Решите задачу:
а задание какое??
Cosπ/12*cosπ/12 - cosπ/6*sinπ/6 =cos²π/12- (1/2)*sin2*π/6 = =(1/2)*(1+cos2*π/12 )- (1/2)*sin(π/3) =(1/2)*(1+cos(π/6) -sin(π/3)) = (1/2)*(1+√3/2 -3/2) =1/2. ************************** Использованы формулы : sinα*cosα =(sin2α)/2 и cos²α =(1+cos2α)/2 .