cos(π/6-α)=m,m绝对值小于等于1,求cos(5π/6+α)+sin(2π/3-α）
问题描述：
cos(π/6-α)=m,m绝对值小于等于1,求cos(5π/6+α)+sin(2π/3-α）
答
cos(5π/6+α)=cos[π-(π/6-α)]=-cos(π/6-α)=-m
sin(2π/3-α)=cos[π/2-(2π/3-α)]=cos(-π/6+α)=cos(π/6-α)=m
所以原式=0