Transcript Hs6 Goniometrische formules

Diagnostische Toets Wis B VWO Hoofdstuk 6 Goniometrische formules Getal en Ruimte wiskunde uitwerkingen www.uitwerkingensite.nl
6.5
Diagnostische toets
bladzijde 78
1
a sin(-270°) = 1
b sin 135° = sin 45° =
1
2
2
c cos 225° = - cos 45° = − 1 2
2
d cos(-120°) = - cos 60° = - 1
2
1
e sin 240° = - sin 60° = − 2 3
f cos 330° = cos 30° =
2
1
2
3
x
P = cos 205° ≈ - 0,91
yP = sin 205° ≈ - 0,42
x
Q = cos(-37°) ≈ 0,80
yQ = sin(-37°) ≈ - 0,60
y
1
205°
–1
O
1
1
– 37°
x
P
Q
–1
3
a
1
5
r ad = 15
⋅180° = 36°
b 10 rad = 10 ⋅ 180° = 1800°
c - 4 rad = - 4 ⋅ 180° = -720°
180°
d − 4 rad = − 4 ⋅
≈ −229, 2°
4
e
2
3
f
2
3
c
d
e
f
a
b
c
6
270
⋅ rad = 1 1
2
r ad
180
60
60° =
⋅ rad = 1
3
r ad
180
150
−150° = −
⋅ rad = − 6
5
r ad
180
135
−135° = −
⋅ rad = − 4
3
r ad
180
540
540° =
⋅ rad = 3 r ad
180
390
390° =
⋅ rad = 2 6
1
r ad
180
26
26° =
⋅ rad ≈ 0, 45
rad
180
−73
−73° =
⋅ ≈ −1, 27
rad
180
1010
1010° =
⋅ ≈ 17, 63
rad
180
a 270° =
b
5
r ad = 23 ⋅180° = 120°
180°
≈ 38,
2°
rad = 23 ⋅
( ) ≈ 0,
78
sin ( ) ≈ 0,
28
cos (1 ) ≈ 0,
31
a sin
2
7
b
2
7
c
3
5
Goniometrische formules 63
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7
b
c
8
( % )=
cos ( ) = −
cos (1 ) = −
5
6
a sin
1
2
3
4
1
2
1
3
a sin(α) =
α=
2
1
2
1
2
∨ α=
1
6
5
6
b sin(α) = − 12 2
∨ α = 1 43
α = 1 14
c cos(α) =
α=
9
3
∨ α = 1 65
1
6
(
1
2
)=0
1
a sin 2x + 2
2x + 2 = k ⋅2
1
2
2x = − 12 2 + k ⋅2
x = − 14 4 + k ⋅ 124
(
b cos 2x +
)=1
1
6
2x + 2 = k ⋅ 22
1
6
2x = − 61 2 + k ⋅ 22
x = − 121 2 + k ⋅2
( x ) - sin ( x ) = 0
sin ( x ) ( sin ( x ) - 1) = 0
sin ( x ) = 0 ∨ sin ( x ) = 1
1
2
c sin 2
1
2
1
2
1
2
1
2
1
2
x = k ⋅ 2 ∨ 12 x = 12 2 + k ⋅ 22
x=k⋅2 ∨ x= +k⋅4
1
2
10 a sin
(
1
2
)=
x+
1
2
x+
1
2
x = − 2 + k ⋅ 22
=
1
4
1
2
2
+k⋅2
∨
3
4
x = −1 12 4 + k ⋅ 44
(
b cos − 13 x +
− 13 x +
1
2
)=−
1
2
=
2
3
1
2
1
2
1
6
x = − 2 + k ⋅ 62
1
2
x+
=
3
4
x = − 4 + k ⋅ 24
∨ x = − 1
2 4 + k ⋅ 44
1
2
∨ − 13 x +
1
2
= − 23
+ k⋅2
∨ − x = −1 2 + k ⋅ 22
1
3
1
6
1
∨ x = 3 2 2 + k ⋅ 62
( x) = 3
cos ( x ) =
∨ cos ( x ) = −
cos ( x ) =
cos ( x ) = 3 ∨ cos ( x ) = −
1
2
c 4cos 2
1
2
2
+ k ⋅2
1
4
+ k ⋅2
− x = 2 + k ⋅ 22
1
3
∨
3
4
1
2
1
2
3
4
1
2
1
2
1
2
1
2
x=
∨
1
2
1
6
+ k ⋅2
x=−
5
6
∨
1
2
x=−
1
6
3
4
1
2
3
+ k⋅2
∨
1
2
x=
5
6
+ k ⋅2
+ k⋅2
x = + k ⋅ 4 ∨ x = - 13 + k ⋅ 4 ∨ x = 1 23 + k ⋅ 4 ∨ x = -1 23 + k ⋅ 4
1
3
x = 13 + k ⋅ 2 ∨ x = - 13 + k ⋅ 2
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11
