已知cos(π/6-α),则cos(5π/6+α)×sin(2π/3-α)=

问题描述:

已知cos(π/6-α),则cos(5π/6+α)×sin(2π/3-α)=

原式=cos(5π/6+α)*sin(2π/3-α)
=cos[(α+π/3)+π/2]*sin[π-(α+π/3)]
=-sin(α+π/3)*sin(α+π/3)
=-sin²(α+π/3)
=-cos²(α-π/6)不好意思,我打错了,应该是cos(π/6-α)=1/3,其他不变那么答案就应该是-cos²(α-π/6)=-1/9 求采纳...