G&R vwo B deel 3 C. von Schwartzenberg 1
11 Goniometrie en beweging 1/22
spiegelen in de y -as
y = sin(x ) → f (x ) = sin( −x ) ⇒ f (x ) = sin( −x ) heeft dezelfde grafiek als y = − sin(x ). spiegelen in de y -as
y = cos(x ) → g (x ) = cos( −x ) ⇒ g (x ) = cos( −x ) heeft dezelfde grafiek als y = cos(x ). translatie ( − 1 π , 0)
1 1 2 → y = sin(x ) h (x ) = sin(x + 2 π ) ⇒ h (x ) = sin(x + 2 π ) heeft dezelfde grafiek als y = cos(x ). translatie ( − 1 π , 0)
2 → j (x ) = cos(x + 1 π ) ⇒ j (x ) = cos(x + 1 π ) heeft dezelfde grafiek als y = − sin(x ). y = cos(x ) 2 2 translatie ( −π , 0) → y = sin(x ) k (x ) = sin(x + π ) ⇒ k (x ) = sin(x + π ) heeft dezelfde grafiek als y = − sin(x ).
translatie ( −π , 0) y = cos(x ) → l (x ) = cos(x + π ) ⇒ l (x ) = cos(x + π ) heeft dezelfde grafiek als y = − cos(x ).
2a
2c sin(α )
α +π α
α O
O
α − 21 π
cos(α − 1 π ) = sin(α ) 2
cos(α −
1 2
cos(α + π )
cos(α + π ) = − cos(α )
π)
2b
2d sin(α )
sin(α + π ) = − sin(α )
α +π
1
O
cos(α )
De stelling van Pythagoras geeft:
sin(α + π )
sin2 (α ) + cos2 (α ) = 1
sin(x + 1 π ) = cos(x + 1 π − 1 π ) = cos(x − 1 π ). 6
2
3b
3
cos(2x + 1 π ) = sin(2x + 1 π + 1 π ) = sin(2x + 5 π ).
3d
3 3 2 6 2 2 1 1 1 1 − sin(3x − π ) = sin(3x − π + π ) = sin(3x + π ) = cos(3x + π − π ) = cos(3x − π ). 3 3 3 3 2 6 1 1 1 1 1 − cos(4x + 1 π ) = cos(4x + 1 π + π ) = cos(4x + 2 π ) = sin(4x + 2 π + π ) = sin(4x + 2 2 π ) = sin(4x + 2 π ). 6 6 6 6 2 3 3
4a
(sin(x ) − cos(x ))2 = sin2 (x ) − 2 sin(x ) cos(x ) + cos2 (x ) = sin2 (x ) + cos2 (x ) − 2 sin(x ) cos(x ) = 1 − 2 sin(x ) cos(x )
4b
2sin2 (x ) + cos2 (x ) 2 sin2 (x ) cos2 (x ) sin(x ) = + = 2⋅ cos(x ) cos2 (x ) cos2 (x ) cos2 (x )
4c
sin (3x ) 2 2 2 (1 + tan2(3x )) ⋅ cos2(3x ) = 1 + cos ⋅ cos (3x ) = cos (3x ) + sin (3x ) = 1. (3x )
5a
sin2 (x ) + 4 cos(x ) = 1 − cos2 (x ) + 4 cos(x ).
5b
2cos2 (x ) + sin(x ) − 2 = 2 ⋅ 1 − sin2 (x ) + sin(x ) − 2 = 2 − 2 sin2 (x ) + sin(x ) − 2 = −2 sin2 (x ) + sin(x ).
5c
2 sin2 (x ) + cos2 (x ) + cos(x ) = 2 ⋅ 1 − cos2 (x ) + cos2 (x ) + cos(x )
3c
6
sin(α )
α
α O
3a
cos(α )
(
2
)
+ 1 = 2 tan2 (x ) + 1.
2
2
(
)
(
)
= 2 − 2cos2 (x ) + cos2 (x ) + cos(x ) = 2 − cos2 (x ) + cos(x ).
6
sin(2x − 1 π ) = − cos(x + 1 π )
3 3 cos(2x − 1 π − 1 π ) = cos(x + 1 π + π ) 3 2 3 cos(2x − 5 π ) = cos(x + 1 1 π ) 6 3 hiernaast gaat het verder
2x − 5 π = x + 1 1 π + k ⋅ 2π
∨ 2x − 5 π = −x − 1 1 π + k ⋅ 2π
x
∨ 3x = − 1 π + k ⋅ 2π
x
6 3 = 2 1 π + k ⋅ 2π 6 = 2 1 π + k ⋅ 2π 6
6
3
2
∨ x = − 1 π +k ⋅ 2π 6
3
x op [0, 2π ] geeft x = 61 π ∨ x = 21 π ∨ x = 1 61 π ∨ x = 1 56 π .
G&R vwo B deel 3 C. von Schwartzenberg 7a
7b
11 Goniometrie en beweging 2/22
sin(x + 1 π ) = cos(2x )
2 + 1 π − 1 π ) = cos(2x ) 2 2
7d
cos(x x = 2x + k ⋅ 2π −x = k ⋅ 2π
∨ x = −2x + k ⋅ 2π ∨ 3x = k ⋅ 2π
x = k ⋅ 2π
∨ x = k ⋅ 2π
x = −2 − π + k ⋅ 2π ∨ x = − 31 π + k ⋅ 23 π x op [0, 2π ] ⇒ x = −2 + π ∨ x = 31 π ∨ x = π ∨ x = 1 23 π .
3
x op [0, 2π ] ⇒ x = 0 ∨ x = 32 π ∨ x = 1 31 π ∨ x = 2π . sin(3x ) = − cos(x ) cos(3x − 1 π ) = cos(x + π ) 2
∨ 3x − 1 π = −x − π + k ⋅ 2π
2x = 1 1 π + k ⋅ 2π
∨ 4x = − 1 π + k ⋅ 2π
2
sin(2x + π ) = 1 − 2 sin(2x ) − sin(2x ) = 1 − 2 sin(2x ) sin(2x ) = 1 2x = 1 π + k ⋅ 2π
7e
3x − 1 π = x + π + k ⋅ 2π 2
cos(x − 1) = − cos(2x + 1) cos(x − 1) = cos(2x + 1 + π ) x − 1 = 2x + 1 + π + k ⋅ 2π ∨ x − 1 = −2x − 1 − π + k ⋅ 2π −x = 2 + π + k ⋅ 2π ∨ 3x = −π + k ⋅ 2π
2
2
x = 41 π + k ⋅ π x op [0, 2π ] ⇒ x = 41 π ∨ x = 1 41 π .
2
∨ x = − 1π +k ⋅ 1π x = 34 π + k ⋅ π 8 2 x op [0, 2π ] ⇒ x = 34 π ∨ x = 1 34 π ∨ x = 38 π ∨ x = 78 π ∨ x = 1 38 π ∨ x = 1 78 π . 7c
sin2 (x ) + 1 cos(x ) = 1
7f
2
(
2
2
2 − 2cos (x ) + cos2 (x ) + cos(x ) = 0
− cos (x ) + 1 cos(x ) = 0 2 − cos(x ) ⋅ cos(x ) − 1 = 0 2 cos(x ) = 0 ∨ cos(x ) = 1 2 x = 21 π + k ⋅ π ∨ x = 31 π + k ⋅ 2π ∨ x = − 31 π + k ⋅ 2π x op [0, 2π ] ⇒ x = 21 π ∨ x = 1 21 π ∨ x = 31 π ∨ x = 1 23 π .
(
8a
9a
9b
)
2 ⋅ 1 − cos2 (x ) + cos2 (x ) + cos(x ) = 0
1 − cos2 (x ) + 1 cos(x ) = 1 2
2 sin2 (x ) + cos2 (x ) + cos(x ) = 0
− cos2 (x ) + cos(x ) + 2 = 0
)
cos(2πt ) = sin( 1 πt )
2 cos(2πt ) = cos( 1 πt − 1 π ) 2 2 2πt = 1 πt − 1 π + k ⋅ 2π ∨ 2πt = − 1 πt + 1 π + k ⋅ 2π 2 2 2 2 1 1 πt = − 1 π + k ⋅ 2π ∨ 2 1 πt = 1 π + k ⋅ 2π 2 2 2 2 t = − 31 + k ⋅ 34 ∨ t = 51 + k ⋅ 54 t op [0, 3] ⇒ t = 1 ∨ t = 2 31 ∨ t = 51 ∨ t = 1 54 ∨ t = 2 53 .
2 sin(x ) = sin(x ) sin(x ) = 0
8b
cos2 (x ) − cos(x ) − 2 = 0 ( cos(x ) − 2) ⋅ ( cos(x ) + 1) = 0 cos(x ) = 2 (kan niet) ∨ cos(x ) = −1 x = −π + k ⋅ 2π x op [0, 2π ] ⇒ x = π . sin( πt ) = − cos(πt ) 6
cos( πt − 1 π ) = cos(πt + π ) 6
2
πt − 1 π = πt + π + k ⋅ 2π ∨ πt − 1 π = −πt − π + k ⋅ 2π 6
2
6
6
2
6
9d 9e
--sin(2x ) = sin(x + 1 π ) 3
)
x x x
3 3 = 1 π + k ⋅ 2π ∨ 2x = π − x − 1 π + k ⋅ 2π 3 3 = 1 π + k ⋅ 2π ∨ 3x = 2 π + k ⋅ 2π 3 3 = 1 π + k ⋅ 2π ∨ x = 2 π + k ⋅ 2 π 3 9 3
---
10a
verm. in de x -as, −1 2 y = cos(x ) → → g (x ) = − cos(2x ). y = cos(2x )
9f
---
1
10b
Zie de schets hiernaast.
10c
f (x ) = − 21 2 ⇒ sin(x ) = − 21 2
f
x = − 41 π + k ⋅ 2π ∨ x = π − − 41 π + k ⋅ 2π x op [0, 2π ] ⇒ x = 74 π = 1 34 π ∨ x = 54 π = 1 41 π .
10e
g (x ) = 21 ⇒ − cos(2x ) = 21
cos(2x ) = − 1
2
2x = π − 1 π + k ⋅ 2π ∨ 2x = − 2 π + k ⋅ 2π 3
(
2x = x + 1 π + k ⋅ 2π ∨ 2x = π − x + 1 π + k ⋅ 2π
9c
10d
2
∨ t = − 3 + k ⋅ 12 t = − 59 + k ⋅ 12 5 7 7 t op [0, 3] ⇒ t = 53 ∨ t = 3 ∨ t = 1 72 .
x = k ⋅π sin(2x ) = sin(x ) 2x = x + k ⋅ 2π ∨ 2x = π − x + k ⋅ 2π x = k ⋅ 2π ∨ 3x = π + k ⋅ 2π x = k ⋅ 2π ∨ x = 31 π + k ⋅ 23 π
verm. in de y -as,
2
− 5πt = 1 1 π + k ⋅ 2π ∨ 7πt = − 1 π + k ⋅ 2π
3
x = 31 π + k ⋅ π ∨ x = − 31 π + k ⋅ π x op [0, 2π ] ⇒ x = 31 π ∨ x = 1 31 π ∨ x = 23 π ∨ x = 1 23 π .
g
f (x ) = g (x ) ⇒ sin(x ) = − cos(2x ) cos(x − 1 π ) = cos(2x + π ) 2
x − 21 π = 2x + π + k ⋅ 2π ∨ x − 21 π = −2x − π + k ⋅ 2π −x = 1 1 π + k ⋅ 2π ∨ 3x = − 1 π + k ⋅ 2π 2 2
x = −1 21 π + k ⋅ 2π ∨ x = − 61 π + k ⋅ 32 π x op [0, 2π ] ⇒ x = 21 π ∨ x = 1 61 π ∨ x = 1 56 π . f (x ) ≤ g (x ) (zie de schets) ⇒ x = 21 π ∨ 1 61 π ≤ x ≤ 1 56 π .
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 3/22
11a
f (x ) = sin(2x − 31 π ) heeft evenwichtsstand 0; amplitude 1; periode 22π = π en beginpunt ( 61 π , 0). g (x ) = − cos(x + 61 π ) heeft evenwichtsstand 0; amplitude 1; periode 2π en laagste punt ( − 61 π , − 1).
11b
Gebruik de plot hiernaast voor een schets van de grafieken.
11c
f (x ) = sin(2x − 31 π ) = 0 2x − 1 π = k ⋅ π 3
g (x ) = − cos(x + 61 π ) = 0 cos(x + 1 π ) = 0 6
x + 61 π = 21 π + k ⋅ π x = 31 π + k ⋅ π
2x = 1 π + k ⋅ π 3
x = 61 π + k ⋅ 21 π x op [0, 1 21 π ] ⇒ x = 61 π ∨ x = 32 π ∨ x = 1 61 π . De nulpunten van f zijn 1 π , 2 π en 1 1 π . 6 3 6 11d
f (x ) = 21 ⇒ sin(2x − 31 π ) = 21 2x − 1 π = 1 π + k ⋅ 2π ∨ 2x − 1 π = π − 1 π + k ⋅ 2π 3 6 3 6 2x = 1 π + k ⋅ 2π ∨ 2x = 1 1 π + k ⋅ 2π 2 6
AB = yA − yB = 2 ⋅ yA = 2 sin(α ).
12b
AB 2 = OA2 + OB 2 − 2 ⋅ OA ⋅ OB ⋅ cos ∠AOB 2
(2 sin(α ) )
2
f (x ) = g (x ) ⇒ sin(2x − 31 π ) = − cos(x + 61 π ) cos(2x − 1 π − 1 π ) = cos(x + 1 π + π ) 3 2 6
11e
cos(2x − 5 π ) = cos(x + 1 1 π ) 6
14a
6
6
6
x op [0, 1 21 π ] ⇒ x = 0 ∨ x = 95 π ∨ x = 1 29 π . f (x ) < g (x ) (zie plot) ⇒ 59 π < x < 1 92 π . 13b cos(t + u ) = cos(t ) ⋅ cos(u ) − sin(t ) ⋅ sin(u ) u vervangen door u − 1 π geeft
2
= 1 + 1 − 2 ⋅ 1 ⋅ 1 ⋅ cos(2α )
2
4 sin2 (α ) = 2 − 2 cos(2α )
cos(t + u − 1 π ) = cos(t ) ⋅ cos(u − 1 π ) − sin(t ) ⋅ sin(u − 1 π )
2cos(2α ) = 2 − 4 sin2 (α )
sin(t + u ) = cos(t ) ⋅ sin(u ) − sin(t ) ⋅ − cos(u ) sin(t + u ) = cos(t ) ⋅ sin(u ) + sin(t ) ⋅ cos(u ) sin(t + u ) = sin(t ) ⋅ cos(u ) + cos(t ) ⋅ sin(u ).
2
cos(t − u ) = cos(t ) ⋅ cos(u ) + sin(t ) ⋅ sin(u ) u vervangen door − u geeft cos(t − ( −u )) = cos(t ) ⋅ cos( −u ) + sin(t ) ⋅ sin( −u ) cos(t + u ) = cos(t ) ⋅ cos(u ) + sin(t ) ⋅ − sin(u ) cos(t + u ) = cos(t ) ⋅ cos(u ) − sin(t ) ⋅ sin(u ).
13c
sin(t + u ) = sin(t ) ⋅ cos(u ) + cos(t ) ⋅ sin(u ) t en u beide vervangen door A geeft sin(A + A) = sin(A ) ⋅ cos(A) + cos(A) ⋅ sin(A) sin(2A) = 2 sin(A ) ⋅ cos(A). cos(t + u ) = cos(t ) ⋅ cos(u ) − sin(t ) ⋅ sin(u ) t en u beide vervangen door A geeft cos(A + A ) = cos(A ) ⋅ cos(A ) − sin(A ) ⋅ sin(A )
14b
2
sin(t + u ) = sin(t ) ⋅ cos(u ) + cos(t ) ⋅ sin(u ) u vervangen door − u geeft sin(t + ( −u )) = sin(t ) ⋅ cos( −u ) + cos(t ) ⋅ sin( −u ) sin(t − u ) = sin(t ) ⋅ cos(u ) + cos(t ) ⋅ − sin(u ) sin(t − u ) = sin(t ) ⋅ cos(u ) − cos(t ) ⋅ sin(u ).
cos(2A) = cos2 (A ) − sin2 (A) (zie 14a)
(
2
cos(2A) = cos (A ) − 1 + cos (A ) cos(2A) = 2cos2 (A ) − 1. cos(2A) = cos2 (A ) − sin2 (A) cos(2A) = 1 − sin2 (A) − sin2 (A)
cos(2A) = 1 − 2 sin2 (A)
15b
−2 cos (A ) = −1 − cos(2A )
2 sin2 (A ) = 1 − cos(2A)
cos2 (A) = 1 + 1 cos(2A).
sin2 (A) = 1 − 1 cos(2A).
