© h.hofstede (h.hofstede@hogeland.nl)

       
1. a. cos(2α) = cos2α - sin2α = 1 - sin2α - sin2α = 1 - 2sin2α
       
  b. cos(2α) = cos2α - sin2α = cos2α - (1 - cos2α) = cos2α - 1 + cos2α = 2cos2α - 1
       
  c. (sinα + cosα)2
=
(sinα + cosα)(sina + cosa)
=  sin2α + sinαcos
a + cosαsinα + cos2α
= sin2α + cos2α + 2sinαcosα
= 1 + 2sinαcosα
= 1 + sin2α
       
2. a. sin(3α)
= sin(2α + α)
= sin2αcosα + cos2αsinα
= 2sinαcosα•cosα + (1- 2sin2α)sinα
= 2sinαcos2α + sinα - 2sin3α
= 2sinα(1 - sin2α) + sinα - 2sin3α
= 2sinα - 2sin3α + sinα - 2sin3α
= 3sinα - 4sin3α  
       
  b. sin(4α)
= sin(2α + 2α)
= sin2αcos2α + cos2αsin2α
= 2sin2αcos2α
= 2 • 2sinαcosα • (cos2α - sin2α)
= 4sinαcosα  • cos2α - 4sinαcosα • sin2α
= 4sinαcos3α  - 4sin3αcosα
       
  c. sin(4α)
= sin(2α + 2α)
= sin2αcos2α + cos2αsin2α
= 2sin2αcos2α
= 2sin2α • (1 - 2sin2α)
       
3. a. sin(1/6π)
= sin(2 • 1/12π)
= 2sin(1/12π) • cos(1/12π) = 1/2

4sin2(1/12π) • cos2(1/12π) = 1/4
4sin2(1/12π) • (1 - sin2(1/12π)) = 1/4  noem sin2(1/12π) = p
4p(1 - p) = 1/4
16p - 16p2 - 1 =  0
16p2 - 16p + 1 = 0
p = (16 ±√(256 - 64))/32  = (16 ± √192)/32 = (16 ± 8√3)/32 = 1/2 ± 1/4√3
omdat p kleiner dan 0,25 moet zijn (immers  sin2(1/6π) = 0,25) is p = 1/2 - 1/4√3
sin2(1/12π) = 1/2 - 1/4√3
sin(1/12π)  = √(1/2 - 1/4√3)    (alleen positief omdat de sinus van 1/12π positief is)
       
  b. cos(1/4π)
= cos(2 • 1/8π)
= 2cos2(1/8π) - 1 = 1/2√2

2cos2(1/8π) = 1 + 1/2√2
cos2(1/8π) = 1/2 + 1/4√2
cos(1/8π) = √(1/2 + 1/4√2)
       
  c. cos(1/8π)
= cos(2 • 1/16π)
= 2cos2(1/16π) - 1 = √(1/2 + 1/4√2)

2cos2(1/16π) = 1 + √(1/2 + 1/4√2)
cos2(1/16π) = 1/2 + 1/2√(1/2 + 1/4√2)
cos(1/16π) = √(1/2 + 1/2√(1/2 + 1/4√2))
       
4. Noem de groene hoek x
in driehoek APC:  is PC = 5  (Pythagoras)
Dan is cosx = 4/5

in driehoek ABC:
cos2x = 4/BC
cos2x = 2cos2x - 1 = 2 • (4/5)2 - 1 = 7/25
Dus is 4/BC = 7/25
BC = 100/7
       
5. a. Vermenigvuldig alles met 2sinx:  (sinx ≠ 0)
2sin3x + sin2x = 2sinx
2sin3x + 2sinxcosx - 2sinx = 0
2sinx • (sin2x + cosx - 1) = 0
sin2x + cosx - 1 = 0   (want  sinx ¹ 0)
1 - cos2x+ cosx - 1 = 0
cosx - cos2x = 0
cosx(1 - cosx) = 0
cosx = 0  ∨   cosx = 1
x = 1/2π  x = 11/2π ∨  x = 0  ∨  x = 2π
Die laatste twee vallen af, want dan is sinx = 0, dus blijft over  x = 1/2π  ∨  x = 11/2π.
       
  b. cos2x - sinx = 0
1 - 2sin2x - sinx = 0   noem  sinx = p
-2p2 - p + 1 = 0
ABC- formule:   p = (1 ±√(1 + 8))/-4 = (1 ± 3)/-4 = 1/2  of -1
sinx = 1/2  ∨  sinx = -1
x = 1/6π  ∨  x = 5/6π  ∨  x =  11/2π
       
6. cosinusregel op twee manieren:
42 = 52 + 62 - 2 • 5 • 6 • cosα  en daaruit volgt  cosα = 9/12
62 = 42 + 52 - 2 • 4 • 5 • cosγ  en daaruit volgt  cosγ = 1/8

2 • (9/12)2 - 1 = 1/8
Dus  cosγ = 2 • cos2α - 1
Dan is  γ = 2α
       
7.
       
