 1)
 = cot2x
Clave E
07 • sec2x – secx – 1 = 0
 (1 + tan2x) – secx – 1 = 0
 sen2x
cos2x
 = 1
cosx
  cosx = sen2x
 • M = cosx – cos2x
sen2x
  M = cosx – cos2x
cosx
  M = cosx – cosx = 0
Clave E
08 1
cos2x
 = n senx
cosx
  senx cosx = 1
n 
 
(senx – cosx)(sen2x + senx cosx + cos2x)
(senx – cosx)(sen2x – 2senx cosx + cos2x)
 
 
1 + senx cosx
1 – 2senx cosx
 = 
1 + 1
n
1 – 2
n
 = n + 1
n – 2
Clave C
09 1
cosx
 + 1
senx
 = m  
senx + cosx
senx cosx
 = m (I)
 senx + cosx = n  1 + 2senx cosx = n2 (II)
 De (I): 
n
senx cosx
 = m  senx cosx = n
m
 En (II): 1 + 2n
m
 = n2  2n
m
 = n2 – 1
  2n = m(n2 – 1)
Clave D
10 E = 
a(sen2x )
1
2 – bsenx
a(1 + sen2x + cos2x + 2senx – cos2x)
1
2 – a
1
 E = 
asenx – bsenx
a(1 + senx) – a
 = 
asenx – bsenx
asenx
 = a – b
a
Clave D
TAREA
01 E = senx
1
senx
 + cosx
1
cosx
 + tanx
1
tanx
 E = sen2x + cos2x + tan2x
 1
 E = 1 + tan2x = sec2x 
02 E = (1 + cosx) 

1
senx
 – cosx
senx


 E = (1 + cosx) (1 – cosx)
senx
 E = 1 – cos2x
senx
 = sen2x
senx
 = senx
 
03 1
 E = sen2x + cos2x + 2senx cosx – 1 
senx cosx
 E = 2senx cosx 
senx cosx
 = 2
04 cosx 

1
1 + senx
 + 1
1 – senx


 = 2
k
 cosx 

2
1 – sen2x


 = 2
k
 cosx 

2
cos2x


 = 2
k
  k = cosx
 REFORZANDO
01 4 

senx
cosx
 + cosx
senx


 cosx senx – 4sen2x
 = 4 

1
cosx senx


 cosx senx – 4sen2x
 = 4(1 – sen2x) = 4cos2x
Clave E
02 sec2x – tan2x + 2senx cosx
 1
 = sen2x + cos2x + 2senx cosx = (senx+cosx)2
 = (0,3)2 = 0,09
Clave C
03 sen2x
csc4x
 + cos2x
sec4x
 – sen6x 
 = sen6x + cos6x – sen6x = cos6x
Clave C
04 (tan2x + 1)2
sec4x
 = (sec2x)2
sec4x
 = 1
Clave B
05 cosx 

senx – 1 + senx + 1
sen2x – 1


 = cosx(2senx)
 –cos2x
 = –2tanx = –10
Clave E
06 sec2a – tg2a – 2
1
tga
 + tga
 – cosa
 = 1 + 2
cosa
sena
 + sena
cosa
 – cosa
 = 1 + 2sena ⋅ cosa – cosa
 = (sena + cosa)2 – cosa
 = (sena + cosa) – cosa = sena 
Clave A
07 tan4a + 2tan2a cot2a + cot4a = 34 + 2
  (tan2a + cot2a)2 = 36  tan2a + cot2a = 6
  sec2a – 1 + csc2a – 1 = 6  sec2a csc2a = 8
  sen2a cos2a = 1
8 Clave B
08 2A = tg3β – ctg3β
sec2β ⋅ csc2β – 1 
 
 = (tgβ – ctgβ)(tg2β + tgβ ⋅ ctgβ + ctg2β)
 sec2β ⋅ csc2β – 1
 = (tgβ – ctgβ)(sec2β – 1 + 1 + csc2β – 1)
 sec2β ⋅ csc2β – 1
 = (tgβ – ctgβ)(sec2β ⋅ csc2β – 1)
 sec2β ⋅ csc2β – 1
 = tgβ – ctgβ  2A = B
  B
A
 = 2
Clave E
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
28
09 [(sen10° + cos10°) + 1][(sen10° + cos10°) 
– 1] – sen10° ⋅ cos10°= (sen10° + cos10°)2 
– 12 – sen10° ⋅ cos10° =
 = sen210° + cos210° + 2sen10° ⋅ cos10° – 1 
 – sen10° ⋅ cos10°
 = 1 + 2sen10° ⋅ cos10° – 1 – sen10° ⋅ cos10°
 = sen10° ⋅ cos10°
Clave E
10 E = (1 + ctg2x) – 10ctgx
 = (ctg2x – 10ctgx + 52) – 24 = (ctgx – 5)2 – 24
 Como:
 (ctgx – 5)2 ≥ 0  (ctgx – 5)2 – 24 ≥ – 24
 Min = 0 
 E ≥ – 24  Emin = – 24
Clave D
11 sec225° sen225°+cos225°+2sen25°⋅cos25°
1
cos25°
 + 
1
sen25°
 = sec225° (sen25°+cos25°)2
sen25° + cos25°
sen25° ⋅ cos25°
 = sen25° + cos25°
sen25° + cos25°
 ⋅ sen25° ⋅ cos25° ⋅ sec225°
 = sen25° ⋅ cos25° ⋅ sec225° = sen25°
cos25°
 = tg25° 
 
Clave D
12 senx (1 – secx) = –2cosx
 sen2x (1 – 2secx + sec2x) = 4cos2x
 1 – 2secx + sec2x = 4ctg2x
  sec2x – 4cot2x
1 – 2secx
 = –1
Clave C
13 • k = secx – tanx
1
tanx
 – 1
secx
 = secx – tanx
secx – tanx
secx tanx
 k = secx tanx
 • De la condición:
 1 = tan3x + tanx  1 = tanx (tan2x + 1)
  1 = tanx sec2x  1 = tanx secx
  k = 1 
 Clave D
14 cosx (1 – senx)
secx – tgx ⋅ 
secx + tgx
secx + tgx 
 = cosx (1 – senx)(secx + tgx)
sec2x – tg2x 
 = 
cosx 

1 + senx
cosx


(1 – senx)
1
 = 1 – sen2x = cos2x 
Clave B
15 Sea: secx = a  tan2x = a2 – 1
 Reemplazando: 
 a2 + a4 + a2 (a2 – 1) + (a2 – 1)3
a3 + (a2 – 1)
 = a2 + a4 + a4 – a2 + a6 – 3a4 + 3a2 – 1
a3 + a2 – 1
 = a2 + a
6 – (a4 – 2a2 + 1)
a3 + a2 – 1
 = a4 + a
6 – (a2 – 1)2
a3 + a2 – 1
 = a2 + [a3 + (a2 – 1)][a3 – (a2 – 1)]
a3 + a2 – 1
 = a2 + a3 – a2 + 1 = a3 + 1
 = sec3x + 1 ≡ secM(x) + N
  M + N = 3 + 1 = 4 
Clave C
ACTIVIDADES CAP 16
IDENTIDADES TRIGONOMÉTRICAS AUXILIARES
01 secxcscx = 3
 J = sec2x + csc2x + 2secxcscx
 J = sec2xcsc2x + 2(3)
 J = (3)2 + 6 = 15
Clave D
02 R = 
secqcscq + 2
secqcscq
 – senq
 R = 1 + 2senqcosq – senq
 R = (senq + cosq)2 – senq
 R = senq + cosq – senq = cosq
Clave B
03 sec2x – tan2x = 1 ; secx – tanx = 2
 (secx + tanx)(secx – tanx) = 1
 (secx + tanx)(2) = 1 ⇒ secx + tanx = 1
2
 Sumando: 2secx = 5
2
 ⇒ secx = 5
4
Clave B
04 sec2xcsc2x = 2cot2xsec3x
 1
sen2x
 = 2 cos2x
sen2x
 · 1
cosx
 cosx = 1
2
 x = 60° ∨ 300°
Clave E
05 secxcscx = 4 ⇒ senxcosx = 1
4
 J = 1 – 3sen2xcos2x = 1 – 3

1
4


2
 = 13
16
Clave E
06 sen4β + cos4β = 1 – cos4β
 1 – 2sen2βcos2β = (1 –cos2β)(1 + cos2β)
 1 – sen2β = 3sen2βcos2β
 1
3
 = sen2β
 senβ = 1
3
 
3
2
1
 tanβ = 1
2
 ⇒ tan2β = 1
2
Clave D
07 1 + 2senxcosx = 4
3
 ⇒ senxcosx = 1
6
 J = 1
senxcosx
 = 1
1
6
 ⇒ J = 6
Clave B
08 senx + cosx = 1
3
 1 + 2senxcosx = 1
9
 senxcosx = – 4
9
 J = 1 + senx + cosx + senxcosx
 J = 1 + 1
3
 + 

– 4
9


 = 8
9
Clave E
09 (sec2x –tan2x)2 = (1)2
 sec4x + tan4x – 2sec2xtan2x = 1
 a + bsec2xtan2x = 1 + 2sec2xtan2x
 a = 1 ; b = 2
 2a + b = 2(1) + 2 = 4
Clave C
10 1 – 2sen2xcos2x = m ... (I)
 1 + 2senxcosx = n2
 senxcosx = n
2 – 1
2
 ... (II)
 (II) en (I): 1 – 2 n
2 – 1
2
 
2
 = m
 2(1 – m) = (n2 – 1)2
Clave E
CUADERNO DE TRABAJO
01 J = (tanx + cotx – cotx)cscx
 J = tanx · cscx = senx
cosx
 · 1
senx
 = secx
Clave B
02 E = 
1 – 2sen2x cos2x – 1
1 – 3sen2x cos2x – 1
 = 
–2sen2x cos2x
–3sen2x cos2x
 E = 2
3 Clave C
03 C = (tanx + cotx – cotx)(tanx + cotx – tanx)
 C = tanx · cotx = 1
Clave A
04 secx · cscx = cotx + 1 
  tanx + cotx = cotx + 1
 tanx = 1  x  {45°; 225°}
Clave E
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
29
05 senx
cosx
 + cosx
senx
 = 2 2  
sen2x + cos2x
senx cosx
1
 = 2 2
 
 senx · cosx = 
1
22
 = 
2
4
 J = 1 – 2sen2x · cos2x
 J = 1 – 2




2
4
2
 = 1 – 1
4
 = 3
4 Clave C
06 A = 
csc2x(sec2x – 1)
sec2x(csc2x – 1)
 = 
csc2x · tan2x
sec2x · cot2x
 A = cos2x
sen2x
 · tan2x
cot2x
 = cot2x · tan2x
cot2x
 = tan2x
Clave B
07 senx
cosx
 + cosx
senx
 = n  senx cosx = 1
n
 
 (senx + cosx)2 = m2  1 – 2senx cosx = m2
  1 + 2



1
n
 = m2  2
n
 = m2 – 1
  n(m2 – 1) = 2 Clave A
08 • senx + cosx = m  1 + 2senx cosx = m2
  senx cosx = m
2 – 1
2
 • E = 1 + (senx + cosx) + senx cosx
 E = 1 + m + m
2 – 1
2
 = 0,5(m + 1)2
Clave E
09 [sec8x – sec6x + sec4x – sec2x]csc2x
 [sec6x(sec2x – 1) + sec2x(sec2x – 1)]csc2x
 sec2x(sec2x – 1) [sec4x + 1]csc2x
 sec2x · tan2x [sec4x + 1]csc2x
 sec2x · 
sen2x
cos2x
 [sec4x + 1] · 
1
sen2x
  sec4x(sec4x + 1) = sec8x + sec4x 
 sec8x + sec4x = secMx + secNx
  M + N = 12
Clave B
10 (secx+ tanx)2 = m2
 sec2x + tan2x + 2secx tanx = m2
 (sec2x + tan2x)2 = (m2 – 2n)2
 sec4x + tan4x + 2sec2x tan2x = m4 – 4m2n + 4n2 
 1 + 4sec2x tan2x = m4 – 4m2n + 4n2
 1 + 4n2 = m4 – 4m2n + 4n2  4m2n = m4 – 1
Clave C
TAREA
01 E = (secx ⋅ cscx)cosx
 E = (secx ⋅ cosx)cscx  E = cscx
 1 
02 Si secx – tanx = 2
 secx + tanx = 1
2
 2tanx = – 3
2
↑ (–)
 tanx = – 3
4
03 E = 2(1 – 3sen2x cos2x) – 3(1 – 2sen2xcos2x)
 E = 2 – 6sen2x cos2x – 3 + 6sen2x cos2x
 E = – 1 
04 M = (sec2x + csc2x – sec2x) ⋅ sen2x
 M = csc2x ⋅ sen2x = 1
 
 REFORZANDO
01 sen4x + cos4x = 1 – 2sen2x cos2x
 sen4x + cos4x = 1 – 2

1
4


2
 = 7
8
Clave A
02 sen2x
csc4x
 + cos2x
sec4x
 – 3sen4x
 = sen6x + cos6x – 3sen4x 
 = 1 – 3sen2x cos2x – 3sen4x
 = 1 – 3sen2x (1 – sen2x) – 3sen4x
 = 1 – 3sen2x + 3sen4x – 3sen4x = 1 – 3sen2x
Clave E
03 secx + tanx = 3
4
  secx – tanx = 4
3
 
  1
cosx
 – senx
cosx
 = 4
3
  1 – senx
cosx
 = 4
3
 
Clave D
04 • tanx + cotx = 3
2
  senx
cosx
 + cosx
senx
 = 3
2
 
  1
senx cosx
 = 3
2
  senx cosx = 2
3
 • sen6x + cos6x = 1 – 3 sen2x + cos2x
 = 1 – 3 

2
3


2
= – 1
3
 
Clave A
05 1senx + 1cosx = 2 y 12 + 12 = 2
  senx = cosx = 1
2
  x = 45°
  tanx + cotx = 1 + 1 = 2
Clave C
06 • senx = 2
3
 
 2
3
x
5
  cosx = 5
3
 
 • sen6x + cos6x = 1 – 3 sen2x cos2x
 sen6x + cos6x = 1 – 3 

2
3


2

5
3


2
 = 1 – 20
27
 = 7
27
 
Clave B
07 M = sec2x + 2secx cscx + csc2x
secx cscx
 – secx cscx 
 M = sec2x csc2x + 2secx cscx
secx cscx
 – secx cscx 
 M = secx cscx + 2 – secx cscx = 2
Clave D
08 3sen4x – 2sen6x + 3cos4x – 2cos6x
 = 3(sen4x + cos4x) – 2(sen6x + cos6x)
 = 3(1 – 2sen2x cos2x) – 2(1 – 3sen2x cos2x)
 = 3 – 6sen2x cos2x – 2 + 6sen2x cos2x = 1
Clave B
09 Si cscx + cotx = 3
x + 1
  cscx – cotx = x + 1
3
 
 Del dato: cscx – cotx = 2x – 2
3
  x + 1
3
 = 2x – 2
3
  x = 3
Clave C
10 3 senx + 2 cosx = 5 se observa que: 
 ( 3)2 + ( 2)2 = ( 5)2 
  senx = 3
5
 y cosx = 2
5
 Sea:
 E = 

1 + 3
5


 

1 + 2
5


 (1 – senx – cosx)6 
 E = 

1 + 3
5


 

1 + 2
5


 2(1 – senx)(1 – cosx)
 E = 

1 + 3
5


 

1 + 2
5


 2

1 – 3
5


 

1 – 2
5


 E = 2

1 – 3
5


 

1 – 2
5


 = 12
25
Clave B
11 sec2x + csc2x = sec2x csc2x 
 sec2x + csc2x = 1
(senx cosx)2
 sec2x + csc2x = 
1


1
n


2 = n2
Clave C
12 (senx + cosx)2 = 

1 + 3
2


2
 
  1 + 2senx ⋅ cosx = 4 + 2 3
4
  senx ⋅ cosx = 3
4
 Luego:
 16(sen6x + cos6x) + 3(sec2x + csc2x)
 = 16[1 – 3sen2x ⋅ cos2x] + 3sec2x ⋅ csc2x
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
30
 = 16

1 – 9
16


 + 3

16
3


 = 16

7
16


 + 3

16
3


 = 7 + 16 = 23
Clave D
13 senx + 2cosx = 3  1senx + 2cosx = 3
 Se observa: 12 + ( 2)2 = ( 3)2
  senx = 1
3
 y cosx = 2
3
 sec2x + csc2x = sec2x csc2x
 sec2x + csc2x = 3
2
 + 3 = 9
2
Clave E
14 Multiplicando por senx la condición:
 3senx + 7 = 2cosx
  – 3senx + 2cosx = 7
 Se observa: (– 3)2 + 22 = ( 7)2 
  senx = – 3
7
 y cosx = 2
7
 Reemplazando:
 

– 7
3


4 
+ 2

– 7
3


2 
+
 

7
2


4 
+ 2

7
2


2
 = 91
9
 + 105
16
 = 2401
144
 = 49
12
Clave E
15 k = 1 – 2sen6x
sen2x (1 – sen2x)
  k = 1 – 2sen6x
sen2x cos2x
 De la condición:
 1
cos4x
 – sen2x
cos2x
 = 2cos2x  1 – cos2x sen2x = 
 = 2cos6x
  1 – cos2x sen2x = 2(1–3sen2x cos2x–sen6x)
 1 – 2sen6x = 5sen2x cos2x 
  1–2sen6x
sen2x cos2x
 = 5  k = 5 
Clave E
ACTIVIDADES CAP 17
IDEN. TRIGON. FUNDAMENTALES DE SUMA Y DIFERENCIA
01 L = senxcosa
senxcosa
 = tanx
Clave A
02 a = ω – q
 tana = tanω – tanq
1 + tanωtanq
 
q a ω
a
1
23
2
 tana = 
1 + 

3
2




1
5


3
2
 – 1
5
 = 1
Clave A
03 3q = 2q + q
 sen2q + cos2qtanq + cos2qcotq – sen2q
 cos2q(tanq + cotq)
 cos2q(2csc2q)
 2cot2q
Clave A
04 cosxcosy + senxseny = 3senxseny
 cosxcosy = 2senxseny
 tanxtany = 1
2 Clave C
05 y = cos(a + 45°) + [–sen(45° – a)]
 (a + 45°) + (45° – a) = 90°
 y = cos(a + 45°) – cos(a + 45°)
 y = 0
Clave D
06 L = 1 + tan40°tan5° + tan40° + tan5°
 tan(45° – 5°) = tan45° – tan5°
1 + tan45°tan5°
 tan40° = 1 – tan5°
1 + tan5°
 tan40°tan5° + tan40° + tan5° = 1
 Reemplazando: L = 1 + 1
 L = 2
Clave B
07 a – 3q = (a + 2β) – (3q + 2β)
 tan(a – 3q) = 2 – 7
1 + 2(7)
 = – 1
3 Clave D
08 a + 53°
2
 = 45°
 a = 37°
2
 tana = 1
3
 
45° 53°/2
5
2
E 5 5
B
5
a
D C
Clave B
09 Propiedad: x + y + z = 180°
 tanx + tany + tanz = tanxtanytanz
 Aplicando:
 1 + tanatanβtanγ = 1 + tana + tanβ + tanγ
 1 + tanatanβtanγ = 1+ 2013 = 2014
Clave A
10 E = cotAcotB + 1 + cotBcotaC + 1 + cotAcotC + 1
 E = 3 + (cotAcotB + cotBcotC + cotAcotC)
 1
 E = 3 + 1 = 4
Clave E
CUADERNO DE TRABAJO
01 
sena cosb + cosa senb + sena cosb – cosa senb
cosa cosb – sena senb + cosa cosb + sena senb
 
2sena cosb
2cosa cosb
 = tana
Clave A
02 
 
sen50°sen10°x – sen50°cos10°y = –1
sen50°sen10°x + sen10°cos50°y = 1 


(–)
 y(sen10°cos50° + sen50°cos10°) = 2
 ysen(10° + 50°) = 2  y 
3
2
 = 2 
  y = 
3
4
  y = 
3
34
Clave A
03 tan(q + a) = 
tanq + tana
1 – tanq tana
 