2
sin(x ) ⋅ cos(x ) = 1 sin(x − 1) 2
1 ⋅ 2 ⋅ sin(x ) ⋅ cos(x ) = 1 sin(x − 1) 2 2 1 sin(2x ) = 1 sin(x − 1) 2 2
sin(2x ) = sin(x − 1) 2x = x − 1 + k ⋅ 2π ∨ 2x = π − x + 1 + k ⋅ 2π x = −1 + k ⋅ 2π ∨ 3x = π + 1 + k ⋅ 2π
x = −1 + k ⋅ 2π ∨ x = 31 π + 31 + k ⋅ 23 π .
)
2
cos(2A) = 1 − 2 sin2 (A) (zie ook 12b).
cos(2A) = 2cos2 (A ) − 1
2
2
cos(2A) = cos2 (A ) − 1 − cos2 (A )
2
16a
6
x = 2π + k ⋅ 2π ∨ 3x = − 31 π + k ⋅ 2π x = 2π + k ⋅ 2π ∨ x = − 91 π + k ⋅ 23 π
cos(2A ) = cos2 (A) − sin2 (A ). 15a
6
2x − 5 π = x + 1 1 π + k ⋅ 2π ∨ 2x − 5 π = −x − 1 1 π + k ⋅ 2π
cos(2α ) = 1 − 2 sin2 (α ).
13a
f
x op [0, 1 21 π ] ⇒ x = 31 π ∨ x = 1 31 π . De nulpunten van g zijn 1 π en x = 1 1 π . 3 3
7 π + k ⋅π x = 41 π + k ⋅ π ∨ x = 12 7 π. x op [0, 1 21 π ] ⇒ x = 41 π ∨ x = 1 41 π ∨ x = 12 7 π ∨ 1 1 π < x ≤ 1 1 π. f (x ) > 21 (zie plot) ⇒ 41 π < x < 12 2 4
12a
g
2
16c
2
sin2 ( 1 x ) = cos(x ) + 1 1 2
4
Gebruik: cos(2A) = 1 − 2 sin2 (A) ⇒ sin2 (A ) = 1 − 1 cos(2A) 1 − 1 cos(x ) = cos(x ) + 1 1 2 2 4 −1 1 cos(x ) = 3 2 4 cos(x ) = − 1 2 x = 23 π + k ⋅ 2π ∨ x = − 23 π + k ⋅ 2π .
2
2
G&R vwo B deel 3 C. von Schwartzenberg 16b
11 Goniometrie en beweging 4/22
cos2 (2x ) = cos(4x ) + 1
16d
2
cos2 (2x ) = 2cos2 (2x ) − 1 + 1
sin2 (x ) + 2 sin(x ) cos(x ) + cos2 (x ) = 1 1
2
− cos2 (2x ) = − 1
2
sin2 (x ) + cos2 (x ) + sin(2x ) = 1 1
2
cos2 (2x ) = 1
( sin(x ) + cos(x ) )2 = 1 21
2
sin(2x ) = 1
2
2
cos(2x ) = ± 1 = ± 1 ⋅ 2 = ± 1 2
2x = 1 π + k ⋅ 2π ∨ 2x = π − 1 π + k ⋅ 2π
2x = 1 π + k ⋅ 1 π
x
2
4
2 2
2
2
6 6 = 1 π + k ⋅π ∨ x = 5 π + k ⋅π. 12 12
x = 81 π + k ⋅ 41 π . 17a
Evenwichtsstand 1 ; amplitude 1 ; periode π en beginpunt (hoogste punt bij cosinus) (0, 1). 2
2
Dus y = 1 + 1 cos(2x ). 2
17b
2
cos(2A ) = 1 − 2 sin2 (A) ⇒ sin2 (A) = 1 − 1 cos(2A ) 2
2
y = sin2 (x ) + cos(2x ) = 21 − 21 cos(2x ) + cos(2x ) = 21 + 21 cos(2x ). 18
sin(3x ) = sin(2x + x ) = sin(2x ) ⋅ cos(x ) + cos(2x ) ⋅ sin(x )
19b
(
cos(3x ) = cos(2x + x ) = cos(2x ) ⋅ cos(x ) − sin(2x ) ⋅ sin(x )
)
(
= 2 sin(x ) ⋅ cos(x ) ⋅ cos(x ) + 1 − 2 sin2 (x ) ⋅ sin(x ) 2
3
3
= 2cos (x ) − cos(x ) − 2 sin2 (x ) cos(x )
= 2 sin(x ) ⋅ cos (x ) + sin(x ) − 2 sin (x )
(
)
(
= 2 sin(x ) ⋅ 1 − sin2 (x ) + sin(x ) − 2 sin3(x ) 3
19a
)
= 2cos3 (x ) − cos(x ) − 2 ⋅ 1 − cos2 (x ) ⋅ cos(x )
3
3
= 2 sin(x ) − 2 sin (x ) + sin(x ) − 2 sin (x )
= 2cos (x ) − cos(x ) − 2cos(x ) + 2 cos3 (x )
= 3sin(x ) − 4 sin3(x ).
= 4 cos3 (x ) − 3cos(x ).
cos(2A ) = 1 − 2 sin2 (A) ⇒ sin2 (A ) = 1 − 1 cos(2A ).
y
)
= 2cos2 (x ) − 1 ⋅ cos(x ) − 2 sin(x ) cos(x ) ⋅ sin(x )
2 2 = 1 − cos(x ) − sin2 ( 1 x ) = 1 − cos(x ) − 1 − 1 cos(x ) = 1 − 1 cos(x ). 2 2 2 2 2
(
20a
− cos(A) = cos(A + π ).
20b
− 1 π ). 2
sin(A ) = cos(A
19b staat hierboven uitgewerkt.
)
cos(2A ) = 1 − 2 sin2 (A) of sin2 (A) = 1 − 1 cos(2A).
20c
2 2 cos(2A ) = 2cos (A ) − 1 of cos (A ) = 1 + 1 cos(2A ). 2 2 1 cos(A ) = sin(A + π ). 2 2
20d 20e
2
21
Voor B geldt: xB = −xA en yB = yA . Voor C geldt: xC = −xA en yC = −yA .
22a
Voor elke p geldt: f ( − p ) = − p cos( − p ) = − p ⋅ cos( p ) = −f ( p ). f ( − p ) = −f ( p ) ⇒ f ( − p ) + f ( p ) = 0 ⇒ f is (punt)symmetrisch in O .
22b
Voor elke p geldt: g ( − p ) = − p sin( − p ) = − p ⋅ − sin( p ) = p sin( p ) = g ( p ) ⇒ g is (lijn)symmetrisch in de y -as.
23a
Voor elke p geldt: f ( − p ) = cos2 ( − p ) sin( − p ) = cos2 ( p ) ⋅ − sin( p ) = − cos2 ( p ) sin( p ) = −f ( p ). f ( − p ) = −f ( p ) ⇒ f ( − p ) + f ( p ) = 0 ⇒ f is symmetrisch in O .
23b
f ( 21 π − p ) = cos2 ( 21 π − p ) sin( 21 π − p ) = cos2 ( 21 π + p ) sin( 21 π + p ) = f ( 21 π + p ).
1 2
1 π −p 2
π +p
Dus f is symmetrisch in de lijn x = 1 π . 2
Gebruik: cos( 1 π − p ) = − cos( 1 π + p ) (kwadr.) 2
2
cos2 ( 1 π − p ) = − − cos2 ( 1 π + p ) ⇒ cos2 ( 1 π − p ) = cos2 ( 1 π + p ). 2
24a
2
2
f ( − 41 π − p ) = 2 sin( − 41 π − p ) − 2 cos( − 41 π − p )
( = 2(− 1 2
2
O
) (
= 2 sin( − 1 π ) cos( p ) − cos( − 1 π ) sin( p ) − 2 cos( − 1 π ) cos( p ) + sin( − 1 π ) sin( p ) 4
4 4 2 ⋅ cos( p ) − 1 2 ⋅ sin( p ) − 2 1 2 ⋅ cos( p ) + − 1 2 ⋅ sin( p ) 2 2 2
) (
4
)
= − 2 ⋅ cos( p ) − 2 ⋅ sin( p ) − 2 ⋅ cos( p ) + 2 ⋅ sin( p ) = −2 2 ⋅ cos( p ).
)
G&R vwo B deel 3 C. von Schwartzenberg 24b
11 Goniometrie en beweging 5/22
f ( − 41 π + p ) = 2 sin( − 41 π + p ) − 2cos( − 41 π + p )
( = 2(− 1 2
) (
= 2 sin( − 1 π ) cos( p ) + cos( − 1 π ) sin( p ) − 2 cos( − 1 π ) cos( p ) − sin( − 1 π ) sin( p ) 4
4 4 2 ⋅ cos( p ) + 1 2 ⋅ sin( p ) − 2 1 2 ⋅ cos( p ) − − 1 2 ⋅ sin( p ) 2 2 2
) (
4
)
)
= − 2 ⋅ cos( p ) + 2 ⋅ sin( p ) − 2 ⋅ cos( p ) − 2 ⋅ sin( p ) = −2 2 ⋅ cos( p ). Voor elke p geldt: f ( − 1 π − p ) = f ( − 1 π + p ) ⇒ f is symmetrisch in de lijn x = − 1 π . 4
25a
4
4
f ( 41 π − p ) = cos( 41 π − p ) + sin( 41 π − p ) + 1
= cos( 1 π ) cos( p ) + sin( 1 π ) sin( p ) + sin( 1 π ) cos( p ) − cos( 1 π ) sin( p ) + 1 4
4
4
4
= 1 2 ⋅ cos( p ) + 1 2 ⋅ sin( p ) + 1 2 ⋅ cos( p ) − 1 2 ⋅ sin( p ) + 1 = 2 ⋅ cos( p ) + 1. 2
2
2
2
f ( 41 π + p ) = cos( 41 π + p ) + sin( 41 π + p ) + 1
= cos( 1 π ) cos( p ) − sin( 1 π ) sin( p ) + sin( 1 π ) cos( p ) + cos( 1 π ) sin( p ) + 1 4
4
4
4
= 1 2 ⋅ cos( p ) − 1 2 ⋅ sin( p ) + 1 2 ⋅ cos( p ) + 1 2 ⋅ sin( p ) + 1 = 2 ⋅ cos( p ) + 1. 2
2
2
2
1 4
Er geldt: f ( 1 π − p ) = f ( 1 π + p ) ⇒ f is symmetrisch in de lijn x = 1 π . 4
4
Alternatieve uitwerking
1 4
f ( 41 π + p ) = cos( 41 π + p ) + sin( 41 π + p ) + 1 (gebruik de eenheidscirkel hiernaast) = sin( 1 π − p ) + cos( 1 π − p ) + 1 4
4
= cos( 1 π − p ) + sin( 1 π − p ) + 1 = f ( 1 π − p ).
f(1 π − 4
25b
f ( 34 π −
f ( 34 π +
f ( 43 π −
π +p
4
O
4 4 4 p ) = f ( 1 π + p ) ⇒ f is symmetrisch in de lijn x = 1 π . 4 4 p ) = cos( 34 π − p ) + sin( 34 π − p ) + 1 = cos( 3 π ) cos( p ) + sin( 3 π ) sin( p ) + sin( 3 π ) cos( p ) − cos( 3 π ) sin( p ) + 1 4 4 4 4 = − 1 2 ⋅ cos( p ) + 1 2 ⋅ sin( p ) + 1 2 ⋅ cos( p ) − − 1 2 ⋅ sin( p ) + 1 = 2 ⋅ sin( p ) + 1. 2 2 2 2 p ) = cos( 34 π + p ) + sin( 34 π + p ) + 1 = cos( 3 π ) cos( p ) − sin( 3 π ) sin( p ) + sin( 3 π ) cos( p ) + cos( 3 π ) sin( p ) + 1 4 4 4 4 = − 1 2 ⋅ cos( p ) − 1 2 ⋅ sin( p ) + 1 2 ⋅ cos( p ) + − 1 2 ⋅ sin( p ) + 1 = − 2 ⋅ sin( p ) + 1. 2 2 2 2 p ) + f ( 34 π + p ) = 2 ⋅ sin( p ) + 1 − 2 ⋅ sin( p ) + 1 = 2 ⇒ f is symmetrisch in het punt ( 43 π , 1).
26a
26b
f (x ) = sin(x ) ⇒ waarschijnlijk is f '(x ) = cos(x ).
26c
f (x ) = cos(x ) ⇒ waarschijnlijk is f '(x ) = − sin(x ).
27a
Zie de plot van y 2 hiernaast.
27b
f (x ) = sin( 2x ) ⇒ f '(x ) = 2cos(2x ).
27c
f (x ) = cos( 3x ) ⇒ f '(x ) = −3sin(3x ).
28
f (x ) = cos x = sin( x + 21 π ) ⇒ f '(x ) = cos(x + 21 π ) = − sin(x ).
29
f (x ) = sin( ax + b ) ⇒ f '(x ) = cos(ax + b ) ⋅ a = a cos(ax + b ). g (x ) = cos( ax + b ) ⇒ g '(x ) = − sin(ax + b ) ⋅ a = −a sin(ax + b ).
30a
f (x ) = 3 + 4 sin( 2x − 31 π ) ⇒ f '(x ) = 4 cos(2x − 31 π ) ⋅ 2 = 8cos(2x − 31 π ).
30b
g (x ) = 10 + 16cos( 21 ( x − 1 ) ) ⇒ g '(x ) = −16 sin( 21 (x − 1)) ⋅ 21 ⋅ 1 = −8 sin( 21 (x − 1)).
30c
h (x ) = x cos(x ) ⇒ h '(x ) = 1 ⋅ cos(x ) + x ⋅ − sin(x ) = cos(x ) − x sin(x ).
30d
j (x ) = x cos( 2x ) ⇒ j '(x ) = 1 ⋅ cos(2x ) + x ⋅ −2 sin(2x ) = cos(2x ) − 2x sin(2x ).
π −p
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 6/22
30e
k (x ) = x 2 ⋅ sin( 3x ) ⇒ k '(x ) = 2x ⋅ sin(3x ) + x 2 ⋅ 3cos(3x ) = 2x sin(3x ) + 3x 2 cos(3x ).
30f
l (x ) = 2x ⋅ sin( 3x − 1 ) ⇒ l '(x ) = 2 ⋅ sin(3x − 1) + 2x ⋅ 3cos(3x − 1) = 2 sin(3x − 1) + 6x cos(3x − 1).
31a
f (x ) = 3tan( 2x ) ⇒ f '(x ) = 3 ⋅
31b
g (x ) = tan2 (x ) = tan(x )
31c
h (x ) = cos(x ) ⋅ tan(x ) = cos(x ) ⋅
32
I
f (x ) = sin2 (x ) = sin(x ) ⋅ sin(x ) ⇒ f '(x ) = cos(x ) ⋅ sin(x ) + sin(x ) ⋅ cos(x ) = 2 sin(x ) cos(x ).
II
f (x ) = sin2 (x ) = sin(x )
2
(
)
6 1 ⋅2 = . cos2 (2x ) cos2 (2x )
⇒ g '(x ) = 2 tan(x ) ⋅
sin(x ) = sin(x ) ⇒ h '(x ) = cos(x ). cos(x )
2
(
sin(x ) 2 sin(x ) 1 1 = 2⋅ ⋅ = . cos(x ) cos2 (x ) cos3 (x ) cos2 (x )
)
⇒ f '(x ) = 2 sin(x ) ⋅ cos(x ).
III f (x ) = sin2 (x ) = 1 − 1 cos( 2x ) ⇒ f '(x ) = − 1 ⋅ − sin( 2x ) ⋅ 2 = sin(2x ). 2
2
2
Mijn persoonlijke voorkeur gaat uit naar II omdat in dit geval f (x ) niet hoeft te worden herschreven. 2
33a
f (x ) = cos2 (x ) = cos(x )
33b
g (x ) = 2 sin2 (x ) = 2 sin(x )
33c
h (x ) = 1 + 2cos2 (x ) = 1 + 2 cos(x )
33d
j (x ) = x + 3 sin2 (x ) = x + 3 sin(x )
34a
f (x ) = sin3(x ) = sin(x )
34b
g (x ) = x ⋅ sin2 (x ) = x ⋅ sin(x )
34c
h (x ) = cos2 (2x ) = cos( 2x )
34d
j (x ) = cos2 (x 2 ) = cos( x 2 ) ⇒ j '(x ) = 2 cos(x 2 ) ⋅ − sin(x 2 ) ⋅ 2x = −4x sin(x 2 ) cos(x 2 ).
