e9mwm-Exos 08 FonctionsTrigoReciproques

∀x∈[−1; 1] ,arccos(x)+arccos (−x) = π
0,π
2
∀x∈R∗
+,arctan (x) + arctan 1
x=π 2
arctan (x) + arctan 1
xx < 0
∀x∈[−1; 1] ,arccos(x) + arcsin (x) = π
2
y=π
4
∀x>0,arcsin(x)>x
y=xR+
∀x>0,arctan(x)6x
y=xR+
sin (arccos(x))
cos (arcsin(x))
cos (2 arccos(x))
cos (2 arcsin(x))
sin (2 arccos(x))
cos (2 arctan(x))
sin (2 arctan(x))
tan (2 arcsin(x))
π
4= arctan 1 2+ arctan 1 3
A= 4 arctan 1
5
x y 0< x < y arctan x
y+ arctan y−x y+x=π 4
4 arctan 1
5−arctan 1 239=π 4
π

fR∀x∈R, f(x) = arctan x3
fRJ
narctan 1
n+ 2−arctan 1 n+ 1= arctan 1 n2+ 3n+ 1
SN=
N
n=0
arctan 1
n2+ 3n+ 1Nlim
N→+∞
SN
n
arccos 1
n+ arccos 1 n+ 1= arccos 1−(n2−1)n(n+ 2) n(n+ 1) 
A(x) = cos (arcsin(x)) x A(x)D
A(x)x D
arcsin 1
3+ arcsin 1 4= arcsin √8 + √15 12 
f I =0; π
4
x I f(x) = 1
cos(x)
f I J f−1
f−1
x∈J, 
cos f−1(x)=1
x
sin f−1(x)=1−1
x2
f−1J\{1}
∀x∈J\{1},f−1′(x) = 1
x√x2−1
R
: 1 + tan(x) + tan2(x) + tan3(x) = 0
1 + X+X2+X3