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Trigonometriese Identiteite

Afleiding van Waardes vir Trigonometriese Funksies vir : 3030, 4545 and 6060

Hou in gedagte dat die trigonometriese funksies slegs van toepassing is op reghoekige driehoeke. Dus kan ons waardes aflei vir trigonometriese funksies vir 3030, 4545 en 6060. Ons sal begin met 4545 omdat dit die maklikste is

Neem enige reghoekige driehoek met een hoek 4545. Dus omdat een hoek gelyk is aan 9090, moet die derde hoek ook gelyk wees aan 4545. Ons het dus 'n gelyksydige reghoekige driehoek soos aan gedui in figuur 1.

Figuur 1: 'n Gelyksydige reghoekige driehoek
Figuur 1 (MG11C17_019.png)

As die twee sye gelyk is in lengte aan aa, dan kan die skuinssy hh, soos volg bereken word:

h 2 = a 2 + a 2 = 2 a 2 h = 2 a h 2 = a 2 + a 2 = 2 a 2 h = 2 a
(1)

Dus het ons:

sin ( 45 ) = teenoorstaande ( 45 ) skuinssy = a 2 a = 1 2 sin ( 45 ) = teenoorstaande ( 45 ) skuinssy = a 2 a = 1 2
(2)
cos ( 45 ) = aanliggende ( 45 ) skuinssy = a 2 a = 1 2 cos ( 45 ) = aanliggende ( 45 ) skuinssy = a 2 a = 1 2
(3)
tan ( 45 ) = teenoorstaande ( 45 ) aanliggende ( 45 ) = a a = 1 tan ( 45 ) = teenoorstaande ( 45 ) aanliggende ( 45 ) = a a = 1
(4)

Ons kan iets soortgelyks probeer vir 3030 en 6060. Ons begin met 'n gelyksydige driehoek en halveer een hoek soos aangedui in figuur 2. Dit gee ons die verlangde reghoekige driehoek met een hoek gelyk aan 3030 en een hoek gelyk aan 6060.

Figuur 2: 'n Gelyksydige driehoek met een hoek gehalveer.
Figuur 2 (MG11C17_020.png)

As die gelyke sye se lengte gelyk is aan aa, dan is die basis gelyk aan 12a12a en die lengte van die vertikale sy vv kan dan soos volg bereken word:

v 2 = a 2 - ( 1 2 a ) 2 = a 2 - 1 4 a 2 = 3 4 a 2 v = 3 2 a v 2 = a 2 - ( 1 2 a ) 2 = a 2 - 1 4 a 2 = 3 4 a 2 v = 3 2 a
(5)

Dus het ons:

sin ( 30 ) = teenoorstaande ( 30 ) skuinssy = a 2 a = 1 2 sin ( 30 ) = teenoorstaande ( 30 ) skuinssy = a 2 a = 1 2
(6)
cos ( 30 ) = aanliggende ( 30 ) skuinssy = 3 2 a a = 3 2 cos ( 30 ) = aanliggende ( 30 ) skuinssy = 3 2 a a = 3 2
(7)
tan ( 30 ) = teenoorstaande ( 30 ) aanliggende ( 30 ) = a 2 3 2 a = 1 3 tan ( 30 ) = teenoorstaande ( 30 ) aanliggende ( 30 ) = a 2 3 2 a = 1 3
(8)
sin ( 60 ) = teenoorstaande ( 60 ) skuinssy = 3 2 a a = 3 2 sin ( 60 ) = teenoorstaande ( 60 ) skuinssy = 3 2 a a = 3 2
(9)
cos ( 60 ) = aanliggende ( 60 ) skuinssy = a 2 a = 1 2 cos ( 60 ) = aanliggende ( 60 ) skuinssy = a 2 a = 1 2
(10)
tan ( 60 ) = teenoorstaande ( 60 ) aanliggende ( 60 ) = 3 2 a a 2 = 3 tan ( 60 ) = teenoorstaande ( 60 ) aanliggende ( 60 ) = 3 2 a a 2 = 3
(11)

Jy hoef nie hierdie identiteite te memoriseer nie as jy weet hoe om hulle af te lei.

leidraad: Twee bruikbare driehoeke om te onthou.:

Figuur 3
Figuur 3 (MG11C17_021.png)

Alternatiewe definisie vir tanθtanθ

Ons weet dat tanθtanθ soos volg gedefinieer word: tanθ= teenoorstaande aanliggende tanθ= teenoorstaande aanliggende Dit kan soos volg geskryf word:

tan θ = teenoorstaande aanliggende × skuinssy skuinssy = teenoorstaande skuinssy × skuinssy aanliggende tan θ = teenoorstaande aanliggende × skuinssy skuinssy = teenoorstaande skuinssy × skuinssy aanliggende
(12)

Maar ons weet ook dat sinθsinθ soos volg gedefinieer word: sinθ= teenoorstaande skuinssy sinθ= teenoorstaande skuinssy en dat cosθcosθ soos volg gedefinieer word: cosθ= aanliggende skuinssy cosθ= aanliggende skuinssy