35a
f (x ) = sin3(x ) + sin(x ) = sin(x )
(
)
(
⇒ f '(x ) = 2cos(x ) ⋅ − sin(x ) = −2 sin(x ) cos(x ). 2
)
⇒ g '(x ) = 4 sin(x ) ⋅ cos(x ).
(
(
(
3
)
2
)
⇒ h '(x ) = 4 cos(x ) ⋅ − sin(x ) = −4 sin(x ) cos(x ).
2
)
⇒ j '(x ) = 1 + 6 sin(x ) ⋅ cos(x ).
⇒ f '(x ) = 3sin2 (x ) ⋅ cos(x ).
(
2
)
2
(
)
⇒ g '(x ) = 1 ⋅ sin2 (x ) + x ⋅ 2 sin(x ) ⋅ cos(x ) = sin2 (x ) + 2x sin(x ) cos(x ).
⇒ h '(x ) = 2cos(2x ) ⋅ − sin(2x ) ⋅ 2 = −4 sin(2x ) cos(2x ).
2
(
3
)
+ sin(x ) ⇒ f '(x ) = 3 sin2 (x ) ⋅ cos(x ) + cos(x )
(
)
= 3cos(x ) ⋅ 1 − cos2 (x ) + cos(x ) = 4 cos(x ) − 3cos3(x ). 35b
g (x ) = sin2 (x ) ⋅ cos(x ) = sin(x )
(
2
)
⋅ cos(x ) ⇒
g '(x ) = 2 sin(x ) ⋅ cos(x ) ⋅ cos(x ) + sin2 (x ) ⋅ − sin(x ) = 2 sin(x ) ⋅ cos2 (x ) − sin3(x )
(
)
= 2 sin(x ) ⋅ 1 − sin2 (x ) − sin3(x ) = 2 sin(x ) − 2 sin3(x ) − sin3(x ) = 2 sin(x ) − 3sin3(x ). tan( x ) sin(x ) = ⋅ 1 = 1 = sin( x ) cos(x ) sin( x ) cos( x )
( cos(x ) )
−1
35c
h (x ) =
36a
f (x ) = 1 + 2 sin(x − 31 π ) heeft evenwichtsstand 1; amplitude 2; periode 2π en beginpunt ( 1 π , 1). 3
⇒ h '(x ) = −1 ⋅ ( cos(x ) )
−2
x
= 5 π + 1 ⋅ 2π = 1 5 π . 6 2 6
36b
Horizontale raaklijnen in de toppen bij x
37a
f (x ) = −2 + 2 sin(3x − 21 π ) = −2 + 2 sin(3(x − 61 π )) heeft evenwichtsstand − 2; amplitude 2; periode 2π = 2 π en beginpunt ( 1 π , − 2). 3 3 6 Hoogste punten zijn ( 1 π + 1 ⋅ 2 π + k ⋅ 2 π , − 2 + 2) = ( 1 π + k ⋅ 2 π , 0). 6
4 3
3
3
3
6
4 3
3
3
3
Laagste punten zijn ( 1 π + 3 ⋅ 2 π + k ⋅ 2 π , − 2 − 2) = ( 2 π + k ⋅ 2 π , − 4). 3
3
3
sin( x ) . cos2 ( x )
f
= 1 π + 1 ⋅ 2π = 5 π en 3 6 4
De toppen zijn ( 1 π + k ⋅ 2 π , 0) en (k ⋅ 2 π , − 4).
⋅ − sin(x ) =
G&R vwo B deel 3 C. von Schwartzenberg 37b
11 Goniometrie en beweging 7/22
g (x ) = 2 + cos( 31 x + 21 π ) = 2 + cos( 31 (x + 23 π )) heeft evenwichtsstand 2; amplitude 1; periode 21π = 6π en beginpunt (hoogste punt) ( −1 1 π , 2 + 1) = ( −1 1 π , 3). 2 2 3
Hoogste punten zijn ( −1 1 π + k ⋅ 6π , 2 + 1) = ( −1 1 π + k ⋅ 6π , 3). 2
2
Laagste punten zijn ( −1 1 π + 1 ⋅ 6π + k ⋅ 6π , 2 − 1) = (1 1 π + k ⋅ 6π , 1).
37c
37d
2 2 2 1 h (x ) = 1 − 3 sin(x + 6 π ) heeft evenwichtsstand 1; amplitude 3; periode 21π = 2π en beginpunt ( − 61 π + π , 1) = ( 56 π , 1). Hoogste punten zijn ( 5 π + 1 ⋅ 2π + k ⋅ 2π , 1 + 3) = (1 1 π + k ⋅ 2π , 4). beginpunt bij een sinus-grafiekis een punt waar de 6 3 4 grafiek STIJGEND door de evenwichtsstand gaat Laagste punten zijn ( 5 π + 3 ⋅ 2π + k ⋅ 2π , 1 − 3) = ( 1 π + k ⋅ 2π , − 2). 3 6 4
j (x ) = −2 − cos(2x ) heeft evenwichtsstand − 2; ampl. 1; periode 22π = π en beginpunt (0 + 21 π , − 2 + 1) = ( 21 π , − 1).
Hoogste punten zijn ( 1 π + k ⋅ π , − 2 + 1) = ( 1 π + k ⋅ π , − 1). 2
beginpunt van een cosinusgrafiek is een hoogste punt
2
Laagste punten zijn ( 1 π + 1 ⋅ π + k ⋅ π , − 2 − 1) = (k ⋅ π , − 3). 2
38a
2
f (x ) = cos( 2x ) − 2 sin(x ) + 2 ⇒ f '(x ) = −2 sin(2x ) − 2 cos(x ). f '(x ) = 0 ⇒ −2 sin(2x ) − 2 cos(x ) = 0 sin(2x ) = − cos(x ) cos(2x − 1 π ) = cos(x + π ) 2 2x − 1 π = x + π + k ⋅ 2π ∨ 2x − 1 π = −x − π + k ⋅ 2π 2
2
x = 1 21 π + k ⋅ 2π ∨ 3x = − 21 π + k ⋅ 2π x = 1 21 π + k ⋅ 2π ∨ x = − 61 π + k ⋅ 32 π . Dus x A = 1 π ; xB = 1 1 π ; xC = 1 1 π ; en xD = 1 5 π . 2
6
2
6
yA = f ( 21 π ) = cos(π ) − 2 sin( 21 π ) + 2 = −1 − 2 + 2 = −1 ⇒ A ( 21 π , − 1);
yB = f (1 61 π ) = cos(2 31 π ) − 2 sin(1 61 π ) + 2 = 21 − 2 ⋅ − 21 + 2 = 21 + 1 + 2 = 3 21 ⇒ B (1 61 π , 3 21 ); yC = f (1 21 π ) = cos(3π ) − 2 sin(1 21 π ) + 2 = −1 − 2 ⋅ −1 + 2 = −1 + 2 + 2 = 3 ⇒ C (1 21 π , 3) en
yD = f (1 56 π ) = cos(3 23 π ) − 2 sin(1 56 π ) + 2 = 21 − 2 ⋅ − 21 + 2 = 21 + 1 + 2 = 3 21 ⇒ D (1 56 π , 3 21 ). 38b
f (0) = f (2π ) = cos(0) − 2 sin(0) + 2 = 1 − 0 + 2 = 3 Dus f (x ) = p heeft vier oplossingen (zie ook figuur 11.14) voor 3 ≤ p < 3 1 . 2
39a
f (x ) = 21 x + cos(x ) ⇒ f '(x ) = 21 − sin(x ).
f '(x ) = 0 ⇒ 21 − sin(x ) = 0
f '(x ) = 1 ⇒ 21 − sin(x ) = 1
− sin(x ) = 1 ⇒ sin(x ) = − 1 2
2
x = 1 61 π + k ⋅ 2π ∨ x = π − 1 61 π + k ⋅ 2π x op [0, 7] ⇒ x = 1 61 π ∨ x = 1 56 π .
− sin(x ) = − 1 ⇒ sin(x ) = 1 2
39b
2
x = 61 π + k ⋅ 2π ∨ x = π − 61 π + k ⋅ 2π x = 61 π + k ⋅ 2π ∨ x = 56 π + k ⋅ 2π .
x op [0, 7] ⇒ x = 61 π ∨ x = 56 π ∨ x = 2 61 π .
40
f (x ) = cos3(x ) = cos(x )
(
3
)
⇒ f '(x ) = 3cos2 (x ) ⋅ − sin(x ) = −3 sin(x ) cos2 (x ).
f '(x ) = 0 ⇒ −3sin(x ) cos2 (x ) = 0 sin(x ) = 0 ∨ cos(x ) = 0 x = k ⋅ π ∨ x = 21 π + k ⋅ π
x op [0, 2π ] ⇒ x = 0 ∨ x = 21 π ∨ x = π ∨ x = 1 21 π ∨ x = 2π . De punten zijn (0, 1); ( 1 π , 0); (π , − 1); (1 1 π , 0) en (2π , 1). 2 2
41a
(2 − sin(x ) ) ⋅ −3sin(x ) − 3cos(x ) ⋅ −cos(x ) −6 sin(x ) + 3sin2 (x ) + 3cos2 (x ) −6 sin(x ) + 3 3cos(x ) ⇒ f '(x ) = = = . 2 − sin( x ) (2 − sin(x ) )2 (2 − sin(x ) )2 (2 − sin(x ) )2 3cos(x ) f (x ) = 0 ⇒ = 0 (teller = 0) ⇒ cos(x ) = 0 ⇒ x = 1 π + k ⋅ π . Nu x op [0, 2π ] ⇒ x = 1 π ∨ x = 1 1 π . 2 2 2 2 − sin( x ) x = 21 π (en y = 0) ⇒ S 1 ( 21 π , 0) y = −3x + b 1 1 1 −6 sin( 21 π ) + 3 ⇒ door S ( 1 π , 0) ⇒ 0 = −3 ⋅ 2 π + b ⇒ b = 1 2 π , dus k : y = −3x + 1 2 π . 1 6 1 3 3 − ⋅ + − rc = f '( π ) = = = 1 2 2 2 1 (2 − 1 )2 2 − sin( 1 π ) f (x ) =
(
2
)
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 8/22
y =x +b 1 1 1 +3 −6 ⋅ −1 + 3 = 9 ⇒ door S (1 1 π , 0) ⇒ 0 = 1 2 π + b ⇒ b = −1 2 π , dus l : y = x − 1 2 π . rc = f '(1 1 π ) = = 2 2 2 2 2 9 (2 − sin(1 21 π ) ) (2 − −1)
x = 1 21 π (en y = 0) ⇒ S 2 (1 21 π , 0) −6 sin(1 21 π )
41b
f '(x ) =
−6 sin(x ) + 3
(2 − sin(x ) )2
= 0 (teller = 0) ⇒ −6 sin(x ) + 3 = 0 ⇒ −6 sin(x ) = −3 ⇒ sin(x ) = 1 ⇒ x = 1 π ∨ x = 5 π (op [0,2π ]). 2
6
6
1 ⇒ B = [− 3, 3]. maximum (zie plot) is f ( 1 π ) = = = = 3 > 1 f 6 2 2 − sin( 61 π ) 2 − 21 1 21 5 1 1 − − 3cos( ) 3 ⋅ 3 1 3 π 6 2 minimum (zie plot) is f ( 5 π ) = = = 21 = − 3. 6 2 − 21 12 2 − sin( 56 π ) 3cos(2π ) = 3 ⋅1 = 1 1 2 2 − sin(2π ) 2 − 0 3cos( 61 π ) 3 ⋅ 21 3 1 21 3
randmaximum (zie plot) is f (2π ) =
42a
F (x ) = − 31 cos( 3x ) ⇒ F '(x ) = − 31 ⋅ − sin(3x ) ⋅ 3 = sin(3x ) = f (x ).
42b
G (x ) = 51 sin( 5x ) ⇒ G '(x ) = 51 ⋅ cos(5x ) ⋅ 5 = cos(5x ) = g (x ).
43a
f (x ) = 4 sin( 31 x ) ⇒ F (x ) = 4 ⋅ 11 ⋅ − cos( 31 x ) + c = −12cos( 31 x ) + c .
43b
g (x ) = x 2 − 5 cos( 2x ) ⇒ G (x ) = 31 x 3 − 5 ⋅ 21 ⋅ sin(2x ) + c = 31 x 3 − 25 sin(2x ) + c .
43c
h (x ) = sin( 2x + 31 π ) ⇒ H (x ) = 21 ⋅ − cos(2x + 31 π ) + c = − 21 cos(2x + 31 π ) + c .
43d
j (x ) = 3cos( 21 x − 61 π ) ⇒ J (x ) = 3 ⋅ 11 ⋅ sin( 21 x − 61 π ) + c = 6 sin( 21 x − 61 π ) + c .
3
2
1 π 3
44a
∫ (2x + cos( 21 x ) ) dx
0
1 π 3
44b
∫
1 π 6
(
1
π
3 = x 2 + 2 sin( 1 x ) = ( 1 π )2 + 2 sin( 1 π ) − 02 + 2 ⋅ 0 = 1 π 2 + 2 ⋅ 1 = 1 π 2 + 1. 2 3 6 9 2 9 0
(
1
π x 2 − 2 sin( x − 61 π ) dx = 31 x 3 + 2cos(x − 61 π ) 31 = 31 ⋅ ( 31 π )3 + 2cos( 61 π ) − 31 ⋅ ( 61 π )3 + 2 cos(0) π 6
)
(
= 1 π3 + 2⋅ 1 3 − 81
45
)
2
7 π3 + ( 6481 π 3 + 2 ⋅ 1) = 648
)
3 − 2.
f (x ) = 1 + 2cos( 21 x − 56 π ) = 0 ⇒ 2cos( 21 x − 56 π ) = −1 ⇒ cos( 21 x − 56 π ) = − 21 ⇒ 1 x − 5 π = 2 π + k ⋅ 2π ∨ 1 x − 5 π = − 2 π + k ⋅ 2π ⇒ 1 x = 9 π + k ⋅ 2π ∨ 1 x = 1 π + k ⋅ 2π ⇒ 2 6 3 2 6 3 2 6 2 6
V
x = 3π + k ⋅ 4π ∨ x = 31 π + k ⋅ 4π . Er geldt: x op [0,4π ] ⇒ x = 3π ∨ x = 31 π . 3π
O (V ) =
∫
1 π 3
(1 + 2 cos(
1 x − 5 π ) dx = x + 4 sin( 1 x − 5 π ) 3π = 3π + 4 sin( 2 π ) − 1 π + 4 sin( − 2 π ) 2 6 2 6 1π 3 3 3
)
(
)
3
(3
)
= 3π + 4 ⋅ 1 3) − 1 π + 4 ⋅ − 1 3 = 2 2 π + 4 3. 2
46a
g (x ) = 31 sin3(x ) = 31 sin(x )
(
3
)
2
2
⇒ g '(x ) = 1 ⋅ 3 sin2 (x ) ⋅ cos(x ) ≠ f (x ). Dus g (x ) = 1 sin3(x ) is geen primitieve van f . 3
3
2
2
46bc
cos(2A ) = 1 − 2 sin (A ) ⇒ 2 sin (A) = 1 − cos(2A) ⇒ sin (A) = 1 − 1 cos(2A ). 2 2 f (x ) = sin2 (x ) = 21 − 21 cos( 2x ) ⇒ F (x ) = 21 x − 21 ⋅ 21 ⋅ sin(2x ) + c = 21 x − 41 sin(2x ) + c .
47a
cos(2A ) = 2cos2 (A ) − 1 ⇒ cos(2A ) + 1 = 2cos2 (A ) ⇒ 1 cos(2A ) + 1 = cos2 (A ). 2
2
f (x ) = cos2 (x ) = 21 cos( 2x ) + 21 ⇒ F (x ) = 21 ⋅ 21 ⋅ sin(2x ) + 21 x + c = 41 sin(2x ) + 21 x + c . 47b cos(2A ) = 1 − 2 sin2 (A ) ⇒ 2 sin2 (A) = 1 − cos(2A) ⇒ sin2 (A) = 1 − 1 cos(2A ). 2 2
1 sin(6x ) + c . g (x ) = sin2 (3x ) = 21 − 21 cos( 6x ) ⇒ G (x ) = 21 x − 21 ⋅ 61 ⋅ sin(6x ) + c = 21 x − 12
47c
sin(2A ) = 2 sin(A ) cos(A ) ⇒ 1 sin(2A) = sin(A) cos(A ). 2
3
h (x ) = sin( 21 x ) cos( 21 x ) = 21 sin(x ) ⇒ H (x ) = 21 ⋅ − cos(x ) + c = − 21 cos(x ) + c .