 2 =
tan
tan
θ
θ
+
−
1
2
1
2
 
2
1
3
  tanq = 3
4 Clave D
04 
 tan x
7
 =
tan tan
tan tan
4 3
4 3
x x
x x
7 7
1
7 7
−
+ ⋅
= a – b
1 + ab
 tanx =
tan tan
tan tan
4 3
4 3
x x
x x
7 7
1
7 7
+
− ⋅
= a – b
1 – ab
 
 E = (1 – a2b2) a – b
1 + ab




a – b
1 – ab




 = a2 – b2
Clave B
05 N = 






sen4a
sen2a
cos4a
cos2a
–
2
 – tan22a
 N = 






sen4a cos2a – cos4a sen2a
sen2a cos2a
2
 – tan22a
 N = 






sen(4a – 2a)
sen2a cos2a
2
 – tan22a
 N = sec22a – tan22a = 1
  N2 – 1 = 0 
Clave C
06 • tan(x – y) = a – b
a + b
 • tan(y – z) = 1
 Siendo x – y + y – z = x – z
 tan(x – z) = 
tan(x – y) + tan(y – z)
1 – tan(x – y) · tan(y – z)
 
 tan(x – z) =
a b
a b
a b
a b
−
+
+
− −
+
1
1
= a
b
  cot(x – z) = b
a
Clave B
07 C = 3senx + 4cosx
32 + 42Máx =
 + 5 = 10
 
 L = 1senx + 3cosx
312 + 2 Mín = –
 + 1= –1  10 + (–1) = 9
Clave C
08 tana = 2k; tanb = 3k; tanq = 5k
 Como a + b + q = 90°
 tana tanb + tanb tanq + tanq tana = 1
 6k2 + 15k2 + 10k2 = 1
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
31
 k = 
31
1  tanq = 
31
5  cotq = 31
5
Clave D
09 k = 2cotA
tanB
 + tanB
tanB
 + 2cotB
tanC
 + tanC
tanC
 + 2cotC
tanA
 + tanA
tanA
 k = 3 + 2(cotA cotB + cotB cotC + cotC cotA)
 k = 3 + 2(1) = 5
Clave B
10 	 Del	gráfico:
 tanb = 1
7
; 
 tan(a + b) = 2
7
; 
53°
43
3
1
1
1
 tan(a + b + q) = 3
4
 
 tan(a + b – b) = 
2
7
 – 1
7
1 + 2
7
 · 1
7
 = 7
51
  tana = 7
51
 
 tan(a + b + q – a – b) = 
3
4
 – 2
7
1 + 3
4
 · 2
7
 = 13
34
 
  tanq = 13
34
  
1 – tanq
3tana
 = 
21
51
21
34 = 51
34
 = 3
2
Clave A
TAREA
01 R = sen30°cosx + senxcos30° + sen30°cosx
 – senxcos30°
 R = 2sen30°cosx = 2

1
2


cosx
 R = cosx 
02 V = cos45°cosx – sen45°senx + cos45°cosx
 + sen45°senx 
 V = 2

2
2


cosx = 2cosx
03 tan(x + y) = tanx + tany
1 – tanxtany
 tan(x + y) = 2 + 3
1 – 2(3)
 = –1
04 tan(45° – 37°) = tan45° – tan37°
1 + tan45°tan37°
 tan8° = 
1 + 1

3
4


1 – 3
4
 = 1
7
 REFORZANDO
01 senx = 1
2
 1
2
x
3
 cosy = 1
3
 
3
1
y
8
 sen(x + y) = senx cosy + seny cosx
 sen(x + y) = 1
2
 ⋅ 1
3
 + 8
3
 ⋅ 3
2
 = 1 + 2 6
6
Clave E
02 tan(30° – x) = tan30° – tanx
1 + tan30° tanx
 tan(30° – x) = 
3
3
 – 3
1 + 3
3
 ⋅ 3
 = – 2 3
6
 = – 
3
3
Clave E
03 sen(x + y) sen(x – y) + cos2x + sen2y
 sen2x – sen2y + cos2x + sen2y = 1 
Clave B
04 sen(x – y) 
cosx cosy
 + tany + tanx
 tanx – tany + tany + tanx = 2 tanx
Clave C
05 cosx = cosy cosz
 y + z = p – x  cos(y + z) = cos(p – x)
  cos(y + z) = –cosx
  cosy cosz – seny senz = –cosy cosz
  2cosy cosz = seny senz coty cotz = 1
2
 
 Clave D
06 tanφ = – 3
2
 
X
(2; –3)
(–2; 1)
β
aa – φ
Y
 tanβ = – 1
2
 
 φ – a = β  a = φ – β
 
 tana = tan(φ – β) = 
– 3
2
 – 

– 1
2


 
1 + 

– 3
2




– 1
2


 = – 4
7
 
 Clave C
07 • tana 
tanβ 
 = 7
18
  sena cosβ 
cosa senβ 
 
 = 7
18
 
sena cosβ = 7k
cosa senβ = 18k

 • sen(a + β) = 4
5
  sena cosβ + senβ cosa = 4
5
  k = 4
125 7k 18k
 • 25sen(β – a) = 25(senβ cosa – sena cosβ)
 18k 7k
 25sen(β – a) = 25 ⋅ 11k = 25 ⋅ 11 ⋅ 4
125
 = 8,8 
 
Clave D
08 • tan70° – tan20° = 2a  cot20° – tan20° = 2a
  2cot40° = 2a  tan40° = 1
a
 
 • sen5° = sen(45° – 40°)
 sen5° = sen45° cos40° – sen40° cos45°
 sen5° = 1
2
 (cos40° – sen40°)
 = 1
2
 

a
a2 + 1
 – 1
a2 + 1


 = a – 1
2a2 + 2
Clave D
09 • Valor máximo: (2n)2 + (n2 – 1)2 = 5
  4n2 + n4 – 2n2 + 1 = 25  n4 + 2n2 + 1 = 25
  (n2+1)2 = 25  n2 + 1 = 5  n2 = 4  n = 2
 • Máximo valor pedido:
 = (12 n–1)2 + 102
n2 = 144 2–1 + 100
4
 
 = 169 = 13
Clave C
10 M = (cot40° – tan40°) – tan10°
tan(180° + 10°)
 M = (2cot80°) – tan10°
tan10°
 = 2tan10° – tan10°
tan10°
 = 1 
Clave E
11 Aplicamos la identidad auxiliar:
 sen(a – β)
cosa cosβ
 = tana – tanβ
 k = (tan27° – tan2x) + (tanx – tan27°) 
 +(tan2x – tanx)
 k = 0
Clave A
12 	 Del	gráfico:
 tana = 2
x
 tanβ = 8
x
 tan(a + β) = 12
x
 En la identidad: 
 tan(a + β) = tana + tanβ
1 – tana tanβ
 12
x
 = 
2
x
 + 8
x
1 – 2
x
 ⋅ 8
x
  12
x
 

1 – 16
x2


 = 10
x
  16
x2 = 1
6
  x = 4 6
Clave B
13 De i) cota cotβ = 2cosa cotβ (cota + cotβ)
 cosa
sena
 ⋅ cosβ
senβ
 = 2cosa cosβ sen(a + β)
sena senβ
  sen(a + β) = 1
2
 De ii) (1 – sen2a) – (1 – sen2β) = 1
3
  sen2β – sen2a) = 1
3
 sen(β – a) sen(β + a) = 1
3
1
40°
a
a2 + 1
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
32
  sen(β – a) ⋅ 1
2
 = 1 
3
  sen(β – a) = 2
3
  sen(a – β) = – 2
3
Clave A
14 Obsérvese que: 

p
5
 – q 

 + 

p
20
 + q 

 = p
4
  p
5
 – q = p
4
 – 

p
20
 + q 

 Dato: tan

p
20
 + q 

 = 2
3
 
  tan

p
5
 – q 

 = tan p
4
 – 

p
20
 + q 

 tan

p
5
 – q 

 = 
tan p
4
 – tan

p
20
 + q 

 
1 + tan p
4
 tan 

p
20
 + q 

 
 tan

p
5
 – q 

 = 
1 – 2
3
1 + 1 ⋅ 2
3
 = 1
5
Clave A
15 De la condición:
 sen(q–β)
tana
 = 2cosq cosβ  sen(q–β)
cosqcosβ
 = 2tana 
  tanq = tanβ + 2tana (1) 
 En la expresión:
 k = 
sen(q – a)
cosq cosa
sen(a + β)
cosa cosβ
 = tanq – tana
tana + tanβ
 k = tanβ + 2tana – tana
tana + tanβ
 = tanβ + tana
tana + tanβ
 = 1
Clave D
ACTIVIDADES CAP 18
IDENTIDADES TRIGONOMÉTRICAS DE ÁNGULO DOBLE
01 sena = – 3
5
 ; a ∈ IIIC
 3cos2a = 3(1 – 2sen2a)
 3cos2a = 3

1 – 2 – 3
5
 
2 

 = 21
25 Clave B
02 E = 1
2


2sen
p
12
 cos
p
12


 + 
3
2


2sen
p
6
 cos
p
6


 E = 1
2
sen
p
6
 + 
3
2
sen
p
3
 E = 1
2


1
2


 + 
3
2


3
2


 = 1
Clave C
03 2cos22a – 1 + 1 – cos2a = 0
 2cos22a = cos2a
 cos2a = 1
2
 2cos2a – 1 = 1
2
 ⇒ cos2a = 3
4
Clave E
04 M = 2[2senxcosx](cos2x – sen2x)
 M = 2sen2xcos2x = sen4x
Clave C
05
 a + 2q = 90°
 cosa = sen2q
 
q
q
a 2q
4
A
C
M B
2
2
6 6 cosa = 2senqcosq
 cosa = 2

2
6




2
6


  cosa = 

2 2
3


Clave D
06 E = – 2cot20°
cot20°
 = –2
Clave B
07 tany
2
 = 2x ⇒ 
1 – cosy
1 + cosy
 = 2x
 1 – cosy
1 + cosy
 = 2x ⇒ 1 + cosy
1 – cosy
 = 1
2x
 Se pide:
 1 + cosy
1 – cosy
 + 1 + 1
2x
 = 1
2x
 + 1 + 1
2x
 = x + 1
x
 Clave B
08 E = 2sen2q + 2senqcosq
2cos2q + 2senqcosq
 E = 2senq(senq + cosq)
2cosq(senq + cosq)
 E = senq
cosq
 = tanq
Clave A
09 P = 2cos2x
cos2x
 + 2sen2x
sen2x
 P = 2 + 2 = 4
Clave A
10 M = 4(2sen20°cos20°)cos40°cos80°
sen20°
 M = 2(2sen40°cos40°)cos80°
sen20°
 M = 2sen80°cos80°
sen20°
 = sen160°
sen160°
 M = 1
Clave A
CUADERNO DE TRABAJO
01 Siendo 0 < q < 
p
8
 , 0 < 2q < 
p
4
 sen2q = 
26
1 , cos2q = 
26
5
 Piden: 13sen4q + 
5
sen4q
 
1
5
26
 = 26sen2q · cos2q + 
5
2sen2q cos2q
 Pero: sen2q · cos2q = 
26
1 · 
26
5 = 5
26
  26



5
26
 + 
2



5
26
5
 = 5 + 13 = 18 
Clave E
02 E = 
2sen2x cos2x
2cos22x
 = sen2x
cos2x
 = tan2x
 E = 2tanx
1 – tan2x
 = 2(2)
1 – 22 = –4
3
Clave A
03 • x  IV cuadrante
 senx = 5–
6
, cosx = 
6
1
 56
1
x
 • 18cos4x = 18[1 – 2sen22x]
 Siendo sen2x = 2senx cosx
  sen2x = 2




5–
6



 6
1
 = 
6
5–2
 = – 
5
3
  18cos4x = 18



10
9
1 – = –2
Clave D
04 tanq = 2
x
 tan2q = 5
x
 
3
2
x
 Sabemos que tan2q = 
2tanq
1 – tan2q
  5
x
 =
2 2
1 2 2
x
x
( )
− ( )
  x = 2 5 
  tanq = 2
x
 = 
2
52
 = 
5
1
Clave C
05 Por propiedad:
 1 – sen40° = (sen20° – cos20°)2
 C = 1 – sen40° + sen20°
 C = (sen20° – cos20°)2 + sen20°
 C =|sen20° – cos20°| + sen20°
 Por C.T. sen20° < cos20°
  C = cos20° – sen20° + sen20° = cos20°
Clave A
06 
1
sena
 – 
cosa
sena
 = 1
5
  csca – cota = 1
5
 (1)
 Sabemos csc2a – cot2a = 1
  (csca – cota)
1/5
(csca + cota) = 1
  csca + cota = 5 (2)
 (1) y (2): csca = 13
5
  sena = 5
13
, cosa = 12
13
  sen2a = 2



5
13




12
13
 = 120
169
 
  cos2a = 


12
13
5
 – 


5
13
5
 = 119
169
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
33
  26(sen2a – cos2a) = 2
13 Clave B
07 E = tan55° – tan35° = cot35° – tan35°
 E = 2cot2(35°) = 2cot70° = 2tan20° = 2b
Clave E
08 sen15° = 1
x
 
CA
1
M
D
B
15°15°
120°
30°
30°
x
x
 Sabemos: 
 cos2a = 1 – 2sen2a 
  sena = 
2
1 – cos2a
  sen15° = 
2
1 – cos30°
 = 
2
32 – 
  1
x
 = 
2
32 –  x = 2 32 + 
Clave A
09 Área = 1
2
 



1
2
1
2
+ cos2a 



1
2
sen2a
 Área = 1
8
 (1 + cos2a)sen2a
 Área = 1
8
 2cos2a 2sena cosa
 Área = 1
2
 cos3a sena
 
1
2
1
2
1
2
1
2
Clave D
10 Usando cos2A = 1 + cos2A
2
 = 1
2
 + 1
2
 cos2A
 1
2
 + 1
2
 cos p
50
 = cos2 p
100
 E = 
1
2
 + 1
2
 cos2 p
100
 = 1
2
 + 1
2
 cos p
100
 E = cos2 p
200
 = cos p
200 Clave C
TAREA
01 sen2q = 2senqcosq
 sen2q = 2
 13
2 


 13
3 

 = 12
13 
q
3
213
02 cos2q = cos2q – sen2q
 cos2q = 

6
7


2
 – 

1
7


2
 = 5
7
03 cotq = 4 ⇒ tanq = 1
4
 tan2q = 
2tanq
1 – tan2q
 = 
1 – 

1
4


2
2

1
4


 = 8
15
 
04 cos2q = cos2q – sen2q
 
5
x
 = 

5
3


2
 – 

2
3


2
 2
3
x
q q
5
  x = 9 5
 REFORZANDO
01 cos74° = cos2(37°) = 1 – 2sen237°
 = 1 – 2 

3
5


2
 = 1 – 18
25
 = 7
25
 
Clave B
02 tan2a = 2tana
1 – tan2a
 
 x + 1
3
 = 
2 

1
3


 
1 – 

1
3


2
 x + 1
3
 = 3
4
 ⇒ 4x + 4 = 9  x = 5
4
 
Clave A
03 • cotx = 5
12
 • Construimos 
	 	 la	figura
 con los datos:
 donde tanx
2
 = 12
13 + 5
 = 12
18
 = 2
3
Clave B
04 	 Construimos	la	figura	con	sena = 3
10
 sen2a = 2sena cosa
 sen2a = 2 3
10
 ⋅ 7
10
 = 21
5
 
Clave E
05 	 Construimos	la	figura	con	tan2a = 2 2
	 De	la	figura:
 sena = 2 2
2 6
 = 3
3
 
a 2a
2 2
3
3
1
2 6
Clave D
06 2 tanq + 3tanq cotq = 1 + 2cotq 
 2 tanq – 2cotq = 1 – 3
 cotq – tanq = 3 – 1
2
 2cot2q = 3 – 1
2
  cot2q = 3 – 1
2 2
 
Clave C
x
a
a
3
1
13
x/2 x
13 12
5
a
7
3
10
07 • cosβ = – 3
5
 • tanβ
2
 = – 1 – cosβ
1 + cosβ
 
 • tanβ
2
 = – 
1 + 
3
5
1 – 
3
5
= –2
 • tan2β = 2tanβ
1 – tan2β
 = 
2 

4
3


 
1 – 

4
3


2 = – 24
7
  3tanβ
2
 – 7tan2β = 3(–2) – 7 

– 24
7


 = 18
Clave C
08 • sena = – 0,2 = – 1
5
 • cota
2
 = csca + cota 
 cota
2
 = –5 – 2 6
 cota
2
 + 5 = – 2 6 
Clave D
09 tan2a = a + bsec2a
bcota sec2a
 bsec2a(cota tan2a– 1) = a 
 bsec2a

tan2a
tana
 – 1 

 = a
 bsec2a(sec2a) = a  cos22a = b
a
 cos4a = 2cos22a – 1 = 2b
a
 – 1 = 2b – a
a
 
Clave A
10 Halle el máximo valor de:
 2(cos5a sena– sen5a cosa)sen2a
cos2a
 = 2cosa sena (cos4a – sen4a)sen2a
cos2a
 = sen2a (cos2a – sen2a)sen2a
cos2a
 = sen22a cos2a
cos2a
 = sen22a  Máx = 1
Clave C
11 K = [(sena–cosa)2]2 + 2sen2a + 1
2
(1–2 sen22a)
 K = (1 – sen2a)2 + 2sen2a + 1
2
 – sen22a
 K = 1–2sen2a + sen22a + 2sen2a + 1
2
 – sen22a
 K = 1 + 1
2
 = 3
2 Clave D
12 3(cos4x + 8cos2x – 8cos4x)
 = 3[2cos22x – 1 + 8cos2x(1 – cos2x)]
 = 3[2cos22x – 1 + 8cos2x sen2x)]
 = 3[2cos22x – 1 + 2(2cosx senx)2]
X
Y
(–3; –4)
5 β
X
Y
(2 6 ; –1)
5a
b c
ota
 se
c2
a
b sec2a
b cota
b a
a
a
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
34
 = 3[2cos22x – 1 + 2sen22x]
 = 3[2(cos22x + sen22x) – 1] = 3
Clave E
13 • tana cot a
2
 = tana 1 + cosa
1 – cosa
 
 = tana 1 + cosa
1 – cosa
 ⋅ 1 – cosa
1 – cosa
 = tana sena
1 – cosa
 = tana sena
1 – cosa
 ⋅ 1
 + cosa
1 + cosa
 = tana sena(1 + cosa)
sen2a
 
