高一数学 求第二问!
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高一数学 求第二问!
2π/3<2α-π/3<5π/3
所以cos(2α-π/3)<0
cos(2α-π/3)=-12/13
sin2α=sin(2α-π/3+π/3)=sin(2α-π/3)cosπ/3+cos(2α-π/3)sinπ/3
=5/13*1/2-12/13*√3/2
=(5-12√3)/26
采纳我
高一数学 求第二问!
2π/3<2α-π/3<5π/3
所以cos(2α-π/3)<0
cos(2α-π/3)=-12/13
sin2α=sin(2α-π/3+π/3)=sin(2α-π/3)cosπ/3+cos(2α-π/3)sinπ/3
=5/13*1/2-12/13*√3/2
=(5-12√3)/26
采纳我