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 9/22
(
)
48a
f (x ) = tan2 (x ) = 1 + tan2 (x ) − 1 ⇒ F (x ) = tan(x ) − x + c .
48b
g (x ) = x + tan2 (x ) = x + 1 + tan2 (x ) − 1 ⇒ G (x ) = 21 x 2 + tan(x ) − x + c .
49a
sin(2A) = 2 sin(A) cos(A) ⇒ 1 sin(2A ) = sin(A) cos(A ).
(
1 π 6
∫
1 π 6
∫
sin(2x ) cos(2x ) dx =
0
49b
0
)
2
1
1 sin( 4x ) dx = − 1 cos(4x ) 6 π = − 1 cos( 2 π ) − − 1 cos(0) = − 1 ⋅ − 1 + 1 ⋅ 1 = 1 + 1 = 3 . 2 8 3 8 8 2 8 16 8 16 8 0
cos(2A ) = 1 − 2 sin2 (A ) ⇒ cos(2A ) − 1 = −2 sin2 (A ) ⇒ 1 cos(2A ) − 1 = − 1 sin2 (A). π
∫ (2 − 21 sin
2
)
( x ) dx =
1 π 3
4
π
∫ ( 41 cos( 2x ) + 1 43 ) dx
1 π 3
2
4
π
= 1 sin(2x ) + 1 3 x 1 = 1 sin(2π ) + 1 3 π − 1 sin( 2 π ) + 1 3 ⋅ 1 π 8 8 3 4 π 4 4 3 8 3
(8
(
)
= 1 ⋅0 +13π − 1 ⋅ 1 3 + 7 π =13π − 1 8
50
4
2
12
4
16
3 − 7 π = 7π − 1 12
6
)
3.
16
cos(2A ) = 1 − 2 sin2 (A ) ⇒ 2 sin2 (A) = 1 − cos(2A ) ⇒ sin2 (A ) = 1 − 1 cos(2A ). 1 π 2
I (L ) =
2
∫ π ⋅ (f (x ) )
1 π 2
dx =
0
∫ π ⋅ sin
0
1
2
2
1 π 2
(2x ) dx =
∫π
0
π
2
(
V
)
⋅ 1 − 1 cos( 4x ) dx 2 2
2 = π ⋅ 1 x − 1 sin(4x ) = π ⋅ 1 π − 1 sin(2π ) − π ⋅ 0 − 1 sin(0) = π ⋅ 1 π − 0 − π ⋅ ( 0 − 0 ) = 1 π 2 . 2 8 8 8 4 4 4 0
(
51a
(
f (x ) = 2 sin(x )
)
2
)
(
)
(
(
)
)
+ sin(x ) − 1 ⇒ f '(x ) = 4 sin(x ) ⋅ cos(x ) + cos(x ).
f '(x ) = 0 ⇒ 4 sin(x ) ⋅ cos(x ) + cos(x ) = 0 ⇒ cos(x ) ⋅ ( 4 sin(x ) + 1 ) = 0 cos(x ) = 0 ∨ 4 sin(x ) = −1 ⇒ x = 1 π + k ⋅ π ∨ sin(x ) = − 1 ⇒ x = 1 π ∨ x = 1 1 π ∨ sin(x ) = − 1 . 2 2 2 4 4 x = 21 π ⇒ f ( 21 π ) = 2 sin2 ( 21 π ) + sin( 21 π ) − 1 = 2 ⋅ 12 + 1 − 1 = 2 + 0 = 2. x = 1 21 π ⇒ f (1 21 π ) = 2 sin2 (1 21 π ) + sin(1 21 π ) − 1 = 2 ⋅ ( −1)2 + −1 − 1 = 2 − 2 = 0. sin(x ) = − 1 ⇒ f (x ) = 2 ⋅ ( − 1 )2 + − 1 − 1 = 2 ⋅ 1 − 1 − 1 = 1 − 1 − 1 = 1 − 2 − 1 = −1 1 . 16 4 8 4 8 8 8 4 4 4
Randextreem: f (0) = f (2π ) = 2 ⋅ 02 + 0 − 1 = −1. Dus Bf = [ −1 1 , 2]. 8 51b
f (x ) = 0 ⇒ 2 sin2 (x ) + sin(x ) − 1 = 0 (stel sin(x ) = t ) 2t 2 + t − 1 = 0 ⇒ D = b 2 − 4 ⋅ a ⋅ c = 12 − 4 ⋅ 2 ⋅ −1 = 1 + 8 = 9 ⇒ D = 3.
t = sin(x ) = −21 ⋅±23 ⇒ t = sin(x ) = −1 ∨ t = sin(x ) = 21 (met x op [0, 2π ]) ⇒ x = 1 21 π (zoeken we niet) ∨ x = 61 π ∨ x = 56 π . cos(2A ) = 1 − 2 sin2 (A ) ⇒ 2 sin2 (A ) − 1 = − cos(2A ). 5 π 6
5 π 6
∫ (2 sin (x ) + sin(x ) − 1 ) dx = ∫ ( − cos( 2x ) + sin(x ) ) dx = − 21 sin(2x ) − cos(x ) 2
1 π 6
1 π 6
(
)
5 π 6 1 π 6
(
= − 1 sin( 5 π ) − cos( 5 π ) − − 1 sin( 1 π ) − cos( 1 π ) = − 1 ⋅ − 1 3 − − 1 3 − − 1 ⋅ 1 3 − 1 3 2
3
6
2
3
6
2
2
= 1 4 52
2
3+1 2
3+ 1 4
1 = 0 ⇒ sin(x ) + 1 2 = 0 ⇒ sin(x ) = − 1 ⇒ x = 1 1 π ∨ x = 1 5 π . 2 2 6 6 4 2 1 1 1 5 f (0) = 0 + 0 + 4 = 4 > 0 ⇒ het ingesloten gebied loopt van x = 1 6 π tot x = 1 6 π . cos(2A) = 1 − 2 sin2 (A) ⇒ 2 sin2 (A ) = 1 − cos(2A ) ⇒ sin2 (A) = 1 − 1 cos(2A ). 2 2 1 56 π 1 56 π 1 5π 1 − 1 cos( 2x ) + sin(x ) + 1 dx = 3 x − 1 sin(2x ) − cos(x ) 6 sin2 (x ) + sin(x ) + 1 dx = 2 2 4 4 4 4 11π 6 1 61 π 1 61 π
f (x ) = 0 ⇒ sin2 (x ) + sin(x ) +
∫
(
)
(
∫
)
(
)
(
= 3 ⋅ 1 5 π − 1 sin(3 2 π ) − cos(1 5 π ) − 3 ⋅ 1 1 π − 1 sin(2 1 π ) − cos(1 1 π ) 4 6 = 33 π − 24 = 33 π + 24
4 3 6 4 6 4 3 1 ⋅ − 1 3 − 1 3 − 21 π − 1 ⋅ 1 3 − − 1 3 2 2 2 4 24 4 2 1 3 − 1 3 − 21 π + 1 3 − 1 3 = 12 π + 2 3 − 8 2 24 8 2 24 8
(
6
)
)
3 = 1 π − 3 3. 2
4
)
2 2 2 3 + 1 3 = 1 1 3. 2 2
G&R vwo B deel 3 C. von Schwartzenberg 53a
11 Goniometrie en beweging 10/22
f (x ) = 1 21 − 3sin( 21 x ) heeft
evenwichtsstand 1 1 ; amplitude 3; periode 21π = 4π en beginpunt (2π , 1 1 ). 2
53b
2
2
Zie de grafiek van f (x ) = 1 1 − 3sin( 1 x ) hiernaast. 2 2 1 1 f (x ) = 0 ⇒ 1 2 − 3sin( 2 x ) = 0 ⇒ −3sin( 21 x ) = −1 21 ⇒ sin( 21 x ) = 21 ⇒ 1 x = 1 π + k ⋅ 2π ∨ 1 x = 5 π + k ⋅ 2π ⇒ x = 1 π + k ⋅ 4π ∨ x = 5 π + k ⋅ 4π . 2 6 2 6 3 3 x [0, 4π ] ⇒ x = 31 π ∨ x = 35 π . (het gevraagde gebied ligt ONDER de x -as) 5 π 3
O (V ) =
∫
5 π 3
−f (x ) dx =
1 π 3
5
π
5 π 3 3 ∫1 −1 21 + 3sin( 21 x ) dx = −1 21 x + 321 ⋅ − cos( 21 x ) 1 = −1 21 x − 6cos( 21 x ) 31 π π π 3 3
(
)
(
)
(
= −1 1 ⋅ 5 π − 6cos( 1 ⋅ 5 π ) − −1 1 ⋅ 1 π − 6cos( 1 ⋅ 1 π ) = − 5 π − 6cos( 5 π ) − − 1 π − 6cos( 1 π ) 2 3 2 3 2 3 2 3 2 6 2 = −2 1 π − 6 ⋅ − 1 3 − − 1 π − 6 ⋅ 1 3 = −2 1 π + 3 3 + 1 π + 3 3 = 6 3 − 2π . 2 2 2 2 2 2
(
5 π 3
53c
I (L ) =
2
∫ π ⋅ (f (x ) )
5 π 3
dx =
1 π 3
∫π
1 π 3
6
)
)
2 ⋅ 1 1 − 3sin( 1 x ) dx = 2 2
(
)
5 π 3
∫ π ⋅ (2 41 − 9 sin( 21 x ) + 9 sin
1 π 3
)
2 1 ( x ) dx 2
Nu is: cos(2A) = 1 − 2 sin2 (A) ⇒ 2 sin2 (A ) = 1 − cos(2A) ⇒ sin2 (A) = 1 − 1 cos(2A). 2
2
5 π 3
=
∫ π ⋅ (2 41 − 9 sin( 21 x ) + 9 ⋅ ( 21 − 21 cos(x ) ) ) dx
1 π 3
5
π
5
π 3 3 = π ⋅ 2 1 x − 91 ⋅ − cos( 1 x ) + 9 x − 9 sin(x ) = π ⋅ 27 x + 18cos( 1 x ) − 9 sin(x ) 2 2 2 2 2 1 4 4 π 2 1 π 3
(
)
3
(4 3 = π ⋅ ( 45 π + 18 ⋅ − 1 2 4
) (
(
= π ⋅ 27 ⋅ 5 π + 18cos( 5 π ) − 9 sin( 5 π ) − π ⋅ 27 ⋅ 1 π + 18cos( 1 π ) − 9 sin( 1 π ) = 11 1 π 2 − 9π 4 54a
6
2 3 6 2 4 3 3 − 9 ⋅ − 1 3 − π ⋅ 9 π + 18 ⋅ 1 3 − 9 ⋅ 1 3 2 2 2 2 2 4 3 + 2 1 π 3 − 2 1 π 2 − 9π 3 + 2 1 π 3 = 9π 2 − 13 1 π 3. 2 4 4 4
)
y P = sin(ct ) en de periode is 5 = 2cπ ⇒ c = 25π . Of voor t = 5 is y P = sin( 2π ⋅ 5) = sin(2π ) 5
3
)
)
54b
Formule II: x P = cos( 2π t ). 5
(de sinus heeft dan precies één periode doorlopen)
55
rcy =−x +3 = −1 ⇒ ∠AMB = 45° ⇒ (∆AMB is een 1-1- 2 driehoek) AB = MB .
AM = 4 ⇒ AB = MB = 4 = 4 ⋅ 2 = 422 = 2 2. 2 2 2 xA = x M − AB = 2 − 2 2 en yA = y M + AB = 1 + 2 2. 56a
t = 0 ⇒ P (1, 3). P draait linksom. t op [0, 1 21 π ] ⇒ driekwartcirkel. De baan van P is driekwartcirkel met middelpunt M ( −1, 3) en straal 2.
56b
x = 0 ⇒ −1 + 2 cos(t ) = 0 ⇒ 2cos(t ) = 1 ⇒ cos(t ) = 21 (t op [0, 1 21 π ]) ⇒ t = 31 π . t = 31 π ⇒ yA = 3 + 2 sin( 31 π ) = 3 + 2 ⋅ 21 3 = 3 + 3 ⇒ A(0, 3 + 3).
56c
rcy =x + 4 = 1 ⇒ bij B hoort t = 1 π en bij C hoort t = 1 1 π .
t t 56d
4 4 = 1 π ⇒ xB = −1 + 2cos( 1 π ) = −1 + 2 en yB = 3 + 2 sin( 1 π ) = 3 + 2. 4 4 4 = 1 1 π ⇒ xC = −1 + 2cos(1 1 π ) = −1 − 2 en yC = 3 + 2 sin(1 1 π ) = 3 − 2. 4 4 4
x = −2 ⇒ −1 + 2cos(t ) = −2 (intersect of) ⇒ 2cos(t ) = −1 ⇒ cos(t ) = − 1 ⇒ t = 2 π + k ⋅ 2π ∨ t = − 2 π + k ⋅ 2π . 2 3 3 t op [0, 1 1 π ] ⇒ t = 23 π ≈ 2, 09 ∨ t = 34 π ≈ 4,19. 2 x < −2 (zie driekwartcirkel of plot) ⇒ 2, 09 < t < 4,19.
y =x +4
t = 21 π A B
t =π
M
C
t = 1 21 π
t =0
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 11/22
57a
x P = 5 + 3cos(2t ) en yP = 2 + 3sin(2t ). (met t in seconden)
57b
De eerste rondgang van t = 0 tot t = π . y = 0 (x -as) ⇒ 2 + 3sin(2t ) = 0 (intersect) ⇒ t ≈ 1, 94 ∨ t ≈ 2, 78 . y < 0 (zie plot) ⇒ 1, 94 < t < 2, 78. Dus 0,84 seconden per rondgang.
58a
t = 0 ⇒ P (1 21 , − 3).
x = −1 21
t op [0, 34 π ] ⇒ 2t op [0, 1 21 π ].
P draait linksom.
A
De baan van P is driekwartcirkel met middelpunt M ( − 1 , − 3) en straal 2.
t = 41 π
2
58b
y = 0 (x -as) ⇒ − 3 + 2 sin(2t ) = 0 ⇒ 2 sin(2t ) = 3 ⇒ sin(2t ) = 21 3 ⇒
2t = 1 π + k ⋅ 2π ∨ 2t = π − 1 π + k ⋅ 2π 3
3
t = 21 π
t = 61 π + k ⋅ π ∨ t = 31 π (deze zoeken we) + k ⋅ π . xA = − 21 + 2cos( 23 π ) = − 21 + 2 ⋅ − 21 = −1 21 ⇒ A( −1 21 , 0). 58c
x = −1 21 ⇒ − 21 + 2cos(2t ) = −1 21 ⇒ 2cos(2t ) = −1 ⇒ cos(2t ) = − 21 ⇒
2t = 2 π + k ⋅ 2π ∨ 2t = − 2 π + k ⋅ 2π ⇒ t = 1 π + k ⋅ π ∨ t = − 1 π + k ⋅ π . 3
58d
3
3
3 < −1 1 (zie de baan bij 58a) ⇒ 1 π < t < 2 π . 2 3 3
M
y = −2 t = 0
t = 34 π
t op [0, 34 π ] ⇒ t = 31 π ∨ t = 32 π . Dus x y = −2 ⇒ − 3 + 2 sin(2t ) = −2 (intersect) ⇒ t ≈ 1, 64. Dus y < −2 (zie 58a of de plot) ⇒ 1, 64 < t ≤ 3 π . 4
59a
y = x + 1 ⇒ 2 sin(t ) = 2cos(t ) + 1 (met 0 ≤ t < 2π ) intersect geeft dan t ≈ 1,15 en y ≈ 1, 82 ⇒ x = y − 1 = 0, 82 ⇒ snijpunt (0, 82; 1, 82) of t ≈ 3, 57 en y ≈ −0,82 ⇒ x = y − 1 = −1, 82 ⇒ snijpunt ( −1, 82; − 0,82).