 = 1 + cosa
cosa
 = seca + 1 ≅ Aseca + B
  A = 1; B = 1  A2 + B + 1 = 3
Clave E
14 


cos x
4
 + sen x
4




cos x
4
 – sen x
4


 
2sen x
4
 

1 – 2sen2 x
8


 = 
cos2 x
4
 – sen2 x
4
2sen x
4
 cos x
4
 = 
cos x
2
sen x
2
 = cot x
2
Clave D
15 (1 + cos2a) + sen2a
(1 – cos2a) + sen2a
 = 12
5
 2cos2a + 2sena cosa
2sen2a + 2sena cosa
 = 12
5
 2cosa (cosa + sena) 
2sena (sena + cosa) 
 = 12
5
  cota = 12
5
	 En	la	figura:	
 tan x
2
 = 5
25
 = 1
5
Clave D
ACTIVIDADES CAP 19
IDENTIDADES TRIGONOMÉTRICAS DE ÁNGULO TRIPLE
01 C = 3senx – 4sen3x + senx
cos2x
 C = 4senx(1 – sen2x)
(1 – sen2x)
 = 4senx
Clave C
02 L = senx(2cos2x + 1)
senx
 + cosx(2cos2x – 1)
cosx
 L = 2cos2x + 1 + 2cos2x – 1 = 4cos2x
Clave C
03 
 BHT: cos2x = m – n
2n
 
n
n
xx
m
m – n
B
T
H
A
2x
2x
C
m – n
2
Clave C
13
13
12
5
a/2 a
04 C = 4
4cos20°cos(60° – 20°)cos(60° + 20°)
 = 4
cos3(20°)
 C = 
1
2
4
 = 8
Clave A
05 K = sen2x(2cos4x + 1)
sen2x
 – cos2x(2cos4x – 1)
cos2x
 K = 2cos4x + 1 – 2cos4x + 1 = 2
Clave B
06 M = 
4cos3a
3senacosa
sena
 – sena(2cos2a + 1)cosa
sena
 + 1
 M = 3 – 2cos2a – 1
4cos2a
 + 1 = 1 – cos2a
2cos2a
 + 1
 M = 2sen2a
2cos2a
 + 1 = tan2a + 1 = sec2a
Clave E
07 1
cot2q
= 4 ⇒ tan2q = 4 ⇒ tanq = 2
 tan3q = 3tanq – tan3q
1 – 3tan2q
 = 3(2) – (2)3
1 – 3(2)2
 ∴ 11tan3q = 2
Clave C
08 E = cos2a
cos2a
 + cos3a
cos3a
 – 8tana
2tana
1 – tan2a
 E = cos2a
cos2a
 – sen2a
cos2a
 + cosa(2cos2a – 1)
cos3a
 – 4
 + 4tan2a
 E = 1 – tan2a + 4cos2a – 3
cos2a
 – 4 + 4tan2a 
 E = 1 + 3tan2a – 3sec2a
 E = 1 – 3(sec2a – tan2a) = 1 – 3 = –2
Clave E
09 K = csc2Asen2A(2cos2A + 1)2 +
 sec2Acos2A(2cos2A – 1)2 + 2(2cos22A – 1)
 K = 8cos22A + 2 + 4cos22A – 2 = 12cos22A
Clave E
10 sen33°1
2
 + cos33°
3
2
 = a
2
 sen33°sen30° + cos33°cos30° = a
2
 cos(33° – 30°) = a
2
 ⇒ cos3° = a
2
 Se pide: cos9° = cos3(3°)
 cos9° = 4cos33° – 3cos3°
 cos9° = 4

a
2


3
 – 3

a
2


 cos9° = a3 – 3a
2 Clave C
CUADERNO DE TRABAJO
01 A = 
4cos3x – 3cosx – cosx
sen2x
 = 
4cos3x – 4cosx
sen2x
 A = 
4cosx(cos2x – 1)
sen2x
 = 
4cosx(–sen2x)
sen2x
 A = –4cosx
Clave E
02 M = 
senx(2cos2x + 1)
senx
 – 
cosx(2cos2x – 1)
cosx
 M = (2cos2x + 1) – (2cos2x – 1) = 2 
Clave D
03 Del triángulo: H
m
 = tan3x, H
n
 = tanx
  mtan3x = ntanx
  
msenx(2cos2x + 1)
cosx(2cos2x – 1)
 = nsenx
cosx
  m
n
 = 
2cos2x – 1
2cos2x + 1
  n + m
n – m
 = 
4cos2x
2
  2cos2x = n + m
n – m Clave B
04 S = cos20° · cos(60° – 20°) · cos(60° + 20°)
 S = 1
4
 · cos(3(20°)) = 1
4
 · cos60° = 1
4
 · 1
2
 = 1
8
Clave D
05 tanx = tan12° · tan72° · cot42°
 tanx = tan12° · tan72° · tan48°
 tanx = tan12° · tan(60° + 12°) · tan(60° – 12°)
 tanx = tan3(12°) = tan36°
  x = 36° 
Clave A
06 E = 1
cos80°
 + 8cos280° = 1 + 8cos380°
cos80°
 Siendo cos3(80°) = 4cos380° – 3cos80°
  4cos380° = cos240° + 3cos80°
 4cos380° = – 1
2
 + 3cos80°
 
 E = 
cos80°




1
2
1 + 2 + 3cos80°–
 
 = 1 – 1 + 6cos80°
cos80°
  E = 6
Clave B
07 J = 3sen(60° + x) – sen(180° + 3x) – sen3x
 J = 3sen(60° + x) + sen3x – sen3x
 J = 3sen(60° + x)
Clave D
08 4J = 4cos3x + 4cos3(120° – x) + 4cos3(120° + x)
 4J = 3cosx + cos3x + 3cos(120° – x) + cos3x
 + 3cos(120° + x) + cos3x
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
35
 4J = 3cosx + 3cos3x + 3[cos(120° – x) + 
 cos(120° + x)]
 4J = 3cosx + 3cos3x + 3[2cos120°cosx]
 4J = 3cosx + 3cos3x + 3 · 2



1
2
– cosx
  J = 3
4
 cos3x
Clave E
09 tan20° + tan40° + tan220° tan240° cot10°
 tan20° + tan40° + tan220° tan240° tan80°
 Por propiedad: tan20° tan40° tan80° = tan60°
  tan20° + tan40° + 3tan20° tan40°
  tan(20° + 40°) = 3
Clave B
10 DHC: tan2x = 11
5
 NHC: cotx = 11
1
 Luego: 
 tan2x · cotx = 11
5
 
2
2
2
2
1
H
B N
C2k
2x
2k
k
x
x
x
xx
2x 6
D
A
E
11
Clave A
TAREA
01 Dato: sena = 1
5
, piden sen3a
 sen3a = Sen 3a – 4sen3a
 sen3a = 3



1
5
 – 4



1
5
3
 = 3
5
 – 4
125
 = 71
125
02 cosq = 1
3
 , piden: cos3q
 cos3q = 4 cos3q – 3cosq
 cos3q = 4



1
3
3
 – 3



1
3
 = 4
27
 – 1 = – 23
27
03 Tan = 3q = 3tanq – tan3q
1 – 3tan2q
 = 3(2) – 23
1 – 3(2)2
 tan3q = 2
11
04 M = 
cos3x
cosx
+1
 m = cosx(2cos2x – 1)
cosx
 + 1
 M = 2cos2x – 1 + 1 = 2cos2x
REFORZANDO
01 senx = 0,5
  senx = 1
2
 
 sen3x = 3senx – 4sen3x
 sen3x = 3



1
2
 – 4



1
2
3
 = 1
Clave E
02 Se sabe que:
 sen3x = 3senx – 4sen3x
 sen3x + 4sen3x = 3senx
 sen3x + 4sen3x
3
 = senx
Clave A
03 Se sabe:
 I. sen3x = senx(2cos2x + 1)
  sen3x
senx
 = 2cos2x + 1
 II. cos3x = cosx(2cos2x – 1)
  cos3x
cosx
 = 2cos2x – 1
 Reemplazando en la expresión:
 (2cos2x + 1) + (2cos2x – 1) – 4cos2x = 0
Clave D
04 Se sabe:
 tan3x = tanx



2cos2x + 1
2cos2x – 1
 
 
  tan3x
tanx
 = 




1
5
2 + 1




1
5
2 – 1
 = – 7
3
Clave E
05 tanx = 11
 tan3x = 3tanx – tan3x
1 – 3tan2x
 = 3(11) – (11)3
1 – 3(11)2
 tan3x = 1298
362
 = 3,58
Clave C
06 csc3x = 5
3
  sen3x = 3
5
 
 sen3x = sen(2x + x)
 = sen2xcosx + senxcos2x
 = 2senxcosxcosx + senxcos2x
 = senx(2cos2x + cos2x)
 = senx(1 + cos2x + cos2x)
 = senx + 2senxcos2x = 3
5
Clave A
07 cotx = 2 2 ,	construimos	la	figura:
 cos3x = 4cos3x – 3cosx
 cos3x = 4



2 2
3
3
 – 3



2 2
3
 cos3x = 
10 2
27
Clave D
13
x
22
08 tan2a = 2tana
1 – tan2a
 x + 1
3
 = 








1
3
1
3
2
1 –
2 
  x = 5
4
 tan3a = 3tana – tan3a
1 – 3tan2a
 
 1 + x + y
3
 = 




1
3
1
27
1
9
3 –
1 – 3 .
  1 + x + y
3
 = 13
9
  1 + 5
4
 + y = 13
3
  y = 25
12
Clave C
09 Se sabe que:
 cos3x = cosx(2cos2x – 1) 
 cos3x = 6
2 15
 (2 . 1
5
 – 1)
 cos3x = 3
15
 . 



–3
5
 = – 3 15
25
Clave B
10 4cos3x – 3cosx
cosx
 = m  4cos2x – 3 = m
  2(1 + cos2x) = m + 3  cos2x = m + 1
2
Clave C
11 3senx – 4sen3x
senx
 + 4cos3x – 3cosx
cosx
 = 3 – 4sen2x + 4cos2x – 3
 = –4(1 – cos2x) + 4cos2x = –4 + 8cos2x
 = – 4 + 4(1 + cos2x) = 4cos2x ≡ 2kcoskx
  k = 2
Clave C
12 cos3a – (4cos3a – 3cosa)
cosa
 + 
 sen3a + (3sena – 4sen3a)
sena
 
 = – 3cos2a + 3 + 3 – 3sen2a
 = – 3(cos2a + sen2a) + 6 = 3
Clave C
13 sen2x = cos3x
 2senxcosx = 4cos3x – 3cosx
 2senx = 4(1 – sen2x) – 3
 4sen2x + 2senx – 1 = 0
 senx = –2 ± 4 – 4(4)(–1)
2(4)
 , x∈〈0; p
2
〉
  senx = 
5 – 1
4 Clave A
1
x
y
3
15
5
x
x
2x
62152
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
36
14 tan2φ = 2tanφ
1 – tan2φ
 
 7
y
 = 








3
y
3
y
2
1 –
2
  y = 3 7
 tan3φ = 3tanφ – tan3φ
1 – 3tan2φ
 
 x + 7
3 7
 = 












3
3 7
3
3 7
3
3 7
3
3
2
–
1 – 3
 
 x + 7
3
 = 5  x = 8 
Clave E
15 BH = sen2x = 2cos3x
 2senxcosx = 2(4cos3x – 3cosx) 
 senx = 4(1 – sen2x) – 3
 4sen2x + senx – 1 = 0
 senx = –1 ± 12 – 4(4)(–1)
2(4)
  senx = 17 – 1
8 Clave A
ACTIVIDADES CAP 20
TRANSFORMACIONES TRIGONOMÉTRICAS I
01 E = 2sen60°cos10°
cos10°
 = 2
3
2


 = 3
Clave B
02 M = 2sen6xcos2x + 2 3cos2x
cos2x
 M = 2sen6x + 2 3; 9x – 4(180°) = 0 ⇒ x = 80°
 M = 2sen(+480°) + 2 3
 M = +2sen120° + 2 3 = +2

3
2


 + 2 3
 M = 3 3
Clave D
03 R = 1 + cos4a + 2sen4acos2a
(1 + 2sen2a)(1 + cos4a)
 R = 2cos22a + 2(2sen2acos2a)cos2a
(1 + 2sen2a)(2cos22a)
 R = 2cos22a(1 + 2sen2a)
(1 + 2sen2a)2cos2a
 = 1
Clave A
04 3cosa = sen70° – sen10° –(–cos20°)
 3cosa = 2sen30°cos40° + cos20°
 3cosa = 2

1
2


cos40° + cos20°
 3cosa = 2cos30°cos10°
3
4
x
y
φ
φφ
1 2
A H C
B
2x
3x
 3cosa = 2

3
2


cos10°
 a = 10°
Clave C
05 2E = 2sen30°cos10° – cos10°
 2E = 2

1
2


cos10° – cos10°
  E = 0
Clave D
06 2M sen4°
cos8°
 = 2sen8°cos1° + 2sen8°cos8°
2cos8°cos1° + cos2(8°) + 1
 2M sen4°
cos8°
 = 2sen8°(cos1° + cos8°)
2cos8°cos1° + 2cos28°
 2M sen4° = sen8°(cos1° + cos8°)
(cos1° + cos8°)
 2M sen4° = 2sen4°cos4°
 M = cos4°
Clave E
07 ksen40°cos40°cos100° = cos100° + cos40°
 – k
2
sen80°cos80° = 2cos70°cos30°
 – k
4
sen160° = 2sen20°

3
2


 – ksen20° = 4 3sen20°
  k = –4 3
Clave E
08 Propiedad:
 F = 4senAsenBsenC
senAsenBsenC
 = 4
Clave D
09 E = sen40° + cos70°
cos10°
 = sen40° + sen20°
cos10°
 tana = 2sen30°cos10°
cos10°
 = 2

1
2


 = 1
 tana = tan45° ⇒ a = 45° = 
p
4
 Ca = 45° = 
p
4 Clave E
10 x = –1 ; y = –2
 r = 5
 E = 2sen3acos2a
 
X
Y
O
(–1; –2)
(–2; 1)
a
5
 E = 2(3sena – 4sen3a)(1 – 2sen2a)
 E = 2 3

–2
5


 – 4

–2
5


3
 1 – 2

–2
5


2
 
 25 5E = –12
Clave C
CUADERNO DE TRABAJO
01 Por transformación de suma a producto:
 
2sen2x cosx
–2sen2x senx
 = –cotx
Clave C
02 Transformando de suma a producto:
 
2sen3x · cos2x
2cos3x · cos2x
 = 3  tan3x = 3 
  3x = 60°  x = 20°
Clave B
03 Usando 0,5 como sen30°.
 sen30° + 2sen20° + sen10° = Asen20°cos25°
 2sen20°cos10° + 2sen20° = Asen20°cos25°
 2sen20°(1 + cos10°)
2cos25°
 = Asen20°cos25°
 4sen20°cos25° = Asen20°cos25°  A = 4
Clave A
04 2csc18° + 4sen6°csc18°
 4csc18°



1
2
+ sen6° 
 4csc18°(sen30° + sen6°)
 4csc18°(2sen18°cos12°)
 8csc18°sen18°
1
cos12°  8cos12°
Clave B
05 ksen82° = sen8° + 2cos38°
 ksen82° = sen8° + cos38° + cos38°
 ksen82° = sen8° + sen52° + cos38°
 ksen82° = 2sen30°
1/2
cos22° + cos38°
 ksen82° = cos22° + cos38°
 kcos8° = 2cos30°cos8° 
  k = 2




3
2
 = 3
Clave C
06 E = sen2130° – sen210°
 2E = 2sen2130° – 2sen210°
 2E = (1 – cos260°) – (1 – cos20°)
 2E = cos20° – cos260° = 2sen140° sen120°
 E = sen40° sen60° = 
3
2
 sen40° = 
2csc40°
3
 Piden:
  2Ecsc40° = 3 
Clave C
07 Factorizando 8sen3a:
 8sen3a (2sen4a – 3sen2a + 1)
 8sen3a (2sen2a – 1)(sen2a – 1)
 8sen3a (1 – 2sen2a)(1 – sen2a)
 8sen3a · cos2a · cos2a
 2sena · cos2a (4sen2a · cos2a)
sen22a
 sena · sen2a (2sen2a · cos2a)
  sena · sen2a · sen4a
Clave E
08 Llevando senos y cosenos
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
37
 
 = 
 
 
 = senA senB
cosA cosB
A B A B
A B A B
+
+
=
+( ) −( )
+( ) −( )
2
2 2
2
2 2
sen cos
cos cos
 
 = tan



A + B
2
 = tan60° = 3
Clave D
09 cos(2a + b) – cos(2b + a) = m
 –2sen



3
2
(a + b) sen



a – b
2
 = m
 –2sen







3
2
5p
3
 sen



a – b
2
 = m
 – 2
m
 = csc



a – b
2
  – 2
m
 = –csc



b – a
2
  csc



b – a
2
 = 2
m Clave D
10 2sen5a – sen3a = 0
 sen5a + sen5a – sen3a = 0
 sen(4a + a) + 2cos4a sena = 0
 sen4a cosa + cos4a sena + 2cos4a sena = 0
 Dividiendo entre cos4a cosa:
 tan4a + tana + 2tana = 0  tan4a = –3tana
 Piden: a = tan4a – 2 + 3tana
0
 = –2
  aa = (–2)–2 = 1
4 Clave C
TAREA
01 J = 2sen4xcos3x
cos3x
 J = 2sen4x
02 H = 2cos2xcosx
sen2x
 = 2cot2xcosx
03 M = 2sen16°cos64°
sen26°
 = 2sen16°sen26°
sen26°
 M = 2sen16°
04 P = 2sen45°sen25°
sen25°
 P = 2

2
2


 = 2 
REFORZANDO
01 
2sen



5x + 3x
2
 cos



5x – 3x
2
sen4xcosx
 = 
2sen4xcosx
sen4xcosx
 = 2
Clave B
02 
2cos



3x + x
2
 cos



3x – x
2
2cosxcos2x
 = 
2cos2xcosx
2cosxcos2x
 = 1
Clave A
03 cos165° + cos105°
 = cos



165° + 105°
2
cos



165° – 105°
2
 
 = cos135°cos30° 
 = – 2
2
 . 3
2
 = – 6
4
Clave E
04 sen7x + senx
sen4x
 = 2sen4xcos3x
sen4x
 = 2cos3x
Clave D
05 (sen20° + cos20°)sec25°
 = (cos70° + cos20°) 1
cos25°
 
 = 
2cos45°cos25°
cos25°
 = 2cos45° = 2 2
2
 = 2
Clave B
06 
(cos3a+ cosa) + cos2a
(sen3a + sena) + sen2x
 = 
2cos2acosa + cos2a
2sen2acosa + sen2a
 = 
cos2a (2cosa + 1)
sen2a (2cosa + 1)
 = 
cos2a
sen2a
 = cot2a
Clave C
07 (cos40° + cos40° + cos80°)csc80°
 = (cos40° + 2cos60°cos20°)csc80°
 = (cos40° + cos20°)csc80°
 = 2cos30°cos10°csc80°
 = 2 3
2
 . sen80°csc80° = 3
Clave C
08 M = 2cos60°senx = senx
 N = 2cos30°senx = 3 senx
  MN = senx( 3 senx) = 3 sen2x
Clave A
09 M = 
sen10° + sen70° + sen50°
sen70° + sen10° – sen50°
 M = 
2sen 30°cos20° + sen70°
2cos60°sen10° + sen10°
 M = 
2sen70°
2sen10°
 = 
cos 20°
sen 10°
 Msen10° = cos20°
Clave B
10 [6csc10°cos65° – 3 2csc10°] sen35°
 = [6csc10°(cos65° – 2
2
)] sen35°
 = [6csc10°(cos65° – cos45°)] sen35°
 = [6csc10°(–2sen55°sen10°)] sen35°
 = –6csc10°sen10° (2cos35°sen35°)
 = –6sen70° = –6cos20°
Clave D
11 




1
2
+ sen10°
2cos10°
2 sen40°
+ sen30°
 = 
2(sen30° + sen10°)sen40°
2cos10°
 + sen30°
 = 
2(2sen20°cos10°)sen40°
2cos10°
 + sen30°
 = 2sen40°sen20° + sen30°
 = cos20° – cos60° + sen30° = cos20°
Clave C
12 • cos220° – 
1
4
 = cos220° – sen230°
 = 
2cos220° - 2sen230°
2
 
 = 
(1 + cos40°) – (1 – cos60°)
2
 = 
cos40° + cos60°
2
 = 
2cos50°cos10°
2
 = cos50° . cos10°
 A = 50° , B= 10°
 • 4 3 sen(A + B) = 4 3 sen60°
 = 4 3




3
2
 = 6
Clave B
13 Tenemos :
 4csc25° . sen5°(2sen25°.cos10° + sen25°)
 = 4csc25° . sen5° . sen25°(2cos10° + 1)
 = 4csc25° . sen5° . sen25°



sen15°
sen5°
 = 4sen15° = 4



6 - 2
4
 = 6 - 2
Clave C
14 cos42x – cos44x
(cos6x.cos2x + 1)sen6x.sen2x
 = 
(cos22x + cos24x)(cos2x + cos4x)(cos2x – cos4x)
(cos6x.cos2x + 1)sen6x.sen2x
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
38
 = 
(cos6xcos2x + 1)(2sen3xcos3x)(2senxcosx)
1 + cos4x
2
1 + cos8x
2
+ (2cos3xcosx)[–2sen3xsenx]



