Fonctions trigonométriques - Sebjaumaths

O,~
i,~ j
O
M~ i, ~ OM
~
i, ~ OM_
−1−0.5 0.5 1
−1
−0.5
0.5
1
0
π
6
π
4
π
3
2π
3
3π
4
5π
6
−π
6
−π
4
−π
3
−2π
3
−3π
4
−5π
6
√2
2
−√2
2
√2
2
−√2
2
√3
2
√3
2
−√3
2
−√3
2

M x ~ i, ~ OMk∈Z
x+ 2kπ ~ i, ~ OM
2π
M~ i, ~ OM
x x ∈]−π, π]
M M0x~ i, ~ OMx0
~
i, ~ OM0  ~ OM, ~ OM0x0−x+ 2kπ
k∈Z
M[R, θ]M R =OM θ
~
i, ~ OM
M(x, y)O[R, θ]M
R=px2+y2
cos (θ) = x
R
sin (θ) = y
R
M[R, θ] (x, x)M
x=Rcos (θ)
y=Rsin (θ)

M x ~ i, ~ OM
Mcos(x)x
Msin(x)x
Rxcos(x)
Rxsin(x)
x0π
6
π
4
π
3
π
2π
cos(x) 0 √3
2
√2
2
1
20−1
sin(x) 1 1
2
√2
2
√3
21 0
x∈Rk∈Z
• −16cos(x)61
• −16sin(x)61
•cos2(x) + sin2(x) = 1
•cos(x+ 2kπ) = cos(x)
•sin(x+ 2kπ) = sin(x)
•cos(−x) = cos(x)
•sin(−x) = −sin(x)
•cos(π−x) = −cos(x)
•sin(π−x) = sin(x)
•cos(π+x) = −cos(x)
•sin(π+x) = −sin(x)
•cos(π−x) = −cos(x)
•sin(π−x) = sin(x)
•cos π
2−x= sin(x)
•sin π
2−x= cos(x)
•cos π
2+x=−sin(x)
•sin π
2+x= cos(x)

∀α∈R
•cos(x) = cos(α)⇔x=α+ 2kπ
x=−α+ 2kπ k∈Z
•sin(x) = sin(α)⇔x=α+ 2kπ
x=π−α+ 2kπ k∈Z