59b
x = 1 ⇒ 2cos(t ) = 1 ⇒ cos(t ) = 21 ⇒ t = 31 π + k ⋅ 2π ∨ t = − 31 π + k ⋅ 2π . Er geldt nu: x > 1 voor − 1 π < t < 1 π . De baan van P is een cirkel met middelpunt (0, 0) en straal 2. 3 3 Dus ligt
2 π 3
2π
= 1 deel van de cirkel rechts van de lijn x = 1. (omtrek van een cirkel = 2π r ) 3
De lengte van het deel rechts van de lijn x = 1 is 1 ⋅ 2π ⋅ 2 = 4 π . 3
59c
3
P (2cos(t ), 2 sin(t )) en Q (cos(2t ), sin(2t )). Nu de stelling van Pythagoras: 2
2
PQ 2 = ( cos(2t ) − 2cos(t ) ) + ( sin(2t ) − 2 sin(t ) )
= cos2 (2t ) − 4 cos(2t ) cos(t ) + 4 cos(t )2 + sin2 (2t ) − 4 sin(2t ) sin(t ) + 4 sin2 (t ) = cos2 (2t ) + sin2 (2t ) + 4 cos(t )2 + 4 sin2 (t ) − 4 cos(2t ) cos(t ) − 4 sin(2t ) sin(t )
(
)
= cos2 (2t ) + sin2 (2t ) + 4 ⋅ cos(t )2 + sin2 (t ) − 4 ⋅ ( cos(2t ) cos(t ) + sin(2t ) sin(t ) ) = 1 + 4 ⋅ 1 − 4 ⋅ cos(2t − t ) = 5 − 4 cos(t ). 59d
PQ = PQ =
Dus PQ = 5 − 4 cos(t ).
5 − 4 cos(t ) = 1 1 (intersect) ⇒ t ≈ 0,81 ∨ t ≈ 5, 47. 2 5 − 4 cos(t ) > 1 1 (zie plot) ⇒ 0,81 < t < 5, 47. 2
Dus gedurende 4,66 seconde per rondgang. 60a
De translatie (2, 0).
61a
De omlooptijd T = 2π1 = 4π = 4 π seconden. 22
60b
5
xQ = 3cos( 21 (t − 2)) en yQ = 3sin( 21 (t − 2)). (met t in seconden)
5
Na 1 ⋅ 4 π = 1 π seconde (voor t = 51 π ) bevindt P zich in (7, − 2). 4 5
t = 51 π
5
Dus xP = 4 + 3cos(2 1 (t − 1 π )) en yP = −2 + 3sin(2 1 (t − 1 π )). (met t in seconden) 2 5 2 5 61b
t = 2 ⇒ xP = 4 + 3cos(2 21 (2 − 51 π )) ≈ 1,12 en yP = −2 + 3sin(2 21 (2 − 51 π )) ≈ −2,85.
61c
Na 3 ⋅ 4 π = 3 π seconde (voor t = 35 π ) bevindt P zich voor het eerst in (1, − 2). 4 5
5
t =0
G&R vwo B deel 3 C. von Schwartzenberg 61d
11 Goniometrie en beweging 12/22
(
)
y = 0 (x -as) ⇒ −2 + 3sin 2 21 (t − 51 π ) = 0 (intersect) t ≈ 0,92 ⇒ x P = 4 + 3cos 2 21 (t − 51 π ) ≈ 6,24 ⇒ snijpunt met x -as (6,24; 0).
( ) 1 1 t ≈ 1,59 ⇒ xP = 4 + 3cos (2 2 (t − 5 π ) ) ≈ 1, 76 ⇒ snijpunt met x -as (1,76; 0).
62a
62b
62c
x = 15 + 6cos 4π (t − 1 ) P 10 (t in seconden). 1) y π t = 23 + 6 sin 4 ( − P 10
( ) ω = 2π = 41π = 4π ( ) (15, 23) x = 15 + 6cos ( 4π (t + 1 − 1 ) ) = 15 + 6cos ( 4π (t + 1 ) ) Q 5 10 10 (t in seconden). 1 1 1 yQ = 23 + 6 sin ( 4π (t + 5 − 10 ) ) = 23 + 6 sin ( 4π (t + 10 ) ) x = 15 + 6cos ( 4πt − π ) = 15 + 6cos ( 4π (t − 1 ) ) = 15 + 6cos ( 4π (t − 1 − 3 ) ) R 10 20 4 (t in seconden). 1 1 3 yR = 23 + 6sin ( 4πt − π ) = 23 + 6 sin ( 4π (t − 4 ) ) = 23 + 6sin ( 4π (t − 10 − 20 ) ) 1 2
Dus R loopt 3 seconde achter op P of (omdat de omlooptijd 20
63a
1 2
seconde is) 7 seconde voor op 20
t =
Omlooptijd is 2π = 1 π seconde. 4
2
(
)
(
(
)
)
(
)
4
(
(
)
5 1 5 yQ = 2 + 3sin 4 ⋅ (t + 12 ⋅ 2 π ) = 2 + 3sin 4 ⋅ (t + 24 π )
(
(
)
) (t in seconden).
)
( −4, 2)
( −1, 2)
O
Op t = 0 heeft P een fasevoorsprong van 1 op (2, 2).
2 Q heeft een faseachterstand van 41 op P ⇒ fasevoorsprong van Q op (2, 2) is 21 − 41 = 41 . x = −1 + 3cos 4 ⋅ (t + 1 ⋅ 1 π ) = −1 + 3cos 4 ⋅ (t + 1 π ) Q 8 4 2 (t in seconden). 1 1 1 yQ = 2 + 3sin 4 ⋅ (t + 4 ⋅ 2 π ) = 2 + 3sin 4 ⋅ (t + 8 π )
(
)
(
(
)
)
(
)
2
1
π π 3 = 1. 64b De faseachterstand van R op P is 2 = 1 . 2π 3 2π 4 7 7 5 1 1 1 is + = ( > 2 ) ⇒ faseverschil tussen Q en R is 1 − = . 3 4 12 12 12
64a
De fasevoorsprong van Q op P is
64c
De fasevoorsprong van Q op R
65a
Omlooptijd in stand I is 1 seconde ⇒ ω = 21π = 2π ⋅ 15 = 30π rad/sec. 15
Q heeft een faseachterstand van
1 op 3
15
P
x = 20 cos 30π ⋅ (t − 1 ⋅ 1 ) = 20 cos 30π ⋅ (t − 1 ) xP = 20 cos(30πt ) Q 3 15 45 en (t in seconden). y = 20 sin(30πt ) 1 ⋅ 1 ) = 20 sin 30π ⋅ (t − 1 ) P 20 sin 30 ( = ⋅ − y t π Q 3 15 45 x = 20 cos 30π ⋅ (t + 1 ) R 45 R heeft een fasevoorsprong van 31 op P , dus (t in seconden). 1 yR = 20 sin 30π ⋅ (t + 45 ) v = 108 km/uur = 30 m/s = 3000 cm/s. Omlooptijd bij II is 2π ⋅ 20 = 4π sec.
( (
(
) )
3 000
300
Dus ω = 2π : 4π = 2π ⋅ 300 = 150 rad/sec. 300
4π
x P = 20 cos(150t ) y = 20 sin(150t ) (t in seconden). P
)
(
( (
65b
( −1, 5)
Op t = 0 heeft P een fasevoorsprong van 1 op (2, 2). 5. Q heeft een fasevoorsprong van 61 op P ⇒ fasevoorsprong van Q op (2, 2) is 41 + 61 = 12 x = −1 + 3cos 4 ⋅ (t + 5 ⋅ 1 π ) = −1 + 3cos 4 ⋅ (t + 5 π ) Q 12 2 24
63c
(21, 23)
P.
1 1 1 xP = −1 + 3cos(4t ) xQ = −1 + 3cos 4 ⋅ (t − 3 ⋅ 2 π ) = −1 + 3cos 4 ⋅ (t − 6 π ) en (t in seconden). y = 2 + 3sin(4t ) 1 1 1 P yQ = 2 + 3sin 4 ⋅ (t − 3 ⋅ 2 π ) = 2 + 3sin 4 ⋅ (t − 6 π )
63b
1 10
)
) )
(2, 2)
G&R vwo B deel 3 C. von Schwartzenberg 66a
66b
11 Goniometrie en beweging 13/22
De diameter van rol II is de helft van rol I, dus de omlooptijd van rol II is de helft van de omlooptijd van rol I. x = 10 cos( − 2π t ) = 10 cos(πt ) xQ = 15 + 5 cos(2πt + π ) P 2 en (t in seconden). 2 π yQ = 5 sin(2πt + π ) y P = 10 sin( − 2 t ) = −10 sin(πt ) Omlooptijd rol II is 1 sec. ⇒ elke seconde loopt 2π ⋅ 5 = 10π cm papier tussen de rollen door. Dat is per uur 10π ⋅ 60 ⋅ 60 = 36 000π cm ≈ 113 097 cm ≈ 1131 m.
67b
x P = −2 + 4 cos( −πt ) y = 1 + 4 sin( −πt ) (t in seconden). P t = 1,2 ⇒ x P = −2 + 4 cos( −1,2π ) ≈ −5,24 en yP = 1 + 4 sin( −1,2π ) ≈ 3,35.
67c
De baan wordt in negatieve richting (met de wijzers van de klok) doorlopen.
67a
1 periode is 1 ⋅ 2 = 1 seconde ⇒ t = 1 , 2 2 4 4
67d
4
M
3
(2, 1) 1 2 N
4
A
−πt = 1 π + k ⋅ 2π ∨ −πt = − 1 π + k ⋅ 2π ⇒ t = − 1 + k ⋅ 2 ∨ t = 1 + k ⋅ 2. 3
2 3
( −2,1)
t = 2 21 en t = 4 21 .
x = 0 (y -as) ⇒ −2 + 4 cos( −πt ) = 0 ⇒ 4 cos( −πt ) = 2 ⇒ cos( −πt ) = 21 ⇒ 3
P
B
O
3
t = − 31 ⇒ y = 1 + 4 sin( 31 π ) = 1 + 4 ⋅ 21 3 = 1 + 2 3 ⇒ P (0, 1 + 2 3). t = 31 ⇒ y = 1 + 4 sin( − 31 π ) = 1 − 4 ⋅ 21 3 = 1 − 2 3 ⇒ Q (0, 1 − 2 3).
( −2, −3)
Alternatieve uitwerking: NP 2 = 42 − 22 = 12 ⇒ NP = 12 = 2 3 ⇒ P (0, 1 + 2 3) en Q (0, 1 − 2 3).
67e
cos ∠AMB = 1 ⇒ ∠AMB ≈ 1,318 (rad) 4
De lengte van de boog onder de x -as is 2 ⋅ Ans ⋅ 2π ⋅ 4 ≈ 10, 54. 2π Alternatieve uitwerking: y = 0 (x -as) ⇒ 1 + 4 sin( −πt ) = 0 (intersect) ⇒ t ≈ 0, 08 ∨ t ≈ 0, 92. y < 0 (zie plot) ⇒ 0, 08 < t < 0, 92. Dus ongeveer 0,84 seconden van de 2 seconden per omwenteling onder de x -as. De lengte van de boog onder de x -as is Ans ⋅ 2π ⋅ 4 ≈ 10,54. 2
68a
π = 1 π rad/min. Op t = 0 zit Frits in het hoogste punt. T = 30 ⇒ ω = 230 15
1 1 1 1 1 1 x Frits = 67 2 cos 15 π ⋅ (t + 4 ⋅ 30) = 67 2 cos( 15 πt + 2 π ) (t in minuten). Voor Frits geldt: 1 1 1 1 1 1 1 1 y Frits = 67 2 + 67 2 sin 15 π ⋅ (t + 4 ⋅ 30) = 67 2 + 67 2 sin( 15 πt + 2 π )
(
)
(
)
68b
Saskia heeft 4 = 1 faseachterstand op Frits.
68c
1 1 1 1 1 1 1 x Saskia = 67 2 cos 15 π ⋅ (t + 4 ⋅ 30 − 8 ⋅ 30) = 67 2 cos 15 π ⋅ (t + 8 ⋅ 30) (t in minuten). 1 1 1 1 1 1 1 1 1 y Saskia = 67 2 + 67 2 sin 15 π ⋅ (t + 4 ⋅ 30 − 8 ⋅ 30) = 67 2 + 67 2 sin 15 π ⋅ (t + 8 ⋅ 30) De omtrek van het reuzenrad wordt afgelegd in 30 minuten
32
8
(
)
(
(
)
)
(
)
Dus 2π ⋅ 67 1 meter in 30 minuten ⇒ 848 m/uur ⇒ de snelheid is ongeveer 0,85 km/uur. 2
68d
1 πt ) = 100 (intersect) ⇒ t ≈ 2, 40 ∨ t ≈ 12, 60 y = 100 ⇒ 67 21 + 67 21 sin( 15 y > 100 (zie plot) ⇒ 2, 40 < t < 12, 60.
Dus gedurende ongeveer 10,2 minuten ≈ 612 seconden boven 100 meter.
35
Alternatieve uitwerking: 32,5 ⇒ α ≈ 1, 068 (rad). 67,5 Dus gedurende 2α ⋅ 30 ≈ 10,2 min. ≈ 612 sec. boven 100 meter. 2π
32, 5 α
cos(α ) =
67, 5
O
67, 5
Q
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 14/22
Diagnostische toets D1a
− cos(3x − 1 π ) = cos(3x − 1 π + π ) = cos(3x + 3 π ) = sin(3x + 3 π + 1 π ) = sin(3x + 1 1 π ). 4
4
2
4
4
2
2
2
4
D1b
( sin(x ) + cos(x ) ) = sin (x ) + 2 sin(x ) cos(x ) + cos (x ) = sin (x ) + cos2 (x ) + 2 sin(x ) cos(x ) = 1 + sin(2x ).
D1c
2 + cos(x ) − 2 sin2 (x ) = 2 + cos(x ) − 2 1 − cos2 (x ) = 2 + cos(x ) − 2 + 2cos2 (x ) = 2cos2 (x ) + cos(x ).
D2a
sin(3x − 1 π ) = cos(2x )
(
cos(3x
2
)
D2c cos( 2 πt ) = − sin( 1 πt )
4 − 1 π − 1 π ) = cos(2x ) 2 4
5
5 2 6 2 πt + 1 π = 1 πt + π + k ⋅ 2π ∨ 2 πt + 1 π = π − 1 πt − π + k ⋅ 2π 5 2 6 5 2 6 7 πt = 1 π + k ⋅ 2π ∨ 17 πt = − 1 π + k ⋅ 2π 30 2 30 2 t = 15 + k ⋅ 60 ∨ t = − 15 + k ⋅ 60 7 7 17 17 t op [0, 10] ⇒ t = 15 ∨ t = 45 ∨ t = 105 ∨ t = 165 . 7 17 17 17
3x − 3 π = 2x + k ⋅ 2π ∨ 3x − 3 π = −2x + k ⋅ 2π
x x x
6
sin( 2 πt + 1 π ) = sin( 1 πt + π )
4 4 = 3 π + k ⋅ 2π ∨ 5x = 3 π + k ⋅ 2π 4 4 = 3 π + k ⋅ 2π ∨ x = 3 π + k ⋅ 2 π 20 5 4 op [0, π ] ⇒ x = 3 π ∨ x = 3 π ∨ x = 11 π ∨ x = 19 π . 20 20 20 4 2
D2b 2 sin (2x ) = sin(2x ) + 1
− sin(2x ) = 1 − 2 sin2 (2x ) sin(2x + π ) = cos(4x ) cos(2x + π − 1 π ) = cos(4x ) 2
2x + 1 π = 4x + k ⋅ 2π ∨ 2x + 1 π = −4x + k ⋅ 2π 2
2
−2x = − 1 π + k ⋅ 2π ∨ 6x = − 1 π + k ⋅ 2π 2
2
1 π +k ⋅ 1π x = 41 π + k ⋅ π ∨ x = − 12 3 7 π ∨ x = 11 π ∨ x = 19 π ∨ x = 23 π . x op [0, 2π ] ⇒ x = 41 π ∨ x = 1 41 π ∨ x = 12 12 12 12
D3a
sin(x + 1 π ) = 2 sin(2x ) ⋅ cos(2x ) sin(x
D3b
3 + 1 π ) = sin(4x ) 3
sin2 (2x ) + 1 = cos(4x ) 4
sin2 (2x ) + 1 = 1 − 2 sin2 (2x ) 4
x + 31 π = 4x + k ⋅ 2π ∨ x + 31 π = π − 4x + k ⋅ 2π
3sin2 (2x ) = 3
−3x = − 1 π + k ⋅ 2π ∨ 5x = 2 π + k ⋅ 2π
sin2 (2x ) = 1
2 π + k ⋅ 2π. x = 91 π + k ⋅ 23 π ∨ x = 15 5
sin 2x = ± 1
3
3
4
4
2
2x = 1 π + k ⋅ π ∨ 2x = − 1 π + k ⋅ π 6
6
1 π + k ⋅ 1 π ∨ x = − 1 π + k ⋅ 1 π. x = 12 2 12 2
D4
2sin(x ) cos(x ) 2sin(x ) 1 sin(2x ) 2sin(x ) cos(x ) 2sin(x ) cos(x ) cos2 (x ) cos2 (x ) cos(x ) 2tan(x ) tan(2x ) = = = ⋅ = = = . 2 1 cos(2x ) cos2 (x ) − sin2 (x ) cos2 (x ) − sin2 (x ) cos2 (x ) sin2 (x ) 1 − tan2 (x ) sin( ) x − 2 1 − 2 2 cos (x ) cos (x ) cos (x ) cos(x )
D5
f (1 21 π − p ) + f (1 21 π + p ) = sin(3π − 2 p ) + cos(1 21 π − p ) + sin(3π + 2 p ) + cos(1 21 π + p )
= sin(2 p ) + sin( p ) − sin(2p ) − sin( p ) = 0 ⇒ f is symmetrisch in het punt (1 1 π , 0). 2
D6a f (x ) = cos( 2x ) + sin( 2x ) ⇒ f '(x ) = −2 sin(2x ) + 2cos(2x ). D6b f (x ) = 2cos3(x ) = 2 cos(x )
(
D6c f (x ) =
3
)
2
⇒ f '(x ) = 2 ⋅ 3 ( cos(x ) ) ⋅ − sin(x ) = −6 sin(x ) cos2 (x ).
