‫اﻟﻌﻼﻗﺎت اﻷﺳﺎﺳﯿﺔ ﺑﯿﻦ اﻟﻨﺴﺐ اﻟﻤﺜﻠﺜﯿﺔ‬

sin θ 1 1 1
sin 2 θ + cos 2 θ = 1 tan θ = = 1 + tan 2 θ = 1 + cot 2
θ =
cos θ cot θ cos 2 θ sin 2 θ
sin tan
cot ‫اﻻرﺟﺎع اﻟﻰ اﻟﺮﺑﻊ اﻷول‬
sin ‫ﻛﻞ‬ ‫ ﻻ ﺗﻐﯿﺮ اﻟﻨﺴﺒﺔ اﻟﻤﺜﻠﺜﯿﺔ‬2π ، π
tan cos cos
cot 3π π
( ‫ﺗﻐﯿﺮ اﻟﻨﺴﺒﺔ اﻟﻤﺜﻠﺜﯿﺔ ) ﻣﻊ اﻻﻧﺘﺒﺎه ﻟﻺﺷﺎرة ﺑﺤﺴﺐ اﻟﺮﺑﻊ اﻟﺬي ﺗﻘﻊ ﻓﯿﮫ اﻟﺰاوﯾﺔ‬
،
2 2
‫ اﻷﺧﺮى‬cot = ‫ إﺣﺪاھﻤﺎ‬tan ، ‫ اﻷﺧﺮى‬cos = ‫ إﺣﺪاھﻤﺎ‬sin : ‫ ﻓﺈن‬90° ‫ إذا ﻛﺎن ﻣﺠﻤﻮع زاوﯾﺘﯿﻦ‬: ‫* ﻣﻼﺣﻈﺔ ھﺎﻣﺔ‬
cos ( −x ) = cos x sin ( −x ) = − sin x tan ( −x ) = −tan ( x ) cot ( −x ) = −cot x
−cos x = cos (π m x ) − sin x = sin ( −x ) − tan x = tan ( −x ) − cot = cot ( −x )
π π
cos x = sin  − x  sin x = cos  − x 
2  2 
: ‫ وﺑﺎﻟﺘﺎﻟﻲ ﯾﻜﻮن‬A + B + C = 180 ° ‫ ﻧﻌﻠﻢ أن ﻣﺠﻤﻮع اﻟﺰواﯾﺎ‬ABC ‫ ﻓﻲ اﻟﻤﺜﻠﺚ‬: ‫* ﺣﺎﻟﺔ ﺧﺎﺻﺔ‬
sin ( A + B ) = sin (180 ° − C ) = sinC , cos ( A + B ) = cos (180 ° − C ) = −cos C

A+B π C  C A+B π C  C
sin = sin  −  = cos , cos = cos  −  = sin
2 2 2  2 2 2 2  2
sin ( A + B ) = sinC cos ( A + B ) = −cos C tan ( A + B ) = −tanC cot ( A + B ) = −cot C
A+B C A+B C A+B C A+B C
sin = cos cos = sin tan = cot cot = tan
2 2 2 2 2 2 2 2
‫اﻟﻨﺴﺐ اﻟﻤﺜﻠﺜﯿﺔ ﻟﻠﺰواﯾﺎ اﻟﺸﮭﯿﺮة‬
π π π π 3π
0 30 ° 45 ° 60 ° 90 ° π 180 ° 270 ° 2 π 360 °
6 4 3 2 2
1 2 3
sin 0 1 0 −1 0
2 2 2
3 2 1
cos 1 0 −1 0 1
2 2 2
1
tan 0 1 3 +∞ 0 −∞ 0
3
1
cot +∞ 3 1 0 −∞ 0 +∞
3
 ‫ﺣﻞ اﻟﻤﻌﺎدﻻت اﻟﻤﺜﻠﺜﯿﺔ اﻟﺒﺴﯿﻄﺔ‬
sin x = sin θ ⇔ {x = θ + 2 π k , x = π −θ + 2π k }: k ∈ Z
π π
sin x = 0 ⇔ x = π k , sin x = 1 ⇔ x = + 2 π k , sin x = − 1 ⇔ x = − + 2 π k : ‫* ﺣﺎﻻت ﺧﺎﺻﺔ‬
2 2
cos x = cos θ ⇔ {x = θ + 2 π k , x = −θ + 2 π k } : k ∈ Z
π
cos x = 0 ⇔ x = + π k , cos x = 1 ⇔ x = 2 π k , cos x = − 1 ⇔ x = π + 2 π k : ‫* ﺣﺎﻻت ﺧﺎﺻﺔ‬
2
π
sin x = cos x ⇔ x = + π k : ‫* ﺣﺎﻟﺔ ﺧﺎﺻﺔ‬
4
(tan x = tan θ , cot x = cot θ ) ⇔ x = θ + π k : k ∈ Z

