Med HAMDANE Med HAMDANE

www.riyadiyat.net
2
J



2


 cos (x) =
+ kπ/k ∈ Z :


2


 sin (x) =
1
1 + tan2 (x)
tan2 (x)
M(α)
sin(α)
1 + tan2 (x)
α
I′
∀x ∈ R\
π
2
+ kπ/k ∈ Z :
t = tan
cos(a) =
1 − t2
a
2
tan(x) =
O
y
cos(x)
2t
J′
‫ﻟﺪﻳﻨﺎ‬
1 + t2
‫ﺍﻟﻌﻼﻗـﺔ ﺑﻴﻦ ﺍﻟﻨﺴﺐ ﺍﻟﻤﺜﻠﺜﻴــﺔ‬
(+)
(+)
π
2
2
2
5π
6
π
−π
π
2
π
3
√
√
3
2
1
2
α
sin(x)
0
cos(x)
1
tan(x)
0
π
6
1
2
√
3
2
√
3
3
[−3π; 3π]
π
4
√
2
2
√
2
2
1
0
Med HAMDANE
−α
π+α
− π2
0
−α
α
π
−π
x
√ √
2 3
2 2
Med HAMDANE
x
π
2
π−α
0
−
+α
π
6
0
3
− 12
2 √
− 22
π
2
π
4
1
2
√
− π2
π
3
√
3
2
1
2
√
3
π
2
.
sin
cos
tan
1
0
‫ﻏﻴﺮ ﻣﻌﺮﻑ‬
π
2
−α
cos(α)
sin(α)
1
tan(α)
(π − α)
sin(α)
− cos(α)
− tan(α)
π
2
+α
cos(α)
− sin(α)
−1
tan(α)
(π + α)
− sin(α)
− cos(α)
tan(α)
‫ ﻋﻠﻰ ﺍﻟﻤﺠﺎﻝ‬x 7−→ cos(x) ‫ ﻭ ﺟﻴﺐ ﺍﻟﺘﻤﺎﻡ‬x 7−→ sin(x) ‫ﺍﻟﺘﻤﺜﻴﻞ ﺍﻟﻤﺒﻴﺎﻧﻲ ﻟﺪﺍﻟﺘﺎ ﺍﻟﺠﻴﺐ‬
1 −
|
−3π
|
− 5π
2
x
cos(α)
sin(x)
‫ﻧﺴﺐ ﻣﺜﻠﺜﻴــﺔ ﺍﻋﺘﻴﺎدﻳــﺔ‬
2π
3
I
‫ﻧــﻀﻊ‬
sin(a) =
‫ﻭ‬
1 + t2
3π
4
(+)
tan(α)
π
y
sin2 (x) + cos2 (x) = 1
(∀x ∈ R) :
∀x 6=
‫ﺛﺎﻧﻮﻳﺔ ﺃﺑﻲ ﺣﻴﺎﻥ ﺍﻟﺘﻮﺣﻴـﺪﻱ‬
‫ ﻣﺤﻤــﺪ ﺣﻤـﺪﺍﻥ‬:‫ﺍﻻﺳﺘﺎﺫ‬
‫ﺍﻟﺤﺴـــﺎﺏ ﺍﻟﻤﺜﻠﺜــــﻲ‬
‫ﻣﻠﺨــﺺ ﺍﻟــﺪﺭﺱ‬
‫ ﺑﺎﻙ ﻉ ﺭﻳﺎﺿﻴﺔ ﻭ ﻉ ﺗﺠﺮﻳﺒﻴﺔ‬2
|
−2π
Document traité par LATEX
|
− 3π
2
|
−π
|
− π2
??
O
−1−
(Csin )
|
π
2
‫ ﻣﻦ‬1 ‫ﺍﻟﺼﻔﺤﺔ‬
|
π
|
3π
2
|
2π
|
5π
2
|
3π
(Ccos )
www.riyadiyat.net
x 7−→ sin(x)
x 7−→ cos(x)
‫دﺍﻟــﺔ ﺍﻟﺠﻴﺐ‬
∀x ∈ R : −1 6 sin(x) 6 1
∀x ∈ R : −1 6 cos(x) 6 1
∀x ∈ R : sin(−x) = − sin(x)
∀x ∈ R : cos(−x) = cos(x)
∀k ∈ Z : sin(x + 2kπ) = sin(x)
∀k ∈ Z : cos(x + 2kπ) = cos(x)
∀k ∈ Z : sin(x + kπ) = (−1)k sin(x)
∀k ∈ Z : cos(x + kπ) = (−1)k cos(x)
∀n ∈ N : sin(n π2 ) = (−1)n
∀n ∈ N : cos(nπ) = (−1)n
sin(x) = sin(α) ⇐⇒

 x = α + 2kπ
x = π − α + 2kπ

cos(x) = cos(α) ⇐⇒
;k ∈ Z
π
2
sin(x) = 1
⇐⇒
x=
sin(x) = 0
⇐⇒
x = kπ,
sin(x) = −1
⇐⇒
x = − π2 + 2kπ,

