若cos(π+α)=-1/2, 3π/2<α<2π, 则sin(2π-α)等于

显示全部楼层 · 2011-2-22 23:07:33

cos(π+α)= sin(-π/2 -α)= -sin(π/2 +α)= -cos(-α)= -cosα= -1/2则cosα＝1/2，sin(2π-α)＝sin(-α)= -sinα3π/2<α<2π第三象限，则sinα为负，则sinα ＝－√3/2则sin(2π-α)＝sin(-α)= -sinα ＝ √3/2，即sin(2π-α)＝sin(2π-α)...

千问 · 2011-2-22 23:07:33

3π/2<α<2πsinα<0cos(π+α)=-1/2-cosα=-1/2cosα=1/2sinα=-√3/2sin(2π-α)=-sinα=-(-√3/2)=√3/2...

千问 · 2011-2-22 23:07:33

∵cos(π+α)=-1/2∴-cosα=-1/2cosα=1/2sin(2π-α)=-sinα=-(-)(1-(1/2)^2)^1/2=2分支根号3不懂发信息问我！...