- 数学问题求cos2/7兀+cos4/7兀+cos6/7兀的值
- 求c2/7兀+cos4/7兀+cos6/7兀的值
- cos2/7兀+cos4/7兀+cos6/7兀
=(cos2/7兀+cos6/7兀)+cos4/7兀
=2cos4/7兀cos2/7兀+2(cos2/7兀)^2-1
=2cos2/7兀(cos4/7兀+cos2/7兀)-1
=4cos1/7兀cos2/7兀cos3/7兀-1
4cos1/7兀cos2/7兀cos3/7兀
=4(sin1/7兀cos1/7兀cos2/7兀cos3/7兀)/sin1/7兀
=2(sin2/7兀cos2/7兀cos3/7兀)/sin1/7兀
=sin4/7兀cos3/7兀/sin1/7兀
=-sin4/7兀cos4/7兀/sin1/7兀
=-sin8/7兀/(2sin1/7兀)
=1/2
===> cos2/7兀+cos4/7兀+cos6/7兀
= 1/2 -1 = -1/2