Rappels de trigonométrie

θ
cos θ= ; sin θ= ; tan θ=sin θ
cos θ=.
(0,0) 1
(cos θ, sin θ)
M θ tan θ
(OM)x= 1
θ0π
6
π
4
π
3
π
2
sin θ01
2
√2
2
√3
21
cos θ1√3
2
√2
2
1
20
tan θ01
√31√3
sin √0
2,√1 2,√2 2,√3 2,√4 2cos
cos π
3=1 2
•cos sin R2π−1 1
•cos sin
•tan R\{π
2+kπ, k ∈Z}π
lim
x→(π
2+kπ)+tan x=−∞ ; lim
x→(π
2+kπ)−
tan x= +∞
•cos(x)0=−sin x; sin(x)0= cos x; tan(x)0= 1 + tan2x=1
cos2x

cos(−x) = cos xcos(π−x) = −cos xcos(π+x) = −cos x
sin(−x) = −sin xsin(π−x) = sin xsin(π+x) = −sin x
tan(−x) = −tan xtan(π−x) = −tan xtan(π+x) = tan x
cos(π
2−x) = sin xsin(π 2−x) = cos xtan(π 2−x) = 1 tan x= cotan x
cos2x+ sin2x= 1
cos sin
cos(a+b) = cos acos b−sin asin b(∗)
cos(a−b) = cos acos b+ sin asin b
sin(a+b) = sin acos b+ sin bcos a(∗)
sin(a−b) = sin acos b−sin bcos a
cos 2x= cos2x−sin2xsin 2x= 2 sin xcos x
= 2 cos2x−1
= 1 −2 sin2x
cos2x=1 + cos 2x
2sin2x=1−cos 2x 2
tan cos sin
tan(a+b) = tan a+ tan b
1−tan atan btan(a−b) = tan a−tan b 1 + tan atan b
cos sin
cos acos b=1
2(cos(a−b) + cos(a+b))
sin asin b=1
2(cos(a−b)−cos(a+b))
sin acos b=1
2(sin(a−b) + sin(a+b))
p=a−b q =a+b
cos sin
cos p+ cos q= 2 cos p+q
2cos p−q 2
cos p−cos q=−2 sin p+q
2sin p−q 2
sin p+ sin q= 2 sin p+q
2cos p−q 2

I J f :I→J f
y∈J x ∈I f(x) = y
f f−1g
∀x∈I, g(f(x)) = x∀y∈J, f(g(y)) = y .
x∈I y ∈J y =f(x)⇔x=f−1(y)
f[a, b]f0(x)>0
x∈]a, b[f[a, b] [f(a), f(b)] f0(x)<0
x∈]a, b[f[a, b] [f(b), f(a)]
f:I→J f0(x)6= 0 x∈I
f−1J
f−1(x)0=1
f0(f−1(x)) .
f(f−1(x)) = x
f0(f−1(x)) ×f−1(x)0= 1
•ln : ]0,+∞[→Rexp : R→]0,+∞[
•x7→ √xR+R+R+
x7→ x2R+∗
•x7→ 3
√xR R x7→ x3
R∗
f−1f
y=x
arccos arcsin arctan
x7→ cos x[0, π] [−1,1]
arccos : [−1,1] →[0, π]
x∈[−1,1] arccos x θ [0, π] cos θ=x
∀x∈[−1,1],cos(arccos x) = x
arccos(cos θ)θ θ
θ∈[0, π]
arccos ] −1,1[
∀x∈]−1,1[,arccos(x)0=−1
√1−x2.
θ∈]0, π[ cos(θ)0=−sin θ6= 0 x∈]−1,1[
arccos(x)0=1
−sin(arccos x)θ= arccos x

θ∈]0, π[ sin θ > 0 sin θ=√1−cos2θ=p1−cos2(arccos x) = √1−x2
x7→ sin x[−π
2,π 2] [−1,1]
arcsin : [−1,1] →h−π
2,π 2i
x∈[−1,1] arcsin x θ [−π
2,π 2] sin θ=x
∀x∈[−1,1],sin(arcsin x) = x
arcsin(sin θ)θ θ
θ∈[−π
2,π 2]
arcsin ] −1,1[
∀x∈]−1,1[,arcsin(x)0=1
√1−x2.
θ∈]−π
2,π 2[ sin(θ)0=−cos θ6= 0 x∈]−1,1[
arcsin(x)0=1
cos(arcsin x)θ= arcsin x θ ∈]−π
2,π 2[ cos θ > 0 cos θ=
p1−sin2θ=p1−sin2(arcsin x) = √1−x2
arcsin(x)0+arccos(x)0= 0
]−1,1[ ∀x∈[−1,1],arcsin x+ arccos x=π
2
x7→ tan x]−π
2,π 2[R
arctan : R→i−π
2,π 2h
x∈Rarctan x θ ]−π
2,π 2[ tan θ=x
∀x∈R,tan(arctan x) = x
arctan(tan θ)θ θ
θ∈]−π
2,π 2[
arctan R
∀x∈R,arctan(x)0=1
1 + x2.
θ∈]−π
2,π 2[ tan(θ)0= 1 + tan2θ6= 0 x∈R
arctan(x)0=1
1 + tan2(arctan x)=1
1 + x2