cos(x ) sin(x ) ⋅ −sin(x ) − cos(x ) ⋅ cos(x ) − sin2 (x ) − cos2 (x ) ⇒ f '(x ) = = = −21 . 2 sin(x ) sin (x ) sin2 (x ) sin (x )
D6d f (x ) = x 2 ⋅ sin( 2x − 1 π ) ⇒ f '(x ) = 2x ⋅ sin(2x − 1 π ) + x 2 ⋅ 2 cos(2x − 1 π ) = 2x sin(2x − 1 π ) + 2x 2 cos(2x − 1 π ). 2
D6e f (x ) = sin(x ) ⋅ tan( 2x
D6f
2
tan( 2x ) f (x ) = ⇒ f '(x ) = sin(x )
OF ..... f (x ) =
2
2
2
2sin(x ) 1 ) ⇒ f '(x ) = cos(x ) ⋅ tan(2x ) + sin(x ) ⋅ ⋅ 2 = cos(x ) ⋅ tan(2x ) + . cos2 (2x ) cos2 (2x )
tan( 2x ) ⇒f sin(x )
2sin(x ) sin(2x ) 1 ⋅ 2 − tan(2x ) ⋅ cos(x ) − ⋅ cos(x ) cos2 (2x ) cos2 (2x ) cos(2x ) cos2 (2x ) = ⋅ 2 2 sin (x ) sin (x ) cos2 (2x ) 2sin(x ) − sin(2x ) cos(2x ) cos(x ) = . sin2 (x ) cos2 (2x ) 2 sin(x ) ⋅ 1 + tan (2x ) ⋅ 2 − tan(2x ) ⋅ cos(x ) 2sin(x ) + 2 sin(x ) tan2 (2x ) − tan(2x ) cos(x ) '(x ) = = . sin2 (x ) sin2 (x )
sin(x ) ⋅
(
)
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 15/22
D7a f (x ) = 3 − 2 sin(x − 1 π ) heeft evenwichtsstand 3; amplitude 2; periode 2π = 2π en beginpunt ( 1 π + π , 3) = (1 1 π , 3). 6
1
Hoogste punten zijn ( 7 π + 1 ⋅ 2π + k ⋅ 2π , 3 + 2) = ( 5 π + k ⋅ 2π , 5). 6
6
6
3
4
Laagste punten zijn ( 5 π + 1 ⋅ 2π + k ⋅ 2π , 3 − 2) = ( 8 π + k ⋅ 2π , 1) = ( 2 π + k ⋅ 2π , 1). 3
D7b
3
2
3
f (x ) = −4 + 3cos(2x − 41 π ) = −4 + 3cos(2(x − 81 π )) heeft evenwichtsstand − 4; amplitude 3; periode 2π = π en beginpunt ( 1 π , − 4 + 3) = ( 1 π , − 1). 2 8 8 Hoogste punten zijn ( 1 π + k ⋅ π , − 4 + 3) = ( 1 π + k ⋅ π , − 1). 8 8 Laagste punten zijn ( 1 π + 1 ⋅ π + k ⋅ π , 3 − 2) = ( 5 π + k ⋅ π , 1). 8 2 8
D8a f (x ) = sin( 2x ) − 2 sin(x ) ⇒ f '(x ) = 2 cos(2x ) − 2 cos(x ). f (π ) = sin(2π ) − 2 sin(π ) = 0 − 2 ⋅ 0 = 0 ⇒ A(π , 0) en rc = f '(π ) = 2 cos(2π ) − 2 cos(π ) = 2 ⋅ 1 − 2 ⋅ −1 = 2 + 2 = 4. y = 4x + b ⇒ 0 = 4π + b ⇒ −4π = b . Dus de raaklijn in A(π , 0) is y = 4x − 4π . door A(π , 0) D8b f '(x ) = 2 cos(2x ) − 2 cos(x ) = 0 ⇒ 2 cos(2x ) = 2 cos(x ) ⇒ cos(2x ) = cos(x ) ⇒ 2x = x + k ⋅ 2π ∨ 2x = −x + k ⋅ 2π ⇒
x = k ⋅ 2π ∨ 3x = k ⋅ 2π ⇒ x = k ⋅ 2π ∨ x = k ⋅ 23 π . Dus (zie ook figuur 11.25) xB = 23 π en xC = 34 π . yB = f ( 23 π ) = sin( 34 π ) − 2 sin( 23 π ) = − 21 3 − 2 ⋅ 21 3 = − 21 3 − 3 = −1 21 3 ⇒ B ( 32 π , − 1 21 3).
yC = f ( 34 π ) = sin( 38 π ) − 2 sin( 34 π ) = 21 3 − 2 ⋅ − 21 3 = 21 3 + 3 = 1 21 3 ⇒ C ( 34 π , 1 21 3). D8c f '(x ) = 2cos(2x ) − 2 cos(x ) = −2 ⇒ cos(2x ) − cos(x ) = −1 ⇒ 2cos2 (x ) − 1 − cos(x ) + 1 = 0 ⇒ 2cos2 (x ) − cos(x ) = 0 ⇒ cos(x ) ⋅ (2cos(x ) − 1 ) = 0 ⇒ cos(x ) = 0 ∨ cos(x ) = 1 ⇒ x = 1 π + k ⋅ π ∨ x = 1 π + k ⋅ 2π ∨ x = − 1 π + k ⋅ 2π . 2 2 3 3
x op [0, 2π ] ⇒ x = 31 π ∨ x = 21 π ∨ x = 1 21 π ∨ x = 1 32 π . f ( 31 π ) = sin( 23 π ) − 2 sin( 31 π ) = 21 3 − 2 ⋅ 21 3 = 21 3 − 3 = − 21 3 ⇒ raakpunt ( 31 π , − 21 3). f ( 21 π ) = sin(π ) − 2 sin( 21 π ) = 0 − 2 ⋅ 1 = −2 ⇒ raakpunt ( 21 π , − 2). f (1 21 π ) = sin(3π ) − 2 sin(1 21 π ) = 0 − 2 ⋅ −1 = 2 ⇒ raakpunt (1 21 π , 2).
f (1 23 π ) = sin(3 31 π ) − 2 sin(1 23 π ) = − 21 3 − 2 ⋅ − 21 3 = − 21 3 + 3 = 21 3 ⇒ raakpunt (1 23 π , 21 3).
D9a f (x ) = − 1 sin( 2x + 1 π ) ⇒ F (x ) = − 1 ⋅ 1 ⋅ − cos(2x + 1 π ) + c = 1 cos(2x + 1 π ) + c . 2
D9b
2
2 2
2
2
4
g (x ) = 3x 2 + cos( 31 x ) ⇒ G (x ) = 3 ⋅ 31 x 3 + 3 ⋅ sin( 31 x ) + c = x 3 + 3sin( 31 x ) + c .
(
)
D9c h (x ) = x − 2 sin2 (x ) = x + 1 − 2 sin2 (x ) − 1 = x + cos( 2x ) − 1 ⇒ H (x ) = 1 x 2 + 1 ⋅ sin(2x ) − x + c .
(
2
2
2
)
2
D9d k (x ) = 2 + tan (x ) = 1 + 1 + tan (x ) ⇒ K (x ) = x + tan(x ) + c . 1 π 3
D10a
∫ ( sin( 2x ) + cos(x ) ) dx
1 π 6
1
π
(
= − 1 cos(2x ) + sin(x ) 31 = − 1 cos( 2 π ) + sin( 1 π ) − − 1 cos( 1 π ) + sin( 1 π ) 2 π 2 3 3 2 3 6
)
6
(
)
= − 1 ⋅ − 1 + 1 3 − − 1 ⋅ 1 + 1 = 1 + 1 3 + 1 − 1 = 1 3. 2
1 π 3
D10b
∫
sin2 (x ) dx =
1 π 6
1 π 3
∫
1 π 6
( 21 − 21 cos( 2x ) ) dx = 21 x − 21 ⋅ 21 sin(2x )
2
2
2 2
2
4
2
4
2
2
1 π 3 1 π 6
cos(2A ) = 1 − 2 sin2 (A ) ⇒ 2 sin2 (A ) = 1 − cos(2A ) ⇒ sin2 (A ) = 1 − 1 cos(2A ) 2 2
(2
)
= 1 ⋅ 1 π − 1 sin( 2 π ) − 1 ⋅ 1 π − 1 sin( 1 π ) = 1 π − 1 ⋅ 1 3 − 1 π + 1 ⋅ 1 3 = 1 π . 2 3
3 π 2
D11
∫
1 π 2
2
π ⋅ (f (x ) ) dx =
4
3 π 2
∫
3
6
4
3
6
4 2
12
4 2
12
3 π 2
2
π ⋅ cos (x ) dx =
1 π 2
∫ ( 21 + 21 cos( 2x ) ) dx = π ⋅ ( 21 x + 21 ⋅ 21 sin(2x ) ) π⋅
1 π 2
3 π 2 1 π 2
cos(2A ) = 2cos2 (A) − 1 ⇒ 1 + cos(2A ) = 2cos2 (A ) ⇒ 1 + 1 cos(2A) = cos2 (A) 2 2 = π ⋅ 1 ⋅ 3 π + 1 sin(3π ) − π ⋅ 1 ⋅ 1 π + 1 sin(π ) = 3 π 2 + 1 π ⋅ 0 − 1 π 2 − 1 π ⋅ 0 = 1 π 2 .
(2
2
4
)
(2
2
4
)
4
4
4
4
2
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 16/22
t =
1 4
π
A
D12a De baan is een driekwartcirkel met middelpunt ( −1, 0) en straal 4. D12b x = 1 ⇒ −1 + 4 cos(2t ) = 1 ⇒ 4 cos(2t ) = 2 ⇒ cos(2t ) = 1 ⇒
t = 21 π
2
Dus de lengte boog CE is
2π
⋅ ( omtrek cirkel ) =
2 π 3
2π
M
t =
3 4
π
⋅ 2π ⋅ 4 = 2 π ⋅ 4 = 8 π . 3
3
D13a De omlooptijd is 3 seconden ⇒ ω = 2π = 2 π . 3
3
xP = 5 + 13cos( 2 π (t − 5)) 3 (met t in seconden). De parametervoorstelling voor de baan van punt P is: 2 yP = 12 + 13sin( 3 π (t − 5)) D13b Punt Q met 1 seconde achterstand op P geeft als parametervoorstelling voor de baan van Q : xQ = 5 + 13cos( 2 π (t − 5 − 1)) xQ = 5 + 13cos( 2 π (t − 6)) 3 3 )⇒ (met t in seconden). 2 2 yQ = 12 + 13sin( 3 π (t − 5 − 1)) yQ = 12 + 13 sin( 3 π (t − 6)) D13c Punt R met een fasevoorsprong van 1 op P geeft een voorsprong van 1 ⋅ 3 = 3 seconde. 4
4
4
De parametervoorstelling voor de baan van R is: xR = 5 + 13cos( 2 π (t − 5 + 3 )) x = 5 + 13cos( 2 π (t − 4 1 )) 3 3 4 ⇒ R 4 (met t in seconden). 2 3 2 1 yR = 12 + 13sin( 3 π (t − 5 + 4 )) yR = 12 + 13 sin( 3 π (t − 4 4 ))
y =2
t =0 D
O
2t = 1 π + k ⋅ 2π ∨ 2t = − 1 π + k ⋅ 2π ⇒ t = 1 π + k ⋅ π ∨ t = − 1 π + k ⋅ π . 3 3 6 6 Dus t = 1 π en yA = 4 sin(2 ⋅ 1 π ) = 4 sin( 1 π ) = 4 ⋅ 1 3 ⇒ A(1, 2 3). 6 6 3 2 D12c In ∆MDC geldt: sin ∠M = CD = 2 = 21 ⇒ ∠M = 61 π . 4 MC ∠CME = π − 2 ⋅ 1 π = 2 π . 6 3 2 π 3
C
E
t op [0, 34 π ] ⇒ 2t op [0, 1 21 π ] ⇒ driekwartcirkel.
x =1
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 17/22
Gemengde opgaven 10. Goniometrie en beweging 2sin(x ) ⋅ cos(x ) sin(2x ) G25a = = tan(2x ). 2 cos(2x )
1 − 2sin (x )
4
(
4
)(
)
G25b cos (x ) − sin (x ) = cos2 (x ) + sin2 (x ) ⋅ cos2 (x ) − sin2 (x ) = 1 ⋅ cos(2x ) = cos(2x ). sin(2x ) 2sin(x ) ⋅ cos(x ) sin(x ) = = = tan(x ). G25c 1 + cos(2x ) cos(x ) 2cos2 (x )
G25d cos(x − y ) ⋅ cos( y ) − sin(x − y ) ⋅ sin(y ) = cos(x − y + y ) = cos(x ). G26a sin(x ) ⋅ cos(x ) = 41
G26c cos(x + 31 π ) = − sin(x ) sin(x + 1 π + 1 π ) = sin(x + π )
2 sin(x ) ⋅ cos(x ) = 1
3
2
sin(2x ) = 1
2
2x = 1 π + k ⋅ 2π ∨ 2x = 5 π + k ⋅ 2π 6
6
1 π + k ⋅π ∨ x = 5 π + k ⋅π. x = 12 12
5 π + k ⋅π. x = − 12
G26d cos(2x ) − sin2 (x ) = 41
G26b cos(x − 31 π ) = sin(2x ) cos(x − 1 π ) = cos(2x − 1 π ) 3
cos(2x ) + 1 cos(2x ) − 1 = 1
2
2
x − 31 π = 2x − 21 π + k ⋅ 2π ∨ x − 31 π = −2x + 21 π + k ⋅ 2π −x = − 1 π + k ⋅ 2π ∨ 3x = 5 π + k ⋅ 2π 6 6
G27a f (x ) = 21
O (V ) =
∫(
1 π 6
)
sin(x ) − 1 dx + 2
1 π 3
2
4
2
3
3
x = 61 π + k ⋅ π ∨ x = − 61 π + k ⋅ π . g (x ) = 21
sin(x ) = cos(x )
cos(x ) = 1
x = 41 π .
x = 31 π .
