Răspuns:
|cos α| = √1+cos2α/2 ( cos2α= 2cos²α-1 => 2cos²α=cos2α+1
deci cos²α= (1+cos2α)/2 => |cos α| = √1+cos2α/2
in particular,
|cosα|= √1+1/2/2
|cosα|= √(3/4)
|cosα|= ± √3/2
α ∈(π/2, π) ( e in cadranul 2, unde cosx<0)
deci, cosα = -√3/2
ai obtinut corect:)