 = 
1
2
(cos4x + cos8x + 2)
cos6x.cos2x + 1
 
 = 
1
2
(2cos6x.cos2x + 2)
cos6x.cos2x + 1
 = 1
Clave A
15 sen2a + sen2β + sen2ω
 = 2sen(a + β)cos(a – β) + sen2ω
 = 2senω . cos(a – β) + 2senω . cosω– = 
2senω[cos(a - β) + cosω]
 = 2senω2cos



a – β + ω
2
cos



a – β – ω
2
 = 4 senω . senβ . sena
 ∴ Si E es el número buscado, 
 E = 
4sena.senβ.senω 
–10senω.senβ.sena
 = – 2
5
Clave C
ACTIVIDADES CAP 21
TRANSFORMACIONES TRIGONOMÉTRICAS II
01 M = sen70° + sen20° – cos20° + cos90°
 M = sen70° + sen20° – sen70° + 0
 M = sen20°
Clave E
02 Ysen10° = sen70° – 2cos80°cos20° –
 2cos80°cos40° – (1)cos80°
 Ysen10° = sen70° – cos100° – cos60° –
 cos120° – cos40° – cos80°
 Ysen10° = sen70° – sen50° = 2sen10°cos60°
 Y = 2

1
2


 = 1
Clave E
03 P = 3cos10° – 6sen20°sen10°
 P = 3cos10° – 3(cos10° – cos30°)
 P = cos30° = 3
2
3
Clave A
04 2Y = sen120° + sen20° + sen540° – sen20°
 2Y = sen60° + 0 ⇒ 2Y = 
3
2
 Y = 
3
4 Clave C
05 2M = 1 + 2(cos10° – cos60°)
 M = cos10°
 3M – 4M3 = 3cos10° – 4sen310° = – cos30°
 3M – 4M3 = – 
3
2 Clave D
06 M = sen7x – senx – sen7x + sen3x
cos11x + cos3x – cos11x – cosx
 M = sen3x – senx
cos3x – cosx
 Pero: x = 
p
24
 ⇒ 12x = 
p
2
 M = cos9x – cos11x
sen9x – sen11x
 = – 2sen10xsenx
2cos10xsenx
 M = –tan10x = – tan10

p
24


 = – tan75°
 M = – 

6 + 2
6 – 2


 = –(2 + 3)
Clave D
07 2H = 
cosx
2
2
 + senx
2
2
 + 
2
2
sen3x – 
2
2
senx
cosx – cos

3x + 
p
2


 2H = 
2
2
(cosx + sen3x)
cosx + sen3x
 = 
2
2
1
 = 2
 H = 
2
2 Clave A
08 Evalunado para: x = 
p6
 = 30°
 16(sen30°)5 = Asen30° + Bsen90° + Csen150°
 16

1
2


5
 = A

1
2


 + B(1) + C

1
2


 A + 2B + C = 1
Clave C
09 L = sen6q – sen3q + sen4q + sen3q
2sen2qsen5q
 L = sen6q + sen4q
2sen2qsen5q
 = 2sen5qcosq
2sen2qsen5q
 L = cosq
2senqcosq
 = cscq
2
Clave D
10 dato: a + γ = 153°30'
 mBB = 26°30' = 53°
2
 a
γ
26°30'
S
A c = cos26° B
C
a = sen27°
 sen1° = 7
400
 S = 1
2
sen27°cos26°sen53°
2
 S = 1
4
(sen53° + sen1°)

1
5


 S = 
5
20


4
5
 + 7
400


 = 327 5
8000
Clave E
CUADERNO DE TRABAJO
01 E = 2cos80° + 4sen70°sen10°
 E = 2cos80° + 2[cos60° – cos80°]
 E = 2cos60° = 2



1
2
 = 1
Clave A
02 Multiplicamos por 2 para generar la 
transformación:
 2E = 2senx sen3x + 2senx sen5x + 2senx sen7x
 + 2senx sen9x
 2E = cos2x – cos4x + cos4x – cos6x + cos6x 
 – cos8x + cos8x – cos10x
 2E = cos2x – cos10x = 2sen6x sen4x
  E = sen6x sen4x
Clave B
03 Factorizando:
 
 = 
2sen20°




3
2
+ sen20°
2sen70°
 + cos60°
 = 
2sen20°(sen60° + sen20°)
2sen70°
 + cos60°
 = 
2sen20° 2sen40° cos20°
2sen70°
 + cos60°
 = 2sen20° sen40° + cos60°
 = cos20° – sen60° + cos60° = cos20°
Clave B
04 M = sen12° cos18° + sen18° cos12°
sen30°
 + 
 cos48°cos12°
 M = 1
2
 + cos48°cos12° = 
1 + 2cos48°cos12°
2
 M = 
1 + cos60° + cos36°
2
 M = 
1 + 1/2 + cos36°
2
  cos36° = 
4M – 3
2
 
Clave C
05 Sea M = csc10° – 4cos20°= 
1
sen10°
 – 4cos20°
 M = 
1 – 4sen10°cos20°
sen10°
 
 M = 
1 – 2(sen30° – sen10°)
sen10°
 M = 
1 – 2sen30° + 2sen10°
sen10°
 = 2
Clave E
06 
(sen6x – sen4x) + sen4x
(cos3x + cosx) – cosx
 = 
2sen3x cos3x
cos3x
  2sen3x
Clave D
07 R = 
cos60° – 2cos(15° + x) sen(15° – x) + sen4x
cos4x + cos2x
 R = 
cos60° – (sen30° – sen2x) + sen4x
2cos3x cosx
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
39
 R = 
sen2x + sen4x
2cos3x cosx
 = 
2sen3x cosx
2cos3x cosx
  tan3x
Clave C
08 senx



1 + cos2x
2
 = asenx + bsen3x
 
senx
2
 + 
senx cos2x
2
 = asenx + bsen3x
 2senx
4
 + 
2senx cos2x
4
 = asenx + bsen3x
 
2senx + sen3x – senx
4
 = asenx + bsen3x
 
senx + sen3x
4
 = asenx + bsen3x
 1
4
 senx + 1
4
 sen3x = asenx + bsen3x
  a + b = 1
4
 + 1
4
 = 1
2 Clave A
09 
2cos2a cosa + cosa – cos3a
2sen3a cosa
 
cos3a + cosa + cosa – cos3a
2sen3a cosa
 
 = 
2cosa
2sen3a cosa
  csc3a
Clave B
10 • 2sen210° = 1 – a  1 – cos20° = 1 – a
  a = cos20°
 • Piden: 
 
1 + 2(2cos40° sen10°)
2
 
 
1 + 2(sen50° – sen30°)
2
  sen50° = cos40° = 2cos220° – 1 = 2a2 – 1
Clave C
TAREA
01 A = cos30° + cos10° – cos10°
 A = 
3
2
02 E = sen50° + sen30° – cos40°
 E = cos40° + 1
2
 – cos40°
 E = 1
2
03 E = sen70° + sen30° – sen90° – sen70°
 E = 1
2
 – 1 = – 1
2
04 P = cos6x – cos8x + cos8x
 P = cos6x
REFORZANDO
01 2
2
sen45°cos15° = 1
2
(sen60° + sen30°)
 = 1
2




3
2
 – 1
2
 = 3 – 1
4
Clave B
02 2sen41°cos12° – sen29°
 = sen(41° + 12°) + sen(41° – 12°) – cos29°
 = sen53° + sen29° – sen29° = 4
5
Clave D
03 2 2
2
 cos8° = 2cos45°cos8°
 = cos53° + cos37°
 = 3
5
 + 4
5
 = 7
5
Clave E
04 2 3
2
 sen7° + sen23°
 = 2cos30°sen7° + sen23°
 = sen37° – sen23° + sen23° = 3
5
Clave C
05 M = 
sen60° + sen20° – sen20°
cos45° + cos25° – cos25°
 
 M = 
sen60°
sen45°
 = 
3
2
2
2
 = 3
2
 = 6
2
Clave B
06 2(2senxcos2x)sen3x + 2sen3xsenx
 2(sen3x – senx) sen3x + 2sen3xsenx
 = 2sen23x – 2senxsen3x + 2sen3xsenx
 = 2sen23

p
6


 = 2sen2 p
2
 = 2
Clave A
07 M = 
2cos40°cos20° + 2sen50°cos80°
–2(2sen60°cos10°)
 M = 
cos60° + cos20° + sen130° – sen30°
–2(sen70° + sen50°)
 M = 
sen70° + sen50°
–2(sen70° + sen50°)
 = – 1
2
 
Clave D
08 1
2
(2sen6xsen2x) + cos24x
 = 1
2
(cos4x – cos8x) + 1
2
(1 + cos8x)
 = 1
2
(1 + cos4x) = 1
2
(2cos22x) = cos22x
Clave B
09 2M = cos(a+β) + cos(a–β) + cos(β+γ) +
 cos(β–γ) + cos(γ+a) + cos(γ–a) –
 [cos(a–β) + cos(β–γ) + cos(γ-a) - 1]
 2M = –cosγ – cosa – cosβ + 1
 2M = –2cos



γ + a
2
 cos



γ – a
2
 + 2sen2 β
2
 2M = –2sen β
2
 cos



γ – a
2
 + 2sen2 β
2
 2M = 2sen β
2
 

sen β
2
 – cos



γ – a
2

 
 = 2sen β
2
 

cos



γ + a
2
 – cos



γ – a
2


 
 2M = 2sen β
2
 

–2sen γ
2
sena
2


 
 M = –2sena
2
sen β
2
sen γ
2
Clave C
10 E = 1
4
+ +
2cos55°sen45°
2
2sen80°sen20°
2
 = 1
4
+ – + –
sen100°
2
cos100°
2
sen10°
2
cos60°
2
 = 1
2
+ – –cos10°
2
sen10°
2
–sen10°
2
 = 1 +cos10°
2
 = 2cos25°
2
 = cos5°
Clave C
11 4cos20° – 3 cot20°
 = 4cos20° – 3
cos20°
sen20°
 = 
2sen40° – 3 cos20°
sen20°
 = 
2(2sen30°sen40° – cos30°cos20°)
sen20°
 = 
2(cos10° – cos70°) – (cos50° + cos10°)
sen20°
 = 
cos10° – cos50° – 2cos70°
sen20°
 = 
2sen30°sen20° – 2sen20°
sen20°
 = 
–sen20°
sen20°
 = –1
Clave B
12 tg x
2
 . tg 5x
2
 = a
 
2sen
2cos
sen
cos
5x
2
5x
2
x
2
x
2
 = a
 
cos2x – cos3x
cos3x + cos2x
 = a
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
40
 cos2x(1 – a) = cos3x(a + 1)
 
cos2x
cos3x
 = 
a+ 1
1 – a Clave D
13 Multiplicando por senx el 1° miembro:
 
senx + 4cos3xcosxsenx
senx
 = 
sen(Ax)
sen(Bx)
 
senx + 2(2senxcosx)cos3x
senx
 = 
sen(Ax)
sen(Bx)
 
senx + 2sen2xcos3x
senx
 = 
sen(Ax)
sen(Bx)
 
senx + (sen5x – senx)
senx
 = 
sen(Ax)
sen(Bx)
 
sen5x
senx
 = 
sen(Ax)
sen(Bx)
  A = 5 ∧ B = 1  AB + A – B = 9
Clave C
14 M = cos2p
7
 + cos4p
7
 + cos6p
7
 2sen3p
7
M = 2sen3p
7
cos2p
7
 + 2sen3p
7
cos4p
7
 +
 2sen3p
7
cos6p
7
 = sen5p
7
 + sen p
7
 + sen7p
7
 – sen p
7
 
 + sen9p
7
 – sen3p
7
 = sen2p
7
 + 0 – sen2p
7
 – sen3p
7
 
 M = – 1
2 Clave B
15 Multiplicando por 2 y transformando: 
 2E = cos2p
7
 – cos6p
7
 + cos2p
7
 – cos10p
7
 – 3cos2p
7
 
 2E = –cos6p
7
 – cos10p
7
 – cos2p
7
 Pero: cos10p
7
 = cos4p
7
 2E = –(cos2p
7
 + cos4p
7
 + cos6p
7
) 
 2E = –(– 1
2
)  E = 1
4 Clave D
ACTIVIDADES CAP 22
FUNCIONES TRIGON.: SENO Y COSENO
01 –2cosx – 1 ≥ 0 ⇒ cosx ≤ – 1
2
 x ∈ 2p
3
 ; 4p
3
 
prob. 14
 –cosx ≥ 0 ⇒ cosx ≤ 0
 x ∈ 
p
2
 ; 3p
2
 
  dom(f) = 2p
3
 ; 4p
3
 
Clave D
02 –1 ≤ senx ≤ 1
 – 
p
4
 ≤ 
p
4
senx ≤ 
p
4
 – 
2
2
 ≤ cos

p
4
 senx 

 ≤ 
2
2
 – 2 ≤ –2cos

p
4
 senx 

 ≤ 2
 1 – 2 ≤ 1 – 2cos

p
4
 senx 

 ≤ 1 + 2
 a = 1 – 2 ; b = 1 + 2
  a + b = 2
Clave D
03 f(x) = senx – 2
2 – senx + 1
 = 1
3 – senx
 – 1
 –1 ≤ senx ≤ 1
 2 ≤ 3 – senx ≤ 4
 1
4
 ≤ 1
3 – senx
 – 1 ≤ 1
2
 – 3
4
 ≤ f(x) ≤ – 1
2
Clave A
04 1 + sen

2px – 
p
2


 = 1
 sen

2px – 
p
2


 = 0
 2px – 
p
2
 = 0 ⇒ x = 1
4
 2px – 
p
2
 = p ⇒ x = 3
4
 Por lo tanto, el número de valores de x en 
el intervalo [0 ; 1] es 2.
Clave E
05 x ∈ 
p
2
 ; p ; f(x) = (senx + 3)2 – 5
 0 < senx ≤ 1
 3 < senx + 3 ≤ 4
 9 < (senx + 3)2 ≤ 16
 4 < (senx + 3)2 – 5 ≤ 11
 ∴fmín = 5
Clave C
06 sen(nx)cos(2mx) + sen(2mx)cos(nx) =
 sen(nx + 2mx) = sen(n + 2m)x
Clave B
07 f(x) = [(cos2x + sen2x)(cos2x – sen2x)] = |cos2x|
 
 Periodo:
 T = 
p
2
 
p
4
p
2
3p
4
p
1
Y
X0
T T
Clave D
08 f(x) = –2sen3x
Clave E
09
 
p
4
p
2
p
2
Y
cosx
X
S 2
2
2
2
p
4
–p
2
–


p
4
 ; 
2
2


 ⇒ S = 

p
2


 

2
2


 ∴ S = 
p 2
4
 u2
Clave E
10 f(x) = sen2x – 1 + sen2x
 sen2x – 1 ≥ 0 ⇒ sen2x ≥ 1 ⇒ sen2x = 1
 senx = ± 1 ⇒ x = (2n – 1) 
p
2
 ; n ∈ 
 
p
2
p
2
3p
2
Y
X0–
Clave C
CUADERNO DE TRABAJO
01 Ran(f) = [–1; 1]
 
Clave A
02 T = 3p
4
 – 



p
4
– = p 
 
 
Y
X
Clave A
03 I. (F) f(x) = cos2x y f(x) = 2cosx tienen el 
 mismo dominio .
 II. (F) El rango de f(x) = cos2x es [–1; 1] y 
 el de f(x) = 2cosx, [–2; 2].
 III. (V) El dominio de ambos es .
Clave B
04 	 El	gráfico	corresponde	a	la	función:	
 f(x) = 5sen3x
Clave E
05 
 
EDITORIALINGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
41
	 Respecto	al	gráfico:
 I. (F) El dominio de f(x) es [–p/6; 3p/2
 II. (F) El rango de f(x) es [–5; 5] 
 III. (V) El periodo T = 2p
3
 – 0 = 2p
3
Clave C
06 
 
	 Según	el	gráfico	es	correcto:
 I. (F) Si a = 2  n = p
4
 II. (F) Si a = 3  n = p
6
 
 III. (V) Si a = 1
2
  n = p
Clave C
07 
 Ran(f) = 1
2
3
2
;




 
1
2
3
2
Clave C
08 f(x) = 3secx + 4 = 
3
cosx
 + 4
 Como –1  cosx  1
  – < 
1
cosx
  –1  1  
1
cosx
 < 
  – < 
3
cosx
 + 4  1  7  
3
cosx
 + 4 < 
  f(x)  –; 1]  [7; +
 No pertenecen: 2; 3; 4; 5; 6
  Piden: 2 + 3 + 4 + 5 + 6 = 20
Clave E
09 Como x  p ; 5p
4
  |senx|= –senx
  f(x) = 1 + 2(–senx)cosx = (senx – cosx)2
  f(x) =|senx – cosx|= senx – cosx
  f(x) = 2sen



p
4
x –
 Siendo p < x < 5p
4
  3p
4
 < x – p
4
 < p
  0 < sen



p
4
x – < 
2
2
 
  0 < 2sen



p
4
x – < 1  0 < f(x) < 1
Clave B
10 
 
A
Y
XB
C
x y
y1
x1
 
 
 
 x1 = p 
4
  x2 = 5p
4
 y1 = 
2
2
  y2 = 
2
2–







A




p
4
;
2
2
B(p; 0)  C



;5p
4 2
2–
  A = 
8
2p (Por determinantes)
Clave D
TAREA
01
 
 
Ran(f) = [–1; 1]
1
–1
Ran(f) = [–1; 1]
02 
 
 –1
1
y = sen3x
2
03 f(x) = sen(x + p/5): x ∈ 
 ⇒ x + p/5 ∈ 〈–∞;∞〉
 ⇒ Domf = R ⇒ Domg = 
 ∴ Domf = Domg = 
04 senxcosx ≠ 0 ⇒ 2senxcosx ≠ 0
 ⇒ sen2x ≠ 0
 senx = 0 ⇔ x = 0; p, 2p; ...; np; n ∈ 
 ⇒ sen2x ≠ 0 ⇔ 2x ≠ np ⇒ x ≠ np
2
 ∴ Domf =  – {np
2
/n∈}
REFORZANDO
01 
 
–3
O
3
f(x) = 3sen2x
Ran(f)
 
 Ran(f ) = [–3;3]
Clave A
02
 
–
y = f(x)
O
–3
3 y = 3senx
T = 
3p
8
 – (– p
8
) = p
2
Clave B
03 
 I. (V)
 II. (F)
 III. (F)
Clave A
04 De la ecuación: y = cos(2x – p) = cos(p –2x)
 y = cos2x ⇒ t = p
Clave C
05 
 
–
–
y = f(x)
–3
3
	 Respecto	al	gráfico:
 I. (V) Donf(x) es 〈–3p/5; 3p/5〉
 II. (V) Ranf(x) es [–3; 3]
 III. (F) T = 3/5
Clave D
06 Sea 2(x + p
5
) = q
 En f(x) = Asenq,
 el rango depende
 solamente
 de A y no de q. 
Clave E
07 I. (F) Elrango de f(x) = 2cos2x es [–2;2]
 II. (F) El dominio de f(x) = cos2x es .
 III. (V) El rango depende de A.
Clave B
08 	 El	gráfico		corresponde	a	la	función
 f(x) = 2cos2x
Clave C
09 El dominio es tal que
T = p
f(x) = cos2x
T =
f(x) = cos3x
T =
f(x) = cosx/2
f(x) = –4cosq
Ran(f)
Ran(f) = [–4;4]
y = senx
–4
4
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
42
 sen x
2
 ≠ 0 ∧ cos x
2
 ≠ 0
 ⇒ x
2
 ≠ np/n ∈  ∧ x
2
 ≠ (2n+1) p
2
 / n ∈ 
 ⇒ x ≠ 2np/n ∈  ∧ x ≠ (2n+1)p/ n ∈ 
 ⇒ x ≠ 2kp/k ∈ 
 ∴ x ∈  – {kp/k ∈ }
Clave A
10 Dado que f(x) = 1 – senx – cos2x 
 = sen2x – senx
 f(x) = sen2x – senx + 1
4
 – 1
4
 f(x) = (senx – 1
2
)2 – 1
4
 Como: – 1 ≤ senx ≤ 1
 – 3
2
 ≤ senx – 1
2
 ≤ 1
2
 0 ≤ (senx – 1
2
 )2 ≤ 9
4
 – 1
4
 ≤ (senx – 1
2
 )2 – 1
4
 ≤ 8
4
 ∴ – 1
4
 ≤ f(x) ≤ 2
Clave C
11 Dando la forma a f(x)
 f(x) = 
senx – 2 + 3
senx – 2
 = 1 + 
3
senx - 2
 
