Baixe Trigonometria: Exercícios Resenha e outras Exercícios em PDF para Matemática, somente na Docsity!
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APPENDIX A
Trigonometry Review
EXERCISE SET A
- (a) 5 π/ 12 (b) 13 π/ 6 (c) π/ 9 (d) 23 π/ 30
- (a) 7 π/ 3 (b) π/ 12 (c) 5 π/ 4 (d) 11 π/ 12
- (a) 12 ◦ (b) (270/π) ◦ (c) 288 ◦ (d) 540 ◦
- (a) 18 ◦ (b) (360/π) ◦ (c) 72 ◦ (d) 210 ◦
- sin θ cos θ tan θ csc θ sec θ cot θ
(a)
(b) 3 / 4
(c) 3 /
- sin θ cos θ tan θ csc θ sec θ cot θ
(a) 1 /
(b) 3 / 5 4 / 5 3 / 4 5 / 3 5 / 4 4 / 3
(c) 1 / 4
- sin θ = 3/
10, cos θ = 1/
10 8. sin θ =
5 /3, tan θ =
- tan θ =
21 /2, csc θ = 5/
21 10. cot θ =
15, sec θ = 4/
- Let x be the length of the side adjacent to θ, then cos θ = x/6 = 0.3, x = 1.8.
- Let x be the length of the hypotenuse, then sin θ = 2. 4 /x = 0.8, x = 2. 4 / 0 .8 = 3.
- θ sin θ cos θ tan θ csc θ sec θ cot θ
(a) 225 ◦^ − 1 /
(b) − 210 ◦^1 / 2 −
(c) 5 π/ 3 −
(d) − 3 π/ 2 1 0 — 1 — 0
- θ sin θ cos θ tan θ csc θ sec θ cot θ
(a) 330 ◦ − 1 / 2
(b) − 120 ◦ −
(c) 9 π/ 4 1 /
(d) − 3 π 0 − 1 0 — − 1 —
Exercise Set A 525
- sin θ cos θ tan θ csc θ sec θ cot θ
(a) 4 / 5 3 / 5 4 / 3 5 / 4 5 / 3 3 / 4
(b) − 4 / 5 3 / 5 − 4 / 3 − 5 / 4 5 / 3 − 3 / 4
(c) 1 / 2 −
(d) − 1 / 2
(e) 1 /
(f ) 1 /
- sin θ cos θ tan θ csc θ sec θ cot θ
(a) 1 / 4
(b) 1 / 4 −
(c) 3 /
(d) − 3 /
(e)
(f ) −
- (a) x = 3 sin 25 ◦ ≈ 1. 2679 (b) x = 3/ tan(2π/9) ≈ 3. 5753
- (a) x = 2/ sin 20 ◦ ≈ 5. 8476 (b) x = 3/ cos(3π/11) ≈ 4. 5811
- sin θ cos θ tan θ csc θ sec θ cot θ
(a) a/ 3
9 − a^2 / 3 a/
9 − a^2 3 /a 3 /
9 − a^2
9 − a^2 /a
(b) a/
a^2 + 25 5 /
a^2 + 25 a/ 5
a^2 + 25/a
a^2 + 25/ 5 5 /a
(c)
a^2 − 1 /a 1 /a
a^2 − 1 a/
a^2 − 1 a 1 /
a^2 − 1
- (a) θ = 3π/ 4 ± 2 nπ and θ = 5π/ 4 ± 2 nπ, n = 0, 1 , 2 ,...
(b) θ = 5π/ 4 ± 2 nπ and θ = 7π/ 4 ± 2 nπ, n = 0, 1 , 2 ,...
- (a) θ = 3π/ 4 ± nπ, n = 0, 1 , 2 ,...
(b) θ = π/ 3 ± 2 nπ and θ = 5π/ 3 ± 2 nπ, n = 0, 1 , 2 ,...
- (a) θ = 7π/ 6 ± 2 nπ and θ = 11π/ 6 ± 2 nπ, n = 0, 1 , 2 ,...
(b) θ = π/ 3 ± nπ, n = 0, 1 , 2 ,...
- (a) θ = π/ 6 ± nπ, n = 0, 1 , 2 ,...
(b) θ = 4π/ 3 ± 2 nπ and θ = 5π/ 3 ± 2 nπ, n = 0, 1 , 2 ,...
- (a) θ = 3π/ 2 ± 2 nπ, n = 0, 1 , 2 ,... (b) θ = π ± 2 nπ, n = 0, 1 , 2 ,...
- (a) θ = 3π/ 4 ± nπ, n = 0, 1 , 2 ,... (b) θ = π/ 6 ± nπ, n = 0, 1 , 2 ,...
- (a) θ = 2π/ 3 ± 2 nπ and θ = 4π/ 3 ± 2 nπ, n = 0, 1 , 2 ,...
(b) θ = 7π/ 6 ± 2 nπ and θ = 11π/ 6 ± 2 nπ, n = 0, 1 , 2 ,...
