Bonjour,
21)
I --> 0
B --> π/3
C --> 2π/3
D --> π
E --> -2π/3
F --> -π/3
22)
I --> 0
A --> 2π/5
B --> 4π/5
C --> 6π/5
D --> 8π/5
23)
a =100π/3 = 102π/3 - 2π/3 = -2π/3 + 34π donc -2π/3
b = 2010π = 1005 x 2π donc 0
c = 41π/4 = 40π/4 + π/4 = π/4 + 10π donc π/4
d = -41π/4 = -40π/4 - π/4 = -π/4 - 10π donc -π/4
e = 20,4π = 20π + 0,4π donc 0,4π
f = 29,2π = 30π - 0,8π donc -0,8π
