Question

已知sin(π+α)=1/2 求sin(2π-α)-1/tan(α-π)·cosα

Answer 1

∵sin(π+α)=1/2 ∴ sinα=--1/2
sin(2π-α)-1/tan(α-π)·cosα
=sin（-α）-1/tanα·cosα
= —sinα-cosα^2/sinα
=-sinα-(1-sinα^2)/sinα
=-1/2-(1-1/4)/1/2
=-2

追问

可答案是2阿- - 是不是你有步骤符号错了

追答

哦，抱歉
∵sin(π+α)=1/2 ∴ sinα=--1/2
sin(2π-α)-1/tan(α-π)·cosα
=sin（-α）-1/tanα·cosα
= —sinα-cosα^2/sinα
=-sinα-(1-sinα^2)/sinα
=1/2+(1-1/4)/1/2
=2

Answer 2

因为sin(π+α)=1/2
所以sina=-1/2
sin(2π-α)-1/tan(α-π)·cosα
=-sinα-1/tanα·cosα
=-（-1/2）-1/sina
=-1/2/-1/2
=1