8. a. Y1 = 2cos(X) en Y2 = cos(2X)
Intersect geeft  X = 1,95   ∨   X = 4,34
 
       
  b. 2cosx = cos(2x)
2cosx = 2cos2x - 1  noem cosx = p
2p2 - 2p - 1 = 0
ABC-formule:  p = (2 ±√(4 + 8))/4 = (2 ±√12)/4 = 1,366 of  -0,366
cosx = 1,366 kan niet.
cosx = -0,366
x = cos-1(-0,366) = 1,95  ∨  x = 2π - 1,95 = 4,34
       
9. Zie de figuur hiernaast.
β = 90 - 2α  (bij hoek D)
∠DEF = 90 - β  dus is ∠AEB = β

In driehoek AEB:
sinb = sin(90 - 2α) = cos(2α) = (16-x)/x = 16/x - 1
cos(2α) = 2cos2α - 1, dus  16/x - 1 = 2cos2α - 1
2cos2α = 16/x  dus  cos2α = 8/x  dus  x = 8/cos2α

In driehoek DBC:  sinα = x/L dus L = x/sinα

Samen geeft dat 
L = 8/cos2α • sinα = 8/(1 - sin2α)sinα8/(sinα - sin3α)

       
10. cos(4x)
= cos(2 • 2x)
= 2cos2(2x) - 1
= 2 • (2cos2x - 1)2 - 1
= 2 • (4cos4x - 4cos2x + 1) - 1
= 8cos4x - 8cos2x  + 1

cos(4x) + 11cos2x = 0
8cos4x - 8cos2x  + 1 + 2cos2x = 0
8cos4x - 6cos2x + 1 = 0   noem nu cos2x = p
8p2 - 6p + 1 = 0
ABC-formule:  p = (6±√(36-32))/16 = (6 ± 2)/16 = 1/2  of  1/4
cos2p = 1/2  ∨  cos2p = 1/4
cosp = 1/2√2 ∨  cosp = -1/2√2  ∨  cosp = 1/2  ∨  cosp = -1/2
p = 1/4π p = 13/4π   p = 3/4π  p =11/4π   p = 1/3π  p = 12/3π   p = 2/3π ∨  p = 11/3π  
       
11. Nulpunten:
sin2x - 2sinx  = 0
2sinxcosx - 2sinx = 0
2sinx(cosx - 1) = 0
sinx = 0  ∨  cosx = 1
x = 0 ∨ x = π  ∨  x = 2π

Extremen:
afgeleide moet nul zijn.
f ' = 2cos2x - 2cosx = 0
 2(2cos2x - 1) - 2cosx = 0
4cos2x - 2 - 2cosx = 0
2cos2x - cosx - 1 = 0   noem nu cos = p
2p2 - p - 1 = 0
ABC-formule:  p = (1 ±√(1 + 8))/4 = (1 ± 3)/4  =  1  of  -1/2
cosx = 1  ∨  cosx = -1/2
x = 0 x = 2π    x = 2/3π  x = 11/3π   
De extremen zijn dan  (0, 0) en (2/3π, -3/23)  en  (11/3π, 3/23)  en  (2π, 0)  
       
12. a. 3sinx1/cosx
3sinxcosx = 1
11/2 • 2sinxcosx = 1
11/2 • sin2x = 1
sin2x = 2/3
2x = sin-1(2/3) = 0,73 + k2π     2x = π - 0,73 = 2,41 + k2π
x = 0,36 + kπ  x = 1,21 + kπ
In [0,1/2π] geeft dat de oplossingen  {0.36, 1.21}
Dat geeft de punten    (0.36, 1.07)  en  (1.21, 2.80)
       
  b. asinx1/cosx
asinxcosx = 1
1/2a • 2sinxcosx = 1
1/2a • sin2x = 1
sin2x = 2/a 
Dat heeft geen oplossing als  2/a > 1  of  2/a < -1
Dat betekent dat a < 2 of a > -2  dus a in het interval  〈-2, 2〉
       
13. sin2x + cos2x = a(sinx + cosx)
2sinxcosx + cos2x - sin2x = a(sinx + cosx)

Deel alles door cosx;
2sinx + cosx - sin²x/cosx = a(sinx/cosx + 1)

Maar sinx/cosx = tanx = 2, dus dat wordt;
2sinx + cosx - 2sinx = a(2 + 1)
cosx = 3a

Omdat  -1 ≤ cosx 1  geldt dus  -1/3   a 1/3
       
14. √(4 - 7sin2x) = cosx - sinx
4 - 7sin2x = (cosx - sinx)2
4 - 7sin2x = cos2x - 2sinxcosx + sin2x
4 - 7sin2x = 1 - sin2x
3 = 6sin2x
sin2x = 1/2
2x = 1/6π + k2π   2x = 5/6π + k2π
x
= 1/12π + kπ    x = 5/12π + kπ   
       
       

© h.hofstede (h.hofstede@hogeland.nl)