PDF - 75.7 ko

0,−→
i , −→ jCO
1C
MCθ
−→
i , −−→ OM
cos θsin θ
M
O
M
sin θ
cos θ
tan θ
cotanθ
θ
tan θ=sin θ
cos θθDtan =R\nπ 2+kπ, k ∈Zo
cotanθ=cos θ
sin θθDcotan =R\{kπ, k ∈Z}
Rx7→ cos x2π
Rx7→ sin x2π
Dtan x7→ tan x π
Rxsin′x= cos x
cos′x=−sin x
i−π
2+kπ;π 2+kπhk
tan′x= 1 + tan2x=1
cos2x
xsin π
2+x= cos x
π
2−→
i
xcos2x+ sin2x= 1
1
2
3
−1
−2
−3
π
2π3π 2
−π
2
−π
−3π
2
−→
i
−→
j

θ0π
6
π
4
π
3
π
2
sin (θ) 0 1
2
√2
2
√3
21
cos (θ) 1 √3
2
√2
2
1
20
tan (θ) 0 √3
3√3
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
2+x=−sin x
sin π
2+x= cos x
cos π
2−x= sin x
sin π
2−x= cos x
cos(a+b) = cos acos b−sin asin b
cos(a−b) = cos acos b+ sin asin b
sin(a+b) = sin acos b+ cos asin b
sin(a−b) = sin acos b−cos asin b
tan(a+b) = tan a+ tan b
1−tan atan b
tan(a−b) = tan a−tan b
1 + tan atan b
cos(2a) = cos2a−sin2a
= 2 cos2a−1
= 1 −2 sin2a
sin(2a) = 2 sin acos a
tan(2a) = 2 tan a
1−tan2a
cos2a=1 + cos(2a)
2
sin2a=1−cos(2a)
2
cos acos b=1
2(cos (a+b) + cos (a−b))
sin acos b=1
2(sin (a+b) + sin (a−b))
sin asin b=1
2(cos (a−b)−cos (a+b))
sin p+ sin q= 2 sin p+q
2cos p−q 2
sin p−sin q= 2 cos p+q
2sin p−q 2
cos p+ cos q= 2 cos p+q
2cos p−q 2
cos p−cos q=−2 sin p+q
2sin p−q 2
θ t = tan θ
2
cos θ=1−t2
1 + t2
sin θ=2t
1 + t2
tan θ=2t
1−t2
x
cos a= cos b⇔
a=b+ 2kπ
ou
a=−b+ 2kπ
, k ∈Z
sin a= sin b⇔
a=b+ 2kπ
ou
a=π−b+ 2kπ
, k ∈Z

~u ~v
A−→
OA =1
k~uk~u B−−→
OB =1
k~vk~v
AB OA =OB = 1
(~u, ~v)
⌢
AB
A B
2π
(~u, ~v) = θ+ 2kπ (~u, ~v)≡θ[2kπ]θ
0 + 2kπ k ∈Z
π
2+ 2kπ −π 2+ 2kπ k∈Z
(~u, ~v)
π
2+kπ k∈Z