Exercise Set A 527
- From the figure, h = x − y but x = d tan β,
y = d tan α so h = d(tan β − tan α).
x
h
y
β α
d
d x
y
h
α β
- From the figure, d = x − y but x = h cot α,
y = h cot β so d = h(cot α − cot β),
h =
d
cot α − cot β
- (a) sin 2θ = 2 sin θ cos θ = 2(
(b) cos 2θ = 2 cos 2 θ − 1 = 2(2/3) 2 − 1 = − 1 / 9
- (a) sin(α − β) = sin α cos β − cos α sin β = (3/5)(1/
(b) cos(α + β) = cos α cos β − sin α sin β = (4/5)(1/
- sin 3θ = sin(2θ + θ) = sin 2θ cos θ + cos 2θ sin θ = (2 sin θ cos θ) cos θ + (cos^2 θ − sin 2 θ) sin θ = 2 sin θ cos^2 θ + sin θ cos^2 θ − sin 3 θ = 3 sin θ cos^2 θ − sin 3 θ; similarly, cos 3θ = cos^3 θ − 3 sin 2 θ cos θ
cos θ sec θ
1 + tan 2 θ
cos θ sec θ
sec^2 θ
cos θ
sec θ
cos θ
(1/ cos θ)
= cos 2 θ
cos θ tan θ + sin θ
tan θ
cos θ(sin θ/ cos θ) + sin θ
sin θ/ cos θ
= 2 cos θ
- 2 csc 2θ =
sin 2θ
2 sin θ cos θ
sin θ
cos θ
= csc θ sec θ
- tan θ + cot θ =
sin θ
cos θ
cos θ
sin θ
sin 2 θ + cos^2 θ
sin θ cos θ
sin θ cos θ
2 sin θ cos θ
sin 2θ
= 2 csc 2θ
sin 2θ
sin θ
cos 2θ
cos θ
sin 2θ cos θ − cos 2θ sin θ
sin θ cos θ
sin θ
sin θ cos θ
= sec θ
sin θ + cos 2θ − 1
cos θ − sin 2θ
sin θ + (1 − 2 sin 2 θ) − 1
cos θ − 2 sin θ cos θ
sin θ(1 − 2 sin θ)
cos θ(1 − 2 sin θ)
= tan θ
- Using (47), 2 sin 2θ cos θ = 2(1/2)(sin θ + sin 3θ) = sin θ + sin 3θ
- Using (47), 2 cos 2θ sin θ = 2(1/2)[sin(−θ) + sin 3θ] = sin 3θ − sin θ
- tan(θ/2) =
sin(θ/2)
cos(θ/2)
2 sin 2 (θ/2)
2 sin(θ/2) cos(θ/2)
1 − cos θ
sin θ
528 Appendix A
- tan(θ/2) =
sin(θ/2)
cos(θ/2)
2 sin(θ/2) cos(θ/2)
2 cos^2 (θ/2)
sin θ
1 + cos θ
- From (52), cos(π/3 + θ) + cos(π/ 3 − θ) = 2 cos(π/3) cos θ = 2(1/2) cos θ = cos θ
- From the figures, area =
hc but h = b sin A
so area =
bc sin A. The formulas
area =
ac sin B and area =
ab sin C
follow by drawing altitudes from vertices B and C, respectively.
A B
C
h
a
c
b
A
E
B
C
D
h 1
h 2
a
c
b
- From right triangles ADC and BDC,
h 1 = b sin A = a sin B so a/ sin A = b/ sin B. From right triangles AEB and CEB, h 2 = c sin A = a sin C so a/ sin A = c/ sin C thus a/ sin A = b/ sin B = c/ sin C.
- (a) sin(π/2 + θ) = sin(π/2) cos θ + cos(π/2) sin θ = (1) cos θ + (0) sin θ = cos θ
(b) cos(π/2 + θ) = cos(π/2) cos θ − sin(π/2) sin θ = (0) cos θ − (1) sin θ = − sin θ
(c) sin(3π/ 2 − θ) = sin(3π/2) cos θ − cos(3π/2) sin θ = (−1) cos θ − (0) sin θ = − cos θ
(d) cos(3π/2 + θ) = cos(3π/2) cos θ − sin(3π/2) sin θ = (0) cos θ − (−1) sin θ = sin θ
- tan(α + β) =
sin(α + β)
cos(α + β)
sin α cos β + cos α sin β
cos α cos β − sin α sin β
, divide numerator and denominator by
cos α cos β and use tan α =
sin α
cos α
and tan β =
sin β
cos β
to get (38);
tan(α − β) = tan(α + (−β)) =
tan α + tan(−β)
1 − tan α tan(−β)
tan α − tan β
1 + tan α tan β
because
tan(−β) = − tan β.
- (a) Add (34) and (36) to get sin(α − β) + sin(α + β) = 2 sin α cos β so
sin α cos β = (1/2)[sin(α − β) + sin(α + β)].
(b) Subtract (35) from (37). (c) Add (35) and (37).
- (a) From (47), sin
A + B
cos
A − B
(sin B + sin A) so
sin A + sin B = 2 sin
A + B
cos
A − B
(b) Use (49) (c) Use (48)
- sin α + sin(−β) = 2 sin
α − β
2
cos
α + β
2
, but sin(−β) = − sin β so
sin α − sin β = 2 cos
α + β
2
sin
α − β
2
