Trigonométrie Flashcards
À quoi ressemble la fonction arctan ?
π/2+. ________
|. ____/
| ___/
| __/
——-—————/———————-
___/ |
___/. |
_______/ |
-π/2+
On a cos(a)=cos(b) alors on a le systeme :
a = -b +2kπ
k c Z
b = b + 2kπ
tan(x)=
sin(x) / cos(x)
cos^2(x) + sin^2(x) =
1
sin(a+b) =
sin(a)cos(b) + sin(b)cos(a)
sin(2a)
2sin(a)cos(a)
cos(2a)
cos^2(a) - sin^2(a)
cos(a+b) =
cos(a)cos(b) - sin(a)sin(b)
tan(a+b)
(tan(a)+tan(b)) / (1-tan(a)tan(b))
À quoi ressemble la fonction arcsin ?
π/2+ |
|. |
| /
| __/
——-—-1———/———1————-
/—|
/—. |
|. |
|.-π/2+
À quoi ressemble la fonction arccos ?
|. π/2+
|. |
—. |
\—-——\
|. —\
|. |
|. |
—_-1-———————1——
On a sin(a) = sin(b) donc ce qui nous donne le systeme :
a = b + 2kπ
k c Z
a = π - b + 2kπ
sin(a)sin(b)
(1/2)(cos(a-b) - cos(a+b))
sin(a)cos(b)
(1/2)(sin(a+b) + sin(a-b))
cos(a)cos(b)
(1/2)(cos(a+b) + cos(a-b))
sin^2(a)
(1-cos(2a))/2
cos^2(a)
(1+cos(2a))/2
sin(p) + sin(q)
2sin((p+q)/2) cos((p-q)/2)
sin(p) - sin(q)
2cos((p+q)/2) sin((p-q)/2)
cos(p) + cos(q)
2cos((p+q)/2) cos((p-q)/2
cos(p) - cos(q)
-2sin((p+q)/2 sin((p-q)/2)
