© h.hofstede (h.hofstede@hogeland.nl)
     
1. sin2(α + 1/4π)
= sin(α + 1/4π) • sin(α + 1/4π)
= (sinαcos1/4π + cosαsin1/4π) • (sinαcos1/4π + cosαsin1/4π)
= sin2α•cos21/4π  + 2sinα•cos1/4π•cosα•sin1/4π + cos2α•sin21/4π
= sin2α • (1/2√2)2 + 2sinαcosα • 1/2√2 • 1/2√2 + cos2α • (1/2√2)2
= 1/2sin2α + sinαcosα + 1/2cos2α
= 1/2(sin2α + cos2α) + sinαcosα
= 1/2 + sinαcosα
       
2. cos(α + 1/6π) + cos(α - 1/6π)
= cosαcos1/6π - sinαsin1/6π + cosαcos1/6π + sinαsin1/6π
= 2cosαcos1/6π
= 2cosα • 1/2√3
= √3 • cosα
       
3. cos2(1/4π - 1/2x)
= cos(1/4π - 1/2x) • cos(1/4π - 1/2x)
= (cos1/4πcos1/2x + sin1/4πsin1/2x) •  (cos1/4πcos1/2x + sin1/4πsin1/2x)
= cos2(1/4π) • cos2(1/2x) + 2cos(1/4π) •  cos(1/2x) •  sin(1/4π) •  sin(1/2x) + sin2(1/4π) •  sin2(1/2x)
= (1/2√2)2 • cos21/2x + 2 • 1/2√2 • cos1/2x • 1/2√2 • sin1/2x + (1/2√2)2 • sin21/2x
= 1/2cos21/2x + cos1/2x •  sin1/2x + 1/2sin21/2x
= 1/2(cos21/2x + sin21/2x) + cos1/2x •  sin1/2x
= 1/2 + 1/2 • sin(2 • 1/2x)
= 1/2 + 1/2sinx
       
4. a. sin(5/12π)
= sin(3/12π + 2/12π)
= sin(3/12π)cos(2/12π) + cos(3/12π)sin(2/12π)
= 1/2√2 • 1/2√3 + 1/2√2 • 1/2
= 1/4√6 + 1/4√2
       
  b. cos(1/12π)
= cos(1/4π - 1/6π)
= cos(1/4π)cos(1/6π) + sin(1/4π)sin(1/6π)
= 1/2√2 • 1/2√3 + 1/2√2 • 1/2
= 1/4√6 + 1/4√2
       
  c. cos(1/12π) = sin(5/12π)
Dat is logisch, want  cos(5/12π) = sin(1/2π - 5/12π)
       
5. Zie de figuur hiernaast.  AD = 5  (Pythagoras)

cos(α + β) = 3/AC    (in driehoek ABC)   ......(1)

cos(α + β) = cosαcosβ - sinαsinβ
cosα = 3/5  en  sinα = 4/5      (in driehoek ABD)
cosβ = AE/5  en sinβ = 1/5  (in driehoek ADE)

invullen geeft dan cos(α + β) = 3/5 • AE/5 - 4/5 • 1/5

Dus moet gelden (met (1)):   3/AC = 3AE/25 - 4/25
AE = √(52 - 12) = √24
3/AC = 3√24/25 - 4/25
75/AC = 3√24 - 4
AC = 75/(3√24 - 4)
EC = AC - AE = 75/(3√24 - 4) - √24

       
6. AP = √(22 + 12) = √5 = AQ

sin(α + θ) =  2/Ö5  en cos(α + θ) = 1/Ö5
sinα = 1/Ö5 en cosα = 2/Ö5

sin(α + θ) = sinαcosθ + cosαsinθ
2/√5 = 1/√5 • cosθ + 2/√5 • sinθ
2 = cosθ + 2sinθ  ...... (1)

 
cos(α + θ) = cosαcosθ - sinαsinθ
1/√5 = 2/√5 • cosθ - 1/√5 • sinθ
1 = 2cosθ - sinθ  .......(2)

 uit (1) volgt  cosθ = 2 - 2sinθ
invullen in (2):   1 = 2(2 - 2sinθ) - sinθ
1 = 4 - 5sinθ
5sinθ = 3
sinθ = 3/5
       
7. a. 2 • sin2x • cos3x  = 0
sin2x = 0  ∨  cos3x = 0
2x = 0 + k2p ∨  2x = p + k2p ∨  3x = 1/2p + k2p ∨   3x = -1/2p + k2p           
x = 0 ∨  x = 1/2p + kp  ∨  x = 1/6p + k2/3  ∨  x = -1/6p + k2/3p
 dat geeft de oplossingen  {0, 1/2p, 3/2p, 1/6p, 5/6p, 7/6p, 2p}
       
  b.

sin(2x + 3x) = sin2xcos3x + cos2xsin3x
sin(2x – 3x) = sin2x cos3x – cos2xsin3x
optellen  geeft de gevraagde formule.

 f(x) = sin(5x) + sin(- x)
De primitieve is dan   F(x) =  -1/5cos(5x) + cos(-x)