TD1 - Anthony Mansuy

p q
sin p+ sin q= 2 sin p+q
2cos p−q 2; cos p+ cos q= 2 cos p+q 2cos p−q 2
sin p−sin qcos p−cos q
D x tan xtan x
2
x∈D t = tan x
2
sin x=2t
1 + t2; cos x=1−t2 1 + t2; tan x=2t 1−t2
tan π
8tan π 12
cos x+ cos(2x) + cos(3x)
√3 cos(5x) = sin(2x) + sin(12x)
(ABC)A
sin(\ ACB) cos(\ ACB) tan(\ ACB)AC BC
AB BC CA AC sin(\ ABC)
(DEF )E
F D cos(\ EDF )
(IJK)K
IK tan([ IJK)
(RST )Ssin( [ RT S) cos( [ RT S) tan( [ RT S)
RS ST
(ABC)Csin(\ BAC) cos(\ BAC) tan(\ BAC)
AB AC
T.S.V.P.

M
S
h
N
®¯Montagne
H
S
h
MN
α β
(MHS)M H h α
NH h β
h MN α β
(ABC)α=\ BAC β =\ CBA γ =\ BCA
sin α
BC =sin β AC =sin γ AB
α β γ
A
B
C
°¯
A
B
C
°¯
H A (BC) (AHB)
(AHC)AH AB β AC γ
sin β
AC =sin γ AB
B(AC)sin α
BC =sin γ AB
α α
AB
OA OB x AB

α β
A
O
B
x
®¯
H B (AO)
AH AB α
BH AB α
◃OH BH \ OBH
◃\ OBH α β
OH AB α β
x OA OB AB α β
(OAB)OA OB
AB α β
α h d
A
B
C
d
h
D
®
AD α h d H
C(AD)
\
DCH d DH
\
BAC h d
\
DCH α \ BAC \ DCH α
h d
DH AD α h d
T.S.V.P.

(3 −2i) + 3 (5 + i) (2 −5i)(1 + 3i) (2 + 3i)2(1 + 2i)62−3i
3+4i
1−2i
2+3i+2−i 3 + i
1
(2 −3i)(5 −i)(1 −i)21 i2010 in(n∈N)
2−3 3 i−5i√3−i1
3+1 3i1
1−i√3
1 + i
√3 + i(1 −i)9(1 + i√3)−5
−1−i√3
42008 2−5i 3i−7tan 3π 7+icos 5π 9−isin 5π 9sin π 9+icos π 9
−√5cos π
11+isin π 11 cos π 3+isin π 6iθ +1 θ∈R) (1 + cos θ+isin θ)nθ∈R)
z z =1
z
z z′z z′+ 1 ̸= 0
z+z′
1 + z z′
z+z′
1 + z z′
2
=1 + Re(z z′)
1 + Re(z z′)
z+z′
1 + z z′= 1 z= 1 z′= 1
C
z2=|z|z2=z|z|=
1
z
= z+1
z|z|=|z2|=|1−z|
M1M2z1z2
M1M2
z1
z2
z1
z2
z∈C\ {1} |z|= 1 1 + z
1−z
z∈C\ {2i}Z=z+ 2i
z−2iA2i B −2i M
z
Z M
Z M [AB]

|Z|= 1 M
|Z|= 3 MC C
θ∈Rcos(3θ) sin(7θ) cos θsin θ
θ∈Rcos(5θ) cos θ
θ=π
10 x= cos θ
x
cos π
10
sin3x; cos5x; cos xsin4x; sin(5x) cos(6x) ; cos4xsin3x; sin3xcos(2x)−sin(2x) cos3(3x)
x∈Rn∈N
n
k=0
cos(kx) ;
n
k=0
sin(kx) ;
n
k=0
cos(x+ky) ;
n
k=0 n
kcos(kx) ;
n
k=0 n
ksin(kx)
(1 + i)nn∈N
062p6n
(−1)pn
2p
062p+16n
(−1)pn
2p+ 1
C
z2= 7
z2=−7
z2=i
z2= 7 + 24i
z2=−5−11i
z2= 32 + 24 i
z2+ 3 z+ 5 = 0
z2+ (1 −4i)z+i−5 = 0
−i z2+ 2(1 + i)z+ 5(2 + i) = 0
(1 −i)z2+ (3 + 6i)z−7 + i= 0
z4−(5 −14i)z2−2 (5i+ 12) = 0
z3+ (1 + i)z2+(8+7i)z+ (18 −4i) = 0
z2−2(1 + 2 cos θ)z+ 5 + 4 cos θ= 0
z4−2z2cos θ+ 1 = 0 θ∈R
T.S.V.P.