 Siendo: – 1 ≤ senx ≤ 1
 – 3 ≤ senx – 2 ≤ –1
 – 1 ≤ 
1
senx – 2
 ≤ – 
1
3
 – 3 ≤ 
3
senx – 2
 ≤ – 1
 – 2 ≤ 1 + 
3
senx – 2
 ≤ 0
 
 ∴ –2 ≤ f(x) ≤ 0
Clave E 
12 Si L = asenx + bcosx
  Lmáx = a2 + b2
  Lmín. = – a2 + b2
 Para f(x) = 3senx + 2cosx
  Ranf = [– 32 + 22 ; + 32 + 22 ]
 Ranf = [– 13; 13]
Clave B
13 – 1 ≤ sen(2px – p
2
) ≤ 1
 0 ≤ 1 + sen(2px – p
2
) ≤ 2  0 ≤ f(x) ≤ 2
Clave C
14 g(x) = sen3x f(x) = cos3x
 Del dato:
 Sen23x + 2cos23x = 1
 1 +cos23x = 1
 cos23x = 0
 cos3x = 0
  3x = p
2
 ; 3p
2
 ; 5p
2
 ; 7p
2
 
 x = p
6
 ; p
2
 ; 5p
6
 ; 7p
6
 
 ∴ p
6
 + p
2
 + 5p
6
 = 3p
2
 
Clave E
15 Siendo –1 ≤ senx ≤ 1 ⇒ –3 ≤ senx – 2 ≤ –1
 ⇒ |senx – 2 |
(–)
 = 2 – senx
 Piden:
 f(x) = 
senx – 2
2 – senx + 1
 = 
senx – 2
3 – senx
 = 
1
3 – senx
 – 1
 Sabemos que:
 –1 ≤ senx ≤ 1
 –1 ≤ –senx ≤ 1
 2 ≤ 3 – senx ≤ 4
 1
4
 ≤ 
1
3 – senx
 ≤ 1
2
 – 3
4
 ≤ 
1
3 – senx
 – 1 ≤ – 1
2
 ∴ – 3
4
 ≤ f(x) ≤ – 1
2 Clave D
ACTIVIDADES CAP 23
FUNCIONES TRIGON.: TANGENTE
01 f(x) = 1
tan2xtanx
 tanx	≠	0				⇒ x ∉ [... ; –p ; 0 ; p ; 2p ; ...]
 tan2x	≠	0		⇒ 2x ∉ [... ; –p ; 0 ; p ; 2p ; ...]
 ∴ Domf =  – { kp
4
/k ∈ }
Clave B
02 f(x) = 3 – tanx + tanx – 1
 3 – tanx ≥ 0 ⇒ tanx ≤ 3
 tanx – 1 ≥ 0 ⇒ tanx ≥ 1
 ∴ Ranf = [1 ; 3]
Clave C
03 f(x) = 1
tanx
 + 1
senx
 = cosx
senx
 + 1
senx
 f(x) = 1 + cosx
senx
 senx	≠	0		⇒ x ∉ {... ; –p ; 0 ; p ; 2p ; ...}
 ∴ Domf =  – {kp/k ∈ }
Clave A
04 f(x) = tan2x – 

1 – tan2x
2tanx


 f(x) = 2tan3x + tan2x – 1
2tanx
	 tanx	≠	0			⇒ x ∉ {... ; –p ; 0 ; p ; 2p ; ...}
 ∴ x ∈  – {kp/k ∈ }
Clave A
05 f(x) = 3tanx ; x ∈ 2p
3
 ; 7p
6
 
 f 

2p
3


 = 3(– 3) = –3
 f 

7p
6


 = 3

3
3


 = 1
 ∴ (–3)(1) = – 3
Clave C
06 
p
4
 ≤ x ≤ 
p
3
 ⇒ 1 ≤ tanx ≤ 3
 0 ≤ tanx – 1 ≤ 3 – 1
 1 – 3 ≤ 1 – tanx ≤ 0
 f(x) = tanx – 1 + –(1 – tanx) = 2tanx – 2
 1 ≤ tanx ≤ 3 ⇒ 2 ≤ 2tanx ≤ 2 3
 0 ≤ 2tanx – 2 ≤ 2 3 – 2
 ∴ Ranf = [0 ; 2 3 – 2] 
Clave D
07 Área: S = p(2 3)
 ∴ S = 2p 3
Clave D
08 Área: S = p(1)
 ∴ S = pu2
Clave D
09 f(x) = 2senxcosxsenx
2senxcosx
 = senx
 f(x) = senx ; x ∈ [0 ; 2p]
Clave B
10 f(x) = senx ; g(x) = tanaa ; x ∈ [0 ; 4p]
 
 
p
4
1
2
3
4
5
6
7
8
9
p 2p 3p 4p
 Por lo tanto hay 9 puntos de intersección.
Clave B
CUADERNO DE TRABAJO
01 i) tanx  x  (2k + 1)p
2
 (1)
 ii) tan2x  2x  (2k + 1)p
2
 
  x  (2k + 1)p
4
 (2)
 iii) tan2x  2x  kp  x  kp
2
 (3)
 De (1), (2) y (3): x  kp
4
  Dom f(x)   – kp
4





 / k  
Clave C
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
43
02 
 
  Dom f(x)   – kp
6 
Y
X
3x  (2k + 1)p
2
 x  (2k + 1)p
6
3x  kp
 x  kp
3 Clave A
03 El dominio de f(x) es tal que:
 3x
2
  (2k + 1)p
2
  x  (2k + 1)p
3
  Dom f(x)   – (2k + 1)p
3 Clave E
04 Como f(x) = sen2x + cos2x + tan2x
 x   x  – (2k + 1)p
2




  Dom f =  – (2k + 1)p
2
 / k ∈ 
Clave B
05 Dando forma f(x) = atan2x + btanx
 Completando a cuadrados:
 f(x) = a


tan2x + b
a
 tanx = a


tanx + b
2a
2
– b
2
4a
 Luego: 0  

tanx + b
2a
2
 < 
 0  a


tanx + b
2a
2
 <  
 –b2
2a
  a


tanx + b
2a
2
– b
2
4a
 <   –b2
4a
  f(x) < 
Clave E
06 g(x) = ( 3tanx)2 + 2( 3tanx) 



1
32
 + 
2



1
32
 – 
2



1
32
 g(x) = 
2



1
32
3tanx + – 1
12
 
 g(x) = 
(6tanx + 1)2
12
 – 1
12
 Como x  – 
p
4
 ; 
p
4
  –1 < tanx < 1
  –6 < 6tanx < 6  –5 < 6tanx + 1 < 7
  0  (6tanx + 1)2 < 49 
  0  
(6tanx + 1)2
12
 < 49
12
  – 1
12
  
(6tanx + 1)2
12
 – 1
12
 < 48
12
  – 1
12
  f(x) < 4 
Clave E
07 Como p
12
  x  p
2
 
  p
6
  2x  p 
  p
3
  2x + p
6
  7p
6
 
Y
X
3
3
3/3
  tan



p
6
2x +  –; 3
3
]  [ 3; +
Clave C
08 Del dato: – p
4
 < x  p
4
  – p
4
  –x < p
4
  0  p
4
 – x < p
2
  f(x)  [0; 
  mínimo = 0
Clave C
09 Cálculo del periodo:
 T = 2p = p
B
  B = 1
2
 (I)
 También 



p
3
; 3  f(x)
  f 



p
3
 = Atan



p/3
2
 = Atan p
6
  3 = A 3
3
  A = 3 
  y = f(x) = AtanBx = 3tan x
2
Clave E
10 
 
h
Y
X
–
y = tan(Mx)
 Se observa T = 2p
3
  x1 = p
6
 También T = p
M
 = 2p
3
  M = 3
2
  f(x) = tan 3
2
 x
  h = f 



p
6
 = tan



p
6
3
2
· = tan



p
4
 = 1
  Área = 1
2
 



4p
3
(1) = 2p
3
 m2
Clave C
TAREA
01 • tan p
3
 = 3
 • tan 



– p
6
 = – 3
3
 ⇒ Ranf = [– 3
3
; 3 〉
02 • tanx = –1 ⇒ x = – p
4
 • tanx = 1 ⇒ x = p
4
 ⇒ Ranf = – p
4
; p
4
 
03 f(x) = 
2tanx
1 – tan2x
 ⇒ f(x)= tan2x
 Dom tan2x =  – (2n + 1) p
4
 , n ∈ 
p
3
p
6
p
4
p
4
–1
1
04 
 Domf(x) = ...; p
4
 ; 5p
4
 ; 9p
4
 ; ...






 Domf(x) = kp + p
4
 , k ∈ 
REFORZANDO
01 • tan 



– p
4
 = 1
 • tan 



p
3
 = 3
 ⇒ Ranf = 〈– 1; 3 ]
Clave B
02 • x ∈ 



p
4
 ; p
2
 ⇒ f ∈ = [1 ; ∞〉
 • x ∈ 



p
2
 ; 5p
4
 ⇒ f ∈ 〈–∞; 1]
 ⇒ Ranf = 〈–∞; 1] ∪ [1 ; ∞〉
 Ranf = 〈–∞; + ∞〉 = 
Clave B
03 • f(x) = cotx ⇒ f(x) = 1
tanx
 ⇒ tanx ≠ 0
 • tanx = 0 ⇔ x ∈ {..., –p; 0; p; 2p;....}
 x = kp, k ∈ 
 ⇒ Domf(x) =  – {kp}; k ∈ 
Clave E
04 • Se observa que el período de
 f(x) = tanAx es 2 p
5
 
 • Se sabe que el período de tanx es p
 ⇒ A



2p
5
 = p ⇒ A = 5
2
 ∴ tanAx = tan 
5
2
 x
Clave C
05 	 En	la	figura:
 a1 = p
3
 ; a2 = p + p
3
 , a3 = 2p + p
3
, ...
 ⇒ ai ∈ np + p
3
/n∈ 






Clave D
p
4
5p
4
9p
4
1
tanx
–1
p
3
p
4
3
p
4
p
2
5p
4
1
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
44
06 • a = 9p
2
 + p
6
 a = 14p
3
 • b = 11p
2
 – p
6
 b = 16p
3
 ⇒ Domf(x) = 14p
3
 ; 16p
3
 
Clave B
07 
 Ranf = [0 ; 3 ]
Clave A
08 I. (V) Ran (tanx) = 
 Ran (tan2x) = 
II. (V) y = tanx e y = |tanx| tienen las
 mismas asíntotas ⇒ tienen el
 mismo dominio.
III. (V) El dominio depende de los va- 
 lores que puede tomar x. En tanx
 y 2tanx, x puede tomar los mis- 
 mo valores.
Clave E
09 f(x) = tg3x – ctg3x + 1 = –2ctg6x + 1
 x ∈ Dom(f) ⇔ x ∈  ∧ 6x ≠ np, n ∈ 
 ⇔ x ∈  ∧ x ≠ np
6
, n ∈ 
 [Dom(f )]° = 
np
6
/n∈ 






Clave D
10 f(x) = sec22x + 4ctg22x
 = 
sen22x + 4cos42x
sen22xcos22x
 = 
4(sen22x + 4cos42x)
sen24x
 Entonces x ∈ Dom(f) ⇔ sen4x ≠ 0
 Entonces Dom(f) =  – 
np
4
/n∈ 






Clave E
11 Como x∈ 3p
10
, 5p
6
 , ⇒ 1
4
 ≤ 1
2
senx ≤ 1
2
 ⇒ p
4
 ≤ p
2
senx ≤ p
2
 ⇒	1	≥	ctg 







p 1
2
senx ≥ 0
 ⇒ Ran(f) = [0,1]
Clave B
9p
2
p
3
p
3
10p
2
11p
2
–
a b
3
3
p
6
p
12
3
3
p
4
–
3
Ranf f(x) = |tan2x|
Tienen el mis-
mo rango
12 f(x) = 
2ctg2x
tgx
 + 2 = 
2
tgx
 



1 – tg2x
2tgx
 + 2 =
 = cot2x + 1 = csc3x ....... (I)
 Como x ∈ p
4
, p
3
 ⇒ 2
3
 ≤ cscx ≤ 2 y de (I)
 ⇒ Ranf(x) = 4
3
, 2 
 ∴ 3a – b2 = 0
Clave D
13 • Período de tanx es p:
 – tan2x: 2T1 = p ⇒ T1 = p
2
 – tan3x: 3T2 = p ⇒ T2 = p
3
 • Período de f(x) = tan2x + tan3x contiene
 a p
2
 y p
3
 ⇒ MCM 



p
2
 , p
3
 = p
 Clave A
14 
 
 Domf =  – (2n + 1) p
4
 /n ∈ 






p
6
2p
6
3p
6
g(x) = tan3x
 Domg =  – (2n + 1) p
6
 /n ∈ 






 Domf ∩ Domg =
 =  – (2n + 1) p
4
 , (2n + 1) p
6
 /n ∈ 






 Clave A
15 f(x) = (cot2x + tan2x)2 – 4 + 5
 = 4csc24x – 4 + 5
 = 2 csc24x – 1 + 5
 = 2|ctg4x|+ 5
 ⇒ Ran(f) = [5, +∞〉 ∧ T = 
p
4
 
 ⇒ secMT = sec 
5p
4
 = – 2
 Clave B
ACTIVIDADES CAP 24
FUNCIONES TRIGONOMÉTRICAS INVERSAS
01 I - V ; II - F ; III - V
Clave E
p
4
p
4
2p
4
3p
4
f(x) = tan2x
–
02 H = –sen 1
2
arcsen

2 2
3


 = –sen 1
2
q 
 

q
 q = arcsen

2 2
3


 ⇒ senq = 2 2
3
 
q
1
2
23
 H = –sen

q
2


 = – 1 – cosq
2
 = – 
1 – 1
3
2
 H = – 3
3 Clave D
03 R = 6cos 2arccos
2
3
 = 6cos2q
 

q
 q = arccos
2
3
 ⇒ cosq = 
2
3
 R = 6cos2q = 6{2cos2q – 1} = 6 2

2
3


 – 1 
 R = 2
Clave B
04 E = (–2)arcsen

cos

p
2
 + 
p
10




 E = 2arcsen

sen 
p
10


 = 2

p
10


 E = 
p
5 Clave E
05 x ∈ [–1 ; 1]
 – 
p
2
 ≤ arcsenx ≤ 
p
2
 ∧ 0 < arccotx < p
 ⇒ arcsenx – arccotx < 
p
2
 Luego: [a ; b] = [–1 ; 1]
 ∴ a2 + b2 = 2
Clave D
06 E = sen(2arccosx) ⇒ E = sen2q
 

q
 q = arccosx ⇒ cosq = 
 E = 2senqcosq = 2( 1 – x2 )x 
1 – x
2
x
1
q
 E2 = 4(1 – x2)x2
Clave D
07 M = tan(arctan(1) + arctan( 26 – 5)) +
 tan(arctan(1) + arctan – ( 26 – 5))
 M = tan arc tan

1 + 26 – 5
1 – 26 + 5


 + 
 tan arc tan

1 – 26 + 5
1 + 26 – 5


 M = 26 – 4
6 – 26
 + 6 – 26
26 – 4
 = 2 26
5
Clave E
08 f(x) = arcsen(2x – 1) + arccos(2x + 1)
 –1 ≤ 2x – 1 ≤ 1 ∧ –1 ≤ 2x + 1 ≤ 1
 0 ≤ 2x ≤ 2 ∧ –2 ≤ 2x ≤ 0
 0 ≤ x ≤ 1 ∧ –1 ≤ x ≤ 0
 ∴ Domf = {0}
Clave D
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
45
09 tan(arcsen 4 – x2 ) = sen(arctan2)
 

q 

ω
 ⇒ tanq = senω
 q = arcsen 4 – x2 ⇒ senq = 4 – x2
 ω = arctan2 ⇒ tanω = 2
 Reemplazando: 
 4 – x2
x2 – 3
 = 2
5
 4 – x2
x2 – 3
 = 4
5
 ⇒ 3|x| = 4 2
Clave D
10 f(x) = |arcsenx|+|arctanx| ; x ∈ [–1 ; 1]
 – 
p
2
 ≤ arcsenx ≤ 
p
2
 ⇒ 0 ≤|arcsenx|≤ 
p
2
 – 
p
4
 ≤ arctanx ≤ 
p
4
 ⇒ 0 ≤|arctanx|≤ 
p
4
 Luego: ranf = 0 ; 3p
4
 
Clave C
CUADERNO DE TRABAJO
01 b = arc sen 3
2
 + arc cos 2
2
 – arc tan 3
3
 b = p
3
 + p
4
 – p
6
 = 5p
12 Clave E
02 Por propiedad: arc cos(cosx) = x, 0  x  p 
 E = arc cos







p
7
–sen = arc cos







p
7
p
2
+cos
 E = arc cos







9p
14
cos  E = 


9p
14
Clave A
03 w = tan 1
2












2p
4
sen3arc cos
 w = tan 1
2








p
2
sen3arc cos

1
 w = tan 1
2







3arc cos

p/3
 = tanp = 0
Clave A
04 cos(arc senx) + sen



p
2
– arc senx = 3
2
 cos(arc senx) + cos(arc senx) = 3
2
 2cos(arc senx) = 3
2
  cos(arc senx) = 3
4
 arc senx = a  sena = x  cosa = 3
4
  x = 1 – cos2a = 1 – 9
16
 = 
7
4
Clave A
05 arc cotx = arc cot 1
1 – x
 arc tan 1
x
 = arc tan 1
1 – x
  1
x
 = 1
1 – x
 
  x2 = 1 – x  x2 + x – 1 = 0
  x = 
2
–1  5 
  x = 
2
–1 + 5
Clave A
06 f(x) = 3arc sen



3x – 4
5
 + 5arc cos



x2 – 2
2
+ p
4
  –1  3x – 4
5
 ≤ 1  –1  x
2 – 2
2
  1
  – 1
3
  x  3  0  x2  4
  – 1
3
  x  3  –2  x  2  x  – 1
3
; 2 
  El menor es – 1
3 Clave A
07 El dominio: –1  4x – 7
10
  1  –3
4
  x  17
4
 x  – 3
4
; 17
4
 
 Calculamos el rango:
 – p
2
  arc sen 



4x – 7
10
  p
2
 – 2p
15
  2
3
 arc sen 



4x – 7
10
 + p
5
  8p
15
  E  – 2p
15
 ; 8p
15
 
  – 3
4
 ; 17
4
 ; – 2p
15
 ; 8p
15
 
Clave A
08 Buscamos el dominio de f(x):
 f(x) = arc cosx + arc cotx
 x  [–1; 1] x   Dom f  [–1; 1]
 Luego: 
 
0  arc cosx  p
p
4
  arc cotx  3p
4 




(+)
 p
4
  f(x)  7p
4
 Comparando: m = p
4
; M = 7p
4
  M
m
 = 7 
 
Clave D
09 Si f(x) = p
2
 arc sen x2 – 8 + p
3
 arc cos x
4
  0  x2 – 8  1  –1  x
4
  1
 8  x2  9  –4  x  4
 x  [–3; –2 2]  [2 2; 3]  x  [–4; 4]
 x  [–3; –2 2]  [2 2; 3] = [–b; –a]  [a; b]
  a2 + b2 = 17
Clave C
10 Calculamos los valores donde exista la 
función:
 f(x) = arc senx + arc cosx + arc tan(1/|x|+ 1)
 x [–1; 1]  x [–1; 1]  x 
 x [–1; 1]
 Luego: f(x) = p
2
 + arc tan 



1
|x|+ 1
 Partiendo: –1  x  1  0  |x|  1
 1  |x|+ 1  2  1
2
  1
|x|+ 1
  1
  arc tan 1
2
  arc tan 1
|x|+ 1
  p
4
  f(x)  p
2
 + arc tan 

