Bonsoir, j'ai pu faire l'exercice 1.
Ex 1 :1) cos(π-x) = -cos x et 3π/5 = π-2π/5Donc cos(3π/5) = cos(2π/5) = (-√5+1)/42) sin²(2π/5)+cos²(2π/5) = 1 ⇒ sin²(2π/5) = 1-((√5-1)/4)² = (5+√5)/8 ⇒ sin(2π/5) = √((5+√5)/8) ou -√((5+√5)/8)Or 2π/5∈[0;π] ⇒ sin(2π/5) ≥ 0 ⇒ sin(2π/5) = √((5+√5)/8)sin(π-x) = sin x donc sin(3π/5) = √((5+√5)/8)3) cos(π/5) = cos(3π/5)*cos(2π/5)+sin(3π/5)*sin(2π/5) = -((3-√5)/8)+((5+√5)/8) = (1+√5)/4