‫دﺳﺎﺗﯿﺮ اﻟﻨﺴﺐ اﻟﻤﺜﻠﺜﯿﺔ ﻟﺠﻤﻊ وﻓﺮق زاوﯾﺘﯿﻦ‬

cos (α ± β ) = cos α ⋅ cos β m sin α ⋅ sin β sin (α ± β ) = sin α ⋅ cos β ± cos α ⋅ sin β
tan α ± tan β
tan (α ± β ) =
1 m tan α ⋅ tan β

π 6 − 2 3 −1
sin15 ° = sin = = = cos75 °
12 4 2 2
5π 6 + 2 3 +1
sin75 ° = sin = = = cos15 °
12 4 2 2
tan15 ° = 2 − 3 = cot75 ° tan75 ° = 2 + 3 = cot15 °
‫دﺳﺎﺗﯿﺮ اﻟﺘﺤﻮﯾﻞ‬
2 cos α ⋅ cos β = cos (α + β ) + cos (α − β ) 2 sin α ⋅ sin β = cos (α − β ) − cos (α + β )
2 sin α ⋅ cos β = sin (α + β ) + sin (α − β ) 2 cos α ⋅ sin β = sin (α + β ) − sin (α − β )

α +β α −β α +β α −β
cos α + cos β = 2 cos cos cos α − cos β = − 2 sin sin
2 2 2 2
α +β α −β α +β α −β
sin α + sin β = 2 sin cos sin α − sin β = 2 cos sin
2 2 2 2
sin (α ± β ) sin ( β ± α )
tan α ± tan β = cot α ± cot β =
cos α ⋅cos β sin α ⋅ sin β
 ‫اﻟﻨﺴﺐ اﻟﻤﺜﻠﺜﯿﺔ ﻟﻀﻌﻔﻲ زاوﯾﺔ‬
2 tan α
sin 2 α = 2 sin α ⋅ cos α tan 2 α =
1 − tan 2 α
cos 2 α = 2 cos 2 α − 1 = 1 − 2 sin 2 α = cos 2 α − sin 2 α = (cos α + sin α )(cos α − sin α )
‫اﻟﻨﺴﺐ اﻟﻤﺜﻠﺜﯿﺔ ﻟﺰاوﯾﺔ ﺑﺪﻻﻟﺔ ﺗﺠﯿﺐ ﺿﻌﻔﯿﮭﺎ‬
1 + cos 2 α 1 − cos 2 α 1 − cos 2 α
cos 2 α = sin 2 α = tan 2 α =
2 2 1 + cos 2 α

‫اﻟﻨﺴﺐ اﻟﻤﺜﻠﺜﯿﺔ ﻟﺜﻼﺛﺔ أﻣﺜﺎل زاوﯾﺔ‬

cos 3 α = 4 cos 3 α − 3 cos α sin 3 α = 3 sin α − 4 sin 3 α
‫ﻣﺘﻄﺎﺑﻘﺘﺎن ھﺎﻣﺘﺎن‬
sin 2 α − sin 2 β = sin (α + β ) ⋅ sin (α − β ) cos 2 α + cos 2 β = cos (α + β ) ⋅ cos (α − β ) + 1