1
2


 ; 3p
4
 
Clave C
TAREA
01 • f(0) = arc sen0 = 0
 • f(0,5) = arc sen(0,5) = p
6
 ⇒ f(0) + f(0,5) = 0 + p
6
 = p
6
02 • f(1) = arc tan(1) = p
4
 • f 3
3




 = arc tan 3
3




 = p
6
 ⇒ f(1) + f 3
3




 = p
4
 + p
6
 = 5p
12
03 
 Ranf = p
4
 , p
3
 
04 • arc sena = p
4
 ⇒ a = sen p
4
 = 2
2
 • arc sen(–1) = b ⇒ senb = –1 ⇒ b = – p
2
 
 ⇒ a – b = 2
2
 – p
2
 = 2 – p
2
REFORZANDO
01 • f 3
2




 = arc sen 3
2




 = p
3
 • f 2
2




 = arc sen 2
2




 = p
4
 ⇒ f 3
2




 + f 2
2




 = p
3
 + p
4
 = 7p
12
Clave E
2
2
3
2
p
3
p
4
f(x)
4–x
2
x2–3
1
q
1
2
5ω
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
46
02 • f(–1) = arc tan(–1) = – p
4
 • f( 3 ) = arc tan( 3 ) = p
3
 ⇒ f(–1) + f( 3 ) = – p
4
 + p
3
 = p
12
 Clave B
03 • f 
1
3




 = arc sen
a

1
3 + arc sen
β

2
3
 
 ⇒ sena = 1
3
 y senβ = 2
3
 2 2
3




⇒ cos a = 5
3




⇒ cos β =
 • sen f 

1
3


 = sen(a + β)
 sen f 

1
3


 = senacosβ + senβcosa
 = 1
3
 · 5
3
 + 2
3
 · 2 2
3
 = 5 + 4 2
9
 
Clave B
04 • arc tan(2) = a ⇒ tana = 2
 • arc tan(–3) = b ⇒ tanb = –3
 • tan(a + b) = 
tana + tanb
1 – tanatanb
 tan(a + b) = 2 – 3
1 – 2(–3)
 = – 1
7
Clave E
05 Sea: a = arc sen 



5
13
 ∧ sena = 5
13
 
 ∧ β = arc cos 



1
5
 ∧ cosβ = 
1
5
 
 Luego: M = tg(a + β) = 
29
2
 Así: csc 



pM
87
 = csc
p
6
 = 2
Clave D
06 Sea: E = sen 1
2
 arc sen 

– 2 2
3


 E = – sen
q
2




 ; q = arc sen 2 2
3
– 



 E = – 1 –
1
3
2
 
 E = – 3
3 Clave D
5
a
13
12
; tga = 5
12
2
β
1
; tgβ = 25
2
q
3
1
2
⇓
07 Sea: q = arc tg
x + 2
2




 ∧ a = arc tg
x
3




 Luego: q – a ≥ p 
4
 ⇒ tg(q – a) ≥ tgp
4
 
 




x + 2
2




x + 2
2




x
3




x
3
–
1 +
≥ 1
 x(x + 1) ≤ 0 ∴ x ∈ [–1 ; 0] 
Clave A
08 – 1 ≤ 1 –|x| ≤ 1 ∧ –1 ≤ x + 2
3 ≤ 1
 |x| ≤ 2 ∧ –3 ≤ x + 2 ≤ 3
 –2 ≤ x ≤ 2 ∧ –5 ≤ x ≤ 1
 Domf = [–2, 2] ∩ [–5, 1] = [–2 ; 1]
Clave B
09 arctanp + arctanq + arctanr = p ... (I)
 φ = arctanp ⇒ tanφ = p
 ω = arctanq ⇒ tanω = q ... (II)
 r = arctanr ⇒ tanr = r
 (II) en (I): φ + ω + r = p
 ⇒ tanq + tanω + tanr = tanqtanωtanr
 Reemplazando: p + q + r = pqr
 ∴ p + q + r
pqr
 = 1
Clave B
10 Como:
 – 1 ≤ 3x – 4
5
 ≤ 1 ∧ – 1 ≤ x
2 – 2
2
 ≤ 1
 –5 ≤ 3x – 4 ≤ 5 – 2 ≤ x2 – 2 ≤ 2
 – 1
3
 ≤ x ≤ 3 – 2 ≤ x ≤ 2
 ∴ x ∈ – 1
3
 ; 2 
Clave A
11 g(x) = p
6
 + arc sen 



x – 1
2 + arc cos(2x + 2)
 – 1 ≤ x – 1
2
 ≤ 1 ∧ –1 ≤ 2x + 2 ≤ 1
 –2 ≤ x – 1 ≤ 2 ∧ –3 ≤ 2x ≤ –1
 –1 ≤ x ≤ 3 ∧ – 3
2
 ≤ x ≤ – 1
2
 
 ⇒ Dom(g) = – 1 ; – 1
2
 
Clave A
12 g(x) = arc tg1
2
 + arc tg1
4
 + arc tg1
8
 tg[g(2)] = 




1
8




1
8




1
4




1
4
1
8
1
8
1
4
1
4
1
2
1
2
1 –
1 –
+
+
+
1 –
























 = 
11
10
 
Clave A
13 Como x ∈ 1 + 3
2
 ; + ∞ ⇒ 2 + 3 ≤ 2x
 
 
5p
12
 ≤ arc tg2x < 
p
2
 
 
25p
12
 ≤ f(x) < 
5p
2
 
 Ran(f) = 
25p
12
 ; 
5p
2
 
Clave A
14 A = 1 + tg2 

arc tg –




1
2


 + 1 – 
 
cos2 



arc cos 7
8
 + 1 – sen2 



arc sen 3
8
 
 = 1 + 1
4
 + 1 – 7
8
 + 1 – 3
8
 = 3 + 1
4
 – 5
4
 A = 2
Clave B
15 Como x ∈ – 1
4
 ; 3
4
 
 ⇒ – 1
4
 ≤ x ≤ 3
4
 ⇒ – 1
2
 ≤ 2x ≤ 3
2
 ⇒ – p
6
 ≤ arc sen2x ≤ p
3
 ⇒ – p
3
 ≤ 2arc sen2x ≤ 2p
3
 ⇒ – p
3
 + p
6
 ≤ 2arc sen2x + p
6
 ≤ 2p
3
 + p
6
 ⇒ – p
6
 ≤ f(x) ≤ 5p
6
 ⇒ f(x) ∈ – p
6
 ; 5p
6
 
  min f(x) = – p
6
 
Clave A
ACTIVIDADES CAP 25
GRÁFICA DE LAS FUNC. TRIGON. INVERSAS
01 A = 1
3
 – 

– 1
3


 ⇒ = 2
3
 2
3
arccosB(0) + C = 0
 2
3
 

p
2


 + C = 0 ⇒ C = – 
p
3
 2
3
 arccosB(8) – 
p
3
 = – 
p
3
 ⇒ B = 1
8
 ∴f(x) = 2
3
 arccos x
8
 – 
p
3
Clave B
02 f(–1) = arccos(–1) = p
 Área de la región sombreada: S
 S = 1
2
 [2(p)]
 S = p u2
Clave D
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
47
03 y = karccosBx ⇒ k = 4
 Área de la región sombreada: S
 S = 1
2
 [8(4p)] ⇒ S = 16p
 Sk = 64p
Clave D
04 f(x) = 4arctanx
Clave C
05 y = arccosx ⇒ cosy = x
 y = arctanx ⇒ tany = x
 cosy = x ⇒ 1
1 + x2
 = x 1
xy1 + x
2
 x 1 + x2 = 1 ⇒ x2(1 + x2) = 1
 Luego:
 x
2
 sen2y + seny = x
2
 2senycosy + seny
 = x

 1
1 + x2
 

 

 1
1 + x2
 

 + x
1 + x2
 = x4
x2(1 + x2)
 + x2
x 1 + x2
 ∴ x
2
 sen2y + seny = x4 + x2 = x2(1 + x2) = 1
Clave B
06 Área de
 la región:
 S = 

p
2


 (2)
 ∴ S = pu2
 
 
Y
X
p
2
p
2
p/2
11
–1
–arc senx
–arc cosx
10
S
p
–
Clave A
07 x ≥ 2 ⇒ 0 < 1
x
 ≤ 1
2
 0 < arcsen 1
x
 ≤ 
p
6
 
p
3
 < 2arcsen 1
x
 + 
p
3
 ≤ 
p
3
 + 
p
3
 
p
3
 < f(x) ≤ 2p
3
 ∴ a + b = 
p
3
 + 2p
3
 = p
Clave A
08 x ≥ 2 ⇒ 0 < 1
x
 ≤ 1
2
 0 < arcsen 1
x
 ≤ 
p
6
 
p
6
 < 3arcsen 

1
x


 + 
p
6
 ≤ 
p
2
 + 
p
6
 ∴ 2(b – a) = 2

4p
6
 – 
p
6


 = p
Clave A
09 f(x) = sen2

p
2
 + arcsenx 

 = sen2(arccosx)
 f(x) = 1 – cos2(arccosx) = 1 – x2
 x ∈ [–1 ; 1]
 
Y
X
O 1
1
–1
Clave D
10 f(x) = arccos(sen4x + cos4x)
 f(x) = arccos(1 – 2sen2xcos2x)
 0 ≤ sen2xcos2x ≤ 1
 –2 ≤ –2sen2xcos2x ≤ 0
 –1 ≤ 1 – 2sen2xcos2x ≤ 1
 p ≤ arccos(1 – 2sen2xcos2x) ≤ 0
 ∴ Ranf = [0 ; p]
Clave B
CUADERNO DE TRABAJO
01 Como – p
2
  arc senBx  p
2
 – Ap
2
  Aarc senBx  Ap
2 
–p p
  A = 2
 Luego: –1  Bx  1  
 
– 1
B
  x  1
B
–2 2
  B = 1
2
 
  y = f(x) = 2arc sen x
2 Clave E
02 Como –1  x – 2
5
  1  –3 < x  7
 También: 
 0  arc cos



x – 2
5
  p 
 0  3arc cos



x – 2
5
  3p
 
10
–3 70
Y
X
 Piden el área: b×h = 10



3p
2
 = 15p
Clave D
03 Trasladando áreas 
 tenemos: 
 Área sombreada:
 b×h = 1(2A) = 4p
  A = 2p 
A
–A
Y
X
–1 1
Clave B
04 y = f(x) = 2arc sen2x + p 
 –1  2x  1  1
2
  x  1
2
 También: 
 – p
2
  arc sen2x  p
2
 
1
2
– 1
2
Y
X
 0  arc sen2x + p  2p
Clave C
05 En el punto P las ordenadas de ambas 
funciones son iguales, entonces:
 g(x) = f(x)  arc senx = arc cosx = p
2
 – arc senx
 2arc senx = p
2
  arc senx = p
4
  y = arc senx = p
4 Clave B
06 Dato: A TOP = p
3
 
(1)(y)
2
 = p
3
  y = 2p
3
 arc cosx = 2p
3
  x = cos 2p
3
 = – 1
2
 
Y
Q R
T
O
P X
y
x–1 11
 En el TQR: 
 A = 1
2
 b×h = 1
2




1
2
(p – y) = 1
4




2p
3
p – = p
12
Clave E
07 Como y = f(x) = arc senx
  y = arc sena (I)
 También: y = 1 – a (II)
 Igualando (I) y (II): 
Y
X
1 – a
y
a 1 1 – a = arc sena 
  a + arc sena = 1
Clave E
08 arc cos



1
2
 = p
3
  arc tan(1) = p
4
 También arc cosx = p
2
 – arc senx
 Luego en: f(x) = p
3
 – p
4
 + 
2p 



p
2
– arc senx
arc senx
 f(x) = p2
arc senx
 – 23p
12
 (I)
 Como x  1
2
; 
3
2
 
  p
6
  arc senx  p
3
 
1
2
Y
X
3
2
  
3
p
  
1
arc senx  
6
p
 Luego: 13p
12
  f(x)  49p
12
  a  f(x)  b 
  b – a = 3p Clave D
09 
 
Y
X
–
x1–x1
y = f(x) =|arctanx|
y = f(x) = arccosx
 Por la simetría se observa que si x1 es so-
lución   – x1 como solución.
  La suma de soluciones en [–8p; 8p] es 
cero.
Clave D
10 Si y = f(x) = x  x  0
  arc tan2x – 5arc tanx + 4  0
 arc tanx –4
 arc tanx –1
  arc tanx  –; 1]  [4; 
O
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
48
 ¡Pero! arc tanx  – 
p
2
 ; 
p
2
 Interceptando se tiene: arc tanx  – p
2
; 1]
 
Y
X
–
x
1
 
  x  –∞; tan1] ∪ [tan4 ; ∞〉
Clave B
TAREA
01 x f(x) = arc senx
–1 –p/2
 0 0
 1 p/2
02 x f(x) = arc sen2x
– 
1
2 – p
 0 0
 
1
2 p
 
03 x f(x) = arc cos
x
2
–2 p
 0 p/2
 2 0
 
04 x f(x) = 2arc senx
–1 2 – p
2




 = –p 
 0 2(0) = 0
 1 2
p
2




 = p
 
REFORZANDO
01 El gráfico de f(x) = arc tanx es:
Clave B
–1
p
2
–
1
p
2
–
–p
p
1
2
1
2
p
2
–2 2
p
–p
p
–1 1
0
p
2
– p
2
02 El gráfico debe estar trasladado 1/4 hacia 
la derecha.
 Entonces: a = 
1
4
 Clave A
03 El gráfico está trasladado p
2
 unidades ha-
cia arriba, entonces la función es:
 f(x) = arc senx + p
2
 
Clave C
04 El gráfico puede corresponder a:
 f(x)= –arc senx
Clave A
05 El gráfico corresponde a la función:
 f(x) = 2arc sen2x
Clave D
06 • arc sena = p
6
 ⇒ a = 1
2
 
 • arc senb = p
4
 ⇒ b = 
2
2
 
 ∴ a + b = 1 + 2
2
 
Clave C
07 El gráfico de f(x) = arc tan2x es:
Clave B
08 • a = 2 arc sen 2
3
 ⇒ a
2
 = arc sen 2
3
 • a = 2 arc cosb ⇒ a
2
 = arc cosb 
 ⇒ cos a
2
 = b ⇒ b = 
5
3
 
 Clave C
09 • f(x) = Aarc tanBx
 • Del gráfico A = 2
 • Del gráfico f(2) = p
2
 ⇒ p
2
 = 2 arc tanB(2) ⇒ p
4
 = arc tan (2B)
 ⇒ 2B = 1 ⇒ B = 1
2
 ∴ A + B = 2 + 1
2
 = 5
2
Clave D
10 
 • f(x) = Aarc cosBx 
f(–1) = p
f(0) = 3p
2
0
p
4
– p
4
2
a/2
3
5
 • 3p
2
 = Aarc cos(0)
 3p
2A
 = arcos0 
 ⇒ 3p
2A
 = p
2
 ⇒ A = 3
 • p = 3arc cos(–B)
 ⇒ cosp
3
 = –B ⇒ B = – 1
2
 ∴ A + B = 3 – 1
2
 = 5
2
Clave D
11 El gráfico de f(x) = arc cot2x es:
Clave C
12 
 Sea: f(x) = Aarc cos(Bx + C) + D
 pA = 11p
6
 + 2p
3
 = 5p
2
 
 ⇒ A = 5/2
 2
B
 = 8 – 2 ⇒ B = 1
3
 
 C
B
 = 2 + 8
2
 ⇒ C = 5
3
 
 D = – 2p
3
 
 ∴ f(x) = 5
2
 arc cos




x
3
– 
5
3
 – 2p
3
Clave C
13 El gráfico de f(x) = –3arc cot2x es:
Clave D
14 La función es f(x) = 2 arc sen3x, pero re-
flejado respecto a Y y trasladado p
6
 hacia 
la izquierda, entonces es:
 f(x) = –2arc sen3



x + p
6
3p
p
2p
3p/2
–1 1 2
0
–4p
4p
2 8C/B
D
pA
2/B
11p
6
– 2p
3
0
–6p
–3p
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
49
 f(x) = –2arc sen



3x + p
2
Clave E
15 S = AB · PH
2
 
 S = 



p + p
2



1 – 2
2
1
2
 S = 3p
8
(2 – 2)
 
A
P
B
H
y = arc cosx
y = arc senx
p
–p/2
p/2
p/4
–1 12
2
Clave D
ACTIVIDADES CAP 26
ECUACIONES TRIGONOMÉTRICAS
01 sen2x = 1
2
 ⇒ VP = 30° = 
p
6
 2x = kp + (–1)k p
6
 x = k 
p
2
 + (–1)k p
12
 k = 0 ⇒ x = 
p
12
 ; k = 1 ⇒ x = 5p
12
 ∴ 
p
12
 + 5p
12
 = 
p
2
 = 90°
Clave B
02 cos(3x – 30°) = 3
2
 ⇒ 3x – 30° = 30°
 3x = 60°
 Luego: 3x = 2kp ± 60°
 x = 2p
3
 k ± 20° = 120°k ± 20°
 k = –1
x = –100°
x = –140°
 ; k = 0
x = 20°
x = –20°
 
 k = 1
x = 140°
x = 100°
 ∴ ∃ 7 soluciones
Clave B
03 tan(3x – 15°) = –1 ⇒ 3x – 15° = –45°
 3x = –30
 Luego: 3x = kp + (–30°)
 x = k 
p
3
 – 10° = 60°k – 10°
 k = –5 ⇒ x = –310° ; k = –4 ⇒ x = –250°
 k = –3 ⇒ x = –190° ; k = –2 ⇒ x = –130°
 k = –1 ⇒ x = –70° ; k = 0 ⇒ x = –10°
 k = 1 ⇒ x = 50° ; k = 2 ⇒ x = 110°
 k = 3 ⇒ x = 170° ; k = 4 ⇒ x = 230°
 k = 5 ⇒ x = 290° ; k = 6 ⇒ x = 350°
 ∴ ∃ 13 soluciones
Clave D
04 tan3x = – 3 ⇒ VP = –60°
 3x = kp + (–60°) ⇒ x = 60°k – 20°
 k = 0 ⇒ x = –20°
Clave B
05 sen3x = 2
2
 ⇒ VP = 
p
4
 Luego:
 3x = kp + (–1)k 
p
4
 ⇒ x = 60°k + 15°(–1)k
 k = 0 ⇒ x = 15° ; k = –1 ⇒ x = – 75°
 Donde: a = –75° ⇒ 2a = –150°
 sen2a = sen(–150°) = –sen30°
 ∴ sen2a = – 1
2 Clave B
06 2sen23x = 2
 1 – cos6x = 2
 cos6x = – 1 ⇒ VP = p
 Luego: 6x = 2pk ± p
 ∴ x = (2k ± 1) 
p
6
 