Daarom kan ons dit soos volg skryf:

tan θ = teenoorstaande skuinssy × skuinssy aanliggende = sin θ × 1 cos θ = sin θ cos θ tan θ = teenoorstaande skuinssy × skuinssy aanliggende = sin θ × 1 cos θ = sin θ cos θ
(13)

leidraad:

tanθtanθ kan ook soos volg gedefinieer word: tanθ=sinθcosθtanθ=sinθcosθ

'n Trignometriese Identiteit

Een van die mees bruikbare resultate van die trigonometriese funksies is dat hulle verwant aan mekaar is. Ons het gesien dat tanθtanθ geskryf kan word in terme van sinθsinθ en cosθcosθ. Net so sal ons wys dat: sin2θ+cos2θ=1sin2θ+cos2θ=1

Ons begin deur te kyk na ABCABC,

Figuur 4
Figuur 4 (MG11C17_022.png)

Ons sien dat: sinθ=ACBCsinθ=ACBC en cosθ=ABBC.cosθ=ABBC.

Volgens die stelling van Pythagoras weet ons dat: AB2+AC2=BC2.AB2+AC2=BC2.

Daarom kan ons die volgende neerskryf:

sin 2 θ + cos 2 θ = A C B C 2 + A B B C 2 = A C 2 B C 2 + A B 2 B C 2 = A C 2 + A B 2 B C 2 = B C 2 B C 2 ( volgens Pythagoras ) = 1 sin 2 θ + cos 2 θ = A C B C 2 + A B B C 2 = A C 2 B C 2 + A B 2 B C 2 = A C 2 + A B 2 B C 2 = B C 2 B C 2 ( volgens Pythagoras ) = 1
(14)

Exercise 1: Trigonometriese Identiteite A

Vereenvoudig deur identiteite te gebruik:

  1. tan 2 θ · cos 2 θ tan 2 θ · cos 2 θ
  2. 1 cos 2 θ - tan 2 θ 1 cos 2 θ - tan 2 θ
Solution
  1. Stap 1. Gebruik bekende identiteite en vervang tanθtanθ :
    = tan 2 θ · cos 2 θ = sin 2 θ cos 2 θ · cos 2 θ = sin 2 θ = tan 2 θ · cos 2 θ = sin 2 θ cos 2 θ · cos 2 θ = sin 2 θ
    (15)
  2. Stap 2. Gebruik bekende identiteite en vervang tanθtanθ :
    = 1 cos 2 θ - tan 2 θ = 1 cos 2 θ - sin 2 θ cos 2 θ = 1 - sin 2 θ cos 2 θ = cos 2 θ cos 2 θ = 1 = 1 cos 2 θ - tan 2 θ = 1 cos 2 θ - sin 2 θ cos 2 θ = 1 - sin 2 θ cos 2 θ = cos 2 θ cos 2 θ = 1
    (16)

Exercise 2: Trigonometriese Identiteite B

Bewys: 1-sinxcosx=cosx1+sinx1-sinxcosx=cosx1+sinx

Solution
  1. Stap 1. Gebruik trig identiteite :
    LHS = 1 - sin x cos x = 1 - sin x cos x × 1 + sin x 1 + sin x = 1 - sin 2 x cos x ( 1 + sin x ) = cos 2 x cos x ( 1 + sin x ) = cos x 1 + sin x = RHS LHS = 1 - sin x cos x = 1 - sin x cos x × 1 + sin x 1 + sin x = 1 - sin 2 x cos x ( 1 + sin x ) = cos 2 x cos x ( 1 + sin x ) = cos x 1 + sin x = RHS
    (17)

Trigonometriese identiteite

  1. Vereenvoudig die volgende met behulp van die basiese trigonometriese identiteite:
    1. cosθtanθcosθtanθ
    2. cos2θ.tan2θ+tan2θ.sin2θcos2θ.tan2θ+tan2θ.sin2θ
    3. 1-tan2θ.sin2θ1-tan2θ.sin2θ
    4. 1-sinθ.cosθ.tanθ1-sinθ.cosθ.tanθ
    5. 1-sin2θ1-sin2θ
    6. 1-cos2θcos2θ-cos2θ1-cos2θcos2θ-cos2θ
  2. Bewys die volgende:
    1. 1+sinθcosθ=cosθ1-sinθ1+sinθcosθ=cosθ1-sinθ
    2. sin2θ+(cosθ-tanθ)(cosθ+tanθ)=1-tan2θsin2θ+(cosθ-tanθ)(cosθ+tanθ)=1-tan2θ
    3. (2cos2θ-1)1+1(1+tan2θ)=1-tan2θ1+tan2θ(2cos2θ-1)1+1(1+tan2θ)=1-tan2θ1+tan2θ
    4. 1cosθ-cosθtan2θ1=11cosθ-cosθtan2θ1=1
    5. 2sinθcosθsinθ+cosθ=sinθ+cosθ-1sinθ+cosθ2sinθcosθsinθ+cosθ=sinθ+cosθ-1sinθ+cosθ
    6. cosθsinθ+tanθ·cosθ=1sinθcosθsinθ+tanθ·cosθ=1sinθ

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