a 2sin(2x) = − 3
sin(2x) = − 12 3
2x = − 13 2 + k ⋅ 22
∨ 2x = 1 1 2 + k ⋅ 22
3
x = − 61 + k ⋅
∨ x=
x op [0, 2 ] geeft x =
(
b 2cos 1 12 x −
(
cos 1 x −
1
2
1 x−
1
6
1 12 x =
11
12
1
2
x=
1
6
=
2
1
2
2
∨ 1 12 x −
+ k ⋅2
3
4
1
6
∨ x=
∨ x = 1 23 2
2
3
= − 43 + k ⋅ 2
∨ 1 12 x = − 127 2 + k ⋅ 22
2 + k ⋅ 22
∨ x = − 187 + k ⋅ 1 13
+ k ⋅ 1 13
11
18
∨ x = 1 65
5
6
)=−
)=−
1
6
+k⋅
2
3
x op [0, 2 ] geeft x =
∨ x = 1 17
18
11
18
∨ x=
17
18
c sin 2 ( x ) - 12 sin( x ) - 12 = 0
(sin( x ) - 1)(sin( x ) + ) = 0
1
2
sin( x ) = 1 ∨ sin( x ) = - 12
x = 12 2 + k ⋅ 22
∨ x = − 61 2 + k ⋅ 22
∨ x = 1 61 2 + k ⋅ 2 2
x op [0, 2 ] geeft
x = 12 2 ∨ x = 1 61
∨ x = 1 65
12 a sin(2x - 1) = sin(x + 2)
2x − 1 = x + 2 + k ⋅ 22 ∨ 2x − 1 =
− (x + 2) + k ⋅ 2
x = 3+ k ⋅2
∨ 2x − 1 = 2 − x − 2 + k ⋅ 2 2
x = 3+ k ⋅2
∨ 3x = −1 + 2 +
k ⋅ 22
x = 3+ k ⋅2
∨ x = − 13 + 13 2 + k ⋅ 232
(
1
b cos x + 3
) = cos ( 2x − )
1
2
(
)
1
1
x + 13 2 = 2x − 12 2 + k ⋅ 22 ∨ x + 3 2 = − 2 x − 2 2 + k ⋅ 22
∨ x + 2 = −2 x + 2 + k ⋅ 2 2
− x = − 2 + k ⋅ 22
5
6
1
3
1
2
x = 2 + k ⋅ 22
∨ 3x = 61 2 + k ⋅ 22
x = 65 2 + k ⋅ 22
∨ x=
5
6
( ( x ) = sin ((
sin ( x ) = sin (
c sin
1
2
1
2
(x + 1)
1
2
x+
2 x =2 x + 2 + k ⋅ 22
)
)
∨
− 2 x =2 + k ⋅ 22 ∨
1
2
1
2
+ k ⋅ 23
1
18
1
2
x = − ( x + ) + k ⋅2
2 x =2 − 2 x −2 + k ⋅ 22
x = -2 + k · 4 ∨ 1 2 x = k ⋅ 22
1
2
1
x = -2 + k · 4 ∨ x = k ⋅1 3
bladzijde 79
13 a y = sin(x)
(
↓ translatie
1
2
,0
y = sin x − 2
)
↓ verm. y-as,
1
3
(
1
2
(
y = sin 3x −
1
2
)
)
↓ verm. x-as, 2
(
f (x) = 2 sin 3x −
1
2
)
Goniometrische formules 65
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b y = cos(x)
↓ verm. y-as, 3
y = cos
( x )
1
3
↓ translatie (-2, 0)
(
y = cos 13
(x + 2)
)
↓ translatie (0, 5)
g(x) = 5 + cos
c y = sin(x)
( (x + 2))
1
3
↓ translatie
(
y = sin x −
1
4
)
↓ verm. y-as,
1
3
(
1
4
�, 0
(
y = sin 3x − 1
4
�
)
↓ verm. x-as, 2
(
y = 2 sin 3x −
)
1
4
)
↓ translatie (0, 1)
(
h(x) = 1 + 2 sin 3x −
14 y = sin(x)
↓ translatie
(
(
1
2
)
)
)
�, 3
y = 3 + sin x − 1
2
2
↓ verm. y-as,
1
4
1
5
)
(
f ( x ) = 3 + sin 5x − 12
�
15 a y = cos(x)
↓ verm. x-as, 2
y = 2 cos(x)
↓ translatie
(
1
3
)
� , −1
(
f ( x ) = −1 + 2cos x − 1
3
�
)
y
y = 2cos(
2cos(xx)
1
y = cos(
cos(xx)
ƒ
π
O
2π
x
y = –1
–1
–2
–3
(
)
b f(x) = -1 geeft −1 + 2 cos x − 1
3 � = −1
(
)
2 cos x − � = 0
(
1
3
)
cos x − 1
3
� = 0
x − 2 = 2 + k ⋅ 2
1
3
x=
5
6
x op [0, 2 ] geeft x =
Dus
(
5
6
) (
1
2
+ k
⋅
5
6
∨ x = 1 6
5
)
� , −1 en 1 6
5
� , −1 .