: ‫* ﺗﺬﻛﺮة ﺑﺎﻟﻤﺘﻄﺎﺑﻘﺎت اﻟﺸﮭﯿﺮة‬

(a + b )2 = a2 + 2 a ⋅b + b 2 , (a − b )2 = a 2 − 2 a ⋅b + b 2 , a 2 − b 2 = (a + b )(a − b )

(a + b )3 = a3 + 3 a 2 ⋅ b + 3 a ⋅ b 2 + b 3 , ( a − b )3 = a3 − 3 a 2 ⋅b + 3 a ⋅ b 2 − b 3

(
a 3 + b 3 = (a + b ) a 2 − a ⋅ b + b 2 ) , (
a 3 − b 3 = (a − b ) a 2 + a ⋅ b + b 2 )
a b c
. ‫اﻟﺘﻲ ﺗﺴﻤﻰ ﻋﻼﻗﺔ اﻟﺠﯿﻮب ﻓﻲ اﻟﻤﺜﻠﺚ‬ = = =2R
sin A sin B sinC
a = 2 R ⋅ sin A b = 2 R ⋅ sin B c = 2 R ⋅ sinC : ‫* ﻧﺘﯿﺠﺔ‬
b2 +c2 −a2
cos A =
2b c
a 2 = b 2 + c 2 − 2b c cos A
a +c −b
2 2 2
cos B = ‫ﺗﺪﻋﻰ ﻗﺎﻋﺪة اﻟﺘﺠﯿﺐ وﻧﺠﺪ‬ b 2 = a 2 + c 2 − 2 a c cos B
2a c
c 2 = a 2 + b 2 − 2 a b cos C
a +b −c
2 2 2
cos C =
2a b
2 p = a + b + c : ‫ أي‬2p ‫ ﺑﺎﻟﺮﻣﺰ‬ABC ‫ ﻧﺮﻣﺰ ﻣﺤﯿﻂ اﻟﻤﺜﻠﺚ‬-

b + a − c = 2 ( p − c ) c + a − b = 2 ( p − b ) b + c − a = 2 ( p − a ) : ‫وﻧﺴﺘﻨﺘﺞ أن‬
 ( ‫) اﻟﻨﺴﺐ اﻟﻤﺜﻠﺜﯿﺔ ﻷﻧﺼﺎف اﻟﺰواﯾﺎ‬
A p ( p −a ) B p ( p −b ) C p ( p −c )
cos = cos = cos =
2 b ⋅c 2 a ⋅c 2 a ⋅b
A ( p − b )( p − c ) B ( p − a )( p − c ) C ( p − a )( p − b )
sin = sin = sin =
2 b ⋅c 2 a ⋅c 2 a ⋅b

A ( p − b )( p − c ) B ( p − a )( p − c ) C ( p − a )( p − b )
tan = tan = tan =
2 p(p −a) 2 p ( p −b ) 2 p ( p −c )

( ‫) ﻗﻮاﻧﯿﻦ ﻣﺴﺎﺣﺔ ﺳﻄﺢ اﻟﻤﺜﻠﺚ‬

1 1 1
S= c ⋅ hc , S= b ⋅ hb , S= a ⋅ ha
2 2 2

1 1 1
S= a ⋅ b ⋅ sin C , S= a ⋅ c ⋅ sin B , S= b ⋅ c ⋅ sin A
2 2 2

a ⋅b ⋅c
S = 2 R 2 ⋅ sin A ⋅ sin B ⋅ sin C ، S =r ⋅p ، S=
4R
S = p ( p − a )( p − b )( p − c )
A r B r C r
tan = tan = tan =
2 p −a 2 p −b 2 p −c

A p −a B p −b C p −c
cot = cot = cot =
2 r 2 r 2 r