∫ ( cos(x ) − 21 ) dx = − cos(x ) − 21 x
1 π 4
(
1 π 4 1 π 6
)
2
1
π
+ sin(x ) − 1 x 31 2 π 4
(
= − cos( 1 π ) − 1 ⋅ 1 π − − cos( 1 π ) − 1 ⋅ 1 π + sin( 1 π ) − 1 ⋅ 1 π − sin( 1 π ) − 1 ⋅ 1 π 2 4 6 2 6 3 2 3 4 =−1 2− 1π + 1 3+ 1 π + 1 3− 1π − 1 2+ 1π =− 2+ 2 8 2 12 2 6 2 8
1 π 4
G27b I (L ) =
∫ π ⋅ sin
1 π 6 1 π 4
=
2
1 π 3
( x ) dx +
2
∫ π ⋅ cos
∫π
1 π 6
(
)
1 π 4
2 4
4 3 − 1 π. 12
)
(x ) dx − π ⋅ ( 1 )2 ⋅ ( 1 π − 1 π ) 2
1 π 4
⋅ 1 − 1 cos( 2x ) dx + 2 2
4
2x = 1 π + k ⋅ 2π ∨ 2x = − 1 π + k ⋅ 2π
f (x ) = g (x )
1 π 4
2
1 1 cos(2x ) = 3 ⇒ cos(2x ) = 1
5 π + k ⋅ 2π. x = 61 π + k ⋅ 2π ∨ x = 18 3
sin(x ) = 1 2 x = 61 π .
2
x + 56 π = x + π + k ⋅ 2π ∨ x + 56 π = π − x − π + k ⋅ 2π geen oplossing ∨ 2x = − 5 π + k ⋅ 2π 6
3
6
1 π 3
∫ π ⋅ ( 21 + 21 cos( 2x ) ) dx − 241 π
1 π 4
1
2
π
3 = π ⋅ 1 x − 1 sin(2x ) 1 + π ⋅ 1 x + 1 sin(2x ) 1 − 1 π 2 2 2 4 4 π π 24 6 4
(
)
(
)
= π ⋅ 1 ⋅ 1 π − 1 ⋅ sin( 1 π ) − π ⋅ 1 ⋅ 1 π − 1 ⋅ sin( 1 π ) + π ⋅ 1 ⋅ 1 π + 1 ⋅ sin( 2 π ) − π ⋅ 1 ⋅ 1 π + 1 ⋅ sin( 1 π ) − 1 π 2
(
)
(
)
(
)
(
2 6 3 2 3 3 2 4 2 4 2 4 4 4 4 = 1 π 2 − 1 π ⋅ 1 − 1 π 2 + 1 π 3 + 1 π 2 + 1 π 3 − 1 π 2 − 1 π ⋅ 1 − 1 π 2 = 1 π 2 − 1 π + 1 π 3. 8 12 8 6 8 8 2 4 4 24 24 4
G27c omtrek = 31 π − 61 π +
= 1π − 1π + 3
(
6
G28a f (x ) = 2 cos(x )
1 π 4
∫
1 π 6 1 π 4
∫
1 π 6
2
)
2
1 π 3
1 + (f '(x ) ) dx +
1 + cos2 (x ) dx +
∫
1 π 4 1 π 3
∫
2
)
24
2
1 + ( g '(x ) ) dx
1 + sin2 (x ) dx (fnInt) ≈ 1,19.
1 π 4
+ sin( 2x ) ⇒ f '(x ) = 4 cos(x ) ⋅ − sin(x ) + 2cos(2x ) = −2 sin(2x ) + 2cos(2x ).
f '(x ) = 0 ⇒ −2 sin(2x ) + 2 cos(2x ) = 0 ⇒ 2cos(2x ) = 2 sin(2x ) ⇒ cos(2x ) = sin(2x ) ⇒ cos(2x ) = cos(2x − 21 π ) ⇒ 2x = 2x − 1 π + k ⋅ 2π (geen oplossing) ∨ 2x = −2x + 1 π + k ⋅ 2π ⇒ 4x = 1 π + k ⋅ 2π ⇒ x = 1 π + k ⋅ 1 π . 2 2 2 8 2 f (x ) = 2cos2 (x ) + sin(2x ) = 2 cos2 (x ) − 1 + 1 + sin(2x ) = cos(2x ) + 1 + sin(2x ). absoluut maximum (zie fig G.13): f ( 1 π ) = cos( 1 π ) + 1 + sin( 1 π ) = 1 2 + 1 + 1 2 = 1 + 2 8 2 2 4 4 ⇒ Bf = [1 − 2, 1 + 2]. absoluut minimum (zie fig G.13): f ( 5 π ) = cos( 5 π ) + 1 + sin( 5 π ) = − 1 2 + 1 − 1 2 = 1 − 2 8 2 2 4 4
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 18/22 1 π 2
2
O (V ) =
G28b f (x ) = g (x ) ⇒ 2cos (x ) + sin(2x ) = 2 sin(2x )
∫
1 π 4 1 π 2
2cos2 (x ) = sin(2x ) 2cos2 (x ) = 2 sin(x ) cos(x )
=
2cos2 (x ) − 2 sin(x ) cos(x ) = 0 2cos(x ) ( cos(x ) − sin(x ) ) = 0 cos(x ) = 0 ∨ cos(x ) = sin(x )
∫
1 π 4 1 π 2
=
x = 21 π + k ⋅ π ∨ cos(x ) = cos(x − 21 π ) x = 21 π + k ⋅ π ∨ x = x − 21 π + k ⋅ 2π ∨ x = −x + 21 π + k ⋅ 2π x = 21 π + k ⋅ π ∨ geen oplossing ∨ 2x = + 1 π + k ⋅ 2π 2
(2 sin(2x ) − (2 cos (x ) + sin(2x ))) dx 2
( sin(2x ) − 2cos2(x ) ) dx
∫ ( sin( 2x ) − cos( 2x ) − 1) dx
1 π 4 1
π
= − 1 cos(2x ) − 1 sin(2x ) − x 21 2 2 π 4
= − 1 cos(π ) − 1 sin(π ) − 1 π + 1 cos( 1 π ) + 1 sin( 1 π ) + 1 π
x = 21 π + k ⋅ π ∨ x = 41 π + k ⋅ π . x op [0, π ] ⇒ x = 1 π ∨ x = 1 π . 2 4
2
2
2
2
= 1 − 0 − 1 π + 0 + 1 + 1 π = 1 − 1 π. 2
2
2
4
2
2
2
4
G28c C het midden van AB als g ( p ) = 1 f ( p ) ⇒ 2 sin(2 p ) = cos2 ( p ) + 1 sin(2 p ) ⇒ 2
2
1 1 sin(2p ) = cos2 ( p ) ⇒ 3sin( p ) cos( p ) = cos( p ) cos( p ) ⇒ 2
cos( p ) = 0(voldoet niet) ∨ 3sin( p ) = cos( p ) ⇒ 3
sin( p ) = 1 ⇒ 3tan( p ) = 1 ⇒ tan( p ) = 1 . 3 cos( p )
G29a De grafiek van f (dezelfde als in G28) is vermoedelijk lijnsymmetrisch in de verticale lijn door de eerste top rechts van de y -as. Vermoedelijk lijnsymmetrisch in de lijn x = 1 π . (zie de berekening in G28a) 8
f ( 81 π + p ) = 2 cos2 ( 81 π + p ) + sin( 41 π + 2p ) = 2 cos2 ( 81 π + p ) − 1 + 1 + sin( 41 π + 2 p ) = cos( 41 π + 2 p ) + sin( 41 π + 2 p ) + 1 = cos( 1 π ) cos(2 p ) − sin( 1 π ) sin(2 p ) + sin( 1 π ) cos(2 p ) + cos( 1 π ) sin(2 p ) + 1 4
4
4
4
= 1 2 ⋅ cos(2 p ) − 1 2 ⋅ sin(2 p ) + 1 2 ⋅ cos(2 p ) + 1 2 ⋅ sin(2 p ) + 1 = 2 cos(2 p ) + 1. 2
2
2
2
f ( 81 π − p ) = 2 cos2 ( 81 π − p ) + sin( 41 π − 2 p ) = 2 cos2 ( 81 π − p ) − 1 + 1 + sin( 41 π − 2 p ) = cos( 41 π − 2 p ) + sin( 41 π − 2 p ) + 1 = cos( 1 π ) cos(2 p ) + sin( 1 π ) sin(2 p ) + sin( 1 π ) cos(2 p ) − cos( 1 π ) sin(2 p ) + 1 4
4
4
4
= 1 2 ⋅ cos(2 p ) + 1 2 ⋅ sin(2 p ) + 1 2 ⋅ cos(2p ) − 1 2 ⋅ sin(2 p ) + 1 = 2 cos(2 p ) + 1. 2
2
2
2
f ( 81 π + p ) = f ( 81 π − p ) (voor elke p ) ⇒ de grafiek van f is symmetrisch in de lijn x = 81 π . G29b Vermoedelijk puntsymmetrisch in A( − 1 π , 1). (A precies midden tussen de toppen bij x = 81 π en x = 81 π − 21 π zie G28a) 8
f ( − 81 π + p ) = 2 cos2 ( − 81 π + p ) + sin( − 41 π + 2 p ) = cos( − 41 π + 2 p ) + sin( − 41 π + 2 p ) + 1
= cos( − 1 π ) cos(2 p ) − sin( − 1 π ) sin(2 p ) + sin( − 1 π ) cos(2 p ) + cos( − 1 π ) sin(2 p ) + 1 4
4
4
4
= 1 2 ⋅ cos(2 p ) + 1 2 ⋅ sin(2 p ) − 1 2 ⋅ cos(2 p ) + 1 2 ⋅ sin(2 p ) + 1 = 2 sin(2 p ) + 1. 2
2
2
2
f ( − 81 π − p ) = 2 cos2 ( − 81 π − p ) + sin( − 41 π − 2 p ) = cos( − 41 π − 2p ) + sin( − 41 π − 2 p ) + 1
= cos( − 1 π ) cos(2 p ) + sin( − 1 π ) sin(2 p ) + sin( − 1 π ) cos(2 p ) − cos( − 1 π ) sin(2 p ) + 1 4
4
4
4
= 1 2 ⋅ cos(2 p ) − 1 2 ⋅ sin(2 p ) − 1 2 ⋅ cos(2p ) − 1 2 ⋅ sin(2 p ) + 1 = − 2 sin(2 p ) + 1. 2
2
2
2
f ( − 81 π + p ) + f ( − 81 π − p ) = 2 sin(2 p ) + 1 + − 2 sin(2 p ) + 1 = 2 ⇒ f is symmetrisch in A ( − 81 π , 1). G30a f (x ) = 0 ⇒ 2 sin2 (x ) + sin(x ) = 0 ⇒ sin(x ) ⋅ (2 sin(x ) + 1 ) = 0 ⇒ sin(x ) = 0 ∨ sin(x ) = − 21 ⇒
x = k ⋅ π ∨ x = − 61 π + k ⋅ 2π ∨ x = 1 61 π + k ⋅ 2π . x op [0, 2π ] ⇒ nulp.: x = 0 ∨ x = π ∨ x = 2π ∨ x = 1 56 π ∨ x = 1 61 π . π
G30b O (V ) =
π
π
2 ∫ (2 sin (x ) + sin(x ) ) dx + ∫ (1 − cos( 2x ) + sin(x ) ) dx = x − 21 sin(2x ) − cos(x ) 0
0
0
(
)
= π − 1 sin(2π ) − cos(π ) − 0 − 1 sin(0) − cos(0) = π − 0 + 1 − ( 0 − 0 − 1 ) = π + 2. 2
G30c f (x ) = 2 sin2 (x ) + sin(x ) = 2 sin(x )
(
2
2
)
+ sin(x ) ⇒ f '(x ) = 4 sin(x ) cos(x ) + cos(x )
f '(x ) = 0 ⇒ 4 sin(x ) cos(x ) + cos(x ) = 0 ⇒ cos(x ) = 0 ∨ 4 sin(x ) + 1 = 0 ⇒ x = 21 π + k ⋅ π ∨ sin(x ) = − 41 . x = 1 21 π ⇒ f (x ) = f (1 21 π ) = 2 sin2 (1 21 π ) + sin(1 21 π ) = 2 ⋅ ( −1)2 + −1 = 2 − 1 = 1. sin(x ) = − 1 ⇒ f (x ) = 2 ⋅ ( − 1 )2 + − 1 = 2 − 1 = − 2 = − 1 . 4
4
4
16
4
16
8
f (x ) = p heeft precies vier oplossingen (zie figuur G.14 en de berekening hierboven) voor − 81 < p < 0 ∨ 0 < p < 1.
4
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 19/22
2π
G30d L(grafiek van f ) =
∫
2π
2
2
∫
1 + (f '(x ) ) dx =
0
1 + ( 4 sin(x ) cos(x ) + cos(x ) ) dx (fnInt) ≈ 11, 07.
0
G31a g (x ) = 2 sin(x ) + cos( 2x ) ⇒ g '(x ) = 2cos(x ) − 2 sin(2x ).
g '(x ) = 0 ⇒ cos(x ) = sin(2x ) ⇒ cos(x ) = cos(2x − 21 π ) x = 2x − 21 π + k ⋅ 2π ∨ x = −2x + 21 π + k ⋅ 2π
A
−x = − 1 π + k ⋅ 2π ∨ 3x = 1 π + k ⋅ 2π 2
x = 21 π + k ⋅ 2π ∨ x x op [0, 2π ] ⇒ x =
C
B
2
= 1 π + k ⋅ 2π.
6 3 1 π ∨ x = 1 π ∨ x = 5 π ∨ x = 1 1 π. 2 6 6 2 Dit geeft toppen: A( 1 π , 1 1 ), B ( 1 π , 1), C ( 5 π , 1 1 ) en 6 2 2 6 2
V
g
f
D
D (1 21 π , − 3).
f ( 61 π ) = 4 cos2 ( 61 π ) − 3sin( 61 π ) = 4 ⋅ ( 21 3)2 − 3 ⋅ 21 = 4 ⋅ 41 ⋅ 3 − 1 21 = 3 − 1 21 = 1 21
⇒ A en C liggen op de grafiek van f .
f ( 56 π ) = 4 cos2 ( 56 π ) − 3sin( 56 π ) = 4 ⋅ ( − 21 3)2 − 3 ⋅ 21 = 4 ⋅ 41 ⋅ 3 − 1 21 = 1 21 5 π 6
G31b O (V ) =
5 π 6
(2 sin(x ) + cos(2x ) − ( 4 cos (x ) − 3sin(x ))) dx = ∫π (2 sin(x ) + cos(2x ) − 4 cos (x ) + 3 sin(x )) dx π 2
∫
1 6 5 π 6
=
5 π 6
∫ (5 sin(x ) + cos(2x ) − 2 ⋅ (cos(2x ) + 1) ) dx
=
1 π 6
2
1 6
∫ (5 sin(x ) − cos( 2x ) − 2) dx
1 π 6
(
)
6 2 3 3 6 2 3 3 = −5 ⋅ − 1 3 − 1 ⋅ − 1 3 − 5 π − −5 ⋅ 1 3 − 1 ⋅ 1 3 − 1 π = 2 1 3 + 1 2 2 2 3 2 2 2 3 2 4
(
G31c omtrek(V ) = 5 π 6
=
∫
2
∫
)
5 π 6
3 − 5 π + 2 1 3 + 1 3 + 1 π = 5 1 3 − 4 π. 3
2
3
4
2
2
∫
1 + (f '(x ) ) dx +
1 π 6
1 + ( g '(x ) ) dx
1 π 6
5 π 6
2
1 π 6
2
∫
1 + ( −8cos(x ) sin(x ) − 3cos(x ) dx +
1 + (2cos(x ) − 2 sin(2x ) ) dx (fnInt) ≈ 11, 72.
1 π 6
(
)
G32a fp (x ) = sin2 (x ) + p cos(2x ) = sin2 (x ) + p ⋅ 1 − 2 sin2 (x ) = p (de andere termen vallen weg) voor p = 1 . 2
(
G32b fp (x ) = sin (x ) + p cos(2x ) = sin(x )
2
)
2
+ p cos(2x ) ⇒
fp '(x ) = 2 sin(x ) cos(x ) − 2p sin(2x ) = sin(2x ) − 2 p sin(2x ) = (1 − 2 p ) ⋅ sin(2x ). fp '(x ) = (1 − 2p ) ⋅ sin(2x ) ≠ 1 ⇒ −1 < 1 − 2 p < 1 ⇒ −2 < −2 p < 0 ⇒ 1 > p > 0 ⇒ 0 < p < 1. G32c
a
a
0
0
∫ fp (x ) dx =
∫(
)
sin2 (x ) + p cos(2x ) dx = a
a
a
0
0
∫ ( 21 − 21 cos(2x ) + p cos(2x ) ) dx = ∫ ( 21 + ( p − 21 ) cos( 2x ) ) dx
= 1 x + 1 ( p − 1 ) sin(2x ) = 1 a + 1 ( p − 1 ) sin(2a ) − 1 ⋅ 0 + 1 ( p − 1 ) sin(0) = 1 a + 1 ( p − 1 ) sin(2a ). 2 2 2 2 2 2 2 2 2 2 2 0 2 Onafhankelijk van p als sin(2a ) = 0 ⇒ 2a = k ⋅ π ⇒ a = k ⋅ 1 π . Gegeven: a op [0, π ] ⇒ a = 0 ∨ a = 1 π ∨ a = π .