Clave D
07 (tan2x – 3)(2senx – 1) = 0
 • tan2x = 3 ⇒ VP = 60°
 2x = kp + 60° ⇒ x = 90°k + 30°
 k = 0 ⇒ x = 30°
 • senx = 1
2
 ⇒ VP = 30°
 x = kp + (–1)k30°
 k = 0 ⇒ x = 30°
Clave E
08 sen5x = 2
2
 ⇒ VP = 
p
4
 = 45°
 5x = k(180°) + (–1)k45° ⇒ x = 36°k + (–1)k9°
 k = –1 ⇒ x = –45° ; k = 0 ⇒ x = 9°
Clave D
09 cotx – 2cotxsenx = 0
 cotx(1 – 2senx) = 0
 • 1 – 2senx = 0 ⇒ senx = 1
2
 ⇒ VP = 30° = 
p
6
 Luego: x = kp + (–1)k p
6
 k = –1 ⇒ x = – 7p
6
 • cotx = 0° ⇒ x = – 
p
2
Clave D
10 (2cos2x + 1)

sen3x – 3
2


 = 0
 • cos2x = – 1
2
 ⇒ VP = 2p
3
 = 120°
 2x = 2pk ± 2p
3
 ⇒ x = kp ± 
p
3
 k = 0 ⇒ x = 
p
3
 ; k = 1 ⇒ x = 2p
3
 • sen3x = 3
2
 ⇒ VP = 
p
3
 3x = kp + (–1)k 
p
3
 ⇒ x = k
p
3
 + (–1)k 
p
9
 k = 0 ⇒ x = 
p
9
 ; k = 1 ⇒ x = 2p
9
 k = 2 ⇒ x = 7p
9
 ; k = 3 ⇒ x = 8p
9
 ∴ 
p
3
 + 2p
3
 + 
p
9
 + 2p
9
 + 7p
9
 + 8p
9
 = 3p
Clave E
CUADERNO DE TRABAJO
01 De la ecuación: sen x
2
 = – 1
2
  Vp = –30°
 Luego x
2
 = 180k + (–1)k(–30°)
 x = 360°k + (–1)k(–60°)
 k = 0  x = –60° Mayor
 k = 1  x = 420° 
 k = 2  x = 660° 
 k = –1  x = –300° Menor
 Piden la diferencia entre la mayor y me-
nor solución:
 –60° – (–300°) = 240°
Clave B
02 Se tiene: cos(2x – 15)° = 
2
2
  Vp = –45°
 Luego aplicamos solución general:
 2x – 15° = 360°k  Vp = 360°k  45°
 k = 0  x 
= 30°
= –15°
 k = –1  x 
= –150°
= –195° 
 k = –2  x 
= –330°
= –375°
 x = { –330°; –195°; –150°; –15°}
  Menor solución negativa = –330°
Clave C
03 De la ecuación: cos2x = 1  Vp = 0
 Luego 2x = 2kp  Vp  2x = 2kp 
  x = kp/k 
Clave C
04 De la ecuación: tan3x = – 3  Vp = – p
3
 Luego 3x = kp + Vp  3x = kp – p
3
 
  x = kp
3
 – p
9
 ; k ∈ 
Clave E
05 De la ecuación: cos x
2
 = 1 + 0 + (–1) = 0
  cos x
2
 = 0  Vp = p
2
 Luego: x
2
 = 2kp  p
2
  x = 4kp  p
 Si k = 0 
x = –p
x = p ( al intervalo pedido)
 Si k = 1 
x = 3p
x = 5p
Clave A
06 De la ecuación: tan25x = 1  tan5x = 1
 i) tan5x = 1  Vp = p
4
 Luego: 5x = kp + Vp 
 5x = kp + p
4
  x = kp
5
 + p
20
 k = 0 ⇒ x = p
20
 k = 1 ⇒ x = p
4
 
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
50
 k = 2 ⇒ x = 9p
20
 k = 3 ⇒ x = 13p
20
 
 k = 4 ⇒ x = 17p
20
 
 ii) tan5x = –1  Vp = – p
4
 Luego: 5x = kp – p
4
  x = kp
5
 – p
20
 k = 1  x = 3p
20
 k = 2  x = 7p
20
 
 k = 3  x = 11p
20
 k = 4  x = 15p
20
 
 k = 5  x = 19p
20
 
Clave E
07 Igualando cada factor a cero:
 i) 2senx + 1 = 0  senx = – 1
2
  Vp = –30°
 Luego: x = 180°n + (–1)n(–30°)
 n = 0  x = –30° 
 n = 1  x = 210° n = 2  x = 330° 
 ii) senx – 1 = 0  senx = 1  x = 90° 
  soluciones:
 210° + 330° + 90° = 630° 
Clave B
08 3tan2x – 3 = 0  tan2x = 
3
3 = 3 
  Vp = 60°
 Luego:
 2x = 180°k + Vp 
  2x = 180°k + 60°  x = 90°k + 30°
 Si k = 0  x = 30° k = 1  x = 120° 
 k = 2  x = 210° k = –1  x = –60° 
 k = –2  x = –150° k = –3  x = –240° 
 Piden: 
 –150° – 60° + 120° + 30° = –60° = – p
3
Clave B
09 Si senx – cosx = 0  senx = cosx
 tanx = 1  Vp = 45°
 Luego: x = 180°n + Vp  x = 180°n + 45°
 n = –2  x = –315° 
 n = –1  x = –135° 
 n = 0  x = 45° 
 n = 1  x = 225° 
   2 soluciones.
Clave C
10 (sen3x – 3)(sen3x + 1) = 0
 i) sen3x = 3 ...(No existe solución)
 ii) sen3x = –1  Vp = – p
2
 Luego: 3x = kp + (–1)k



p
2
–
 x = kp
3
 + (–1)k + 1 p
6
Clave D
TAREA
01 cos2x = 
3
2
 ⇒ Vp = p
6
 
 ⇒ 2x = 2np ± p
6
 ⇒ x = np ± p
12
 , n ∈ 
02 2senxcosx =2



1
4
 ⇒ cos2x = 1
2
 
 ⇒ 2x = p
3
 ⇒ x = p
6
 
03 cos2x – sen2x = 
2
2
 ⇒ cos2x = 
2
2
 
 ⇒ 2x = p
4
 ⇒ x = p
8
04 (1 + sen2x)cos2x = 0
 (2+2sen2x)2cos2x = 0 
 (2 + 1 – cos2x) (1 + cos2x) = 0
 (3 – cos2x) (1 + cos2x) = 0
 ⇒ cos2x = 3 ∨ cos2x = –1 ⇒ VP = p
 descartado ⇒ 2x = 2pk ± p
 ∴ x = pk ± 
p
2
 / k ∈ 
REFORZANDO
01 tan4x = cos p
2
 ⇒ tan4x = 0 ⇒ Vp = 0 
 ⇒ 4x = kp , k ∈ 
 ∴ x = kp
4
 ; k ∈ 
Clave D
02 sen2x = sen36° – cos54° ⇒ sen2x = 0
 0 ⇒ Vp = 0
 Entonces: 2x = {kp/k ∈ }
 ∴ x = kp
2
 ; k ∈  
Clave E
03 (2 + sen2x)
≠ 0
 (2senx – 1) = 0
 ⇒ senx = 1
2
 ⇒ x = p
6
 , 5p
6
 ⇒ La suma es p
Clave C
04 sen2x – sen245° = 0 ⇒ sen2x = 1
2
 ⇒ Vp = p
6
 2x = kp + (–1)k × p
6
 , k ∈ 
 ∴ x = kp
2
 + (–1)k × p
12
 , k ∈ 
Clave A
05 tan2x = 2 · tanx – 1 : x ∈ [0; 2]
 tan2x – 2 · tanx + 1 = 0
 (tanx – 1)2 = 0 → tanx = 1
 tanx = tan



p
4
 ⇒ Vp = p
4
 Luego: x = kp + p
4
 ; k ∈ 
 k = 0 ⇒ x = p
4
 ; k = 1 ⇒ x = 5p
4
 ∴ p
4
 + 5p
4
 = 3p
2 Clave B
06 (2sen2x + 1) (sen2x – 2)
≠0
 = 0
 ⇒ sen2x = – 1
2
 ⇒ 2 = – 5p
6
 ⇒ x = – 5p
12
 
Clave B
07 sen4xcos2x – cos4xsen2x
sen2xcos2x
 = 2
 ⇒ sen(4x – 2x)
sen2xcos2x= 2
 ⇒ sen2x
sen2xcos2x
 = 2 ⇒ sec2x = 2
 ⇒ cos2x = 1
2
 ⇒ 2x = p
3
 ⇒ x = p
6
 
 Luego: a = p
6
 ⇒ tg2a = 3 
Clave C
08 cos4x – cos2x – sen3x = 0
 ⇒ –2sen3xsenx – sen3x = 0
 ⇒ –sen3x(2senx + 1) = 0
 sen3x = 0 ∨ 2senx + 1 = 0 
 ⇒ 



+ p
3
 + 



– p
6
 = p
3
 – p
6
 = p
6
 
 Clave D
09 2cos2x = sen(x + 36°) – sen(x – 36°)
 2cos2x = 2cosx sen36°
 cosx(cosx – sen36°) = 0
 ⇒ cosx = 0 ∨ cosx = sen36° = cos54°
 ⇒ x = (2k + 1)p
2
 ∨ x = 2pk ± 3p
10
 ⇒ La menor solución positiva es 3p
10
Clave E
10 (1 – cos2x) + (1 – cos4x) + (1 – cos8x) = 0
 ⇒ sen2x + sen22x + sen24x = 0
 ⇒ senx = sen2x = sen4x = 0
 ⇒ x = np , 2x = np , 4x = np , n ∈ 
 ⇒ C.S. = {np/n ∈ }
Clave D
sen3x = 0 ⇒ x = ± p
3
senx = – 1
2
 ⇒ x = – p
6
 , x = 7p
6
 
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
51
11 cotx tany = 3 ⇒ tany = 3tanx
 tanx – tany = 1 – 3 ⇒ tanx – 3tanx = 1 – 3 
 ⇒ tanx(1 – 3) = 1 – 3 ⇒ tanx = 1
 ∴ x = kp + p
4
 , k ∈ 
Clave B
12 1 – 2sen2x + 5senx – 5cosx = 0
 Recordemos que (senx – cosx)2 = 1 – sen2x
 Entonces: 
 (senx – cosx)2 + 5(senx – cosx) = 0
 (senx – cosx)2 + 5(senx – cosx) = 0
 (senx – cosx) (senx – cosx + 5)
≠0
 = 0
 De donde solo es posible
 senx – cosx = 0 ⇒ senx = cosx 
 tanx = 1 
 ∴ x = kp + p
4
 k ∈  
Clave B
13 2sen3x – 4sen2x – senx + 2 = 0
 Factorizamos
 2sen2x(senx – 2) – (senx – 2) = 0
 (senx – 2)
≠ 0
(2sen2x – 1) = 0
 ⇒ 2sen2x – 1 = 0 → cos2x = 0
 2x = (2k ± 1) p
2
; k ∈ 
 x = (2k ± 1) p
4
; k ∈  
Clave B
14 senxcos5x – sen5xcosx = 1
8
 
 senxcosx(cos4x – sen4x) = 1
8
 senxcosx(cos2x + sen2x)(cos2x – sen2x) = 1
8
 
 



sen2x
2
 × (cos2x) = 1
8
 
 sen4x
4
 = 1
8
 → sen4x = = 1
2
 Entonces:
 4x = np + (– 1)n  p
6
; n ∈  
 ∴ x = np
4
 + (– 1)n  p
24
; n ∈  
Clave C
15 sen2q + cos2q ≥ 3 + 2 (senq – cosq)
 (sen2q – 2senq) + (cos2q + 2cosq – 3) ≥ 0
 (2senqcosq – 2senq) +
 ((2cos2q – 1) + 2cosq – 3) ≥ 0
 2senq(cosq – 1) + (2cos2q + 2cosq – 4) ≥ 0
 2senq(cosq – 1) + (2cosq + 4)(cosq – 1) ≥ 0
 2(cosq – 1) · (senq + cosq + 2)
(+)
 ≥ 0
 cosq – 1 ≥ 0 → cosq ≥ 1
 Entonces: cosq = 1 ∴ q = 2kp , k ∈ 
Clave A
ACTIVIDADES CAP 27
ECUACIONES TRIGONOMÉTRICAS NO ELEMENTALES
01 2sen2xcos3x = 2cos3x
 cos3x(sen2x – 1) = 0
 sen2x = 1 ⇒ Vp = 
p
2
 2x = kp + 
p
2
 ⇒ x = k 
p
2
 + 
p
4
 ; k ∈ 
Clave D
02 2cos6xcos2x = 0
 cos6x = 0 Vp = 
p
2
 6x = kp ± 
p
2
 ⇒ x = k 
p
3
 ± 
p
12
 ; k ∈ 
 k = 0 ⇒ x = 
p
12
 ⇒ a = 
p
12
 Luego: sen2a = sen 
p
6
 = 1
2
Clave B
03 2senxcosx = cosx
 cosx(2senx – 1) = 0
 senx = 1
2
 ⇒ VP = 
p
6
 x = kp + (–1)k · 
p
6
 k = 0 ⇒ x = 
p
6
 ; k = 1 ⇒ x = 5p
6
 cosx = 0 ⇒ Vp = 
p
2
 x = 2pk ± 
p
2
 k = 0 ⇒ x = 
p
2
 ; k = 1 ⇒ x = 3p
2
 Suma: 
p
6
 + 5p
6
 + 
p
2
 + 3p
2
 = 3p
Clave C
04 2senxcosx + 2cos2xsenx = 0
 2senx(cosx + cosx) = 0
 senx(2cos2x + cosx – 1) = 0
 senx(2cosx – 1)(cosx + 1) = 0
 cosx = 1
2
 ⇒ x = 
p
3
Clave A
05 (sen2x + cos2x)(sen2x – cos2x) = 1
 – cos2x = 1
 cos2x = – 1 ⇒ Vp = p
 2x = 2pk ± p ⇒ x = pk ± 
p
2
 k = 0 ⇒ x = 
p
2
 ; x = – 
p
2
 ⇒ a = – 
p
2
 sen3a – cos2a = – (–1) = 2
Clave D
06 sec2(2x) – tan2(2x) – tan2(2x) – 1 = 0
 1
 tan2(2x) = 0 ⇒ tan2x = 0 ⇒ Vp = 0
 2x = kp + 0 ⇒ x = k 
p
2
 k = 1 ⇒ x = 
p
2
 ; k = 2 ⇒ x = p ;
 k = 3 ⇒ x = 3p
2
 Suma: 
p
2
 + p + 3p
2
 = 3p
Clave B
07 cot4x – 2cos2x – sen2x
sen2x
 = 0
 cot4x – 2cot2x + 1 = 0
 (cot2x – 1)2 = 0
 cot2x = 1 ⇒ 1
tan2x
 = 1 ⇒ tan2x = 1
 tanx = ± 1 ⇒ Vp = ± 
p
4
 Luego: x = kp ± 
p
4
 k = 0 ⇒ x = 
p
4
 ; k = 1 ⇒ x = 3p
4
; x = 
5p
4
 k = 2 ⇒ x = 
7p
4
 ∴ ∃ 4 soluciones
Clave C
08 2senxcos2x + senx = 2coxcos2x – cosx
 sen3x – senx + senx = cos3x + cosx – cosx
 sen3x = cos3x
 ⇒ tan3x = 1 ⇒ Vp = 45° = 
p
4
 
 Luego: 3x = kp + 
p
4
 ⇒ x = k 
p
3
 + 
p
12
 k = 0 ⇒ x = 
p
12
 ; k = 1 ⇒ x = 5p
12
 ; 
 k = 2 ⇒ x = 9p
12
 ; k = 3 ⇒ x = 13p
12
 no cumple
 Suma: 
p
12
 + 5p
12
 + 9p
12
 = 5p
4
Clave E
09 tan3x + tan2x + tan5xtan3x + tan2x = 3
 tan5x
 tan5x = 3 ⇒ 5x = 60°
 ∴ x = 12°
Clave C
10 senpx – cospx > 0
 2sen 

px – 
p
4


 > 0
 sen 

px – 
p
4


 > 0 ; I y II Cuadrante
 x ∈ [0 ; 2] ∧ 0 < px – 
p
4
 < p
 0 ≤ x ≤ 2 ∧ 0 < x – 1
4
 < 1
 1
4
 < x < 5
4
 Luego: [0 ; 2] ∩ 1
4
 ; 5
4
 = 1
4
 ; 5
4
 
Clave B
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
52
CUADERNO DE TRABAJO
01 Aplicando las transformaciones:
 2sen9x cos4x = 3(2cos9x cos4x)
 cos4x(tan9x – 3) = 0
 i) cos4x = 0  4x = p
2
  x = p
8
 ii) tan9x = 3  9x = p
3
  x = p
27
  La menor solución positiva es p
27
.
Clave A
02 De la ecuación: senq + cosq = –1
 2sen



p
4
q + = –1 
  sen



p
4
q + = – 1
2
 
Y
C.T.
X
3
  q + p
4
 = 






3p
4
; 7p
4
  q = 






p
2
; 3p
2
  Suma = 2p
Clave A
03 6sen(2x) – 8cosx + 9senx – 6 = 0
 12senx cosx – 8cosx + 9senx – 6 = 0
 Factorizando: 
 4cosx(3senx – 2) + 3(3senx – 2) = 0
  (3senx – 2)(4cosx + 3) = 0
 Como x  – p
2
 ; p
2
 : senx = 2
3
 Como senx > 0  x  0; p
2

 Por lo tanto, existe un único valor para x.
Clave A
04 sen2x secx = 1  2senx cosx secx = 1
 2senx cosx 1
cosx




 = 1
 Entonces: cosx  0  x  (2n + 1)p
2
 Luego: 2senx = 1  senx = 1
2
  x = 30°
Clave B
05 Cálculo de a:
 sen42x – sen24x = cos24x – cos42x
 sen42x + cos42x = sen24x + cos24x = 1
 3
4
 + 
cos8x
4
 = 1  cos8x = 1  8x = 2p
  x = p
4
 Luego la mayor raíz de:
 x2 – (5tana)x + 6tana = 0, tana = 1
 x2 – 5x + 6 = 0 
 x –3  x = 3 (mayor raíz)
 x –2  x = 2
Clave E
06 Por propiedades:
 (1 + sen2x)2 + cos22x + tan2x = sec2x
 Desarrollando el binomio al cuadrado y 
llevando tan2x al 2° miembro, formamos 
las identidades trigonométricas siguientes:
 1 + 2sen2x + (sen22x + cos22x)
1
 = 1
 1 + 2sen2x = 0 
  sen2x = – 1
2
 
Clave E
07 Factorizando la ecuación:
 2sen3x + sen2x – 2senx – 1 = 0; 0  x  2p
 sen2x(2senx + 1) – (2senx + 1) = 0
 (2senx + 1)(sen2x – 1) = 0
 I. senx = – 1
2
  x = 7p
6
; 11p
6
 II. senx = 1  x = p
2
 III. senx = –1  x = 3p
2
 Entonces la solución es: 
 x1 + x2 + x3 + x4 = 5p
Clave A
08 Usando: 
 cos4q = 3 + 4cos2q + cos4q
8
 En el problema: 
 3 + 4cosx + cos2x
8
 – cos2x
8
 = 7
8
  3 + 4cosx = 7  cosx = 1, x = 2np, n 
  x = {0; 2p}
  Número de soluciones: 2
Clave B
09 Tenemos: (senq)x – y = 0 (1)
 x + (4cosq)y = 0 (2)
 Para que el sistema admita más de una 
solución, estas ecuaciones deben de veri-
ficar	que:
 senq
1
 = 
–1
4cosq
  2senq cosq = – 1
2
 
  sen2q = – 1
2
 Dado que se pide el menor valor positivo:
 2q  IIIC  2q = 180° + 30°  q = 105°
Clave B
10 De sen(px) – cos(px) < 0
 2sen



p
4
px – < 0
  sen



p
4
px – < 0
 Como x  1; 3 sen < 0 sen < 0
Y
X
C.T.
  p < px < 3p
 3p
4
 < px – p
4
 < 11p
4
 ...(I)
 ¡Pero! sen



p
4
px – < 0
  p < px – p
4
 < 2p ...(II)
 Intersectando I y II:
 3p
2
 ; 11p
2
 ∩ 〈p ; 2p 〉  p < px – p
4
 < 2p
  5
4
 < x < 9
4 Clave B
TAREA
01 cos2x + cos2x – 1 = 1
 2cos2x – 1 = 1 ⇒ cos2x = 1 
 ⇒ 2x = p ⇒ x = p
2
 
02 Por legendre:
 4senxcosx = 1 ⇒ 2senxcosx = 1
2
 
 sen2x = 1
2
 ⇒ Vp = p
6
 
 ⇒ 2x = np + (–1)n p
6
 
 ⇒ x = np
2
 + (–1)n p
12
 , n ∈ 
03 En 1 – cos4x
secx
 = 0, como secx = 1
cosx
 ⇒ cosx≠0
 para que secx	esté	definido ⇒ cosx	≠	0	
 y x ∈ 0, 3p
2
 ⇒ x ≠ p
2
 En 1 – cos4x = 0 ⇒ cos4x = 1, pero 0 ≤ x ≤ 3p
2
 ⇒ 0 ≤ 4x ≤ 6p ⇒ 4x = 0 ; 2p ; 4p
 ⇒ x = 0; p
2
; p , pero x ≠ p
2
 ⇒ x = 0; p
04 5sen4x –(1 – 2sen24x) + 3 = 0
 2sen24x + 5sen4x + 2 = 0
 (2sen4x + 1) (sen4x + 2)
≠ 0
 = 0
 ⇒ 2 sen4x + 1 = 0 ⇒ sen4x = –1
2
 ; Vp = –p
6
 
