(sin2π/3+cos11π/6)tan7π/4=
问题描述：
(sin2π/3+cos11π/6)tan7π/4=
答
(sin2π/3+cos11π/6)tan7π/4
=[sin(π-π/3)+cos(2π-π/6)]*tan(2π-π/4)
=(sinπ/3+cosπ/6)*(-tanπ/4)
=[(根号3)/2+(根号3)/2]*(-1)
=-根号3