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c y = 2 cos(x) heeft toppen (0, 2) en ( , -2) dus de toppen van de grafiek van f
)
f(x) = 0 geeft −1 + 2cos ( x − � ) = 0
2cos ( x − ) = 1
cos ( x − ) =
zijn
d
(
1
3
) (
� , 1 en 1 13� , − 3 .
1
3
1
3
1
3
1
2
1
1
x − 13 2 = 132 + k ⋅ 22 ∨ x − 3 2 = − 32 + k ⋅ 22
x=
2
3
x op [0, 2 ] geeft x =
De nulpunten zijn 0,
∨ x = k ⋅ 2�
+ k ⋅2
2
3
∨ x=0 ∨ x=2 .
2
3
en 2 .
16 a evenwichtsstand 2
amplitude 3
periode 2
-3 < 0 dus grafiek dalend door beginpunt − 13 �, 2 .
(
y
)
5
ƒ
4
3
1
(– 3 π, 2)
y=2
2
1
–2π
π
O
–π
2π
x
–1
b evenwichtsstand 4
amplitude 1
2�
=�
periode
2
1 > 0 dus beginpunt
hoogste punt.
y
5
(
5
6
)
5
( 6 π, 5)
g
� , 5 is
y=4
4
3
2
1
O
π
2π
x
Goniometrische formules 67
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17 a evenwichtsstand 30
amplitude 10
22
periode 2
= 7
2
7
V
40
40
V
10 > 0 dus grafiek stijgend door
beginpunt (2, 30).
30
30
V = 30
30
(2, 30)
V = 25
25
20
20
10
10
O
b Voer in y1 = 30 + 10 sin
(
2
7
1,42
4
6,08
8,42
10
12 13,08
t
)
( x − 2) en y2
= 25.
De optie intersect geeft x ≈ 1,42, x ≈ 6,08, x ≈ 8,42 en x ≈ 13,08.
V > 25 geeft 1,42 < t < 6,08 ∨ 8,42 < t < 13,08.
 dV 
c Het beginpunt is (2, 30). De grootste helling is dus   .
 dt  t =2
 dy 
De GR geeft  
≈ 8, 98 dus de grootste helling is 8,98.
 dx
 x=2
65 + -35
= 15
2
b = 65 - 15 = 50
18 a a ==
2
= 1 30
15
beginpunt (25, 15), dus d = 25
periode is 30, dus c =
Dus N = 30 + 10 sin
(
1
15
)
( t − 25) .
b beginpunt (17,5; 65), dus d = 17,5
Dus N = 15 + 50 cos
(
1
15
)
( t − 17, 5) .
19 a = 1 + 0 = 1
De periode is , want de periode van f is
(
en de periode van g is , dus c =
2
= 2.
)
Voer in y1 = 1 + 2 cos(2x) + cos 2 x + 13
.
De optie maximum geeft x ≈ 2,97 en y ≈ 3,65, dus b ≈ 3,65 - 1 = 2,65.
Voer in y2 = 1.
De optie intersect geeft x ≈ 2,19, dus d = 2,19.
Dus h(x) = 1 + 2,65 sin(2(x - 2,19)).
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