Cos Et Sin

  • May 2020
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sin 2 x + cos 2 x = 1 cos ( x + ( 2k + 1) π ) = − cos x k ∈ℤ  sin ( x + ( 2k + 1) π ) = − sin x

cos ( x + 2k π ) = cos x  sin ( x + 2k π ) = sin x cos ( x + π ) = − cos x  sin ( x + π ) = − sin x  π  cos  2 − x  = sin x     sin  π − x  = cos x    2 

cos (π − x ) = − cos x  sin (π − x ) = sin x  π  cos  2 + x  = − sin x     sin  π + x  = cos x    2 

cos ( − x ) = cos x  sin ( − x ) = − sin x   π cos  x − 2  = sin x     sin  x − π  = − cos x    2

cos ( x − y ) = cos x cos y + sin x sin y sin ( x − y ) = sin x cos y − cos x sin y   cos ( x + y ) = cos x cos y − sin x sin y sin ( x + y ) = sin x cos y + cos x sin y sin x 1 cos x tan x = , cotan x = = cos x tan x sin x 1 1 tan 2 x 2 2 x x tan 2 x = 1 , cos , sin − = = cos 2 x tan 2 x + 1 tan 2 x + 1 tan x + tan y tan x − tan y tan ( x + y ) = , tan ( x − y ) = 1 − tan x tan y 1 + tan x tan y cos 2x = cos 2 x − sin 2 x

,

cos 2x = 2 cos 2 x − 1

cos 2x = 1 − 2 sin 2 x

1 − tan 2 x 1 + tan 2 x

,

sin 2x = 2sin x cos x

, tan 2x =

2 tan x 1 − tan 2 x

2 tan x 1 + tan 2 x x +y x −y  x +y x −y  sin x + sin y = 2 sin 2 cos 2 cos x + cos y = 2 cos 2 cos 2   sin x − sin y = 2 cos x + y sin x − y cos x − cos y = −2sin x + y sin x − y  2 2 2 2  sin ( x + y ) sin ( x − y ) tan x + tan y = , tan x − tan y = cos x cos y cos x cos y

cos 2x =

, sin 2x =

k ∈Z k ∈Z  α = β + 2π k  α = β + 2π k     ،  ‫أو‬ ‫أو‬   ‫ ﻤﻌﻨﺎﻩ‬sin α = sin β  ‫ ﻤﻌﻨﺎﻩ‬cos α = cos β  α = π − β + 2π k k ∈ Z   α = − β + 2π k k ∈ Z      .‫ ﻤﻌﺭﻓﻴﻥ‬tan α ‫ و‬tan β ‫( ﻓﻲ ﻫﺫﻩ ﺍﻝﺤﺎﻝﺔ ﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ‬α = β + π k , k ∈ Z ) ‫ ﻤﻌﻨﺎﻩ‬tan α = tan β sin θ =

b a2 + b 2

‫ ﻭ‬cos θ =

a a2 + b 2

  a b ‫ ﺒﻭﻀﻊ‬a cos x + b sin x = a 2 + b 2  cos x + sin x  2 2 a2 + b 2  a +b 

a cos x + b sin x = a 2 + b 2 [ cos θ cos x + sin θ sin x ] = a 2 + b 2 .cos ( x − θ ) :‫ﻨﺠﺩ‬

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