求值(1)cos(兀/11)+cos(3兀/11)+...+cos(9兀/11)
(2)sin(兀/11)+sin(3兀/11)+...+sin(9兀/11)
1)cos(p/11)+cos(2p/11)+……+cos(9p/11)
=[2cos(p/11)sin(p/11)+2cos(3p/11)sin(p/11)+……+2cos(9p/11)sin(p/11)]/[2sin(p/11)]同乘的2sin(π/11)中角π/11是角的公差的一半
={sin(2p/11)+[sin(4p/11)-sin(2p/11)]+[sin(6p/11)-sin(4p/11)]+[sin(8p/11)-sin(6p/11)]+[sin(10p/11)-sin(8p/11)}/[2sin(p/11)]积化和差
=sin(10p/11)/[2sin(p/11)]
...全部
1)cos(p/11)+cos(2p/11)+……+cos(9p/11)
=[2cos(p/11)sin(p/11)+2cos(3p/11)sin(p/11)+……+2cos(9p/11)sin(p/11)]/[2sin(p/11)]同乘的2sin(π/11)中角π/11是角的公差的一半
={sin(2p/11)+[sin(4p/11)-sin(2p/11)]+[sin(6p/11)-sin(4p/11)]+[sin(8p/11)-sin(6p/11)]+[sin(10p/11)-sin(8p/11)}/[2sin(p/11)]积化和差
=sin(10p/11)/[2sin(p/11)]
=sin(p/11)/[2sin(p/11]
=1/2。
2)与前题的方法相同，只是不能得到简单的数值
原式=[2sin(p/11sin(p/11)+2sin(3p/11)sin(p/11)+……+2sin(9p/11)sin(p/11)]/[2sin(p/11)]
={[cos0-cos(2p/11)]+[cos(2p/11)-cos(4p/11)]+……+[cos(4p/11)-cos(6p/11)]+[cos(6p/11)-cos(8p/11)]+[cos(8p/11)-cos(10p/11)]}/[2sin(p/11)]
=[cos0-cos(10p/11]/[2sin(p/11)]
=[1+cos(p/11)]/2sin(p/11)]
=2[cos(p/22)]^2/[4sin(p/22)cos(p/22)]
=(1/2)cos(p/22)/[2sin(p/22)]
=(1/2)cot(p/22)。
收起