 ⇒ 4x = np +(–1)n



– p
6
 ⇒ x = np
4
 + (–1)n+1 p
24
 
 ⇒ x = p
4
 



n + (–1)n+1
6
 , n ∈ 
REFORZANDO
01 cos4x – sen4x = 1
 (cos2x + sen2x)
1
 (cos2x – sen2x)
cos2x
 = 1
 ⇒ cos2x = 1 ⇒ 2x = 2np ⇒ x = np , n ∈ 
Clave A
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
53
02 2cos2x + sen2x = 0
 2cos2x + 2senxcosx = 0
 2cosx(cosx + senx) = 0 
cosx = 0 ⇒ x = p
2
tanx = –1 ⇒ x = 3p
4
 
 ∴ menor solución = p
2
 
Clave B
03 (2sen3x – senx) – 2cos2x = 0
 senx(2sen2x – 1) – 2cos2x = 0
 senx(–cos2x) – 2cos2x = 0
 –cos2x(senx + 2)
≠0
 = 0 ⇒ cos2x = 0
 ⇒ 2x = – p
2
 ⇒ x = – p
4 Clave C
04 sen3x + senx
cos3x + cosx
 = 3 ⇒ 2sen2xcosx
2cos2xcosx
 = 3
 cosx ≠ 0 ⇒ x ≠ (2n + 1) p
2
 ⇒ tg2x = 3 ⇒ x = np
2
 + p
6
 , n ∈ 
 Número de soluciones es 2
Clave B
05 ctgx + tgx – 4ctg2x = 0
 ⇒ 2csc2x – 
4cos2x
sen2x = 0
 ⇒ 
1 – 2cos2x
sen2x = 0
 
 ⇒ 1 – 2cos2x = 0 ⇒ cos2x = 1
2
 Entonces la menor solución positiva es p
6
 
Clave C
06 cos5xcos3x = cosxcos7x
 ⇒ (cos8x + cos2x) – (cos8x + cos6x) = 0
 ⇒ cos2x – cos6x = 0
 ⇒ 2sen4xsen2x = 0
 ⇒ x = np
4
 ∨ x = np
2
 , n ∈  
Clave B
07 tanx = cot(2x + a)
 Si tana = cotq ⇒ a + q = 






(2k + 1)
p
2 / k ∈ 
 ⇒ 3x + a = 






(2k + 1)
p
2 / k ∈ 
 x = 






(2k + 1)
p
6 – a
3
 / k ∈ 
Clave A
08 
sen6xcos3x
cos3x = 0 ⇒ sen6x = 0 si cos3x ≠ 0
 cos3x ≠ 0 ⇒ x = p
6
, p
2
, 5p
6
, 7p
6
, 9p
2
, 11p
6
 cos6x = 0 ⇒ Vp = p
2
 6x = 2pk ± p
2
 ⇒ x = p
3
k ± p
12
 ∴ ∃ 11 soluciones
Clave B
09 sen42x + cos42x – sen22x = 1
 ⇒ 1 - 2sen22xcos22x – sen22x = 1
 ⇒ sen22x (–1–2cos22x) = 0
 ⇒ sen22x (2 + cos4x) = 0
 ⇒ sen2x = 0 ⇒ x = np
2
 ⇒ C.S. = 






np
2
/n ∈ 
Clave A
10 1 – tg2x
1 + tg2x
 = 1 – cos4x
1 + sen4x
 
cos2x – sen2x
cos2x + sen2x = 
1 – cos4x
(sen2x + cos2x)2
 cos4x = 1 – cos4x
 4x = 2np ± p
3
  C.S. = 






np
2
 ± 
p
12 /n ∈ 
Clave C
11 4sen2x
cos2x
 – 8sen2x = 6
 ⇒ 4sen2x – 8sen2xcos2x = 6cos2x
 ⇒ 2(1 –cos2x) - 2sen22x = 3(1 + cos2x)
 ⇒ 2sen22x + 5cos2x + 1 = 0
 ⇒ (cos2x – 3)(2cos2x + 1) = 0
 ⇒ cos2x = – 1
2
 ⇒ x = – p
3
 = a
 Catetos: 10cosa= 5 ∧ –8 3sena = 12
 ⇒ El perímetro es 30 cm.
Clave B
12 tan(x + 45°) + tan(x – 45°) – 2cotx = 0
 cot(45° – x) – tan(45° – x) – 2cotx = 0
 2 . cot(2(45° – x)) – 2cotx = 0
 cot(90° – 2x) = cotx
 tan2x = cotx
 De acuerdo a la nota del problema ante-
rior se tendrá que: 
 3x = 






(2k + 1)
p
2 / k ∈ 
 x = 






(2k + 1)
p
6 / k ∈  
Clave B
13 I. x – y = 53°
2
 II. coty – cotx = 2
 En (II)
 sen(x – y)
senx · seny
 = 2
 sen(x – y) = 2senx · seny 
 sen(x – y) = cos(x – y) – cos(x + y)
 Reemplazamos en (I)
 sen



53°
2
 = cos



53°
2
 – cos(x + y)
 1
5
 = 2
5
 – cos(x + y)
 cos(x + y) = 1
5
 ⇒ x + y = 127°
2
 
 
Clave A
14 cos2x ≠ 0 ⇒ 2x ≠ 
p
2 ; 
3p
2 ⇒ x ± 
p
4 ; 
3p
4 
 sen6x + sen2x
cos2x
 = 0 ⇒ 2sen4xcos2x
cos2x
 = 0
 ⇒ sen4x = 0 ⇒ 4x = 0; p; 2p; 3p; 4p
 ⇒ x = 0, 
p
4
 , p
2
 , 
3p
4
; p ⇒ 3 soluciones
Clave C
15 sen4x
senx
 – sen6x
sen3x
 = 2 . senx
 4 . senx . cosx . cos2x
senx
 – 2 . sen3x . cos3x
sen3x
 = 2 . senx
 4.cosx.cos2x – 2 . cos3x = 2 . senx
 2.cosx.cos2x – cos3x = 
2
2
senx 
 (cos3x + cosx) - cos3x = 
2
2
senx 
 cosx = 
2
2
senx → tanx = 2
 Las soluciones pertenecientes al intervalo 
de 〈0; 2p〉 son:
 ∴ |x2 – x1| = p 
Clave B
ACTIVIDADES CAP 28
RESOLUCIÓN DE TRIÁNGULOS OBLICUÁNGULOS
01 
 
q
2
q
2
2
2
3
n n
n
A H
B
D
C
30° a
 2sen
q
2
 – 3sena = 2

n
2


 – 3

n
3


 ∴ 2sen
q
2
 – sena = 0
Clave A
x ∈ IC : x1 = arc tan( 2)
x ∈ IIIC : x2 = p + arc tan( 2)
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
54
02 El triángulo mostrado es un :
 
60°
xL
L
2xL 
60°
n
3
n
2n
 ⇒ L = xL 3 ∴ x = 3
3
Clave D
03
 
 H
C
A
l h 3l
2a 2q
B
 E = sen2a · tan(a – q)
sen2q · tan(a + q)
 
 E = 
h
l
h
3l
 × 3l – l
3l + l
 = 
1
1
1
3
 × 2
4
 ∴ E = 3
2
Clave D
04
 
60°
C
H
B
H x
ax
ax ax
330°120°
 x2 = (2ax)2 + (ax 3)2 ⇒ 1 = 4a2 + 3a2
 ⇒ a2 = 1
7
 ∴ a = 7
7 Clave B
05
 
2n
3
2n
3
n
13
4n
4n
2n 2n
2n
n
n
A
B
N
T
C
D
 DTM:
 (n 13)2 = (n 3)2 + (2n 3) – 2n 3(2n 3)cosq
 ⇒ cos = 1
6
 ∴ q = arccos

1
6


Clave C
06 5tana = 5 3
 tana = 3 ⇒ a = 60°
 A
12
9
a
C
B
x
 x2 = 122 + 92 – 2(9)(12)cos60° ⇒ x = 3 13 ≅ 11
 Luego: 12 + 9 + 11
3
 = 10 + 2
3
 Por lo tanto, son necesarias 11 estacas 
para cercar el terreno.
Clave D
07 Por lo expuesto anteriormente
 Se concluye que: γ = 2a
 42 = 52 +62 – 2(5)(6)cosa
 
C
A B6
4 5
γ
2a
a ⇒ cosa = 3
4
 
 sen(a + γ)
senγ
 = sen3a
sen2a
 = sena(2cos2a + 1)
sena
 sen(a + γ)
senγ
 = 2(2cos2a – 1) + 1
2cosa
 = 4cos2a – 1
2cosa
 ∴ sen(a + γ)
senγ
 = 
4

3
4


2
 – 1
2

3
4


 = 5
6
Clave A
08
 21sena = 21



5
2
 3
21



 
 ∴ 21sena = 5
2
 3
B
A
A
60°
N
5 C
3
2
5
2
5 3/2
21
Clave C
09
 B 4 C
A
2 3
senB
2
 = 


9
2
 – 4
2


 

9
2
 – 8
2


2(4)
 = 10
8
 ⇒ tanB
2
 = 15
9
senA
2
 = 


9
2
 – 4
2


 

9
2
 – 6
2


2(3)
 = 10
4
 ⇒ tanA
2
 = 15
3
 ⇒ tan

A – B
2


 = 
15
3
 – 15
9
1 + 15
3
 × 15
9
 ⇒ tan

A – B
2


 = 15
7 Clave E
10 b2 + c2 – a2
2abc
 + a
2 + c2 – b2
2abc
 + a
2 + b2 – c2
2abc
 = a
2 + b2 + c2
R3
 ⇒ 2abc = R3 ... (I)
 Se sabe: senA = a
2R
 ; senB = b
2R
 ; senC = c
2R
 ⇒ senAsenBsenC = abc
8R3 ... (II)
 (I) en (II): senAsenBsenC = abc
8(2abc)
 ∴ senAsenBsenC = 1
16
Clave D
CUADERNO DE TRABAJO
01 Aplicando ley de senos:
 
10
sena
 = 
sen2a
10 3  
1
sena
 = 
2sena cosa
3 
 cosa = 
3
2
  a = 30°
 En el triángulo: 30° + 60° + x = 180°
  x = 90°
Clave C
02 Por ley de cosenos:
 52 = (8)2 + (6)2 – 2(8)(6)cosA
 96cosA = 75
  cosA = 75
96
 
  secA = 96
75
 A
B
6 5
8 C
 Luego: 25secA = 25



96
75
 = 32
Clave B
03 Ley de senos:
 
a – 1
sena
 = 
a + 1
sen2a
 
a – 1
a + 1
a
(Lado mayor)
(Lado
menor)
 
a – 1
sena
 = 
a + 1
2sena cosa
 2cosa = a + 1
a – 1
(Lado menor)
(Lado mayor)
Clave A
04 Ley de cosenos:
 x2 = (x + 1)2 + (x – 2)2 – 2(x + 1)(x – 2)1
2
 x = 7
  Perímetro = 3x – 1 = 3(7) – 1 = 20
Clave D
05 Del dato: a
b + c
 = c
b – a
 – 1 = c + a – b
b – a
 ab – a2 = bc + ab – b2 + c2 + ac – bc
 b2 = a2 + c2 + ac ...dando forma de ley de cosenos
 b2 = a2 + c2 + 2ac



1
2
– cosb  b = 120°

 Piden: 3(cotb – tanb) = 3 – 
3
1 – (– 3)



 = –1 + 3 = 2
Clave D
06 Por ley de senos:
 b = 2RsenB
 32 = 2R(0,8) = 2R 8
10
  R = 20 cm 
32 = b
R R
A
COB
Clave C
07 Como A + B + C = 180°
  senA = sen(180° – (B + C))
  senA = sen(B + C)
 Ley de cosenos:
 ( 28)2 = 42 + 36 – 2(6)(4)cosA
 cosA = 16 + 36 – 28
2(6)(4)
 = 24
48
 = 1
2
  A = 60°
 Piden:
 sen(B + C) = senA = 
3
2
Clave A
2ax
n 3
EDITORIAL INGENIOSOLUCIONARIO - TRIGONOMETRÍA 5°
55
08 Ley de senos: 
 
a + 1
sen2a
 = 
a – 1
sena
; 
A
B
C
a – 1
a + 1
a
 cosa = a + 1
2(a – 1)
 ...(1)
 Ley de cosenos: 
 (a – 1)2 = (a + 1)2 + a2 – 2(a + 1)acosa ...(2)
 Reemplazando (1) en (2): 
 a2 – 5a = 0  a = 5
  Suma de lados: 3a = 15
Clave C
09 Aplicando ley de tangente, b = 3  c = 2
 




B + C
2
tan




B – C
2
tan
 = b + c
b – c
  
tan(60°)




B – C
2
tan
 = 5
1
 
 tan



B – C
2
 = 
3
5
  
B – C
2
28
5
3
 Piden: 28cos2



B – C
2
 = 28×
28
5 



2
 
  28×25
28
 = 25
Clave A
10 ACB (Isósceles)
 Sea AB = 2  BC = sec80°
 BPC (Ley de senos)
 2
senx
 = sen80°
sen(160° – x)
  2
senx
 = sen80°
sen(20° + x)
 
 2sen(160° – x) = senx sec80°
 2sen(20° + x) = sena
cos80°
 2sen(20° + x)cos80° = senx
 sen(100° + x) – sen(60° – x) = senx
 sen(80° – x) = senx + sen(60°– x)
 sen(80° – x) = 2sen30° cos(30° – x)
 sen(80° – x) = cos(30° – x)
  80° – x + 30° – x = 90°  x = 10°
 Piden: sen3x = sen30° = 1
2
Clave B
TAREA
01 
 
x
sen45° = 
3 2
sen30°
 ⇒ x = 6
02 
a
senA
 = 
b
senB
 = 
c
senC
 = 2R ⇒ senA = 
a
2R
 ADC: 
8
senA
 = 
b
sen90°
 
 ⇒ 8 = bsenA ⇒ 8 = b



a
2R
 ⇒ R = 
ab
8
A C
x 105°
30°
B
3 2
03 
 (I) Ley de senos: 
3 + 1
senβ = 
1
sen15°
 ⇒ senβ = 
2
2 ⇒ β = 45°
 (II) a = 120° y tga = – 3 
04 (I) Ley de senos: sena
senq
 = 3
2
 (II) Ley de tangentes: 
tg



a + q
2
tg



a – q
2
 = 5
 De (I) y (II):
 senq
sena
 . 
tg



a + q
2
tg



a – q
2
 = 


2
3
(5) = 


10
3
REFORZANDO
01 m
n
 = ?
	 Del	gráfico	por	el	teorema	de	senos
 m
n
 = sen2a
sena
 → m
n
 = 2.sena.cosa
sena
 ∴ m
n
 = 2cosa
Clave B
02 
 Por teorema de Pitágoras:
 d2 = 92 + 402 = 41
Clave C
03 asenB = bsenA
Ley de senos:
asenC = csenA
A D
C
b a
8
8
B
CA
B
1cm
15° a
β
( 3 + 1)cm
A C
2k 3k
a q
B
A C
n m
2a a
B
S
N
O E
20°
70°
40
d
Barco 1
9Barco 2
 E = asenB + bsenA
asenC + csenA
 = bsenA + bsenA
csenA + csenA
 
 E = 2bsenA
2csenA
 = 3c
c
 = 3
Clave A
04 mBAPB = ?
 Del	gráfico	se	tiene	que	x	=	5q
 Pero 8q = 90° → q = 11°15'
 Reemplazamos en x:
 x = 5x(11°15') ∴ x = 56°15'
Clave D
05 Por ley de senos:
 
d
sen135°
 = 
b
sen15°
 ⇒ d = 
sen45°
sen15°
 ⇒ d = 2
2



6 – 2
4
2
 ⇒ d = 2 2( 6 + 2)
4
2
 
 ⇒ d = 3 + 1 
Clave C
06 
a2 = b2 + c2 – 2bccosA
Ley de cosenos:
b2 = a2 + c2 – 2accosB
 
 ⇒ k = –(2bccosA) senA
cosA
 + (2accosB) senB
cosB
 k = –2c[bsenA – asenB] = –2c(0) = 0
 ⇒ k + 1 = 1
Clave D
07 E = 
senA – senB
senA + senB + 
b – a
b + a
 ⇒ E = 
 ⇒ E = tgA – B
2
ctgA + B
2
 – tgA – B
2
ctgA + B
2
 ⇒ E = 0
Clave C
P
E O AB
Persona (B) Persona (A)
4q
x
N
O
N
1/
4N
E
4q
3q 4qq
Oeste
Este1km A
d
B
45°30°
15°
+
sen A – B
2
tg A – B
2
cos A – B
2
tg A + B
2
2cos A + B
2
2sen A + B
2
EDITORIAL INGENIO SOLUCIONARIO - TRIGONOMETRÍA 5°
56
08 Ley de senos: 
 a = ksenA b = ksenB c = ksenC 
 E = 
acosA + bcosB
2cosAcosB + cosC 
 E = 
k(2senAcosA + 2senBcosB)
2(2cosAcosB + cosC) 
 E = 
k(sen2A + sen2B)
2[cos(A + B) + cos(A – B) + cosC] 
 E = 2ksen(A + B)cos(A – B)
2cos(A – B)
 = ksenC = c
 Clave E
09 E = a3 . c2(acosC – ccos(B + C))sen2B . senC
 = a3 . c2(acosC + ccosA)sen2B . senC
 = a3 . c2 . b . sen2B . senC
 = a3(c2 . sen2B) (b . senC)
 = a3(c2 . sen2B) (c . senB)
 = a3 . c3 . sen3B
 = 8



acsenB
2
3
 = 8 . ( 73 )3 = 56
Clave C
10 H = ? 
	 Del	gráfico:		Hcot30°	=	10 3cos15°
 H 3 = 10 3 32 +
2
 ∴ H = 5 32 +
Clave B
15°
E
H
aO
S
10 3
30°
Diana
Alejandro
11 3  cot2a = ? 
 
 cota = n 2
n 3
 ⇒ cota = 2
3
 ⇒ 3cot2a = 2
Clave B
12 A + B + C = 180° ⇒ C = 30°
 
a
senA
 = 
b
senB
 = 
c
sen30°
 = 2c
 ⇒ 
a = 2csenA
b = 2csenB
 
a2 = 4c2sen2A
⇒
b2 = 4c2sen2B
 ⇒ a2 + b2 = 4c2[2sen2A + 2sen2B]
 = 2c2[2 – (cos2A + cos2B)]
 4c2(cos2A + cos2B) = 4c2 – a2 – b2
Clave C
13 Ley de cosenos: c2 = a2 + b2 – 2abcosC
 ⇒ 0 = a2 + b2 – (a2 + b2 – 2abcosC) + 1
2
ab
 ⇒ cosC = – 1
4
 
 ctg2



C
2
 = 
1 +cosC
1 – cosC = 
1 – 1
4
1 – 



– 1
4
 = 
3
4
5
4
 = 3
5
Clave B
a
n 2
n 3
n
60°
Posición (1) Posición (2)
Poste
14 d = ?
d
A
B
T
75m
S
O
30°
60°
Torre
 • AT = 25 3 • BT = 75 3
 d = (25 3)2 + (75 3)2
 ⇒ d = 25 30
Clave A
15 tana  cotβ = ? 
 Del gráfico se tiene que
 tana = 4h
n
 y cotβ = n
6h
 
 Multiplicamos:
 tana  cotβ = 4h
n
  n
6h
 
 ∴ tana  cotβ = 2
3 Clave B
h
h
h
h
h
a
a
β
β h
n
n