sin2π/3=?;cos2π/3=?;sin7π/6=?;cos7π/6=?说下
问题描述:
sin2π/3=?;cos2π/3=?;sin7π/6=?;cos7π/6=?说下
答
sin2π/3=√3/2;cos2π/3=-1/2;sin7π/6=-1/2;cos7π/6=-√3/2
答
(根号3)/2
-1/2
-1/2
-(根号3)/2
答
sin2π/3=sinπ/3=√3/2
cos2π/3=-cosπ/3=-1/2
sin7π/6=-sinπ/6=-1/2
cos7π/6=-cosπ/6=- √3/2