(
)
2
3cos(x )
G33a f (x ) = 2 + sin(x ) ⇒ f '(x ) =
(2 + sin(x ) ) ⋅ −3sin(x ) − 3cos(x ) ⋅ cos(x ) = (2 + sin(x ) )2
2
−6sin(x ) − 3sin2 (x ) − 3cos2 (x )
(2 + sin(x ) )2
=
−6sin(x ) − 3
(2 + sin(x ) )2
f '(x ) = 0 (teller = 0) ⇒ −6 sin(x ) − 3 = 0 ⇒ sin(x ) = − 21 ⇒ x = 1 61 π + k ⋅ 2π ∨ x = π − 1 61 π + k ⋅ 2π . Gegeven: x op [ −π , π ] ⇒ x = − 5 π ∨ x = − 1 π . 6 6 minimum (zie plot): f ( − 5 π ) = 6
maximum (zie plot): f ( − 1 π ) = 6
3cos( − 56 π ) 2 + sin( − 56 π ) 3cos( − 61 π ) 2 + sin( − 61 π )
= − 3 ⇒ Bf = [ − 3, 3]. 3 ⋅ 21 3 1 21 3 = = 1 = 3 2 − 21 12
=
3 ⋅ − 21 3 2−
1 2
=
−1 21 3 1 21
2 9 ⇒ 3cos(x ) ⋅ 3cos( −x ) = 9 ⇒ 3cos(x ) ⋅ 3cos(x ) = 9 ⇒ cos (x ) = 1 ⇒ G33b f (x ) ⋅ f ( −x ) = 7 2 + sin(x ) 2 + sin( −x ) 7 2 + sin(x ) 2 − sin(x ) 7 4 − sin2 (x ) 7
(
)
7 cos2 (x ) = 4 − 1 − cos2 (x ) ⇒ 6cos2 (x ) = 3 ⇒ cos2 (x ) = 1 ⇒ cos(x ) = ± 1 = ± 1 ⋅ 2 = ± 1 ⋅ 2
x
= 1 π + k ⋅ 1 π. 4 2
π
6
= −5 cos( 5 π ) − 1 sin( 5 π ) − 5 π − −5 cos( 1 π ) − 1 sin( 1 π ) − 1 π
5 π 6
5
= −5 cos(x ) − 1 sin(2x ) − 2x 61 2 π
2
2
2 2
2
Gegeven: x op [ −π , π ] ⇒ x = − 3 π ∨ x = − 1 π ∨ x = 1 π ∨ x = 3 π . 4
4
4
4
.
3
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 20/22
G34a t op [0, 3 π ] ⇒ 2t op [0, 1 1 π ]. De baan is driekwartcirkel met middelpunt (1, 1) en straal 2.
2 t = 21 π G34b y = −x + 2 ⇒ 1 + 2 sin(2t − 21 π ) = −1 − 2cos(2t − 21 π ) + 2 B sin(2t − 1 π ) = − cos(2t − 1 π ) 2 2 cos(2t − 1 π − 1 π ) = cos(2t − 1 π + π ) 2 2 2 y = −x + 2 3 2t − π = 2t + 1 π + k ⋅ 2π ∨ 2t − π = −2t − 1 π + k ⋅ 2π t = 41 π t = π 2 2 4 (1,1) 4t = 1 π + k ⋅ 2π geen oplossing 2 t = 81 π + k ⋅ 21 π . A t op [0, 43 π ] ⇒ t = 81 π ∨ t = 58 π . t =0 t = 81 π ⇒ x P = 1 + 2cos( − 41 π ) = 1 + 2 ⋅ 21 2 = 1 + 2 en yP = 1 + 2 sin( − 41 π ) = 1 + 2 ⋅ − 21 2 = 1 − 2 ⇒ A(1 + 2, 1 − 2). t = 58 π ⇒ x P = 1 + 2cos( 34 π ) = 1 + 2 ⋅ − 21 2 = 1 − 2 en yP = 1 + 2 sin( 34 π ) = 1 + 2 ⋅ 21 2 = 1 + 2 ⇒ B (1 − 2, 1 + 2). 4
G34c x = 0 (y -as) 1 + 2cos(2t − 1 π ) = 0
y = 0 (x -as) 1 + 2 sin(2t − 1 π ) = 0 2
2
2cos(2t − 1 π ) = −1
2 sin(2t − 1 π ) = −1
2
2
cos(2t − 1 π ) = − 1
sin(2t − 1 π ) = − 1
2t − 1 π = 2 π + k ⋅ 2π ∨ 2t − 1 π = − 2 π + k ⋅ 2π
2t − 1 π = − 1 π + k ⋅ 2π ∨ 2t − 1 π = 1 1 π + k ⋅ 2π
2t = 7 π + k ⋅ 2π ∨ 2t = − 1 π + k ⋅ 2π
2t = 1 π + k ⋅ 2π ∨ 2t = 1 2 π + k ⋅ 2π
2
2
2
3
2
6
2
3
2
6
6
2
3
7 π + k ⋅ π ∨t = − 1 π + k ⋅ π t = 12 12 7 π x > 0 (zie de baan van P ) ⇒ 0 ≤ t < 12
6
3
t = 61 π + k ⋅ π ∨ t = 56 π + k ⋅ π y > 0 (zie de baan van P ) ⇒ 61 π < t ≤ 3 π 4
Uit bovenstaande regel volgt dan: x > 0 en tevens y G35a v = 0,5 m/s ⇒ per seconde
2
> 0 (zie de baan van P ) ⇒ 1 π < t < 7 π . 6 12
0,5 0,5 gedeelte van de cirkel ⇒ ⋅ 2π rad/s = 5 rad/s. 2π ⋅ 1,1 2π ⋅ 1,1 11
5t + π) G35b x = 1,1 cos( 11 (t in seconden en x en y in meters). 5 y = 1,1 sin( 11 t + π ) G35c De volgende koker loopt 1 cirkel = 1 ⋅ 2π = 1 π radialen achter. 6
6
t =0 ( −1, 1; 0)
3
O wateroppervlak
x = 1,1cos( 5 t + 2 π ) 11 3 (t in seconden en x en y in meters). 5 2 y = 1,1 sin( 11 t + 3 π )
y = −0, 8
G35d y = −0,8 ⇒ 1,1 sin( 5 t ) = −0, 8 (intersect) ⇒ t ≈ 8, 70 ∨ t ≈ 12, 03. 11
y < −0,8 (zie plot) ⇒ 8, 70 < t < 12, 03. Dus gedurende 3,3 seconde. G36c f 4 (x ) = 1 + sin2 (x ) + cos(4x ) = 1 + 1 − 1 cos(2x ) + cos(4x )
G36a f2 (x ) = 1 + sin2 (x ) + cos(2x ) = 1 + 1 − 1 cos(2x ) + cos(2x ) 2
2
2
2
= 1 1 + 1 cos(2x )
= 1 1 − 1 cos(2x ) + cos(4x ).
= 1 1 + 1 sin(2x + 1 π )
2π
2
2
2
2
2
2
2
2
G36d O (V ) =
= 1 1 + 1 sin(2(x + 1 π )). 4
2
4
=
G36b fn ( 1 π ) = 1 ⇒ 1 + sin2 ( 1 π ) + cos(n ⋅ 1 π ) = 1 6 6 6 4 4 2 1 + 1 + cos(n ⋅ 1 π ) = 1 2 6 4 cos(n ⋅ 1 π ) = −1 6 n ⋅ 61 π = π + k ⋅ 2π
∫ ( 4 − (1 21 − 21 cos(2x ) + cos(4x ) ) ) dx
0 2π
( )
n = 6 + k ⋅ 12. 0 < n < 50 ⇒ n = 6 ∨ n = 18 ∨ n = 30 ∨ n = 42.
∫ ( 4 − f 4 (x ) ) d x
0 2π
Dit geeft a = 1 1 , b = 1 , c = 2 en d = − 1 π . 2
2
=
∫ (2 21 + 21 cos( 2x ) − cos( 4x ) ) dx
0
V
2π = 2 1 x + 1 sin(2x ) − 1 sin(4x ) 4 4 2 0
(
= 5π + 1 sin(4π ) − 1 sin(8π ) − 0 + 1 sin(0) − 1 sin(0) 4
4
4
4
= 5π + 0 − 0 − ( 0 + 0 − 0 ) = 5π . O (rechthoek ABCD ) = 2π ⋅ 5 = 10π . Dus de grafiek van f 4 verdeelt de rechthoek in twee gebieden met dezelfde oppervlakte.
)
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 21/22
G 37a x = 3sin(2πt ) en y = 3cos(2πt ) geeft de ciirkel met middelpunt (0, 0) en straal 3. x = 2 sin( 1 πt ) en y = 2cos( 1 πt ) geeft de ciirkel met middelpunt (0, 0) en straal 2. 6
6
3 rondgang gemaakt ⇒ het is 18 minuten over 1. t = 1,3 de grote wijzer heeft 1 10
G37b Wijzers (niet de eindpunten) vallen over elkaar ⇒ sin(2πt ) = sin( 61 πt ) en cos(2πt ) = cos( 61 πt ).
(2πt = 61 πt + k ⋅ 2π ∨ 2πt = π − 61 πt + k ⋅ 2π ) en tevens (2πt = 61 πt + k ⋅ 2π ∨ 2πt = − 61 πt + k ⋅ 2π ) ( 116 πt = k ⋅ 2π ∨ 136 πt = π + k ⋅ 2π ) en tevens ( 116 πt = k ⋅ 2π ∨ 136 πt = k ⋅ 2π ) ( 116 t = k ⋅ 2 ∨ 136 t = 1 + k ⋅ 2) en tevens ( 116 t = k ⋅ 2 ∨ 136 t = k ⋅ 2) (t = k ⋅ 1211 ∨ t = 136 + k ⋅ 1312 ) en tevens (t = k ⋅ 1211 ∨ t = k ⋅ 1312 )
t = k ⋅ 12 ⇒ het eerste tijdstip na t = 0 is dus t = 12 . 11 11
(3sin(2πt ) − 2 sin(
G37c afstand =
1 πt ) 6
2
) + (3cos(2πt ) − 2cos(
1 πt ) 6
2
)
= 9 sin2 (2πt ) − 12 sin(2πt ) sin( 1 πt ) + 4 sin2 ( 1 πt ) + 9 cos2 (2πt ) − 12cos(2πt ) cos( 1 πt ) + 4 cos2 ( 1 πt ) 6
6 6 6 2 1 2 1 1 1 9 sin (2πt ) + cos (2πt ) + 4 sin ( πt ) + cos ( πt ) − 12 cos(2πt ) cos( πt ) + sin(2πt ) sin( πt ) 6 6 6 6 1 11 9 ⋅ 1 + 4 ⋅ 1 − 12cos(2πt − πt ) = 13 − 12cos( πt ). 6 6
(
= =
2
2
G37d Een gelijkbenige driehoek als ∨ afstand = 3
13 − 12cos( 11 πt ) = 3 6 13 − 12 cos( 11 πt ) = 9 6 −12 cos( 11 πt ) = −4 6 cos( 11 πt ) = 1 6 3
) (
) (
)
afstand = 2 13 − 12cos( 11 πt ) = 2
∨
6
∨
13 − 12cos( 11 πt ) = 4
∨
−12cos( 11 πt ) = −9
∨
cos( 11 πt ) = 3
6
6
6
4
Het eerste moment na t = 0 (cos(0) = 1) volgt uit cos( 11 πt ) = 3 ⇒ 11 πt ≈ 0, 723 ⇒ t ≈ 0,125. 6
4
6
G38a f (x ) = sin(x ) ⇒T = ( 21 π , 1) en A(π , 0). g (0) = −42 ⋅ 0 ⋅ (0 − π ) = 0 ⇒ de grafiek van g gaat door O . π
−4 ⋅ 1 π ⋅ ( 1 π − π ) = −4 ⋅ 1 π ⋅ − 1 π = −4 ⋅ − 1 π 2 = 1 ⇒ de grafiek van 2 2 4 π2 2 π2 − 4 − 4 g (π ) = 2 ⋅ π ⋅ (π − π ) = 2 ⋅ π ⋅ 0 = 0 ⇒ de grafiek van g gaat door A.
g ( 21 π ) =
π2 2
π
g gaat door T .
π
G38b f (x ) = sin(x ) ⇒ f '(x ) = cos(x ). 4
π2
x ⋅ (x − π ) = −
π
G38c
4
x 2 + π4 x ⇒ g '(x ) = −
x + π4 . f '(0) = cos(0) = 1 en g '(0) = − 82 ⋅ 0 + π4 = π4 > 1 ⇒ g '(0) > f '(0). π g (x ) = −
π2
π
π
∫ ( g (x ) − f (x ) ) dx = ∫ (ax (x 0
0
8
π2
π
− π ) − sin(x ) ) dx ∫ ax 2 − a π x − sin(x ) dx = 1 ax 3 − 1 a π x 2 + cos(x ) 2 3 0
(3
(
)
0
)
= 1 a π 3 − 1 a π ⋅ π 2 + cos(π ) − 1 a ⋅ 03 − 1 a π ⋅ 02 + cos(0) = 1 a π 3 − 1 a π 3 − 1 − ( 0 − 0 + 1 ) = − 1 a π 3 − 2. 3
π
∫ ( g (x ) − f (x ) ) d x 0
2
2
= 0 ⇒ − 1 a π 3 = 2 ⇒ a π 3 = −12 ⇒ a = − 123 . 6
π
3
2
6
G&R vwo B deel 3 C. von Schwartzenberg
11 Goniometrie en beweging 22/22
1 2
hoek
0
sinus
1 6
1 2
0
cosinus
1
1 2
1 4
π
1 2
3
1 2
π 2 2
1 3
π
1 2
3 1 2
1 2
π
1 2
+ +
π
teken van de SINUS
π
1
0 en 2π
- -
0
1 21
π
- +
π
teken van de COSINUS
0 en 2π
- +
π
1 21 π
sin(A ) = sin(B ) ⇒ A = B + k ⋅ 2π ∨ A = π − B + k ⋅ 2π cos(A ) = cos(B ) ⇒ A = B + k ⋅ 2π ∨ A = −B + k ⋅ 2π sin( −A ) = − sin(A ) − sin(A ) = sin(A + π ) sin(A ) = cos(A − 1 π )
cos( −A ) = cos(A ) − cos(A ) = cos(A + π ) cos(A ) = sin(A + 1 π )
sin2 (A ) + cos2 (A ) = 1
tan(A ) =
2
sin(A ) cos(A )
2
sin(2A ) = 2 sin(A ) cos(A ) cos(2A ) = cos2 (A ) − sin2 (A ) = 2cos2 (A ) − 1 = 1 − 2 sin2 (A ) cos(t + u ) = cos(t ) ⋅ cos(u ) − sin(t ) ⋅ sin(u ) cos(t − u ) = cos(t ) ⋅ cos(u ) + sin(t ) ⋅ sin(u ) sin(t + u ) = sin(t ) ⋅ cos(u ) + cos(t ) ⋅ sin(u ) sin(t − u ) = sin(t ) ⋅ cos(u ) − cos(t ) ⋅ sin(u )
Deze vier formules worden op het eindexamen gegeven.
De grafiek van de functie f is symmetrisch in de lijn x = a als voor elke p geldt: f (a − p ) = f (a + p ).
De grafiek van de functie f is symmetrisch in het punt (a , b ) als voor elke p geldt:
f (a − p ) + f (a + p ) 2
= b.
Dus de grafiek van de functie f is symmetrisch in het punt (a , b ) als voor elke p geldt: f (a − p ) + f (a + p ) = 2b .
f (x )
afgeleide f '(x )
sin(x ) cos(x )
cos(x ) − sin(x )
tan(x )
1 of 1 + tan2 (x ) cos2 (x )
f (ax + b )
a ⋅ f '(ax + b )
f (x )
primitieven F (x )
sin(x ) cos(x )
− cos(x ) + c sin(x ) + c
1 + tan2 (x ) of
f (ax + b )
1 cos2 (x )
tan(x ) + c 1 ⋅ F (ax + b ) + c
a
