Resolvendo livro de matemática - Parte 275

Описание: b)
y = 1 + sen(x)
p = 2𝝿
D = 𝕽
Im = [0, 2]

c)
y = sen(x - 𝝿/2)
Im = [-1, 1]
D = 𝕽
p = 2𝝿
𝝿 - 𝝿/2 = π/2
3𝝿/2 - 𝝿/2 = 𝝿
2𝝿 - 𝝿/2 = 3𝝿/2

d)
y = 2sen(x/4)
D = 𝕽
Im = [-2, 2]
p = (2π/1)/(1/4) = 2π/1*4/1
p = 8𝝿

2
a)
y = sen(x)/2
p = 2𝝿
D = 𝕽
Im = [-1/2, 1/2]

b)
y = sen(x + 3𝝿/2)
p = 2𝝿
D = 𝕽
Im = [-1, 1]
𝝿/2 + 3𝝿/2 = 2𝝿
𝝿 + 3𝝿/2 = 5𝝿/2
5𝝿/2 = 450°

3
a)
y = sen(6x)
p = 𝝿/3

b)
y = sen(x/3)
p = (2π/1)/(1/3) = 2π/1*3/1
p = 6𝝿

c)
y = 5sen(4x)
p = 𝝿/2

d)
y = 4sen(2x + 𝝿/6)
p = 𝝿

4
a)
f(x) = √(sen(x -π/2) )
sen(x -π/2) ≥ 0
0 ≤ x - π/2 ≤ 𝝿
π/2 ≤ x ≤ 3π/2
D = {x ∈ 𝕽| π/2 ≤ x ≤ 3π/2}

b)
f(x) = √(sen(x -π/3) )
sen(x -π/3) ≥ 0
0 ≤ x - π/3 ≤ 𝝿
π/3 ≤ x ≤ 4π/3
D = {x ∈ 𝕽| π/3 ≤ x ≤ 4π/3}

5
sen(⍺) = (2x - 1)/3
-1 ≤ (2x - 1)/3 ≤ 1
-1 ≤ x ≤ 2
S = {x ∈ 𝕽| -1 ≤ x ≤ 2}

6
sen(x) = 5m + 1
-1 ≤ 5m + 1 ≤ 1
-2/5 ≤ m ≤ 0
S = {x ∈ 𝕽| -2/5 ≤ x ≤ 0}

b)
sen(x) = -5 + 8m
-1 ≤ -5 + 8m ≤ 1
1/2 ≤ m ≤ 3/4

c)
sen(x) = 7m - 20
-1 ≤ 7m - 20 ≤ 1
19/7 ≤ m ≤ 3

d)
sen(x) = 1/2 - 3m
-1 ≤ 1/2 - 3m ≤ 1
1/2 ≥ m ≥ -1/6
-1/6 ≤ m ≤ 1/2

Função cosseno

cos(x) = OM'
Ex1
cos(1830°) = cos(30°)
cos(1830°) = √3/2

Ex2
cos(13𝝿) = cos(𝝿)
13𝝿 = 𝝿 + 6 * 2𝝿
cos(13𝝿) = -1

1
a)
cos(810°) = cos(90°)
cos(810°) = 0

b)
cos(-900°) = cos(-180°)
cos(-900°) = -1

c)
cos(1980°) = cos(180°)
cos(1980°) = -1

d)
cos(11𝝿) = cos(𝝿)
11𝝿 = 𝝿 + 5 * 2𝝿
cos(11𝝿) = -1

e)
cos(9𝝿/2) = 0
9π/2*2π/2π
9π/4π*2π/1
9π/4π=2+1/4
(2+1/4)*2π
9π/2=π/2 + 2 * 2𝝿

f)
cos(13𝝿/6)
13π/6=(π + 12π )/6
13π/6=π/6+2π
cos(13𝝿/6) = √3/2

2
cos(-535°)
cos(190°)

3
A = √2/2 + cos(3x) + cos(x/2)
x = 𝝿/2
A = √2/2 + cos(3𝝿/2) + cos(𝝿/4)
A = √2

4
y = (sen(7𝝿/2))cos(31𝝿)
7π/2*2π/2π
7π/4π*2π/1
7π/4π=1+3/4
(1+3/4)*2π
7π/2=3π/2 + 1 * 2𝝿
31𝝿 = 𝝿 + 15 * 2𝝿
y = 1