Bonjour,
1.a. -11π/12 = π/12 - π correspond au point E symétrique de A(π/12) par rapport à O
b. l'antécédent de D qu'on note d = π/12 + 3π/4 = 10π/12 = 5π/6
l'antécédent de H qu'on note h = π/12 - π/4 = -2π/12 = -π/6
2. π/12 + π/4 = 4π/12 = π/3
B(cos(π3) ; sin(π/3)) = B(1/2 ; √3/20
3.a. On a cos²(π/12) = (6 + 2 + 2√12)/16 = (8+4√3)/16 = (2+√3)/4
sin (π/12) = √(1 - cos²(π/12)) = √((2-√3)/4) = √(2-√3) / 2
b. cos(11π/12) = cos(π - π/12) = -cos(π/12) = (√6 + √4) / 4
sin(11π/12) = sin(π - π/12) = sin(π/12) = √(2-√3) / 2
cos(13π/12) = cos(π + π/12) = -cos(π/12) = (√6 + √4) / 4
sin(13π/12) = sin(π + π/12) = -sin(π/12) = -√(2-√3) / 2
