问题是不是错了
a∈(π,2π),tan(a+π/6)=2,则cos(π/6-2a)=?
tan(a+π/6)=2
∴
sin(2a+π/3)
=sin2(a+π/6)
=2sin(a+π/6)cos(a+π/6)
=2sin(a+π/6)cos(a+π/6)/[sin²(a+π/6)+cos²(a+π/6)]
=2tan(a+π/6)/[1+tan²(a+π/6)]
=2*2/(1+2²)
=4/5
∴ cos(π/6-2a)
= cos[π/2-(2a+π/3)]
=sin(2a+π/3)
=4/5