利用等差数列化简三角函数

sin（π/11）+sin（3π/11）+sin（5π/11）+sin（7π/11）+sin（9π/11)
= sin(2π/11)[sin（π/11）+sin（3π/11）+sin（5π/11）+sin（7π/11）+sin（9π/11)]/sin(2π/11)
= -1/2{[cos(3π/11)-cos(-π/11)]+[cos(5π/11)-cos(π/11)]+[cos(7π/11)-cos(3π/11)]+[cos(9π/11)-cos(5π/11)]+[cos(11π/11)-cos(7π/11]}/sin(2π/11)
=-1/2[cosπ + cos(9π/11)-2cos(π/11)]/sin(2π/11)
=-1/2[2cos(10π/11)cos(π/11)-2cos(π/11)]/sin(2π/11)
=[cos(π/11)-cos(10π/11)cos(π/11)]/[2sin(π/11)cos(π/11)]
=[1-cos(10π/11)]/2sin(π/11)
=[cos0-cos(π/11)]/2sin(π/11)
=2sin(π/22)sin(π/22)/4sin(π/22)cos(π/22)
=1/2tan(π/22)

想必楼主是参加高中数学联赛的。。

同感同感..