Résumé
Fonctions Période Angles f(x) = sin x
g(x) = cos x
h(x) = tan x
P 2 b
P 2 b
P b
) ( sin 1
1 k
1
2
k
sin
) ( cos 1
1 k
1
2 2
k
cos
) ( tan1 k
k
tan
Problèmes variés
1 1
4 P
P 2b
b
4 2
2
b
2 min max f f
A
2 3 1
A
A 2
) 1 , 0 ( : int Départ Po
0 a
2 1 cos 2
)
(x x
f
(h, k) = (0, -1)
1
1 1
cos 2 2
)
(x x
f
2 2 1
cos
2 x
2 1 cos 2 x
À l’aide de pythagore u = ±12/13
Deuxième quadrant: u = -12/13 tan(t)= sin(t)/cos(t)
12 5 13
12 13
5
tan
t
12 3 5 5
12 2 5
2
12 3 25 144
50
144 782 144
432 144
300 144
50
72
391
3 3 2
sin 3 x
3
0,72973 x
3
3,87133 x
3032 ,
2
x x 6 , 6968
S = {2,3023 + 6n; 6,6968 + 6n}, n ε z
7 2 2
cos 5
x
2
1,86055
x
9611 ,
4
x x 9 , 0389
S = {-4,9611 + 10n; -9,0389 + 10n}, n ε z
2
4,42275
x
ac b
2 4
a x b
2
0 1
2 u
2 u
x u sin
) 1 )(
2 ( 4 )
1
(
2
9
4 3 1 x
5 ,
1
0
u u
2 1
1
sin x sin x 0 , 5
ac b
2 4
a x b
2
0 6
7
2 u
2 u
x u sin
) 6 )(
2 ( 4 )
7
(
2
1
4 1 7 x
5 ,
1
1
u u
2 2
5 , 1
sin x sin x 2
(h, k) = (0, 0) P=π/2
ab > 0 croissant P/2 = π/4
(h, k) = (2π/3, 0) P=2π
ab > 0 croissant P/2 = π
f(x) = 2 tan 0,5(x-2π/3)
f(x) = 2 tan -2(x + π/8) + 1
(h, k) = (-π/8, 1) P=π/2
ab < 0 décroissant P/2 = π/4
A = 2
(h, k) = (-π/2; 2)
P = 4π b = 1/2
f(x) = 2 sin 1/2(x + π/2) + 2
ab > 0 croissant
A = 2
(h, k) = (-π/2; 0)
P = 3π b = 2/3
f(x) = -2 cos 2/3(x + π/2)
a < 0 par le bas
A = 2 k = 0 P = 4π/3
b = 3/2 h = π/3
f(x) = 2 cos 3/2(x- π/3)
A = 20 k = 0 P = 4π b = 1/2
h = 3π/2
f(x) = 20 cos 1/2(x- 3π/2) f(x) = -20 cos 1/2(x + π/2)
A = 1
(h, k) = (π/3; 0,5) P = 4π
b = 1/2
f(x) = sin 1/2(x- π/3) + 0,5
A = 1
(h, k) = (π/4, 1) P = π
b = 2
f(x) = -sin 2(x- π/4) + 1
x x 2 x
2 2
sin
sin cos x
x x
x cos
sin 1 cos
sin
2
2
x x 2 x
2 2
sin tan cos
x x x
x
2 2 2
2
sin cos cos
sin cosec2xsin x x
x sin sin
1
2
x x
ec2 cos2 cos
x x
2
2 cos
sin 1
x x
x cos cos
sin
t
t t
t
2
2 2
2
cos
cos sin
sin
Le standard, développer le terme de gauche pour obtenir celui de droite.
Le standard, développer le terme de gauche pour obtenir celui de droite.
2 2 1
sin x
) 6 (
2 x
6 ) 5
(
2 x
12
x 12
5
x 12
13
x 12
17
x
[ 2
, 0
[
x P = π
12
x 12
5
x 12
12 12
13
x 12
12 12
17
x
12 , 17 12
, 13 12
, 5 12
x
2 2 1
sin x
[ 2
, 0
[
x
12 , 17 12
, 13 12
, 5 12
x
2
2
3 2
2 2 1
sin )
( x x
f
3 sin 2 ( ) 1 0
2 sin
2
2x x
6 ) 7
(
2 x 2 ( x ) 11 6
12 19
x 12
23
x
[ 2
, 0
[
x
P = π
1 )(sin 2 ( ) 1 ) 0
2 sin 2
( x x
1 0
2 sin
2 x sin 2 ( x ) 1 0
2 2 1
sin x
sin 2 ( x ) 1
2 ) 3
(
2 x
4 7
x
12 7
x 12
11
x 4
3
x
A = 2 k = 0 P = 4π/3
b = 3/2 h = π/3
f(x) = 2 cos 3/2(x- π/3)
A = 20 k = 0 P = 4π b = 1/2
h = 3π/2
f(x) = 20 cos 1/2(x- 3π/2) f(x) = -20 cos 1/2(x + π/2)
A = 1
(h, k) = (π/3; 0,5) P = 4π
b = 1/2
f(x) = sin 1/2(x- π/3) + 0,5
A = 1
(h, k) = (π/4, 1) P = π
b = 2
f(x) = -sin 2(x- π/4) + 1
A = 3
(h, k) = (2, -1) P = 2
f(x) = -3 sin π(t-2) -1
1
π/2 π
3
π/22
π5
π/23
π 2 1
cos
2 x
3 , 4 3
2
[
x
2 , 0
[
x
2 )
cos(
2 )
(x x f