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Trigonometriese identiteite

Afleiding van waardes vir trigonometriese funksies vir : 30 , 45 And 60

Hou in gedagte dat die trigonometriese funksies slegs van toepassing is op reghoekige driehoeke. Dus kan ons waardes aflei vir trigonometriese funksies vir 30 , 45 en 60 . Ons sal begin met 45 omdat dit die maklikste is

Neem enige reghoekige driehoek met een hoek 45 . Dus omdat een hoek gelyk is aan 90 , moet die derde hoek ook gelyk wees aan 45 . Ons het dus 'n gelyksydige reghoekige driehoek soos aan gedui in [link] .

'n Gelyksydige reghoekige driehoek

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

h 2 = a 2 + a 2 = 2 a 2 h = 2 a

Dus het ons:

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

Ons kan iets soortgelyks probeer vir 30 en 60 . Ons begin met 'n gelyksydige driehoek en halveer een hoek soos aangedui in [link] . Dit gee ons die verlangde reghoekige driehoek met een hoek gelyk aan 30 en een hoek gelyk aan 60 .

'n Gelyksydige driehoek met een hoek gehalveer.

As die gelyke sye se lengte gelyk is aan a , dan is die basis gelyk aan 1 2 a en die lengte van die vertikale sy v 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

Dus het ons:

sin ( 30 ) = teenoorstaande ( 30 ) skuinssy = a 2 a = 1 2
cos ( 30 ) = aanliggende ( 30 ) skuinssy = 3 2 a a = 3 2
tan ( 30 ) = teenoorstaande ( 30 ) aanliggende ( 30 ) = a 2 3 2 a = 1 3
sin ( 60 ) = teenoorstaande ( 60 ) skuinssy = 3 2 a a = 3 2
cos ( 60 ) = aanliggende ( 60 ) skuinssy = a 2 a = 1 2
tan ( 60 ) = teenoorstaande ( 60 ) aanliggende ( 60 ) = 3 2 a a 2 = 3

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

Twee bruikbare driehoeke om te onthou.

Alternatiewe definisie vir tan θ

Ons weet dat tan θ soos volg gedefinieer word: tan θ = teenoorstaande aanliggende Dit kan soos volg geskryf word:

tan θ = teenoorstaande aanliggende × skuinssy skuinssy = teenoorstaande skuinssy × skuinssy aanliggende

Maar ons weet ook dat sin θ soos volg gedefinieer word: sin θ = teenoorstaande skuinssy en dat cos θ soos volg gedefinieer word: cos θ = 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 kan ook soos volg gedefinieer word: tan θ = sin θ cos θ

'n trignometriese identiteit

Een van die mees bruikbare resultate van die trigonometriese funksies is dat hulle verwant aan mekaar is. Ons het gesien dat tan θ geskryf kan word in terme van sin θ en cos θ . Net so sal ons wys dat: sin 2 θ + cos 2 θ = 1

Ons begin deur te kyk na A B C ,

Ons sien dat: sin θ = A C B C en cos θ = A B B C .

Volgens die stelling van Pythagoras weet ons dat: A B 2 + A C 2 = B C 2 .

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

Vereenvoudig deur identiteite te gebruik:

  1. tan 2 θ · cos 2 θ
  2. 1 cos 2 θ - tan 2 θ
  1. = tan 2 θ · cos 2 θ = sin 2 θ cos 2 θ · cos 2 θ = sin 2 θ
  2. = 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

Bewys: 1 - sin x cos x = cos x 1 + sin x

  1. 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

Trigonometriese identiteite

  1. Vereenvoudig die volgende met behulp van die basiese trigonometriese identiteite:
    1. cos θ tan θ
    2. cos 2 θ . tan 2 θ + tan 2 θ . sin 2 θ
    3. 1 - tan 2 θ . sin 2 θ
    4. 1 - sin θ . cos θ . tan θ
    5. 1 - sin 2 θ
    6. 1 - cos 2 θ cos 2 θ - cos 2 θ
  2. Bewys die volgende:
    1. 1 + sin θ cos θ = cos θ 1 - sin θ
    2. sin 2 θ + ( cos θ - tan θ ) ( cos θ + tan θ ) = 1 - tan 2 θ
    3. ( 2 cos 2 θ - 1 ) 1 + 1 ( 1 + tan 2 θ ) = 1 - tan 2 θ 1 + tan 2 θ
    4. 1 cos θ - cos θ tan 2 θ 1 = 1
    5. 2 sin θ cos θ sin θ + cos θ = sin θ + cos θ - 1 sin θ + cos θ
    6. cos θ sin θ + tan θ · cos θ = 1 sin θ

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Source:  OpenStax, Siyavula textbooks: wiskunde (graad 11). OpenStax CNX. Sep 20, 2011 Download for free at http://cnx.org/content/col11339/1.4
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