 x = α + 2kπ

;k ∈ Z
x = −α + 2kπ
‫ﻣﻌــــﺎدﻻﺕ ﺧـﺎﺻــــــــــﺔ‬
‫ﻣﻌــــﺎدﻻﺕ ﺧـﺎﺻــــــــــﺔ‬
+ 2kπ,
k∈Z
k∈Z
k∈Z
cos(x) = 1
⇐⇒
x = 2kπ,
cos(x) = 0
⇐⇒
x=
cos(x) = −1
⇐⇒
x = x = π + 2kπ,
π
2
k∈Z
+ kπ,
k∈Z
k∈Z
‫ﻣﺠﻤــــــــــــــﻮﻉ‬
‫ﺻﻴﻎ ﺗﺤﻮﻳﻞ‬
Med
HAMDANE
‫ﻣﺠﻤــــــــــــــﻮﻉ‬
‫ﺻﻴﻎ ﺗﺤﻮﻳﻞ‬
Med
HAMDANE
sin(a + b) = sin(a) cos(b) + cos(a) sin(b)
cos(a + b) = cos(a) cos(b) − sin(a) sin(b)
sin(a − b) = sin(a) cos(b) − cos(a) sin(b)
cos(a − b) = cos(a) cos(b) + sin(a) sin(b)
cos(2a) = cos2 (a) − sin2 (a)
sin(2a) = 2 sin(a) cos(a)
sin(a) = 2 sin a2 cos a2
sin2 (a) =
= 2 cos2 (a) − 1
:‫ﻧﺘﺎﺋـــﺞ‬
1−cos(2a)
2
sin(p) + sin(q) = 2 sin
sin(p) − sin(q) = 2 cos
p+q
2
cos2 (a) =
p+q
2
cos
sin
:‫ﺗﺤﻮﻳﻞ ﺟﺪﺍﺀ ﺇﻟﻰ ﻣﺠﻤـــﻮﻉ‬
sin(a) sin(b) =
− 3π
;
2
3π
2
h
n
\ − π2 ; π2
1
2
o
p−q
2
p−q
2
cos(a) cos(b) =
(Ctan )
−π
O
π
(a, b) 6= (0; 0)
a
a2 +b2
√ 2b 2
a +b
cos(α) = √
sin(α) =
Document traité par LATEX
:‫ﺣﻴﺚ‬
π
2
cos (a + b) + cos (a − b)
‫دﺍﻟــﺔ ﺍﻟﻈـــﻞ‬
tan(−x) = − tan(x)
∀k ∈ Z, tan(x + kπ) = tan(x)
⇐⇒
x = α + kπ, k ∈ Z
tan(a + b) =
tan(a)+tan(b)
1−tan(a) tan(b)
tan(a − b) =
tan(a)−tan(b)
1+tan(a) tan(b)
Med HAMDANE
x
tan(2a) =
a cos(x) + b sin(x)
=
=
??
+ kπ|k ∈ Z :
tan(x) = tan(α)
a cos(x) + b sin(x)
1
2
x 7−→ tan(x)
∀x ∈ R\
3π
2
p+q
2
:‫ﺗﺤﻮﻳﻞ ﺟﺪﺍﺀ ﺇﻟﻰ ﻣﺠﻤـــﻮﻉ‬
cos (a + b) − cos (a − b)
π
2
cos p−q
2
sin p−q
cos(p) − cos(q) = −2 sin p+q
2
2
cos(p) + cos(q) = 2 cos
‫ﺍﻟﺘﻤﺜﻴﻞ ﺍﻟﻤﺒﻴﺎﻧﻲ ﻟﺪﺍﻟﺔ ﺍﻟﻈﻞ ﻋﻠﻰ‬
− π2
1+cos(2a)
2
:‫ﺗﺤﻮﻳﻞ ﻣﺠﻤﻮﻉ ﺇﻟﻰ ﺟــــﺪﺍﺀ‬
y
− 3π
2
:‫ﻧﺘﺎﺋـــﺞ‬
= 1 − 2 sin2 (a)
:‫ﺗﺤﻮﻳﻞ ﻣﺠﻤﻮﻉ ﺇﻟﻰ ﺟــــﺪﺍﺀ‬
i
‫دﺍﻟــﺔ ﺟﻴﺐ ﺍﻟﺘﻤــﺎﻡ‬
2 tan(a)
1−tan2 (a)
tan2 (a) =
tan(a) + tan(b) =
sin(a+b)
cos(a) cos(b)
tan(a) − tan(b) =
sin(a−b)
cos(a) cos(b)
1−cos(2a)
1+cos(2a)
‫ﺗﺤﻮﻳﻞ ﺍﻟﺼﻴﻐــﺔ‬
√
b
√
a2 + b2 √ a
cos(x)
+
sin(x)
a2 +b2
a2 +b2
√
2
2
a + b cos(x − α)
‫ ﻣﻦ‬2 ‫ﺍﻟﺼﻔﺤﺔ‬
www.riyadiyat.net