Question

4 cos (3d-n/6)+ ctg (d+n/12); если d=n/6

Answer 1

4cos(3n/12-n/6)+ctg(n/6-n/12)

4cos(n/12)+ctg(n/12)

cos(n/12)(4+1/sin(n/12))

Answer 2

4 cos (3d-n/6)+ ctg (d+n/12)=4 cos (3п/6-n/6)+ ctg (п/6+n/12) =

4 cos (n/3)+ ctg (n/4)=4*0,5+1=3