Répondre :
cos(90°+α)*cos(180°-α) =(-sin(a)*cos(a)/(-cos(a)*sin(a))=tan(a)
sin(90°-α)*sin(α-360°)
sin(180°-α)*cos(180°+α) =(sin(a)(-cos(a))/(cos^2(a)=-tan(a)
sin(90°+α)*cos(270°+α)
bonjour c'est TRES IMPORTANT car c'est pour demain, simplifier :
cos(90°+α)*cos(180°-α) sin(180°-α)*cos(180°+α)
sin(90°-α)*sin(α-360°) + sin(90°+α)*cos(270°+α)
ce qui est souligné veut dire divisé par... merci pour votre aide
cos(90°+α)*cos(180°-α) =(-sin(a)*cos(a)/(-cos(a)*sin(a))=tan(a)
sin(90°-α)*sin(α-360°)
sin(180°-α)*cos(180°+α) =(sin(a)(-cos(a))/(cos^2(a)=-tan(a)
sin(90°+α)*cos(270°+α)